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273 | Why is the Hubble Telescope in space? Do we get enhanced clarity and range by placing it in space? What can it achieve from space that it could not achieve from Earth? | As @astromax observed, one of the primary factors that makes a space based telescope better than an equivalent telescope of the same size is scattering . Along with scattering, there is also refraction which can be especially problematic when combined with atmospheric turbulence. In the modern era, this problem can be remedied to a certain extent using adaptive optics , but since the Hubble was designed, built, and launched before AO became practical in the 1990's, a space telescope represented the pinnacle of optical clarity for that time. However, there is another important optical property of the atmosphere , absorption. Although the Hubble is primarily a visible light telescope, it does have instruments that cover both the near UV and near IR, both of which are absorbed by the atmosphere more than visible light. Furthermore, there are practical advantages of space based telescopes. There is neither weather, nor light pollution in space. | {
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321 | Jupiter has a great many moons - in the hundreds, and they're still being discovered. What is the current theory for where all these moons came from? Are they rocks flying through space captured by Jupiter's gravity? | Mass. The more massive a body, the larger the gap between its lowest and highest orbit; the range of speeds at which a random body entering its gravity is likely to remain as its satellite. Sun has millions of satellites if you count all the asteroids; smaller planets tend to have one or two moons at most (Pluto with five being a notable and not fully explained exception) To a lesser degree there's a matter of shape too. A regularly round body will have more regular and stable orbit than a potato-shaped one. Jupiter, being a gas giant is perfectly round. This doesn't play that much of a role though, especially with higher orbits. And last but not least, no destabilizing influence of other bodies. It's very hard to maintain a lunar orbit - artificial satellites around our Moon last only a couple years each, because relatively close neighborhood of Earth tends to destabilize orbit of anything orbiting the Moon. Jupiter being a single massive planet with relatively tiny (relatively to its mass) moons doesn't have them influence each other all that much. | {
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349 | Since the moon has gravity, it's almost impossible that there aren't some gasses trapped on the surface by the moon's gravity. Has any free-floating oxygen been found on the Moon? If so, in what concentration? | The Moon's atmosphere is very thin compared to the Earth, so thin that it is usually said to have no atmosphere. The Moon's gravity is not strong enough to retain lighter elements, so they escape into space. Apollo 17 carried an instrument called the Lunar Atmosphere Composition Experiment (LACE). Oxygen is not listed as one of the elements it found on this NASA web page . The principle gases found were neon, helium and hydrogen. Others included methane, carbon dioxide, ammonia, and water. Additionally, ground instruments have detected sodium and potassium in the lunar atmosphere. The recently launched Lunar Atmosphere and Dust Environment Experiment (LADEE - pronounced laddie, not lady) should be able to tell us more soon. Update:
According to this article at Spaceflight Now , LADEE did detect oxygen in the Moon's atmosphere. | {
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537 | Is there a reason that the Earth has the tilt that it does (~23°)? How do we know which way is supposed to be 0°? Does this tilt have major consequences on the planet? Has it changed and will it change in the future? (If so, would there be any consequences?) | First up, the tilt is exactly 23.45 degrees. The reason for Earth's tilt is still not yet really proven, but scientists at Princeton stated on August 25, 2006 that planet Earth may have 'tilted' to keep its balance. Quote: By analyzing the magnetic composition of ancient sediments found in the remote Norwegian archipelago of Svalbard, Princeton University's Adam Maloof has lent credence to a 140-year-old theory regarding the way the Earth might restore its own balance if an unequal distribution of weight ever developed in its interior or on its surface. The theory, known as true polar wander, postulates that if an object of sufficient weight -- such as a supersized volcano -- ever formed far from the equator, the force of the planet's rotation would gradually pull the heavy object away from the axis the Earth spins around. If the volcanoes, land and other masses that exist within the spinning Earth ever became sufficiently imbalanced, the planet would tilt and rotate itself until this extra weight was relocated to a point along the equator. Same goes for the lack of proof about the exact consequences on the planet or if it'll happen again. All that is still being researched and debated, but those Princeton scientists have thrown in some interesting perspectives which you'll discover when you visit the link I provided and completely read what they wrote. Besides "consequences" for Earth as a planet, it should be noted that the tilt of Earth is the reason why we have seasons. So even when we yet have to find out what the consequences for Earth (as a planet) have been and/or will be, we do know that the tilt surely has consequences for all beings that live on planet Earth… the seasons that influence us all: summer, fall, winter, and spring. And - last but not least - as for your question how they know where 0 degrees would be, you just have to look at the Earth's rotation around the sun. Take a 90° angle of the planet's orbit and you know where 0° would be. But I guess a picture explains more than a thousand words, so I quickly created the following graphic: | {
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539 | So, in school (that's a long time ago) they have been teaching us there are 9 planets in our solar system. Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto But every now and then I keep reading stories about another "dwarf planet" (Eris, discovered in 2005) that - depending on what source tells the story - is another planet according to the astronomical definition, while other sources say that it isn't a planet. Some even say Pluto isn't a planet anymore either. The result: I'm confused due to the contradicting stories. Even Wikipedia isn't clear about Eris and only writes (emphasis mine): NASA initially described it as the Solar System's tenth planet. Initially? So, is it a 10th planet or not? Fact is, there is another "something" out there and it surely seems to look like a planet. Yet, some people keep stating there are 9 planets in our solar system, while others say there are more than 9 planets, and then again there are people stating that the latest definition of "planet" has kicked out Pluto too so there are actually fewer than 9 planets in our solar system. Trying to get a definite, official, and astronomically correct answer I can actually rely on, I'm therefore asking: How many planets are there in this solar system? EDIT The "Definition of planet" at Wikipedia doesn't really help either, as it states: Many astronomers, claiming that the definition of planet was of little scientific importance, preferred to recognize Pluto's historical identity as a planet by "grandfathering" it into the planet list.* * Dr. Bonnie Buratti (2005), "Topic — First Mission to Pluto and the Kuiper Belt; "From Darkness to Light: The Exploration of the Planet Pluto"", Jet Propulsion Laboratory. Retrieved 2007-02-22. So, if you link somewhere to provide proof, it would be great if you could point me to a more trusted source than Wikipedia. Ideally, an astronomical trusted source and/or paper. | Since we're talking about terminology, we need to remember that none of this really matters, outside of clarity when communicating. Still, some people tend to have rather strong opinions on it, thus confusion about how many planets are really in the solar system arises. The people The most trusted source in Astronomy would have to be the people that set the generally accepted rules. The IAU (International Astronomical Union) has been in existence since 1919 and is comprised of 10814 Individual Members in 93 different countries worldwide. Of those countries, 73 are National Members. The key activity of the IAU is the organization of scientific meetings. Every year the IAU sponsors nine international IAU Symposia. The IAU Symposium Proceedings series is the flagship of the IAU publications. Every three years the IAU holds a General Assembly, which offers six IAU Symposia, some 25 Joint Discussions and Special Sessions, and individual business and scientific meetings of Divisions, Commissions, and Working Groups. The proceedings of Joint Discussions and Special Sessions are published in the Highlights of Astronomy series. The reports of the GA business meetings are published in the Transactions of the IAU - B series. The definition At the 2006 IAU General Assembly in Prague, the accepted definition of a planet was debated vigorously. The outcome of the meeting was the currently accepted definition of a planet: A celestial body that (a) is in orbit around the Sun, (b) has sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape, and (c) has cleared the neighbourhood around its orbit. What about Pluto? With this in mind, the group decided on Pluto's fate. From page 2 of this official resolution document: The IAU further resolves: Pluto is a "dwarf planet" by the above definition and is recognized as the
prototype of a new category of Trans-Neptunian Objects. The reason given in the FAQ on this page : Q: Why is Pluto now called a dwarf planet? A: Pluto now falls into the dwarf planet category on account of its size and the fact that it resides within a zone of other similarly-sized objects known as the transneptunian region. Basically, they decided that it isn't officially a planet anymore because it didn't match criteria (c): has cleared the neighbourhood around its orbit. It hasn't done this, because it 'resides within a zone of other similarly-sized objects '. Therefore, it hasn't cleared its neighborhood. Soo... what about the number of planets? Q: Based on this new definition, how many planets are there in our Solar System? A: There are eight planets in our Solar System; Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune. Mnemonic: My Very Educated Mother Just Served Us Nachos. But that's if you don't count Dwarf planets - if you do count them, you end up with five more : Ceres Pluto Eris Makemake Haumea So there are 8 planets in the solar system if you don't count Dwarfs, 13 if you do. | {
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639 | This answer on Space Exploration to a question about Mars says that one reason Mars has such a thin atmosphere is because it lacks a magnetic field to protect it from the effect of double solar winds. Here MBR explains that Venus does not have a magnetic field. Image credit: ESA If this is the case, then why has Venus's atmosphere not been stripped away by solar wind like that of Mars? | Venus has a strong ionosphere that protects it against violent solar winds. So, even though Venus has no intrisic magnetic field, it has an effective, induced magnetic field due to the interaction between the solar winds and the atmosphere, that protects it against solar winds. Venus atmosphere is thick enough to have a consequent ionosphere, that would be the difference between Mars and Venus (and Venus was able to keep a thicker atmosphere due to its greater mass, contrary to Mars). Sources: Luhmann & Russell (1997) , on Venus Luhmann & Russell (1997) , on Mars | {
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646 | Our nearest star Proxima Centauri is 4.243 light years away from Earth. Does that mean we are seeing light that is 4.243 years old everyday? | Yes, the speed of light in vacuum (or c ) is 299,792,458 m/s and one light-year is the distance the light travels in one Julian year (365.25 days), which comes out to 9.46073 × 10 15 meters. The exact value is 9,460,730,472,580,800 meters. Since c is the maximum speed at which all energy, matter, and information in the Universe can travel, it is the universal physical constant on which the light-year ( ly ) as one of the astronomical units of length is based. That means that visible light as an electromagnetic radiation cannot travel faster than c and in one Julian year it can traverse a maximum distance of d = t * c d is distance in meters t time in seconds c the speed of light in vacuum in meters per second If we calculate this distance for a 4.243 ly distant object, that comes out as 4.243 * 365.25 * 86,400 s * 299,792,458 m * sˉ¹ or exactly 40,141,879,395,160,334.4 meters (roughly 40 trillion kilometers or 25 trillion miles). That is the distance the light traveled since it was last reflected of (or in our case emitted from, since Proxima Centauri is a red dwarf star) the surface of a celestial object to be 4.243 Julian years later visible at our observation point, in this case our planet Earth from where the distance to Proxima Centauri you quoted was measured. The more powerful the telescope, the further into the past we can see because the light is much older! This goes the same regardless of the distance of the object you're observing, but astronomy is particularly neat in this regard and we can observe objects that are so distant that we see them from the time when they were still forming. For further reading on other units used to measure far away objects you might be interested in reading this question on the parsec. | {
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666 | This arose from a comment posted against a question elsewhere on the stackexchange How powerful a telescope/binoculars would allow me to view the astronauts aboard the ISS do a space-walk? Aperture? Magnification? | This has been done before, so I don't have to go through all the heavy calculations using Rayleigh criterion accounting for atmospheric diffraction and visible light wavelength. Ralf Vandebergh, a Dutch astronomer, professional photographer and veteran satellite spotter has been busy trying to do exactly this since the 2007 and has indeed succeeded on several occasions by now using a 10 inch (25.4 cm) Newtonian reflecting telescope that has a resolving power (angular resolution on the CCD sensor) of roughly one pixel per meter at the distance to the International Space Station (ISS) that is currently in a 230 miles (370 km) orbit above the Earth: Ralf Vandebergh’s detail of an image he took on Mar. 21, 2009 showing astronauts working outside the ISS. Credit: R. Vandebergh Vandebergh's personal page also hosts all kinds of other successful observations of the ISS through his telescope and recorded in both photographs as well as some short videos. Why short? Because targeting the ISS as it moves at the speed of 4.8 mi/s (7.7 km/s) is rather tricky, and the atmospheric conditions and times at which the ISS passes over some area on the surface of the Earth don't make it any easier either . But perseverance and hard work have paid well for this individual astronomer. Raw video of the ISS as seen through the air turbulence. Note the good visibility of the Lira antenna at the Russian Zvezda Module in the lower part of the image. Credit: R. Vandebergh So again, skipping the math to calculate the required angular resolution of a telescope and apply that to some arbitrarily selected image sensor size and resolution, we can see that using a well collimated 10 inch Newtonian or a Dobsonian telescope on a clear night can, with some near perfect targeting, produce a direct proof of a spacewalker doing his or her job 230 miles (370 km) high during an EVA. More powerful telescopes would of course produce better resolution images, but the atmospheric effects limit their use and are of course much harder to target an object moving fast over the skies with. | {
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748 | We know the spectacular explosions of supernovae, that when heavy enough, form black holes. The explosive emission of both electromagnetic radiation and massive amounts of matter is clearly observable and studied quite thoroughly. If the star was massive enough, the remnant will be a black hole. If it wasn't massive enough, it will be a neutron star. Now there's another mode of creation of black holes: the neutron star captures enough matter, or two neutron stars collide, and their combined mass creates enough gravity force to cause another collapse - into a black hole. What effects are associated with this? Is there an explosive release of some kind of radiation or particles? Is it observable? What physical processes occur in the neutrons as they are subjected to the critical increase of pressure? What is the mass of the new black hole, comparing to its neutron star of origin? | A neutron star must have a minimum mass of at least 1.4x solar masses (that is, 1.4x mass of our Sun) in order to become a neutron star in the first place.
See Chandrasekhar limit on wikipedia for details. A neutron star is formed during a supernova , an explosion of a star that is at least 8 solar masses. The maximum mass of a neutron star is 3 solar masses. If it gets more massive than that, then it will collapse into a quark star , and then into a black hole. We know that 1 electron + 1 proton = 1 neutron; 1 neutron = 3 quarks = up quark + down quark + down quark; 1 proton = 3 quarks = up quark + up quark + down quark; A supernova results in either a neutron star (between 1.4 and 3 solar masses), a quark star(about 3 solar masses), or a black hole(greater than 3 solar masses), which is the remaining collapsed core of the star. During a supernova, most of the stellar mass is blown off into space, forming elements heavier than iron which cannot be generated through stellar nucleosynthesis, because beyond iron, the star requires more energy to fuse the atoms than it gets back. During the supernova collapse, the atoms in the core break up into electrons, protons and neutrons. In the case that the supernova results in a neutron star core, the electrons and protons in the core are merged to become neutrons, so the newly born 20-km-diameter neutron star containing between 1.4 and 3 solar masses is like a giant atomic nucleus containing only neutrons. If the neutron star's mass is then increased, neutrons become degenerate, breaking up into their constituent quarks, thus the star becomes a quark star; a further increase in mass results in a black hole. The upper/lower mass limit for a quark star is not known (or at least I couldn't find it), in any case, it is a narrow band around 3 solar masses, which is the minimum stable mass of a black hole. When you talk about a black hole with a stable mass (at least 3 solar masses), it is good to consider that they come in 4 flavors: rotating-charged, rotating-uncharged , non-rotating-charged, non-rotating-uncharged. What we would see visually during the transformation would be a hard radiation flash.
This is because during the collapse, the particles on/near the surface have time to emit hard radiation as they break up before going into the event horizon; so this could be one of the causes of gamma ray bursts (GRBs). We know that atoms break up into protons, neutrons, electrons under pressure. Under more pressure, protons and electrons combine into neutrons. Under even more pressure, neutrons break down into quarks. Under still more pressure, perhaps quarks break down into still smaller particles. Ultimately the smallest particle is a string : open or closed loop, and has a Planck length, which is many orders of magnitude smaller than a quark. if a string is magnified so it is 1 millimeter in length, then a proton would have a diameter that would fit snugly between the Sun and Epsilon Eridani, 10.5 light years away; that's how big a proton is compared to a string, so you can imagine there are perhaps quite a few intermediate things between quarks and strings. Currently it looks like several more decades will be needed to figure out all the math in string theory, and if there is anything smaller than strings then a new theory will be required, but so far string theory looks good; see the book Elegant Universe by Brian Greene. A string is pure energy and Einstein said mass is just a form of energy, so the collapse into a black hole really breaks down the structure of energy that gives the appearance of mass/matter/baryonic particles, and leaves the mass in its most simple form, open or closed strings, that is, pure energy bound by gravity. We know that black holes (which are not really holes or singularities, as they do have mass, radius, rotation, charge and hence density, which varies with radius) can evaporate , giving up their entire mass in the form of radiation, thus proving they are actually energy. Evaporation of a black hole occurs if its mass is below the minimum mass of a stable black hole, which is 3 solar masses; the Schwarzschild radius equation even tells you what the radius of a black hole is given its mass, and vice versa. So you could transform anything you want, such as your pencil, into a black hole if you wanted to, and could compress it into the required size for it to become a black hole; it is just that it would immediately transform itself (evaporate) completely into a flash of hard radiation, because a pencil is less than the stable black hole mass (3 solar masses). This is why the CERN experiment could never have created a black hole to swallow the Earth - a subatomic black hole, even one with the mass of the entire Earth, or the Sun, would evaporate before swallowing anything; there is not enough mass in our solar system to make a stable (3 solar mass) black hole. A simple way for a neutron star to become more massive in order to be able to turn into a black hole is to be part of a binary system, where it is close enough to another star that the neutron star and its binary pair orbit each other, and the neutron star siphons off gas from the other star , thus gaining mass. Here is a nice drawing showing exactly that. Matter falling into a black hole is accelerated toward light speed. As it is accelerated, the matter breaks down into subatomic particles and hard radiation, that is, X-rays and gamma rays. A black hole itself is not visible, but the light from infalling matter that is accelerated and broken up into particles is visible. Black holes can also cause a gravitational lens effect on the light of background stars/galaxies. | {
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752 | My question is about furthest object in the observable universe except Сosmic microwave background . | A neutron star must have a minimum mass of at least 1.4x solar masses (that is, 1.4x mass of our Sun) in order to become a neutron star in the first place.
See Chandrasekhar limit on wikipedia for details. A neutron star is formed during a supernova , an explosion of a star that is at least 8 solar masses. The maximum mass of a neutron star is 3 solar masses. If it gets more massive than that, then it will collapse into a quark star , and then into a black hole. We know that 1 electron + 1 proton = 1 neutron; 1 neutron = 3 quarks = up quark + down quark + down quark; 1 proton = 3 quarks = up quark + up quark + down quark; A supernova results in either a neutron star (between 1.4 and 3 solar masses), a quark star(about 3 solar masses), or a black hole(greater than 3 solar masses), which is the remaining collapsed core of the star. During a supernova, most of the stellar mass is blown off into space, forming elements heavier than iron which cannot be generated through stellar nucleosynthesis, because beyond iron, the star requires more energy to fuse the atoms than it gets back. During the supernova collapse, the atoms in the core break up into electrons, protons and neutrons. In the case that the supernova results in a neutron star core, the electrons and protons in the core are merged to become neutrons, so the newly born 20-km-diameter neutron star containing between 1.4 and 3 solar masses is like a giant atomic nucleus containing only neutrons. If the neutron star's mass is then increased, neutrons become degenerate, breaking up into their constituent quarks, thus the star becomes a quark star; a further increase in mass results in a black hole. The upper/lower mass limit for a quark star is not known (or at least I couldn't find it), in any case, it is a narrow band around 3 solar masses, which is the minimum stable mass of a black hole. When you talk about a black hole with a stable mass (at least 3 solar masses), it is good to consider that they come in 4 flavors: rotating-charged, rotating-uncharged , non-rotating-charged, non-rotating-uncharged. What we would see visually during the transformation would be a hard radiation flash.
This is because during the collapse, the particles on/near the surface have time to emit hard radiation as they break up before going into the event horizon; so this could be one of the causes of gamma ray bursts (GRBs). We know that atoms break up into protons, neutrons, electrons under pressure. Under more pressure, protons and electrons combine into neutrons. Under even more pressure, neutrons break down into quarks. Under still more pressure, perhaps quarks break down into still smaller particles. Ultimately the smallest particle is a string : open or closed loop, and has a Planck length, which is many orders of magnitude smaller than a quark. if a string is magnified so it is 1 millimeter in length, then a proton would have a diameter that would fit snugly between the Sun and Epsilon Eridani, 10.5 light years away; that's how big a proton is compared to a string, so you can imagine there are perhaps quite a few intermediate things between quarks and strings. Currently it looks like several more decades will be needed to figure out all the math in string theory, and if there is anything smaller than strings then a new theory will be required, but so far string theory looks good; see the book Elegant Universe by Brian Greene. A string is pure energy and Einstein said mass is just a form of energy, so the collapse into a black hole really breaks down the structure of energy that gives the appearance of mass/matter/baryonic particles, and leaves the mass in its most simple form, open or closed strings, that is, pure energy bound by gravity. We know that black holes (which are not really holes or singularities, as they do have mass, radius, rotation, charge and hence density, which varies with radius) can evaporate , giving up their entire mass in the form of radiation, thus proving they are actually energy. Evaporation of a black hole occurs if its mass is below the minimum mass of a stable black hole, which is 3 solar masses; the Schwarzschild radius equation even tells you what the radius of a black hole is given its mass, and vice versa. So you could transform anything you want, such as your pencil, into a black hole if you wanted to, and could compress it into the required size for it to become a black hole; it is just that it would immediately transform itself (evaporate) completely into a flash of hard radiation, because a pencil is less than the stable black hole mass (3 solar masses). This is why the CERN experiment could never have created a black hole to swallow the Earth - a subatomic black hole, even one with the mass of the entire Earth, or the Sun, would evaporate before swallowing anything; there is not enough mass in our solar system to make a stable (3 solar mass) black hole. A simple way for a neutron star to become more massive in order to be able to turn into a black hole is to be part of a binary system, where it is close enough to another star that the neutron star and its binary pair orbit each other, and the neutron star siphons off gas from the other star , thus gaining mass. Here is a nice drawing showing exactly that. Matter falling into a black hole is accelerated toward light speed. As it is accelerated, the matter breaks down into subatomic particles and hard radiation, that is, X-rays and gamma rays. A black hole itself is not visible, but the light from infalling matter that is accelerated and broken up into particles is visible. Black holes can also cause a gravitational lens effect on the light of background stars/galaxies. | {
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782 | At this point in time, evidence for the existence of dark matter has accumulated in many ways: it affects galactic rotation curves plays a major role in cosmology, and the evolution of structure in the universe is predicted in copious amounts by gravitational lensing on a wide range of scales influences the dynamics of galaxy clusters to name a few. There are many known candidates for dark matter particles: WIMPs , axions , WISPs , neutrinos, etc (in fact, even bricks, though some other considerations would exclude them). The question then is :
Why do we expect that only one type of dark matter particles is responsible for phenomenological dark matter? For example, $\Lambda$CDM cosmology, the standard cosmological model, requires dark matter to be cold (slow, non-relativistic), which is used to constrain the possible properties of dark matter particles. However, this doesn't actually imply, that dark matter is cold for all the astrophysical systems. For example, galactic halos could be made of warm dark matter, and halos of dwarf galaxies could be made of cold dark matter. One might of course say that one-species model is the simplest one. The counter-argument would be that in reality there well may be many species. This in turn might have profound implications for astrophysical models. To summarize the question: Is there any good reason, preferably supported by observations, to think that only one species of dark matter is present in all the models currently used? | Hot dark matter would be made from very light, fast moving particles. Such particles could not possibly be gravitationally bound to any structure, but rather would be dispersed all across the universe. But dark matter is always "found" (or "inferred") either gravitationally bound to some visible structure (e.g. weak lensing detection of dark matter associated with colliding galaxy clusters / flat rotation curves of spiral galaxies / abnormal velocity dispersion in galaxy clusters ) or not associated to anything visible but nevertheless forming clumps (weak lensing detection of galaxy clusters previously unseen ). That is why dark matter is thought to be cold . Additionally, there is a clear distinction between both types: there is not such thing as dark matter that is "not too cold but not too hot either" (see footnote as well). Dark matter is either made of particles with less than ~10 eV (hot dark matter, made of light particles, mostly dispersed everywhere) or particles with more than ~2 GeV (heavier, slower particles gravitationally bound to some structure). Both limits are found when imposing the maximum amount in which the candidate particles (neutrinos or something more exotic) can possibly contribute to the actual value of the density parameter due to matter in our expaning Universe. Thus, either DM appears gravitationally bound (cold DM) or dispersed (hot DM), and both types are clearly distinct (10 ev vs 2 Gev). Observations favour the first case. However, Cold Dark Matter is not the ultimate solution, and still faces some problems. Regarding the possibility of mixed solutions, many of them have been already ruled out. Microlensing has ruled out the possibility of unseen compact objects (brown dwarfs, stars, stellar black holes) in galactic halos, in our galactic neighbourhood as well as in the extragalactic domain . Ordinary matter (stones, bricks, dust) cannot possibly be, otherwise they would become hot and re-radiate. Any exotic mix of known particles doesn't work. All we think we know is that DM must be made of some heavy particles yet to be discovered. In order to introduce a more complex model (e.g. different types of particles depending on the structure they appear attached to) one needs a justification (i.e. some predictions that better agree with reality) and nobody has been able to do that yet. Remark Note that Dark Matter particles, either from the hot or the cold type, cannot possibly "slow down" and clump too much (e.g. forming planets) because they don't interact electromagnetically like ordinary matter, that is why DM is said to be collisionless . Wherever infalling ordinary matter forms any structure (e.g. protostars or accretion disks ), a very important part of the process is thermalisation , i.e. the redistribution of energy of the infalling particles by means of numerous collisions. This cannot happen with Dark Matter. | {
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856 | Have we discovered any natural satellites of natural satellites of planets or dwarf planets? Even very small, or relatively short-lived - e.g. ringlets around Saturn's moons, some meteorites orbiting Jupiter moons, or something to orbit Charon? Or is the Star-Planets-Moons the deepest naturally ocurring orbital recursion level? | I don't think there are any in the Solar system. We do have around 250 asteroids with moons . Rhea's ring seems to be the only exception. Edit: Originally I said "a moon with a moon would be an unstable system , due to the gravitational influence of the planet." @Florian disagrees with this. However, the answer is more complex than the Hill sphere alone. At first approximation, the Hill sphere gives a radius in which orbits around a moon could be stable. Our Moon's Hill radius is 64000 km. For our own Moon, we know that most low orbits are unstable due to mascons : mass concentrations below the surface which make the Moon's gravitational field noticeably uneven. There are only four inclinations where an object orbiting the Moon avoids all mascons and would be stable: 27º, 50º, 76º, and 86º. High orbits above the Moon aren't all safe either: above 1200 km and inclinations of more than 39.6º, Earth's gravity disrupts the satellite's orbit. Note that these orbits are comfortably within the Moon's Hill sphere. There are stable orbits at high inclinations and high eccentricity: As for other moons in the solar system: most of them are smaller and orbit around larger planets, so their Hill spheres are small, and the planet's gravity will disrupt much of the volume inside the Hill sphere too. Moons below the limit where their gravity is strong enough to make them spherical, will have problems with uneven gravitational fields. Mascons may also be present. | {
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874 | If you rewind the universe back 14,7 billion years, all matter were in one spot, and then started expanding. Do we know where this is in reference to our own solar system? And is there anything there? Or are all matter in the universe just blown inn all different directions and none is left at the "original" spot where it originated | You're envisioning the Big Bang as a cosmological "explosion" in space where the universe is the resulting material expanding in outward all directions. The problem is, the universe doesn't work like that. The universe isn't a region of space that is expanding outward into another thing where you can (even hypothetically) fly out to the border and say, "Yup, that's about all there is. Let's use these borders to find the center of this expanding sphere. " The Universe is literally everything there is: space, time, and everything it contains and ever will contain. And the word contains is even a bit of a misnomer in that it suggests a container with boundaries. But the universe is simultaneously both unbounded and finite at the same time. "Finite" refers to the fixed amount of stuff this universe contains (matter, energy, etc)… and unbounded because this thing we know as "the universe" has no borders in a sense that we point to or even experience. So how would you even define a "center"? There isn't one. There's a famous way to help visualize an expanding universe that is both finite and unbounded without a center or borders; it's called "The Balloon Analogy." Imagine our friends in Flatland living in two-dimensional space… where everything they know (their entire universe) exists on the surface of a balloon. If you start to inflate that balloon (the expanding universe), all the little astronomers on the surface will observe that all the surrounding galaxies are moving away from them — and the farther away those galaxies are, the faster they seem to be moving away. That's pretty much what most cosmologists believe is happening to us. Going back to that balloon analogy, picture yourself existing on the surface of that balloon-universe and ask these questions again — In my expanding universe (the surface of the balloon), where is it's center? Everything is expanding outward in all directions, so if we rewind back in time, certainly everything should converge somewhere, right? Can we just find the borders of this outward expansion and calculate the exact center? So basdically… "Do we know the exact spot where big bang took place?" Be careful taking the balloon analogy too far, because it starts to break down in many ways. But the answer to where the expansion of the universe emanates from really is that it emanates from everywhere; everything we know is contained within the origin of our own existence, and there is no center and there are no borders by any definition of existence that we can experience. | {
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891 | We know that nothing can have proper velocities larger than the speed of light in vacuum. But are there any objects in space that get close to it? Any comets, or other objects thrown by gravity or supernova explosions that were hurled to incredible speeds? | The answer to this is surprising: We are. And many (if not all) other galaxies. And they move faster than light. See, the universe is expanding, at an accelerating rate . The fabric of spacetime itself stretches out, so that galaxies seem to move away from each other. The interesting thing is that relativity does not forbid these from moving away faster than light. While local space is flat and the local speed of light must be upheld, this need not hold at a global scale, so it is possible to have frames which move away from each other faster than $c$. Indeed, there are some galaxies that are moving away from us faster than light ( the only reason we see them is that they used to be closer and moving at a slower speed ). Any pair of galaxies that are 4200 Mpc away from each other (that is, with a redshift of 1.4), are moving away from each other faster than light in each other's frames (numbers stolen from the linked page). Since the only consistent way to talk about motion is relative, one can say that we are moving away from other galaxies faster than light, since the reverse is also true. This can put galaxies in the bucket of the fastest moving objects in the universe. As for which is the fastest, I don't know, we would have to find a pair of galaxies which are the farthest apart (distance measured in the frame of the galaxy, of course), but since the universe is probably more than what we observe 1 , we can't pinpoint the pair of galaxies for which this is true. For those who think that it is cheating 2 to short-circuit the question with space expansion, there are other objects that go faster than light (they are not the fastest objects in the universe though), and these can be found on good 'ol Earth. Electrons : In nuclear reactor cooling pools 3 , we have a phenomenon known as Cerenkov radiation . Basically, emitted beta particles move faster than the speed of light in water. This creates an effect of similar origin as the sonic boom, where strong light emanates from the medium. Saywhat? You think I'm cheating again 2 by putting everything relative to the speed of light in a medium? Alright, fine. Here are some fast objects that don't require space expansion to be fast, nor do they involve any trickery of semantics where the medium in which they are being measured is not mentioned. Many have already been mentioned by astromax. Tachyons :These are particles which go faster than $c$ — this does not violate relativity as long as they never decelerate to subluminal speeds. However, there isn't much (any?) experimental evidence for these. A lot of BSM models do predict their existence, though. So there's still some cheating here, on to bradyonic matter: Gluons : These are massless, and though they don't occur freely (except possibly in glueballs , though these most probably have mass) they do travel at $c$. But these can't move at any other speed, so again, this is slightly cheating. On to fermionic matter: Neutrinos : Now these are viable candidates. The electron neutrino is known to have very, very little mass (we have an upper bound for it, which gives ), and as a result it can easily attain very high speeds. Put it in a gravitational field, and it goes even faster. If you want macroscopic objects, however: Stuff spiraling around spinning black holes : Black holes have a strong gravitational field, and when rotating, they can impart angular momentum (lots of it) to nearby objects like accretion disks. Objects close to a black hole are accelerated to pretty high speeds. In fact, if an object is within the ergosphere , it moves faster than light from the point of view of certain frames of reference. Stuff falling into black holes : From the faraway frame, an object speeds up and approaches the speed of light as it approaches the horizon of a black hole. Arbitrarily large speeds bounded by $c$ can be attained here. Black hole plasma jets : Jets being flung out of black holes can get pretty fast relative to each other. 1. Due to cosmic expansion, there can be galaxies that are no longer visible to us. Some galaxies may never have been visible to us, if we start watching from when galaxies started forming. 2. I, for one, agree with you. 3. And other places where you have massive particles being emitted really fast into a medium | {
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1,001 | The age of the Universe is estimated at 13.8 billion years, and current theory states nothing can exceed the speed of light, which can lead to the incorrect conclusion that the universe can't have a radius of more than 13.8 billion light years. Wikipedia deals with this misconception as follows: This reasoning would only make sense if the flat, static Minkowski spacetime conception under special relativity were correct. In the real Universe, spacetime is curved in a way that corresponds to the expansion of space , as evidenced by Hubble's law . Distances obtained as the speed of light multiplied by a cosmological time interval have no direct physical significance. → Ned Wright, "Why the Light Travel Time Distance should not be used in Press Releases" That doesn't clear the matter up for me, and having no science or maths background beyond high school, further reading into Hubble's law isn't helping much either. One layman's explanation I've seen offers explanation that the Universe itself isn't bound by the same laws as things within it. That would make sense – insofar as these things can – but the above quote ( "Distances obtained as the speed of light multiplied by a cosmological time interval have no direct physical significance" ) seems more general than that. Can anyone offer (or direct me to) a good layman's explanation? | The easiest explanation for why the maximum distance one can see is not simply the product of the speed of light with the age of the universe is because the universe is non-static. Different things (i.e. matter vs. dark energy) have different effects on the coordinates of the universe, and their influence can change with time. A good starting point in all of this is to analyze the Hubble parameter, which gives us the Hubble constant at any point in the past or in the future given that we can measure what the universe is currently made of: $$ H(a) = H_{0} \sqrt{\frac{\Omega_{m,0}}{a^{3}} + \frac{\Omega_{\gamma,0}}{a^{4}} + \frac{\Omega_{k,0}}{a^{2}} + \Omega_{\Lambda,0}} $$
where the subscripts $m$, $\gamma$, $k$, and $\Lambda$ on $\Omega$ refer to the density parameters of matter (dark and baryonic), radiation (photons, and other relativistic particles), curvature (this only comes into play if the universe globally deviates from being spatially flat; evidence indicates that it is consistent with being flat), and lastly dark energy (which as you'll notice remains a constant regardless of how the dynamics of the universe play out). I should also point out that the $0$ subscript notation means as measured today . The $a$ in the above Hubble parameter is called the scale factor, which is equal to 1 today and zero at the beginning of the universe. Why do the various components scale differently with $a$? Well, it all depends upon what happens when you increase the size of a box containing the stuff inside. If you have a kilogram of matter inside of a cube 1 meter on a side, and you increase each side to 2 meters, what happens to the density of matter inside of this new cube? It decreases by a factor of 8 (or $2^{3}$). For radiation, you get a similar decrease of $a^{3}$ in number density of particles within it, and also an additional factor of $a$ because of the stretching of its wavelength with the size of the box, giving us $a^{4}$. The density of dark energy remains constant in this same type of thought experiment. Because different components act differently as the coordinates of the universe change, there are corresponding eras in the universe's history where each component dominates the overall dynamics. It's quite simple to figure out, too. At small scale factor (very early on), the most important component was radiation. The Hubble parameter early on could be very closely approximated by the following expression: $$H(a) = H_{0} \frac{\sqrt{\Omega_{\gamma,0}}}{a^{2}}$$ At around: $$ \frac{\Omega_{m,0}}{a^{3}} = \frac{\Omega_{\gamma,0}}{a^{4}} $$
$$ a = \frac{\Omega_{\gamma,0}}{\Omega_{m,0}} $$
we have matter-radiation equality, and from this point onward we now have matter dominating the dynamics of the universe. This can be done once more for matter-dark energy, in which one would find that we are now living in the dark energy dominated phase of the universe. One prediction of living in a phase like this is an acceleration of the coordinates of universe - something which has been confirmed (see: 2011 Nobel Prize in Physics ). So you see, it would a bit more complicating to find the distance to the cosmological horizon than just multiplying the speed of light by the age of the universe. In fact, if you'd like to find this distance (formally known as the comoving distance to the cosmic horizon), you would have to perform the following integral: $$ D_{h} = \frac{c}{H_{0}} \int_{0}^{z_{e}} \frac{\mathrm{d}z}{\sqrt{\Omega_{m,0}(1+z)^{3} + \Omega_{\Lambda}}} $$ where the emission redshift $z_{e}$ is usually taken to be $\sim 1100$, the surface of last scatter. It turns out this is the true horizon we have as observers. Curvature is usually set to zero since our most successful model indicates a flat (or very nearly flat) universe, and radiation is unimportant here since it dominates at a higher redshift. I would also like to point out that this relationship is derived from the Friedmann–Lemaître–Robertson–Walker metric , a metric which includes curvature and expansion. This is something that the Minkowski metric lacks. | {
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1,007 | I often hear that nothing can escape a black hole because its "escape velocity" is greater than c . If that is accurate, what about the following? I know that the following has a lot of most likely impossible assumptions, but I'm trying to understand whether the escape velocity explanation makes sense. Suppose you've got some sort of space ship that's inside the event horizon, and it hasn't been destroyed by the gravity or tidal forces. Also suppose that you've got a whole lot of reaction mass and a fantastic power supply. Could this space ship, through applying enough constant force escape the event horizon? | One of the clearest way to see what's going on is to look at a Penrose diagram of a Schwarzschild black hole. A Penrose diagram is like a map of the spacetime drawn in such a way as to preserve angles and put the light rays diagonally at $45^\circ$ angles, forming the light cones. Since we're mapping all of an infinite space(time) into a finite drawing, distances are necessarily distorted, but that's a small price to pay. Time goes up on the diagram, and a typical infalling trajectory is in blue. Because every massive object must be locally slower than light, it must stay within those light cones. Hence, no matter how you accelerate, at every point of your trajectory you must go in a direction that stays within those $45^\circ$ diagonals at that point. But once you're inside the horizon, every direction that stays within the light cones leads to the singularity. Accelerating this way or that just means that you get to choose where you hit the singularity--a little on the left on the diagram or a little to the right. Trying to escape using high acceleration brings you closer to the light's $45^\circ$ lines, which because of time dilation will actually shorten your lifespan. You'll actually get to the singularity sooner if you struggle in that manner. This particular image came from Prof. Andrew Hamilton . Note that it pictures an eternal Schwarzschild black hole, i.e. one that has always and will exist. An actual black hole is formed by stellar collapse and will eventually evaporate, so its diagram will be slightly different (in particular, there will be no "antihorizon"). However, in all respects relevant here, it's the same situation. | {
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1,062 | When I stand outside looking at the night sky, to my untrained eye, everything except the moon looks like a star. I know intellectually that some are planets circling our sun, and some are entire galaxies far away, but they all look basically the same. How far out do you have be be from our sun for it to appear the same as any other 'star' in the sky? Edit to clarify As we migrate out through the solar system, when we look up to the sky the sun will get dimmer the farther away we get. On earth there is no doubt which star is our sun. As we occupy bodies in the solar system farther from the sun, where will we be when the sun appears to be the same brightness as any other star in the sky? | One way to answer would be to consider the brightest star in our sky (other than the Sun), which is Sirius. Then determine how far you would have to be from our Sun for it to be as bright as Sirius is from here. That turns out to be 1.8 light years. That's not even halfway to the nearest star, so if you're in any other star system, then our Sun is just another star. If you're anywhere in our solar system, even way out in the Oort cloud, then our Sun is way brighter than anything else. | {
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1,184 | Due to Earth's elliptical orbit, its distance from Sun varies by almost 5 million Kilometers (147 million Kilometers at closest point & 152 million Kilometers at farthest point, i.e. almost 3% of the average distance). As evident from the fact that that Venus has hotter environment than Mars due to their respective distances from the sun. Why then Earth does not observe two winters (at farthest points) and two summers (at closest points)? Additional Note :
I know that Earth's seasonal climate change is caused by its 23 degrees tilt that causes the sunlight density variations for the hemispheres. But to me this 5 million Km distance seems more relevant than the 23 degrees tilt. | There are a few incorrect assumptions in your post, so it is difficult to answer as asked. But I can address the misconceptions. 1. The seasons are not caused by our distance from the sun The seasons are caused by the 23.5° tilt in Earth's axis. When the Northern Hemisphere is tilted towards the sun (summer), the Southern Hemisphere is simultaneously tilted away from the sun (winter). So the seasonal temperature difference has little to do with the Earth's position in its elliptical orbit. Without this tilt, there would be no seasons and the temperature day to day across the globe would be relatively uniform. 2. Even the GLOBAL temperature is NOT consistent with our change in distance As a matter of fact, the average temperature of the Earth globally is hottest when it is the furthest from the sun — hotter by about 2.3°C ( ref ). That's because there is a lot more landmass in the Northern Hemisphere facing the sun (when Earth is farthest away in its orbit). So even though there is less intensity of sunlight, the land is able to be heated up much faster than the vast oceans which have to be heated at perihelion. This distance-temperature inconsistency isn't unique to the Earth. Look at the average temperature of the other inner planets as we move away from the sun: Mercury (167°C) Venus (460°C) ← farther, but hotter than Mercury? Earth (14.0°C) Mars (-60°C) Venus is actually warmer than Mercury because of the thick carbon dioxide atmosphere causing runaway global warming. So it isn't simply the distance from the sun that determines the average temperature of a planet. 3. There's only ONE aphelion/perihelion The closest point of the Earth's orbit (perihelion) and the farthest (aphelion) only happens once per year; not twice. That is because the elliptical orbit of the Earth is such so the sun is at one of the foci, not the center (as illustrated below). Note that the size of the bodies and the eccentricity of the orbits are greatly exagerated here. | {
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1,225 | From Earth, we can distinguish the type of another galaxy very easily simply by observing the shape, colour, and structure of the galaxy when we image it. But since we are located inside the Milky Way, we cannot get an "outside view". Thus, how can we tell if we are located in either a spiral or an elliptical galaxy? | There are several clues. The Milky Way is a flat disk The first one, and the simplest one, is that we live in a disk. As you can see on images like the ones from the 2MASS survey in the infrared range: We clearly see that we are inside a disk, since what we see when we look around us is a disk seen edge-on. It does not look like an ellipsoid or any other shape; it looks like a disk. The Milky Way has arms The other strong clue we have is the velocity of stars of the Milky Way, we can measure. First, we see an overall rotational motions, and, when you look closely to the patterns (like in the following image, that is observations in the radio range of the Milky Way at Green Bank Observatory), you can deduce that stars are even ordered into spirals: | {
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1,269 | In reference to the question, "How can we tell that the milky way is a spiral galaxy?" The answers there clearly sum up the question asked. But Milky Way is not just a spiral galaxy. It is further classified as a barred spiral galaxy. Question: Which particular feature in the distribution of stars, or in general a feature in observations led us to believe that it is a barred galaxy? Note: The edge-on picture is not sufficient to establish it, since a dense distribution at smaller radii could also arise from a non-uniform density on the spiral disk if modeled to fit the observations. We don't have any data from other angles or orientations. | There are several different lines of evidence which together form a coherent picture: that of a barred galaxy. Moreover, as most disc galaxies are barred, we should expect the same from the Milky Way. The various evidences are: The observed light distribution (2MASS) shows a left-right asymmetry in brightness and the vertical height. This is explained by the near end of the bar being located on that side. The distribution of magnitudes of red-clump stars (which have very nearly the same luminosity) is split towards the Galactic centre, as expected from a boxy/peanut bulge (which is always associated with a bar). The observed gas velocities show velocities which are "forbidden" in an axisymmetric or near-axisymmetric (spiral arms only) galaxy. These velocities occur naturally from the orbits of gas in a barred potential. The velocity distribution of stars in the Solar neighbourhood shows some asymmetries and clumping which is most naturally explained by orbital resonance with the bar rotation. The extent, pattern speed, and orientation of the bar is consistent between all of these. | {
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1,509 | Planets form from disks of matter orbiting around a star; some moons form from disks of matter orbiting planets. If this were going to happen around Saturn, approximately how much time would it take? | The answer to the headline question is: No. Most of Saturn's rings are below the Roche limit of about 2.5 Saturn radii. Hence tidal forces will prevent that part of the rings to form a (large) moon. Actually, part of the rings may be caused by loss of material from some of Saturn's moons, as suspected from observations of Enceladus . Accretion of Earth is still going on. So any number between millions of years and billions of years of accretion time for a planet can be justified.
Half of Earth's mass should have accreted within 10 million years, see this paper . | {
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1,518 | We know that the orbits of Pluto and Neptune overlap.
This means that pluto sometimes crosses the orbit of Neptune; will Pluto hit Neptune in any circumstance? | No, Pluto is a so called resonant trans-neptunian object; the orbital period of Pluto is almost exactly 3:2 (1.5) times that of Neptune. This means that every time Pluto nears perihelion and is therefore closest to the Sun and also closest to the orbit of Neptune, Neptune is always at a specific angle (50° according to Wikipedia) in front or behind Pluto.
(See for instance Figure 3 and surrounding text in Jewitt, D., Morbidelli, A., & Rauer, H. (2007). Trans-Neptunian Objects and Comets. Springer.) | {
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1,911 | It is said that the Andromeda and Milky Way galaxies are coming close to each other with a speed of approximately 400000 km/hour. They will be together in the next 4 billion years. What will happen to life on Earth or human beings on Earth? If we are about to collide in the next 4 billion years then how long before we should take action for interstellar voyage? Are scientists working on such projects for an interstellar voyage? Will we be able to get the something like Earth to go away from this galaxy/Earth/ solar system and, considering speed of human instruments/spaceships, how long will it take to go to a safe place? | What will happen to life on earth or human beings on earth? Assuming that human beings, or life, still exists on Earth at that time, they will have survived so much due to the ongoing death of the sun, that the gravitational pertubations due to the galactic collision will be nothing. Keep in mind that in about 1-2 billion years, the sun will be so hot and large that all the water will have boiled off the earth into space. About 3 billion years from now, the surface of the Earth will be so hot that metals will be melting. Any life that has survived those events and still lives on Earth will surely take a galactic collision in stride. I imagine, though, that most humans will have fled Earth - if not for distant star systems, then at least for planets in our own system that are going to be warming up enough for human habitation. If we are about to collide in next 4 billion year then how long before we should take action for inter-stellar voyage? As soon as possible. When interstellar voyage becomes possible, we should start sending out ships to colonize other planets and star systems. This will likely take a long time, but if we are to survive more than a billion years, it is necessary. Keep in mind that the Sun and Andromeda are events we can predict. We don't know, and can't predict, the next cataclysmic asteroid strike, which is likely to happen in a shorter time than a billion years. There are lots of reasons to exit the planet, we should be worried about the ones we can't predict or see, not the ones we can predict. Are scientists working on such projects for inter-stellar voyage? Yes, but in small steps. Manned missions to space, to the moon, and living aboard the ISS have provided significantly valuable information that will be used in such interstellar missions. As we continue to push the boundaries of our ability to survive in space we eventually will be able to live in space, perhaps whole lifetimes will be spent in space. As engine technology progresses beyond simply lifting people out of Earth's gravitational well, we will eventually be sending people on long voyages outside our solar system. It's a very, very long way off, but each advance takes us closer to that eventual goal. | {
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2,240 | So because I can only really think of space-time in 2-dimensions like a sheet of something, my assumptions might be wrong to begin with. I was watching a YouTube video on black-holes and there was a quip regarding black holes and their warping of space time, to the point where the narrator said something along the lines of “there is nothing at the center of the black hole but gravity has warped space time so much that it's this ’pinhole’ like affect that causes the blackhole behavior” . My questions are: When a star dies and collapses into a black hole, what is at its center? The star's mass compacted into the size of the plank length, or something similarly small? Is there really nothing at the center of a black hole? Surely, the core collapsed into something really small right? What is meant by singularity? Is it just the warping of space-time that makes it this way? | This is more of a question for the Physics stack, but I'll give it a shot, since it's fairly basic. You need to understand something before we begin. The theoretical framework we have to gauge and answer this sort of thing is called General Relativity, which was proposed by Einstein in 1915. It describes things such as gravity, black holes, or just about any phenomena where large densities of mass or energy are involved. There's another chapter in Physics called Quantum Mechanics. This describes, usually, what happens at very small scales - things that are super-tiny. Both GR and QM are fine in their own way. Both are tested against reality and work very well. But they are not compatible with each other. Meaning: you cannot describe a phenomenon from a GR and a QM perspective, both at once. Or meaning: we don't have a coherent set of equations that we could write down, and then "extract" out of them either a GR-like view of reality, or a QM-like view. The problem is, the center of a black hole is both very high mass density and very high gravity (and therefore right in the field of GR), and very small (and therefore "quantum-like"). To properly deal with it, we'd have to reconcile GR and QM and work with both at once. This is not possible with current physics. We pretty much have to stick to GR only for now, when talking about black holes. This basically means that anything we say about the center of a black hole is probably incomplete, and subject to further revision. A star dies, collapses into a black hole, what is at the center? The
star's mass compacted into the size of the plank length or something
similarly small? Is there really nothing at the center of a black
hole?, surely the core collapsed into something, just really small
right? According to General Relativity, it collapses all the way down to nothing. Not just "very small", but smaller and smaller until it's exactly zero in size. Density becomes infinite. You can't say "Plank length" because, remember, we can't combine GR and QM, we just don't know how. All we have here is GR, and GR says it goes all the way down. It is quite possible that the singularity is not physical, but just mathematical - in other words, whatever is at the center is not actually zero size. Quantum mechanics in particular would be offended by zero-size things. But we can't say for sure because our knowledge here is incomplete. I'm using words such as "size" (which implies space) and "becomes" (which implies time). But both space and time in the context of a black hole are very seriously warped. The "becoming" of a black hole all the way down to the zero-size dot is a reality only for the unlucky observer that gets caught in it. But for a distant, external observer, this process is slowed and extended all the way to plus infinity (it's only complete after an infinitely long time). Both observers are correct, BTW. EDIT: So, when we are saying "density is infinite and size is zero at the singularity", this language applies to the unfortunate observer being dragged down in the middle of the initial collapse of the star. But from the perspective of the distant observer, a black hole is still a chunk of mass (the original star) in a non-zero volume (the event horizon of the BH). To this observer, the density of that object is finite, and its size is definitely not zero. From this perspective, anything falling into the BH never quite finishes falling, but just slows down more and more. Both observers are correct. So, keep in mind, when I talk about "infinite density", that's the inside observer point of view. What is a singularity? Is it just the warping of space time that makes
it this way? You get a singularity whenever there's a division by zero in the equations, or when the equations misbehave somehow at that point. There are many different kinds of singularities in science. http://en.wikipedia.org/wiki/Mathematical_singularity In the context of a black hole, the center is said to be a gravitational singularity, because density and gravity are suggested to become infinite, according to the GR equations. GR says: when you have a lump of matter that's big enough, it starts to collapse into itself so hard, there's nothing to stop it. It keeps falling and falling into itself, with no limit whatsoever. Extrapolate this process, and it's easy to see that the size of it tends to zero, and density tends to an infinite value. EDIT: Put another way - if density becomes large enough, gravity is so huge, no other force is strong enough to resist it. It just crushes all barriers that matter raises to oppose further crushing. That lump of matter simply crushes itself, its own gravity pulls it together smaller and smaller... and smaller... and so on. According to current theories, there's nothing to stop it (QM might stop it, but we cannot prove it, because we don't have the math). So it just spirals down in a vicious cycle of ever-increasing gravity that increases itself. Space and time are really pathologic inside the event horizon. If you are already inside, there's no way out. This is not because you can't move out fast enough, but because there's really no way out . No matter which way you turn, you're looking towards the central singularity - in both space and time. There is no conceivable trajectory that you could draw, starting from the inside of the event horizon, that leads outside. All trajectories point at the singularity. All your possible futures, if you're inside the event horizon, end at the central singularity. So, why the center of a black hole is called a "singularity"? Because all sorts of discontinuities and divisions by zero jump out of the equations, when you push math to the limit, trying to describe the very center of a black hole, within a GR frame. http://en.wikipedia.org/wiki/Gravitational_singularity Speaking in general, physicists don't like singularities. In most cases, this is an indication that the mathematical apparatus has broken down, and some other calculations are necessary at that point. Or it might indicate that new physics are taking place there, superseding the old physics. One last thing: just because we don't have a combined GR/QM theory to fully describe the center of black holes, that doesn't mean a pure GR research in this area is "wrong" or "useless". It doesn't mean one could imagine some arbitrary fantasy taking place inside a black hole. Astronomers these days are starting to observe cosmic objects that are very much like black holes , and their observed properties are in very close accord with what GR predicts for such things. So research in this field must continue, because it's clearly on the right track, at least in the ways we can verify today in astronomy. | {
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2,247 | The Sun is huge, when compared to the Moon. Despite the huge difference in their size and distance from Earth, is it purely coincidental that they both look almost the same from Earth? | The coincidence isn't so much that they appear very similar sizes from Earth, but that we are alive to see them at the point in time in which they appear very similar sizes. The moon is slowly moving away from the Earth, and at some point in the future the moon will be unable to totally eclipse the sun and conversely, if you could step far into prehistory, you would be able to see the moon with a much greater angular diameter than you see it now. Most research I've found on the topic seem to be unavailable through my institute, however I did find one paper, "Outcomes of tidal evolution" , which references results from Goldreich's research on the subject. This qualitative description of the eventual disruption of the Earth-Moon system is confirmed by the results of Goldreich's numerical integration, which showed that the moon will recede to 75 Earth radii, when spin-orbit synchronism will be reached; then the Moon's orbit will decay steadily inward because of the influence of the Sun. For reference, the Moon is currently at a distance of approximately 60.3 Earth radii. As such, the moon will steadily move away until synchronism would be reached, and from that point begin to recede towards the Earth due to the tidal affects of the Sun on the Earth disturbing the synchronization. It would seem that at some eventual point in the far distant future, it will return to this coincidental position once again. Counselman III, Charles C. "Outcomes of tidal evolution." The Astrophysical Journal 180 (1973): 307-316. | {
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2,249 | How do the scientists calculate the correct time to launch the Mars mission so that the satellite travel time is less? How they are synchronizing the speed of satellite with respect to earth and mars ?
Refer this link for more understanding: Animation of MOM and Maven launch | The coincidence isn't so much that they appear very similar sizes from Earth, but that we are alive to see them at the point in time in which they appear very similar sizes. The moon is slowly moving away from the Earth, and at some point in the future the moon will be unable to totally eclipse the sun and conversely, if you could step far into prehistory, you would be able to see the moon with a much greater angular diameter than you see it now. Most research I've found on the topic seem to be unavailable through my institute, however I did find one paper, "Outcomes of tidal evolution" , which references results from Goldreich's research on the subject. This qualitative description of the eventual disruption of the Earth-Moon system is confirmed by the results of Goldreich's numerical integration, which showed that the moon will recede to 75 Earth radii, when spin-orbit synchronism will be reached; then the Moon's orbit will decay steadily inward because of the influence of the Sun. For reference, the Moon is currently at a distance of approximately 60.3 Earth radii. As such, the moon will steadily move away until synchronism would be reached, and from that point begin to recede towards the Earth due to the tidal affects of the Sun on the Earth disturbing the synchronization. It would seem that at some eventual point in the far distant future, it will return to this coincidental position once again. Counselman III, Charles C. "Outcomes of tidal evolution." The Astrophysical Journal 180 (1973): 307-316. | {
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2,357 | Recently I boarded a flight and noticed outside air temperature as -53°C at an altitude of 36860ft (11.23km). I don't know what causes such a freezing temperature in that altitude but was wondering higher altitudes (space) may have even freezing temperatures. Here I got a doubt i.e what happens if an ice cube is left in space? Would it be melting or stay as it is? | It depends on where in outer space you are. If you simply stick it in orbit around the Earth, it'll sublimate: the mean surface temperature of something at Earth's distance from the Sun is about 220K, which is solidly in the vapor phase for water in a vacuum, and the solid-vapor transition at that temperature doesn't pass through the liquid phase. On the other hand, if you stick your ice cube out in the Oort Cloud, it'll grow: the mean surface temperature is 40K or below, well into the solid phase, so it'll pick up (or be picked up by) gas and other objects in space. A comet is a rough approximation to an ice cube. If you think of what happens to a comet at various places, that's about what would happen to your ice cube. | {
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2,398 | In doing research on vision, I have learned that "20/20" vision corresponds to a visual acuity of being able to resolve details 1 arcminute in size, that most people have around 20/15 vision , and that due to the limits of physiology basically nobody has vision better than 20/10 vision. This is an upper limit of resolving details about 0.5 arcminutes in size. According to Wikipedia the Moon is around 30 arcminutes wide when seen by the naked eye. Put these together, and it seems to say that when looking at the moon with the naked eye nobody can see more details than would be visible in a 60×60 image of the Moon and that the average person can't see any more details than in a 40×40 version Those seem so small on my monitor. Can that really be all the detail that I can see on the Moon with the naked eye? | It doesn't seem so far-fetched to me. Sure, you might be off by a few pixels, due to differences between the human eye and a computer monitor, but the order of magnitude seems about right — the detail in your images, viewed closely, more or less matches what I see when I look at the full moon. Of course, you could fairly easily test it yourself: go outside on a dark night, when the moon is full, and see if you can spot with your naked eye any details that are not visible (even under magnification) in the image scaled to match your eyesight. I suspect you might be able to see some extra detail (especially near the terminator, if the moon is not perfectly full), but not very much. For a more objective test, we could try to look for early maps or sketches of the moon made by astronomers before the invention of the telescope, which should presumably represent the limit of what the naked human eye could resolve. (You needed to have good eyesight to be an astronomer in those days.) Alas, it turns out that, while the invention of the telescope in the early 1600s brought on a veritable flood of lunar drawings, with every astronomer starting from Galileo himself rushing to look at the moon through a telescope and sketch what they saw, very few astronomical (as opposed to purely artistic) drawings of the moon are known from before that period. Apparently, while those early astronomers were busy compiling remarkably accurate star charts and tracking planetary motions with the naked eye, nobody really though it important to draw an accurate picture of the moon — after all, if you wanted to know what the moon looked like, all you had to do was look at it yourself. Perhaps this behavior may be partly explained by the prevailing philosophical opinions at the time, which, influenced by Aristotle, held the heavens to be the realm of order and perfection, as opposed to earthly corruption and imperfection. The clearly visible "spots" on the face of the moon, therefore, were mainly regarded as something of a philosophical embarrassment — not something to be studied or catalogued, but merely something to be explained away. In fact, the first and last known "map of the moon" drawn purely based on naked-eye observations was drawn by William Gilbert (1540–1603) and included in his posthumously published work De Mundo Nostro Sublunari . It is quite remarkable how little detail his map actually includes, even compared to a tiny 40 by 40 pixel image as shown above: Left: William Gilbert's map of the moon, from The Galileo Project ; Right: a photograph of the full moon, scaled down to 40 pixels across and back up to 320 px. Indeed, even the sketches of the moon published by Galileo Galilei in his famous Sidereus Nuncius in 1610, notable for being based on his telescopic observations, are not much better; they show little detail except near the terminator, and the few details there are appear to be inaccurate bordering on fanciful. They are, perhaps, better regarded as "artist's impressions" than as accurate astronomical depictions: Galileo's sketches of the moon, based on early telescopic observations, from Sidereus Nuncius (1610), via Wikimedia Commons. Few, if any, of the depicted details can be confidently matched to actual lunar features. Much more accurate drawings of the moon, also based on early telescopic observations, were produced around the same time by Thomas Harriott (1560–1621), but his work remained unpublished until long after his death. Harriott's map actually starts to approach, and in some respects exceeds, the detail level of even the 60 pixel photograph above, showing e.g. the shapes of the maria relatively accurately. It is, however, to be noted that it is presumably based on extensive observations using a telescope, over several lunar cycles (allowing e.g. craters the be more clearly seen when they're close to the terminator): Left: Thomas Harriott's lunar map, undated but probably drawn c. 1610-1613, based on early telescopic observations, quoted from Chapman, A. "A new perceived reality: Thomas Harriot's Moon maps" , Astronomy & Geophysics 50(1), 2009; Right: same photograph of the full moon as above, scaled down to 60 pixels across and back up to 320 px. Based on this historical digression, we may thus conclude that the 40 pixel image of the moon, as shown in the question above, indeed does fairly accurately represent the level of detail visible to an unaided observer, while the 60 pixel image even matches the detail level visible to an observer using a primitive telescope from the early 1600s. Sources and further reading: Kopal, Zdeněk (1969). "The Earliest Maps of the Moon" . The Moon , Volume 1, Issue 1, pp. 59–66. Available courtesy of the SAO/NASA Astrophysics Data System (ADS). Van Helden, Al (1995). "The Moon" . The Galileo Project (web site). Wikipedia articles on the Moon and Selenography . | {
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2,441 | My understanding is that time slows and approaches stopping when approaching the event horizon of a black hole. I have seen this explained several places, including a brief explanation in the last paragraph under: http://en.wikipedia.org/wiki/Black_hole#General_relativity , quoted below: Oppenheimer and his co-authors interpreted the singularity at the boundary of the Schwarzschild radius as indicating that this was the boundary of a bubble in which time stopped. This is a valid point of view for external observers, but not for infalling observers. Because of this property, the collapsed stars were called "frozen stars",[17] because an outside observer would see the surface of the star frozen in time at the instant where its collapse takes it inside the Schwarzschild radius. Does this mean then that no matter actually falls into a black hole (except possibly what was there at its formation)? Would this also mean matter is accumulating just outside its event horizon? As I understand it, this would be the perspective from outside the black hole. If this is the case, I wonder if we would observe a tremendous amount of matter surrounding the event horizon, but it would be extremely red shifted? Edit: I noticed an answer to a different question, especially the end portion, provides some insight here as well: https://astronomy.stackexchange.com/a/1009/1386 Edit: These YouTube videos someone put together explain the concept very well, and seem to indicate this idea is gaining traction! https://www.youtube.com/watch?v=yZvgeAbrjgc&list=PL57CC037B74307650&index=118 https://www.youtube.com/watch?v=b1s7omTe1HI Edit: This new YouTube video describes this idea very well, and describes it as the way black holes work! https://youtu.be/mquEWFutlbs | Yes, you are absolutely right, from OUR VIEWPOINT it does. From Kip Thorne's book "Black Holes and Time Warps: Einstein's Outrageous Legacy." “Like a rock dropped from a rooftop, the star’s surface falls downward (shrinks inward) slowly at first, then more and more rapidly. Had Newton’s laws of gravity been correct, this acceleration of the implosion would continue inexorably until the star, lacking any internal pressure, is crushed to a point at high speed. Not so according to Oppenheimer and Snyder’s relativistic formulas. Instead, as the star nears its critical circumference, its shrinkage slows to a crawl. The smaller the star gets, the more slowly it implodes, until it becomes frozen precisely at the critical circumference. No matter how long a time one waits, if one is at rest outside the star (that is, at rest in the static external reference frame) one will never be able to see the star implode through the critical circumference. That is the unequivocal message of Oppenheimer and Snyder’s formulas.” “Is this freezing of the implosion caused by some unexpected, general relativistic force inside the star? No, not at all, Oppenheimer and Snyder realized. Rather, it is caused by gravitational time dilation (the slowing of the flow of time) near the critical circumference. Time on the imploding star’s surface, as seen by static external observers, must flow more and more slowly, when the star approaches the critical circumference, and correspondingly everything occurring on or inside the star including its implosion must appear to go into slow motion and then gradually freeze.” “As peculiar as this might seem, even more peculiar was another prediction made by Oppenheimer and Snyder’s formulas: Although, as seen by static external observers, the implosion freezes at the critical circumference, it does not freeze at all as viewed by observers riding inward on the star’s surface. If the star weighs a few solar masses and begins about the size of the sun, then as observed from its own surface, it implodes to the critical circumference in about an hour’s time, and then keeps right on imploding past criticality and on in to smaller circumferences.” “By looking at Oppenheimer and Snyder’s formulas from the viewpoint of an observer on the star’s surface, one can deduce the details of the implosion, even after the star sinks within its critical circumference; that is one can discover that the star gets crunched to infinite density and zero volume, and one can deduce the details of the spacetime curvature at the crunch.” P217-218 OK, so from our perspective all the matter will be clustered around the critical circumference and no further. That's fine, this shell in theory can exert all the forces required on the external universe such as gravitational attraction, magnetic field etc. The point like singularity which is in the indefinite future of the black hole, (from our point of view) indeed in the indefinite future of the universe itself could not exert such forces on this universe. This singularity is only "reached" as an observer rides in past the critical circumference and, through the process of time dilation, reaches the end of the universe. This is obviously an area of active research and thinking. Some of the greatest minds on the planet are approaching this issue in different ways but so far have not reached a consensus but intriguingly a consensus appears to be beginning to emerge. http://www.sciencealert.com/stephen-hawking-explains-how-our-existence-can-escape-a-black-hole Stephen Hawking said at a conference in August 2015 that he believes that "information is stored not in the interior of the black hole as one might expect, but on its boundary, the event horizon." His comment refers to the resolution of the "information paradox," a long-running physics debate in which Hawking eventually concedes that the material that falls into a black hole isn't destroyed, but rather becomes part of the black hole. Read more at: http://phys.org/news/2015-06-surface-black-hole-firewalland-nature.html#jCp In the mid-90s, American and Dutch physicists Leonard Susskind and Gerard 't Hooft also addressed the information paradox by proposing that when something gets sucked into a black hole, its information leaves behind a kind of two-dimensional holographic imprint on the event horizon, which is a sort of ‘bubble’ that contains a black hole through which everything must pass. What occurs at the event horizon of a black hole is very hard to understand. What is clear, and what proceeds from General Relativity, is that from the viewpoint of an external observer in this universe, any infalling matter cannot proceed past the critical circumference. Most scientists then change the viewpoint to explain how, from the viewpoint of an infalling observer, they will proceed in a very short period of time to meet the singularity at the centre of the black hole.
This has given rise to the notion that there is a singularity at the centre of every black hole. However this is is an illusion, as the time it will take to reach the singularity is essentially infinite to us in the external universe. The fact that the matter cannot proceed past the critical circumference is perhaps not an “illusion” but very real. The matter must from OUR VIEWPOINT become a “shell” surrounding the critical circumference. It will never fall through the circumference while we remain in this universe. So to talk of a singularity inside a black hole is incorrect. It has not happened yet. The path through the event horizon does lead to a singularity in each case, but it is indefinitely far in the future in all cases. If we are in this universe, no singularity has yet been formed. If it has not been formed yet, where is the mass? The mass is exerting pull on this universe, correct? Then it must be IN this universe. From our point of view it must be just this side of the event horizon. ASTONISHINGLY IT MAY BE POSSIBLE TO PROVE THIS. The recent announcement of gravitational waves detected on the merger of 2 black holes was accompanied by an unverified but potentially matching gamma ray burst from the same area of the sky. This is inexplicable from the conventional viewpoint which holds that all the matter would be compressed into a singularity and would be incapable of coming out again. If 2 black holes merge and emit gamma rays… the above is certainly an explanation which is also consistent with General Relativity. The mass never quite made it through the event horizon (from our viewpoint) and was perturbed by the huge violence of the merger, some escaping. It may be a deep gravitational well, but a very powerful gamma ray should just be able to escape given the right kick (attraction by an even larger black hole approaching). Further more refined observations of similar events, which are likely to be reasonably frequent, may provide more evidence. There is not likely to be any other credible explanation. | {
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2,545 | Did we ever actually see the Earth revolving around the Sun? Also, is the geocentric model completely disproved or was it sidelined because the heliocentric model made things easier to understand? (BTW I know Earth revolves around the Sun and am just asking out of curiosity.) Thanks! | What you're asking, basically, is whether there are any proofs for the heliocentric model of the Solar System. A literal naked-eye observation of the Earth revolving around the Sun would be rather difficult, since human beings have never gone to another planet yet, and have only been to the Moon briefly, decades ago. Here are several proofs; some of them are historically relevant also. Kepler's laws of planetary motion This became one of the earliest proofs, as soon as Newton figured out the law of universal gravitation, and the "fluxions" (what we would call today differential equations). When you assume a heliocentric model, and the inverse square law for gravity, then Kepler's laws in a heliocentric model come out of the equations naturally, as soon as you do the math. This is like saying: "if it's heliocentric, and knowing that the law of gravity is correct, then Kepler's laws should be such-and-such". And then: "oh, but the theoretical calculations for Kepler's laws match the actual observations with great precision. Therefore, our hypothesis (heliocentric, inverse square law) must be correct." It was the earliest strong indication that the heliocentric model is in natural accord with the basic laws of science, whereas the geocentric view was becoming more and more contrived as evidence accumulated. Tycho Brahe in late 1500s provided the enormous mass of observations of planetary motion. Johannes Kepler in early 1600s used Brahe's observations to come up with his laws empyrically (and also arguing for the heliocentric model). Isaac Newton in late 1600s said "yes, Kepler is right, because of mathematics and the law of gravity, and here's the proof from calculus". http://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion Stellar parallax An early argument against heliocentrism was that, if the Earth was really revolving around the Sun, then very distant objects, such as the stars, would appear to be ever-so-slightly bobbing back and forth around their average positions. Since that's not the case, it was argued, therefore the Earth must be fixed. You can see this argument in historical archives, proposed by theologians in the late 1600s, in favor of the geocentric model, and against the growing consensus then of "natural philosophers" (what we would call today scientists) that the heliocentric model was correct. In reality, parallax does exist, it's just very small. It was measured experimentally in the 1800s, and was then quickly used to determine, for the first time, the distance to the nearest stars. http://en.wikipedia.org/wiki/Stellar_parallax Aberration of starlight The direction where we see a distant star also changes when Earth's speed vector changes during its revolution around the Sun. This is different from parallax; it's more akin to the way raindrops on the side windows of a car leave diagonal traces when the car starts moving (even though the raindrops fall vertically). It's essentially a relativistic phenomenon (when applied to light), but it can be partially explained in a classic framework. It was actually observed before parallax in the late 1600s (the heyday of Newton), but it went unexplained until early 1700s. http://en.wikipedia.org/wiki/Aberration_of_light Orbital mechanics of interplanetary probes Landing a probe on Mars or Venus would simply not work if you assumed a geocentric model. A geocentric description of the Solar System might remain valid in a purely kinematic perspective (just the geometry of motion) as long as you remain on Earth. But the illusion breaks down quickly as soon as you start to consider dynamics (see Kepler's laws), and/or when you try to actually leave Earth (space probes). Let me reinforce this point, since several answers and comments got it wrong: the geocentric and the heliocentric models are not completely interchangeable, or a matter of relativity. You could build an "explanatory" geocentric model, and it would be "correct", purely in a kinematic fashion (the geometry of motion), and only as seen from Earth. But the model breaks down as soon as you consider dynamics (forces and masses); it would also reveal itself as incorrect even from a kinematic perspective as soon as you leave Earth. This is not just an artifice to simplify calculations. The dynamic calculations are wrong in a geocentric model. In order to compute the very high precision trajectory of the space vehicle carrying the Curiosity rover and successfully place it on Mars, you must operate from a heliocentric perspective. The dynamics are all wrong otherwise. You would not miss the target only a little, in a geocentric approach, you would miss it by a lot - the vehicle would not even go in the general direction of Mars. Miscellaneous When observed in a telescope, Venus has phases like the Moon, and also grows and then shrinks in size, synchronized with its phases (it's large as a thin crescent, it's small when it's gibbous). In a geocentric model, the size changes could be explained by an elliptic orbit of Venus around Earth, but the phase changes synchronized with that are harder to explain. Both phenomena become trivial to explain in a heliocentric model. It should be noted that this does not necessarily prove the heliocentric model, just the fact that Venus is orbiting the Sun , not the Earth . So it's an argument against pure (or strict) geocentrism. Jupiter, when observed in a telescope, clearly has its own satellites. This was an early blow against a strict geocentric model, which assumed that everything must orbit the Earth. It opened the door to the idea that orbits could be centered on other celestial bodies too, and to the idea that things in orbit around larger objects could have their own smaller satellites (and therefore the Earth could orbit the Sun without losing the Moon). The list could continue (and the full list is very long) but these arguments should suffice. You don't necessarily have to see something with your own eyes in order to know with certainty that it's there. In the case of Earth's revolution around the Sun, it was simply a matter of an overwhelming amount of evidence piling up in favor of it. Geocentrism simply doesn't make any sense whatsoever in modern science and space exploration. | {
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3,601 | Interstellar exchange of massive objects is difficult across several light years. But as the stars orbit the galaxy the distances between them change. I don't find data for neighbor star distances which covers more than a couple of a hundred thousand years. I can imagine that it is difficult to trace their orbits further given todays uncertainties in distance measurements. But I'm interested in a purely statistical/geometrical estimation of how frequent nearby Sun-star passages have been. How many times has a star come closer than, for example, 1 light year from the Sun since it formed? How well mixed have the nearest stars become over time? What fraction of them have followed the Sun since formation and made multiple passages? I suppose that 1 light year's distance is enough to cause disturbances in the Oort cloud and likely encourage comets. | TL;DR There is an encounter within 1 light year about once every half a million years. So about 9000 close encounters during the lifetime of the Sun The answer to the second part of your question is that the stars near the Sun are extremely well mixed, with a range of ages from a few million years to 12 billion years. None of them are identifiably "following the Sun". Some details Using the re-reduction of the Hipparcos astrometry, Bailer-Jones (2014) has integrated orbits for 50,000 stars to look for objects that might come or might have come close to the Sun. The K-dwarf Hip 85605 is the winner on that timescale, with a "90% probability of coming between 0.04 and 0.20pc between 240,000 and 470,000 years from now". The next best is GL710 a K-dwarf that will come within about 0.1-0.44pc in 1.3 million years. On a statistical basis, some work has been done by Garcia Sanchez et al. (2001) . They estimate, using the Hipparcos data, that encounters within 1pc occur every 2.3 million years. However, the Hipparcos data isn't complete for low-mass, faint objects. Making a correction for this, the authors estimate a $<1pc$ encounter every 100,000 years. The probability for closer approaches scales as the inverse of the square of the separation. i.e. the timescales for approaches closer than 0.1pc is 100 times longer. There are plenty of big error bars in all this work - recall that an error of 1km/s in velocity leads to a 1pc error in position after 1 million years. All this should be nailed down really well by Gaia results in the next 2-3 years. Update There is a new paper based on Gaia DR3 astrometry by Bailer-Jones (2022) . The star GL710 is now much more firmly constrained, coming within 0.060-0.068 pc of the Sun in 1.26-1.33 million years. A new contender on the scene is HD7977, a G3 dwarf that may have come even closer at 0.02-0.12 pc about 2.72-2.80 million years ago . There are a total of about 50 stars that will or have passed within 1 pc of the Sun in the last $\pm 6$ million years - so about one known encounter every 240,000 years. But note that this cannot possibly be a complete list and the figure is a lower limit even over this period, because faint stars and high velocity stars can move in and out of Gaia's sensitivity range on these sorts of timescales and thus would not be in the current catalogues. Bailer-Jones (2018) (using Gaia DR2 data) attempted to correct for this incompleteness and estimated an encounter rate within 5 pc of $545 \pm 59$ per million years. With the inverse-square scaling mentioned above (and also advocated by Bailer-Jones), this becomes about 1 encounter within 1 pc every 50,000 years or about 1 encounter within 1 light year every 0.5 million years. Assuming this rate is roughly correct over the whole lifetime of the Sun then this would yield 9000 close encounters over its 4.5 billion year lifetime (arguably there might have been a slightly smaller number of stars earlier in the Galaxy's history, but this might be outweighed by the Sun probably gradually migrating outwards from a radial position with a higher stellar density.) Projecting this further, then if you are prepared to wait 50 million years, then you might expect an encounter within 0.1 light years, which is in the Oort cloud and really starting to get interesting as far as the timescales of mass-extinction events go. | {
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4,800 | Is it possible that light itself is dark matter? I am speaking of photons (e.g. visible light, infrared, ultraviolet, etc...). I realize light is understood to be massless, but it is obvious it at least contains energy because we can see with it (e.g. it energizes the cells in our retinas). I wonder if light has a very tiny "net mass" (e.g. 0 mass * relativistically infinite speed). I would think that light at least has a little mass, in proportion to its energy. For example, take E=mc^2, then m = E/c^2 would describe how much mass it has. If this is true, light should have a very little gravity too. Although the effect would seem minimal, light is practically everywhere. Gravity from light would be more concentrated inside galaxies, and even more concentrated in the center of galaxies where there are many stars (like dark matter is). It would be interesting to run the calculations, assuming light does have gravity, and see if this matches the gravitational observations of dark matter in the universe. It would be funny and ironic if dark matter really is light. Edit: Note that it appears light does have gravity as per the discussions here: How does light affect the universe? If that much is true, I wonder if this is significant enough to account for dark matter? | Dark matter, is just a name for something we know nothing of. It was named to account for an extra gravity source for which there have been indirect observations, but yet we cannot explain. The force of gravity exerted by light is negligibly small yet we have measured the gravitational pull of Dark Matter to be big enough to affect whole galaxies; it is what binds galaxies together. Furthermore, we have included everything we can observe (all ordinary matter including photons) when we do the calculations for the amount of gravity there should be. So light is already there. 'Dark matter' is that extra gravity which we cannot account for. | {
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5,869 | When I was in college, I posed to my astronomy professor a thought experiment that had been puzzling my mind for some time: "If all the matter in the Sun magically disappeared instantly, how long would it take its gravity to stop having an effect on us?" His response was that the force of gravity is instant, unlike the speed of light, which appears instant. My big question is: "Do we know it's instant?" We can't possibly move an object large enough to have a noticeable gravitational influence fast enough to measure if it creates (or doesn't create) a doppler-like phenomenon. If he was wrong, how do we know it's not? | The first question as stated has a rather trivial answer: "If the sun magically disappeared, instantly, along with all its influences , how long would it take its gravity to stop having an effect on us?" Since the Sun's gravity is among its influences, it would instantly stop having an effect on us. That's just part of the magical situation, and doesn't even involve any physics. A bit more interesting is the question without the bolded part. In general relativity, changes in the gravitational field propagate at the speed of light. Thus, one might expect that the magical and instant disappearance of the Sun would not affect earth for about eight minutes, since that's how long light from the Sun takes to reach Earth. However, this is mistaken because the instant disappearance of the Sun itself violates general relativity, as the Einstein field equation enforces a kind of local conservation law on the stress-energy tensor analogous to the non-divergence of the magnetic field in the electromagnetism: in any small neighborhood of spacetime, there are no local sources or sinks of stress-energy; it must come from somewhere and go somewhere. Since the magical instant disappearance of the Sun violates general relativity, it does not make sense to use that theory to predict what happens in such a situation. Thus, the Sun's gravity instantly ceasing any effect on the Earth is just as consistent with general relativity as having any sort of time-delay. Or to be precise, it's no more inconsistent. My big question, now, is: "How do we know it's instant?" It's not instant, but it can appear that way. We can't possibly move an object large enough to have a noticeable gravitational influence fast enough to measure if it creates (or doesn't create) a doppler-like phenomenon. We don't have to: solar system dynamics are quite fast enough. An simple calculation due to Laplace in the early nineteenth century concluded that if gravity aberrated, Earth's orbit would crash into the Sun on the time-scale of about four centuries. Thus gravity does not aberrate appreciably--more careful analyses concluded that in the Newtonian framework, the speed of gravity must be more than $2\times10^{10}$ the speed of light to be consistent with the observed lack of aberration. This may seem quite a bit puzzling with how it fits with general relativity's claim that changes in the gravitational field propagate at the speed of light, but it's actually not that peculiar. As an analogy, the electric field of a uniformly moving electric charge is directed toward the instantaneous position of the charge-- not where the charge used to be, as one might expect from a speed of light delay. This doesn't mean that electromagnetism propagates instantaneously--if you wiggle the charge, that information will be limited by $c$, as the electromagnetic field changes in response to your action. Instead, it's just something that's true for uniformly moving charges: the electric field "anticipates" where the change will be if no influence acts on it. If the charge velocity changes slowly enough, it will look like electromagnetism is instantaneous, even though it really isn't. Gravity does this even better: the gravitational field of a uniformly accelerating mass is toward its current position. Thus, gravity "anticipates" where the mass will be based on not just current velocity, but also acceleration. Thus, if conditions are such that the acceleration of gravitating bodies changes slowly (as is the case in the solar system), gravity will look instantaneous. But this is only approximately true if the acceleration changes slowly--it's just a very good approximation under the conditions of the solar system. After all, Newtonian gravity works well. A detailed analysis of this can be found in Steve Carlip's Aberration and the Speed of Gravity , Phys.Lett.A 267 :81-87 (2000) [arXiV: gr-qc/9909087 ]. If he was wrong, how do we know it's not? We have a lot of evidence for general relativity, but the best current evidence that gravitational radiation behaves as GTR says it does is Hulse-Taylor binary . However, there is no direct observation of gravitational radiation yet. The connection between the degree of apparent cancellation of velocity-dependent effects in both electromagnetism and gravity, including its connection with the dipole nature of EM radiation and quadrupole nature of gravitational radiation, can also be found in Carlip's paper. | {
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5,994 | I was reading this article and it says the following: Researchers measured the mass of the Milky Way and found that our galaxy is approximately half the weight of a neighbouring galaxy known as Andromeda which has a similar structure to our own. So I was thinking, if we consider a galaxy neighbor to us, then our galaxy should be having an endpoint and the neighboring galaxy should be having a start point. So how do we know which is the starting and ending point of a galaxy, and how do we calculate it? | Here's a rough sketch of the Milky Way and the Andromeda galaxy , * showing their approximate sizes and distance from each other to scale: What the picture (hopefully) illustrates is the incredibly vast gap of empty space — around 2.5 million light years, to be exact — between the galaxies, each of which has a diameter of only(!) around 100 thousand light years or so. While both galaxies are shaped pretty much like "fuzzy whirlpools" of dust and gas (and, apparently, dark matter), with no precisely defined sharp outer edge, they are still pretty unambiguously separated. While we may not be able to point to a specific line in space and say that "the Milky Way ends right here", it's still clear that the Milky Way is over here , while the Andromeda galaxy is over there , and there's a huge gap of pretty much absolutely nothing at all in between. (Of course, there is some stuff even in intergalactic space , including some very diffuse gas, a little bit of dust and even the occasional stray star . Still, compared to the galaxies themselves — which, from a human viewpoint, are already pretty full of empty space — the intergalactic medium can be pretty well described as empty.) *) The little specks in the picture around each major galaxy are meant to represent their smaller satellite galaxies, such as the lesser and greater Magellanic clouds . Their relative positions and distances, alas, are probably not very accurate. | {
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6,519 | The Sun is roughly 4 light-years away from the closest star system, the Alpha Centauri system. The planets in our Solar System, however, aren't even close to that far away from the Sun. Where does our Solar System end? Is the edge considered to be the orbit of Neptune, the Kuiper Belt, the Oort Cloud, or something else? Note: this question on Physics SE is similar, but the answers posted here go in different directions. | According to the Case Western Reserve University webpage The Edge of the Solar System (2006) an important consideration is that The whole concept of an "edge" is somewhat inaccurate as far as the solar system is concerned, for there is no physical boundary to it - there is no wall past which there's a sign that says, "Solar System Ends Here." There are, however, specific regions of space that include outlying members of our solar system, and a region beyond-which the Sun can no longer hold any influence. The last part of that definition appears to be a viable definition of the edge of the solar system. Specifically, valid boundary region for the "edge" of the solar system is the heliopause. This is the region of space where the sun's solar wind meets that of other stars. It is a fluctuating boundary that is estimated to be approximately 17.6 billion miles (120 A.U.) away. Note that this is within the Oort Cloud. Though the article above is a bit dated, the notion of the heliopause has been still of interest to scientists, particularly how far away it is - hence, the interest in the continuing Voyager missions , which states on the website, that it has 3 phases: Termination Shock Passage through the termination shock ended the termination shock phase and began the heliosheath exploration phase. Voyager 1 crossed the termination shock at 94 AU in December 2004 and Voyager 2 crossed at 84 AU in August 2007. (AU = Astronomical Unit = mean Earth-sun distance = 150,000,000km) Heliosheath the spacecraft has been operating in the heliosheath environment which is still dominated by the Sun's magnetic field and particles contained in the solar wind. As of September 2013, Voyager 1 was at a distance of 18.7 Billion Kilometers (125.3 AU) from the sun and Voyager 2 at a distance of 15.3 Billion kilometers (102.6 AU). A very important thing to note from the Voyager page is that The thickness of the heliosheath is uncertain and could be tens of AU thick taking several years to traverse. Interstellar space, which NASA's Voyager page has defined as Passage through the heliopause begins the interstellar exploration phase with the spacecraft operating in an interstellar wind dominated environment. The Voyager mission page provide the follow diagram of the parameters listed above It is a bit complicated as we do not know the full extent of what the dynamics are like out there, a recent observation reported in the article A big surprise from the edge of the Solar System , reveal that the edge may be blurred by a strange realm of frothy magnetic bubbles, Which is suggested in the article could be a mixing of solar and interstellar winds and magnetic fields, stating: On one hand, the bubbles would seem to be a very porous shield, allowing many cosmic rays through the gaps. On the other hand, cosmic rays could get trapped inside the bubbles, which would make the froth a very good shield indeed. | {
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7,716 | Do all stars have an Oort cloud like ours that will be filled with comets and other objects? If not, why are they not around every star? | Awesome question, especially since we know so little of the answer. Nobody knows for sure how the Oort Cloud formed - I'll put that out there right now - but the current hypothesis is that it was originally part of the Sun's protoplanetary disk . All of the ice and rock coalesced into small bodies - proto-comets, if you will. While these bodies were much closer in to the Sun than they are today, they were tossed far out by gravitational interactions with the gas giants. Other interstellar comets could also have been captured by the Sun, adding to the population. So why is the Oort Cloud spherical? After all, the protoplanetary disk was just a flat disk. Why were the orbits of the objects perturbed? Well, the Oort Cloud objects are only loosely bound to the Sun - relatively, that is. They can be influenced by passing stars or other objects. It appears that galactic-scale tidal forces, combined with the influence of passing stars, molded the Cloud into its current spherical shape. So what does this all tell us? Well, we know other stars have protoplanetary disks , right? Some also have exoplanets - gas giants like Jupiter. They are also subject to tidal forces and the passing of nearby stars. So, theoretically, there's no reason why other stars shouldn't have Oort Clouds. So can we find them? The answer is, most likely, no. Here's why. According to Wikipedia , The outer Oort cloud may have trillions of objects larger than 1 km (0.62 mi), and billions with solar system absolute magnitudes brighter than 11 An absolute magnitude of a solar system object of 11 is very dim. Now, the object's apparent magnitude is how it would look from a given distance; the absolute magnitude is how it looks from a distance of 1 AU (in the case of Solar System objects, this quantity is denoted $H$ ). Oort-cloud objects are 2,000 - 50,000 (or more) AU away-- so these objects, to us in the same solar system, have an apparent magnitude much fainter than 11. The point of that poorly-explained interlude is that these objects are faint. Very faint. And objects in Oort Clouds around other stars would appear even fainter. Using the distance modulus , we can calculate the apparent magnitude of an object if the distance to that object and its absolute magnitude are known: $$m-M=5(\log_{10}d-1)$$ (from here ) where $m$ is apparent magnitude, $M$ is a scaling of $H$ normally used for stars, and $d$ is the distance in AU. Given an Oort Cloud object $x$ light-years away, you can figure out how bright (or dim) it would appear, given that 1 light-year is 63241 AU. Try this with the distances of nearby stars, and you'll realize how dim objects in these stars' Oort Clouds would be. As a final note: We don't know for sure if other Oort Clouds exist. From what I've been able to find, we don't have sufficiently powerful telescopes to observe these hypothetical Clouds, and so we don't (and may never) know if they exist. I hope this helps. This paper was instrumental in this answer. Start at page 38 for the relevant information. This page , too, has some good information. As I found from a link from an answer to this question on Physics, we've found Kuiper-Belt-like disks around other stars. This means it is certainly plausible for these stars to have Oort Clouds, too. And exocomets have been detected, which is another good sign. | {
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7,807 | A naive question. When we look at Jupiter, we see that its features didn't change largely over many years, for instance, the red-spot . If it is composed of gases and liquids, then why aren't the effects of mixing of these fluids visible? My intuition is that due to very low temperatures ($-145\, ^{\circ}$C), diffusion of fluids doesn't occur and therefore the superficial appearence of Jupiter remains the same. | Believe it or not, Jupiter isn't too consistent. Take a look at these pictures, the first taken in 2009 and the second taken in 2010: and Quite the difference, eh? Why? Jupiter's atmosphere is made of zones and belts . Zones are colder and are composed of rising gases; they are dark-colored. Belts are warmer and are composed of falling gases; they are light-colored. The reason the two don't intermix is because of constant flows of wind, similar to the Jetstream . These winds make it hard for bands to mix. There are two types of explanations for the jets. Shallow models say that the jets are caused by local disturbances. Deep models say that they are the byproduct of rotating cylinders comprising the mantle. At the moment, we don't know which explanation is correct. | {
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7,980 | The Great Red Spot is a persistent anticyclonic storm, 22° south of Jupiter's equator. Why is it reddish? From Wikipedia : It is not known exactly what causes the Great Red Spot's reddish color. Are there updated data? | It's red because it's a 'sunburn'. The clouds in the red spot reach to higher altitudes than the surrounding ones and are more exposed to solar UV radiation, which in turn changes the structure of some of the organic molecules etc. This is at least the explanation suggested by recent data from NASA's Cassini mission, see this 5-day-old press release . | {
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8,001 | I've read that the Moon is moving away from the Earth by 1-3cm per year. Is this enough to make the Earth lose the Moon before the Sun goes supernova? I'm asking because I would like to do the calculations for Earth's magnetic pull on this subject. It appears to me that it should be a non-linear function (there are also other factors, like if the Moon could be captured by another planet or having asteroids messing with its orbit on Earth). | As HDE 226868 noted in his answer, the Sun is not going to go supernova. That's something only large stars experience at the end of their main sequence life. Our Sun is a dwarf star. It's not big enough to do that. It will instead expand to be a red giant when it burns out the hydrogen at the very core of the Sun. It will continue burning hydrogen as a red giant, but in a shell around a sphere of waste helium. The Sun will start burning helium when it reaches the tip of the red giant phase. At that point it will shrink a bit; a slight reprieve. It will expand to a red giant once again on the asymptotic red giant branch when it burns all the helium at the very core. It will then burn helium in a shell surrounding a sphere of waste carbon and oxygen. Larger stars proceed beyond helium burning. Our Sun is too small. Helium burning is where things stop. The Sun has two chances as a red giant to consume the Earth. Some scientists say the Sun will consume the Earth, others that it won't. It's all a bit academic because the Earth will be dead long, long before the Sun turns into a red giant. I'll have more to say on this in the third part of my answer. The current lunar recession rate is 3.82 cm/year, which is outside your one to three centimeters per year window. This rate is anomalously high. In fact, it is extremely high considering that dynamics says that
$$\frac {da}{dt} = (\text{some boring constant})\frac k Q \frac 1 {a^{11/2}}$$
Here, $a$ is the semi major axis length of the Moon's orbit, $k$ is the Earth-Moon tidal Love number, and $Q$ is the tidal dissipation quality factor. Qualitatively, a higher Love number means higher tides, and a higher quality factor means less tidal friction. That inverse $a^{5.5}$ factor indicates something seriously funky must be happening to make the tidal recession rate so very high right now, and this is exactly the case. There are two huge north-south barriers to the flow of the tides right now, the Americas in the western hemisphere and Afro-Eurasia in the eastern hemisphere. This alone increases $k/Q$ by a considerable amount. The oceans are also nicely shaped so as to cause some nice resonances that increase $k/Q$ even further. If something even funkier happens and the Moon recedes at any average rate of four centimeters per year over the next billion years, the Moon will be at a distance of 425,000 km from the Earth (center to center). That's less than 1/3 of the Earth's Hill sphere. Nearly circular prograde orbits at 1/3 or less of the Hill sphere radius should be stable. Even with that over-the-top recession rate, the Moon will not escape in the next billion years. What about after a billion years? I chose a billion years because that's about when the Moon's recession should more or less come to a standstill. If the Earth hasn't already died before this billion year mark, this is when the Earth dies. Dwarf stars such as our Sun get progressively more luminous throughout their life on the main sequence. The Sun will be about 10% more luminous than it is now a billion years into the future. That should be enough to trigger a moist greenhouse, which in turn will trigger a runaway greenhouse. The Earth will become Venus II. All of the Earth's oceans will evaporate. Water vapor will reach well up into what is now the stratosphere. Ultraviolet radiation will photodissociate that water vapor into hydrogen and oxygen. The hydrogen will escape. Eventually the Earth will not only be bare of liquid water on the surface, it will be bare of water vapor in the atmosphere. Almost all of the Moon's recession is a consequence of ocean tides. Without oceans, that tunar recession will more or less come to a standstill. | {
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8,004 | What can be said about 67P/Churyumov-Gerasimenko axial tilt? Is it fixed (if so, at what value), only changing with precession (at what rate?) or does 67P "tumble" with axial tilt changing over time? (and if so, at what rate?) I'm interested in estimating the 'polar night' area of the comet, and how it changes over time - but without knowing the rotation properties this is quite hard to do. | As HDE 226868 noted in his answer, the Sun is not going to go supernova. That's something only large stars experience at the end of their main sequence life. Our Sun is a dwarf star. It's not big enough to do that. It will instead expand to be a red giant when it burns out the hydrogen at the very core of the Sun. It will continue burning hydrogen as a red giant, but in a shell around a sphere of waste helium. The Sun will start burning helium when it reaches the tip of the red giant phase. At that point it will shrink a bit; a slight reprieve. It will expand to a red giant once again on the asymptotic red giant branch when it burns all the helium at the very core. It will then burn helium in a shell surrounding a sphere of waste carbon and oxygen. Larger stars proceed beyond helium burning. Our Sun is too small. Helium burning is where things stop. The Sun has two chances as a red giant to consume the Earth. Some scientists say the Sun will consume the Earth, others that it won't. It's all a bit academic because the Earth will be dead long, long before the Sun turns into a red giant. I'll have more to say on this in the third part of my answer. The current lunar recession rate is 3.82 cm/year, which is outside your one to three centimeters per year window. This rate is anomalously high. In fact, it is extremely high considering that dynamics says that
$$\frac {da}{dt} = (\text{some boring constant})\frac k Q \frac 1 {a^{11/2}}$$
Here, $a$ is the semi major axis length of the Moon's orbit, $k$ is the Earth-Moon tidal Love number, and $Q$ is the tidal dissipation quality factor. Qualitatively, a higher Love number means higher tides, and a higher quality factor means less tidal friction. That inverse $a^{5.5}$ factor indicates something seriously funky must be happening to make the tidal recession rate so very high right now, and this is exactly the case. There are two huge north-south barriers to the flow of the tides right now, the Americas in the western hemisphere and Afro-Eurasia in the eastern hemisphere. This alone increases $k/Q$ by a considerable amount. The oceans are also nicely shaped so as to cause some nice resonances that increase $k/Q$ even further. If something even funkier happens and the Moon recedes at any average rate of four centimeters per year over the next billion years, the Moon will be at a distance of 425,000 km from the Earth (center to center). That's less than 1/3 of the Earth's Hill sphere. Nearly circular prograde orbits at 1/3 or less of the Hill sphere radius should be stable. Even with that over-the-top recession rate, the Moon will not escape in the next billion years. What about after a billion years? I chose a billion years because that's about when the Moon's recession should more or less come to a standstill. If the Earth hasn't already died before this billion year mark, this is when the Earth dies. Dwarf stars such as our Sun get progressively more luminous throughout their life on the main sequence. The Sun will be about 10% more luminous than it is now a billion years into the future. That should be enough to trigger a moist greenhouse, which in turn will trigger a runaway greenhouse. The Earth will become Venus II. All of the Earth's oceans will evaporate. Water vapor will reach well up into what is now the stratosphere. Ultraviolet radiation will photodissociate that water vapor into hydrogen and oxygen. The hydrogen will escape. Eventually the Earth will not only be bare of liquid water on the surface, it will be bare of water vapor in the atmosphere. Almost all of the Moon's recession is a consequence of ocean tides. Without oceans, that tunar recession will more or less come to a standstill. | {
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8,100 | I just want to be sure I am visualizing this correctly, because it seems odd. The Moon is tidally locked to the Earth but there are wobbles to its motion due to libration. So from a point on the surface of the near side of the Moon, the Earth would always be near the same place in the sky? It would describe a small circle or a side-to-side wobble over the course of a month, but never move far from that point? That would seem very strange, like it was a gigantic stage prop or something. We are so conditioned that everything rises and sets (except for a few stars near the poles). | Still new at stellarium but here are some quick capture gif lasting one month . Sorry about the quality- limited to 256 colors for smaller gifs. Date on lower left corner. By the way the sun is of course the brightest and i use it as reference for recording (start record when sun is in frame then stop when it appears again in the same position which is roughly one month) Location on Moon : Sea of Tranquility You are looking straight up Yellow lines are azimuth second picture: Zoomed view of earth One whole day 24 hours (give or take a few minutes) Location on Moon: Sea of Tranquility You are looking straight up Yellow lines are azimuth (gif itself rotated to approximately match the orientation of the 260 degree azimuth line in the first picture) Answer for Emilio Pisanty comment on whether the oscillation is detectable by eye stellarium Field of view 60 degrees (default view when stellarium first opened) Picture 3 Shows one end of the oscillation near the 200 azimuth line (see yellow arrows) other end of ellipse near 185 azimuth line you can use Gemini (lower right corner) for reference (see yellow arrows) from head to crotch of one Gemini twin would be a good reference for how wide the oscillation is (not sure if your sundial can detect the difference) picture 4 shows other end of the oscilation near the 185 azimuth line (see yellow arrows) picture 5 shows my stellarium location settings on the moon | {
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8,300 | I was just wondering what are the chances that there is a small object (say less than 1 km but more than few meters) that orbits the Earth but has remained undetected by us? Are we actually scanning the space around the Earth continuously for orbiting bodies? | Not strictly satellites/moons, but certainly companions are 2010 TK 7 with a diameter of ~300 m, an Earth trojan at the L4 point, and the ~5 km 3753 Cruithne in a peculiar orbit locked to the Earth's. | {
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8,460 | Does every star become a black hole? Is there any probability that our sun can become a black hole? If yes then is it on its way to become a black hole? what is the current state of sun as per the black hole life cycle? What will be the effect on all the planetary objects in the solar system if the sun turns into a black hole. Sorry for so many questions but i cannot miss them as these are some questions on my mind. | No, the sun won't ever become a black hole. The choice between the three fates of stars (white dwarf, neutron star, black hole) is entirely determined by the star's mass. A star on the main sequence (like most stars, including our sun) is constantly in a balance between the inward pressure of gravity and the outward pressure of the energy generated by the hydrogen fusion that makes it "burn". 1 This balance stays relatively stable until the star runs out of whatever its current fuel is - at that point, it stops burning, which means there's no longer outward pressure, which means it starts collapsing. Depending on how much mass there is, it might get hot enough as it collapses to start fusing helium together. (If it's really massive, it might continue on to burn carbon, neon, oxygen, silicon, and finally iron, which can't be usefully fused.) Regardless of what its final fuel is, eventually the star will reach a point where the collapse from gravity is insufficient to start burning the next fuel in line. This is when the star "dies". White dwarfs If the star's remains 2 mass less than 1.44 solar masses (the Chandrasekhar limit 3 ), eventually gravity will collapse the star to the point where each atom is pushed right up against the next. They can't collapse further, because the electrons can't overlap. While white dwarfs do shed light, they do so because they are extremely hot and slowly cooling off, not because they're generating new energy. Theoretically, a white dwarf will eventually dim until it becomes a black dwarf, although the universe isn't old enough for this to have happened yet. Neutron stars If the collapsing star is above the Chandraskhar limit, gravity is so strong that it can overcome the "electrons can't overlap" restriction. At that point, all the electrons in the star will be pushed into combining with protons to form neutrons. Eventually, the entire star will composed primarily of neutrons pushed right up next to each other. The neutrons can't be pushed into occupying the same space, so the star eventually settles into being a single ball of pure neutrons. Black holes Black holes are the step beyond neutron stars, although they're worth discussing in a bit more detail. Everything, in theory, has a Schwarzschild radius . That's the radius where a ball of that mass would be so dense that light can't escape. For example, the Schwarzschild radius for Earth is about 9mm. However, for all masses smaller than somewhere between 2-3 times the mass of the sun, it's impossible to squeeze the matter small enough to get it inside that radius. Even a neutron star isn't massive enough. But a star that becomes a black hole is. We don't actually know what happens to a star once it's become a black hole - the edges of the "hole" itself is simply the Schwarzschild radius - the point light can't escape. From outside, it doesn't matter whether the matter collapsed to the point that the neutrons started overlapping, whether it stopped just inside the radius, or whether it continued collapsing until it broke all known physical laws. The edges are still the same, because they're just a cutoff based on the escape velocity. 1 I'm ignoring the red giant phase here, since it's just a delay in the "run out of fuel" step. Basically, the core is helium "ash", while the hydrogen fusion process takes place further and further out. Once that runs out, you get a nova and the collapse continues. 2 Likewise, I'm ignoring the mass that stars shed in their various nova phases. All given masses are based on the remnants left behind. 3 Every source I've found for Chandrasekhar mass, except Wikipedia, gives 1.44 or 1.4 solar masses (which are compatible). Wikipedia gives 1.39, and gives at least one source to back that number. | {
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8,483 | This, of course, is a newbie question, as I am nothing more than a hobbyist. But I was quite surprised to see the recently-released "before and after" pictures of the so-called "Pillars of Creation" in the Eagle Nebula ( here and here ). For something that looks like such a "nebulous" cloud of gas (pardon the pun), I would have expected at least some visible sign of change in the 25 years between the two photos. But at the pixel resolution which I had available, I could not detect even the slightest difference between them. Of course, I am accustomed to terrestrial clouds which are in constant motion, so I (mistakenly?) expected something analogous at the astronomical level. Can someone provide, in layman's terms, how my understanding of this cosmological feature is deficient? In other words, how might I adjust my instinctive intuition that this object should be more dynamic? Is it just the sheer scale that I am not comprehending? (By the way: what is the distanse between the three spires, as compared to, say, our solar system?) | It appears to be static because it's huge beyond your imagination. The distance to the nebula is 7,000 light years. Its apparent size is 7 arc minutes. Therefore its linear size is about 14 light years. Think about that. The whole nebula is so big, it takes light 14 years to cross it. Any motion therein must necessarily be much, much slower. No wonder you're not seeing much change. Data source: wikipedia Calculation using Wolfram Alpha | {
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8,788 | I have a 3inch Newtonian reflector telescope with 300 mm focal length. I can use highest magnification of 75x using a 4mm eyepiece. But in 75x I can't see the details of Jupiter what was expected. Instead I see a little blurry image. Now I would like to know how much magnification is necessary to see a good details of Jupiter and other planets. And one more question: Is there any way to improve the vision of my 3 inch telescope? | You're probably asking the wrong question - which I am going to answer anyway, and after that I am going to answer the question you should have asked instead. As a general rule, there isn't much point in pushing the magnification above 2x the diameter of the instrument, measured in mm. 3 inch, that's 75mm, that's 150x max. Beyond that limit, even under ideal skies the image is large but blurry. After that, seeing (or air turbulence) pushes that limit further down. Your aperture is small enough that it almost never suffers from seeing, but larger instruments are often affected. It varies greatly with time, place and season. There are times when a 12" dobsonian, that in theory could do 600x, is clamped down by seeing to 150 ... 180x. There are times when you could take a 20" dobsonian all the way up to 1000x - but that's very, VERY rare, it's the stuff of legends. Assuming average seeing conditions and instruments of usual size (refractors of 3...4" aperture, reflectors 6" or larger), here are some rules of thumb: Jupiter is seen best under mid-high magnification. It's rare that more than 200x is beneficial. This is because it's a very low contrast object, and additional magnification comes at the cost of less contrast, which makes things worse. Saturn works best at high-ish magnification, bit more than Jupiter but maybe not much more. Around 200 ... 250x usually works. It depends on what you do - if you're trying to see the ring divisions, push it a bit higher. Mars can use the highest magnification that you could generate, given the instrument and the conditions. It's a very small object, contrast is not bad, so crank it all the way up. Most instruments are limited by seeing when observing Mars. Moon is the same as Mars. As you can see, magnification is never an issue for you. More magnification will not make it better. In fact, more magnification always means the image is more blurry, not more crisp - it's always a compromise between size and blurriness that decides the optimal magnification. Don't worry, everyone begins thinking that more is always better. Soon enough, experience shows them what's really going on. That being said, I believe it's not magnification that's giving you trouble, but the general condition of the optical stack that you're using. These are things that are extremely important, and yet are ignored by many, many amateurs - and the results are not optimal. Here are a few things that you should investigate: Collimation Is your scope collimated? In other words, are all optical elements aligned on the same axis? The likely answer is no. It makes a huge difference in the scope's performance, especially for planets. Here's a collimated scope, compared to the same scope out of collimation: Further information on Thierry Legault's site , which is extremely informative. A series of articles and documents regarding collimation: http://www.cloudynights.com/documents/primer.pdf Gary Seronik: A Beginner’s Guide to Collimation Gary Seronik: Collimation Tools: What You Need and What You Don’t Gary Seronik: No-Tools Telescope Collimation Note: Some telescopes (e.g. pretty much all refractors) do not require collimation; they are collimated from factory and hold collimation pretty well. But most reflectors (SCTs, all newtonians including dobsonians, etc) do require this periodic maintenance. Thermal equilibrium At 3" aperture, this is probably not a big issue, but there's no reason why you should add another problem to the existing ones. Your scope should be at the same temperature as the air around it, otherwise its performance decreases. Take it outside 1 hour before you start observing, and that should be enough for you. Larger telescopes (around 10" ... 12" and larger) should use active ventilation for better cooling (a fan on the back of the mirror). More details here: Gary Seronik: Beat the Heat: Conquering Newtonian Reflector Thermals — Part 1 Gary Seronik: Beat the Heat: Conquering Newtonian Reflector Thermals — Part 2 In your case, simple passive cooling for 1 hour should be enough, but it's worth reading those articles. Focal ratio A 3" scope, at 300mm focal length, that's an f/4 instrument. That's a pretty steep f/ ratio. Most eyepieces will not do well with such a blunt cone of light, and will start to exhibit aberrations that blur the image. Only very expensive eyepieces work well at such low focal ratios - things like TeleVue Ethos, or Explore Scientific 82 degree eyepieces. Try and keep the planet in the center - most aberrations are lower there. Even very simple eyepieces do better in the middle of the image. Look at the stars. Are they tiny and round in the center, and large and fuzzy at the edge? Those are aberrations from various sources (eyepiece, primary mirror, etc). Coma Of course, at f/4 even the best eyepieces out there cannot do anything about coma - an aberration coming out of any parabolic mirror, which becomes pretty obvious around f/5, very obvious at f/4, and a major problem at f/3. Again, coma is zero in the center of the image, and increases towards the edge. A coma corrector is used in some cases, such as the TeleVue Paracorr, but I strongly recommend that you DO NOT use one - I suspect your instrument is aberrating in ways that overwhelm coma anyway. Jupiter would not be too blurry even at full f/4 coma at the edge. This paragraph is for informational purposes only. Coma should become a concern with large telescopes, using high quality optics, with a focal ratio of around f/5 and less. E.g., you have a 20" dob with an f/4 mirror, then you should worry about coma - provided that collimation and so on are taken care of. Optics quality An f/4 parabola is not super easy to make at any size. I've made my own optics, and the lower the f/ ratio, the more difficult the process is. Many small, cheap telescopes are made in a hurry, and the difficult focal ratio poses additional problems - as a result, many manufacturers do a poor job. There are even cases where the primary mirror is left spherical, with disastrous results. This is something you can do nothing about. If the primary mirror is bad, then that's just the way things are. An optician might try to correct it, but it's a difficult process, and quite expensive. I only added this here so you are informed. This is what I would do in your case: I would take the scope out 1 hour before observing, every time. I would try and learn how to collimate the scope. I would try to figure out a few simple collimation techniques, and a few simple tests. I would spend a few days / weeks practicing that. I would keep reading about collimation. When collimation is at least partially under control, I would learn how to properly focus the scope. Seems simple, but it can be tricky. Use a bright star, and try and make it as small as possible. Use the Moon when it's visible, and try and make it crisp and clear. Do not try this with a miscollimated scope, since it's pointless. After a few months, when I gain confidence that the scope is in better shape, very well collimated, very well focused, I might try to borrow a better eyepiece from a friend. I said borrow, not buy. Something like a 3 ... 4mm eyepiece, good quality, that would give me a comparison for the existing eyepieces. This ONLY makes sense with a scope that is in perfect collimation, perfect temperature, perfect focus. If an improvement is seen, then get a better eyepiece - but do not spend hundreds of dollars for an expensive eyepiece that will then be used in a tiny cheap scope. Second-hand eyepieces often work exactly as well as new ones. If you know someone in your area who makes mirrors, see if they agree to put your primary mirror on the Foucault tester, and assess its condition. But beware: the results might be very disappointing. Or not. You kind of never know with these little scopes. EDIT: After the scope is collimated and so on, you could try to increase magnification by using a 2x barlow with your eyepieces, but do not expect miracles - the image will be bigger, but probably rather "mushy". More magnification is not always better, there's always a trade-off. Good luck, and clear skies to you! | {
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8,849 | It is amazing that Kepler determined his three laws by looking at data, without a calculator and using only pen and paper. It is conceivable how he proved his laws described the data after he had already conjectured them, but what I do not understand is how he guessed them in the first place. I will focus in particular on Kepler's third law, which states that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of the orbit. I assume that Kepler was working with data about the planets only, plus our own moon, and the sun. I make this assumption because I don't think Kepler had data about other moons, comets, or asteroids, which had not been observed by telescope yet. If this is true, knowing that Neptune, Uranus, and Pluto were not yet discovered when Kepler was alive, this means Kepler had less than 9 data points to work with. My friend claims that it is totally concievably how Kepler guessed this relationship (although he provides no method of how Kepler might have done it), and also that Kepler's observations are "not that hard". As a challenge, I gave my friend a data table with one column labeled $x$, the other $y$, and 9 coordinates $(x,y)$ which fit the relationship $x^4=y^3$. I said "please find the relationship between $x$ and $y$", and as you might expect he failed to do so. Please explain to me how in the world did Kepler guess this relationship working with so few data points. And if my assumption that the number of data points Kepler had at his disposal is small, is wrong, then I still think its quite difficult to guess this relationship without a calculator. | Kepler's third law is trivial (in my opinion) compared to his first law. I am quite impressed that he was able to deduce that the orbits were ellipses. To get that, he had to go back and forth plotting Mars' direction from Earth and Earth's direction from Mars. He knew the length of both planets' years, so observations taken one Mars year apart would differ only because Earth had moved. But maybe not so trivial. He published his first two laws in 1609. The third law didn't come along until ten years later, in 1619. With ten years to work on it, even the most obscure relationship will eventually be found. To discover a ratio-of-powers relationship, plot the logarithms of the numbers. In your example with $x^4 = y^3$, the logs would plot on a straight line with a slope of $3/4$. The timing is right. Napier published his book on logarithms in 1614. Kepler may have on a whim applied this shiny new mathematical tool to his crusty old data. The major hurdle was that at the time there were only six known planets, so he didn't have an abundance of data points, and the ones he had were by no means precise. Kepler's other problem is that none of his laws made any sense to him. They fit the data, but he had no idea why. He didn't have Newton's laws of motion to work from, he had no understanding of force, momentum, angular momentum, and certainly not gravity. So far as he knew, the planets moved the way they did because God decreed it, and angels were tasked with pushing the planets along their orbits. The outer planets moved slower because they were being pushed by lesser angels. (Feynman makes the comment that we understand so much more now. We now know that the angels are on the outside pushing in toward the Sun.) | {
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9,990 | The Sun's rays hit our eyes around 8 minutes after they are emitted from the Sun.
Does this mean that the Sun that we see is always the Sun as it was some 8 minutes before? I strongly think this must be happening; is it really a fact? Do we always see the Sun's past? | Yes , you are right. We don't only see the Sun 8 minutes in the past, we actually see the past of everything in space. We even see our closest companion, the Moon, 1 second in the past. The further an object is from us the longer its light takes to reach us since the speed of light is finite and distance in space are really big. | {
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10,189 | It is often stated that the magnetosphere not only shields the planet from cosmic radiation, but also prevents atmospheric loss. Why then did Venus not lose most of its atmosphere if it doesn't have a strong magnetic field? Is there another mechanism at play, or is the statement about the importance of magnetosphere to atmospheric loss prevention wrong? | There is an interesting article on the magnetosphere of Venus on the ESA Science and Technology site. You can find the article here and it will probably answer your question. The article states, like you did, that some planets, like Earth, Mercury, Jupiter and Saturn, have magnetic fields internally induced by their iron core. These magnetic fields shield the atmosphere from particles coming from solar winds. It also confirms your statement that Venus lacks this intrinsic magnetosphere to shield its atmosphere from the solar winds. The interesting thing, however, is that spacecraft observations, like the ones made by ESA's Venus Express, have shown that the Venusian ionosphere's direct interaction with the solar winds causes an externally induced magnetic field, which deflects the particles from the solar winds and protects the atmosphere from being blown away from the planet. However, the article also explains that the Venusian magnetosphere is not as protective as Earth's magnetosphere. Measurements of the Venusian magnetic field show several similarities, such as deflection of the solar winds and the reconnections in the tail of the magnetosphere, causing plasma circulations in the magnetosphere. The differences might explain the fact that some gasses and water are lost from the Venus atmosphere. The magnetic field of Venus is about 10 times smaller than the earth's magnetic field. The shape of the magnetic field is also different. Earth has a more sharp magnetotail facing away from the Sun and Venus has a more comet-shaped magnetotail. During the reconnections most of the plasma is lost in the atmosphere. The article explains therefore that although Venus does not have an intrinsic magnetic field, the interaction of the thick atmosphere with the solar winds causes an externally induced magnetic field, that deflects the particles of the solar winds. The article suggests, however, that the different magnetic field may mean that lighter gasses are not as protected and therefore are lost into space. I hope this sufficiently answers the question. | {
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10,242 | Where there is fire, there is always smoke. So why there isn't any smoke near the Sun? | The sun isn't on fire. Fire is actually very rare in the solar system. It requires chemical potential energy, which happens on earth because of life. Photosynthesis uses solar energy to build things out of carbon, oxygen, nitrogen and hydrogen (and some other elements). It's these carbon chain structures that are ultimately flammable, and pretty much, only in an Oxygen atmosphere. Oxygen atmosphere is produced by the same photosynthesis. The sun's heat comes from nuclear fusion which doesn't make smoke. Also, the nuclear fusion only happens deep the inside of the sun, the outside is just a thick blanket of molecules - mostly hydrogen. The sun also isn't transparent. The light we see from the sun is like the light we see from a red hot piece of metal or lava. Red hot metal doesn't smoke (unless you drop water on it - which is actually steam.) The sun, like red hot metal or lava, glows bright because of temperature. In a certain sense, the tail of a comet is kind of like steam being made by the heat of the sun. | {
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11,305 | I wonder, how did our ancestors discover the Solar System? They did not have any telescopes to see distant objects, right? Even a planet looks like a star from a distance. They discovered the rotations of different planets without having much technology. | 1. Ancient cultures observed the sky Night skies are naturally dark and there was no light-pollution in ancient times. So if weather permits, you can easily see a lot of stars. No need to tell about the Sun and the Moon. Ancient people had good reasons to study the night skies. In many cultures and civilizations, stars (and also the Sun and the Moon) where perceived to have religious, legendary, premonitory or magical significance (astrology), so a lot of people were interested in them. It did not took long to someone (in reality a lot of different people independently in many parts of the world) to see some useful patterns in the stars that would be useful to navigation, localization, counting hours, counting days and relating days to seasons, etc. And of course, those patterns in the stars were also related to the Sun and the Moon. So, surely all ancient cultures had people who dedicated many nights of their lifes to study the stars in detail right from the stone age. They would also perceive meteorites (falling stars) and eclipses. And sometimes a very rare and spetacular comet. Then there are the planets Mercury, Venus, Mars, Jupiter and Saturn. They are quite easily to notice to be distinct from the stars because all the stars seems to be fixed in the celestial sphere, but the planets don't. They are very easily to notice to be wandering around in the sky with the passage of the days, specially for Venus, which is the brightest "star" in the sky and is also a formidable wanderer. Given all of that, the ancient people surely become very aware of those five planets. About Mercury, initially the Greeks thought that Mercury were two bodies, one that showed up only in the morning a few hours before the sunrise and another only a few hours after the sunset. However, soon they figured out that in fact it was only one body, because either one or the other (or neither) could be seen in a given day and the computed position of the unseen body always matched the position of the seen body. 2. The Earth seems to be round Now, out of the stone age, already into ancient times, navigators and merchants who travelled great distances perceived that the Sun rising and setting points could variate not only due to the seasonal variation, but also accordingly to the location. Also, the distance from the polar star to the horizon line also variates accordingly with the location. This fact denounces the existence of the concept nowadays known as latitude, and this was perceived by ancient astronomers in places like Greece, Egypt, Mesopotamia and China. Astronomers and people who dependend on astronomy (like navigators) would wonder why the distance from the polar star to the horizon varied, and one possibility was that it is because the Earth would be round. Also, registering different Sun angles in different locations of the world on a given same day and at a given same hour, also gives a hint that the Earth is round. The shadow on the Moon during a lunar eclipse also gives a hint that the Earth is round. However, this by itself is not a proof that the Earth is round, so most people would bet on some other simpler thing, or simply don't care about this phenomenon. Most cultures in ancient times presumed that the world was flat. However the idea of the world being round exists since the ancient Greece. Contrary to the popular modern misconception, in the Middle Ages, almost no educated person on the western world thought that the world was flat . About the Earth's size, by observing different Sun positions and shadows angles in different parts of the world, Erasthotenes in ancient Greece calculated the size of Earth and the distance between the Earth and the Sun correctly for the first time as back as the third century B.C. However, due to the confusion about all the different and inconsistent unit measures existent back then and the difficulty to precisely estimate long land and sea distances, confusion and imprecision persisted until the modern times. Ancient cultures also figured out that the shiny part of the Moon was illuminated by the Sun. Since the Full Moon is easily seen even at midnight, this implies that the Earth is not infinite. The fact that the Moon enters in a rounded shadow when exactly in the opposite side of the sky as the Sun also implies that it is the Earth's shadow on the Moon. This also implies that Earth is significantly larger than Moon. 3. Geocentrism So, people observed the Sun, Moon, Mercury, Venus, Mars, Jupiter, Saturn and the fixed sphere of stars all revolving around the sky. They naturally thought that the Earth would be the center of the universe and that all of those bodies revolved around the Earth. This culminated with the work of the phylosopher Claudius Ptolemaeus about geocentrism . Altough we now know that the ptolomaic geocentric model is fundamentally wrong, it could be used to compute the position of the planets, the Sun, the Moon and the celestial sphere of stars, with a somewhat acceptable precision at the time. It accounted to include the observation of planets velocity variations, retrograde motions and also for coupling Mercury and Venus to the Sun, so they would never go very far from it. Further, based on the velocity of the motion of those bodies in the sky, then the universe should be something like: Earth at the center. Moon orbiting the Earth. Mercury orbiting the Earth farther than the Moon. Venus orbiting the Earth farther than Mercury. Sun orbiting the Earth farther than Venus. Mars orbiting the Earth farther than Sun. Jupiter orbiting the Earth farther than Mars. Saturn orbiting the Earth farther than Jupiter. The celestial sphere of stars rotating around the Earth, being the outermost sphere. In fact, the ptolomaic model is a very complicated model, way more complicated than the copernic, keplerian and newtonian models. Particularly, this could be compared to softwares that are based on severely flawed concepts but that are still working due to a lot of complex, tangled and unexplainable hacks and kludges that are there just for the sake of making the thing work. 4. The discovery of the Americas Marco Polo , in the last years of the 1200's, was the first European to travel to China and back and leave a detailed chronicle of his experience. So, he could bring a lot of knowledge about what existed in the central Asia, the East of Asia, the Indies, China, Mongolia and even Japan to the Europeans. Before Marco Polo, very few was known to the Europeans about what existed there. This greatly inspired European cartographers, philosophers, politicians and navigators in the years to come. Portugal and Spain fough a centuries-long war against the invading Moors on the Iberian Peninsula . The Moors were finally expelled in 1492. The two states, were looking for something profitable after so many years of war. Since Portugal ended its part of the war first, it had a head start and went to explore the seas first. Both Portugal and Spain were trying to find a navigation route to reach the Indias and the China in order to trade highly profitable spices and silk. Those couldn't be traded by land efficiently anymore due to the fact that the lands on West Asia and North Africa were dominated by Muslim cultures unfriendly to Christian Europeans, a situation that were just made worse after the fall of Constantinople in 1453. Portugal, were colonizing the Atlantic borders of Africa and eventually they managed to reach the Cape of Good Hope in 1488 (with Bartolomeu Dias ). A Genovese navigator called Cristoforo Colombo believed that if he sailed west from the Europe, he could eventually reach the Indies from the east side. Inspired by Marco Polo and subestimating the size of Earth, he estimated that the distance between the Canary Islands and the Japan to be 3700 km (in fact it is 12500 km). Most navigators would not venture in such voyage because they (rightly) tought that Earth was larger than that. Colombo tried to convice the king of Portugal to finance his journey in 1485, but after submitting the proposal to experts, the king rejected it because the estimated journey distance was too low. Spain, however, after finally expelling the Moors in 1492, were convinced by him. Colombo's idea was far-fetched, but, after centuries of wars with the Muslims, if that worked, then Spain could profit quickly. So, the Spanish king approved the idea. And just a few months after expelling the Moors, Spain sent Colombo to sail west towards the Atlantic and then he reach the Hispaniola island in Central America. After coming back, the news about the discovery of lands in the other side of the Atlantic spread quickly. Portugal and Spain then divided the world by the Treaty of Tordesillas in 1494. In 1497, Amerigo Vespucci reached the mainland America. Portugal would not be left behind, they managed to navigate around Africa to reach the Indies in 1498 (with Vasco da Gama ). And they sent Pedro Álvares Cabral , who reached the Brazil in 1500 before crossing the Atlantic back in order to go for the Indies. After that, Portugal and Spain quickly started to explore the Americas and eventually colonize them. France, England and Netherlands also came to the Americas some time later. 5. The Earth IS round After, the Spanish discovered and settled into the Americas (and Colombo's plan in fact didn't worked). The question that if it was possible to sail around the globe to reach the Indies from the east side remained open and the Spanish were still interested on it. They eventually discovered the Pacific Ocean after crossing the Panama Ishtums by land in 1513. Eager to find a maritime route around the globe, the Spanish crown funded an expedition leadered by the Portuguese Fernão de Magalhães (or Magellan as his name was translated to English) to try to circle the globe. Magellan was an experienced navigator, and had reached what is present day Malaysia traveling through the Indian Ocean before. They departed from Spain in September 20th, 1519. It was a long and though journey that costed the lives of most of the crew. Magellan himself did not survived, having died in a battle in the Phillipines on 1521. At least, he lived enough to be aware that they in fact reached East Asia by traveling around the globe to the west, which also proves that the Earth is round. The journey was eventually completed by the leadership of Juan Sebatián Elcano , one of the crewmen of Magellan. They reached Spain back through the Indian and Atlantic Oceans on September 6th, 1522 after traveling for almost three years a distance of 81449 km. 6. Heliocentrism There were some heliocentric or hybrid geo-heliocentric theories in ancient times. Notably by the Greek philosopher Philolaus in the 5th century BC. By Martianus Capella around the years 410 to 420. And by Aristarchus of Samos around 370 BC. Those models tried to explain the motion of the stars as rotation of the Earth and the position of the planets, specially Mercury and Venus as translation around the Sun. However those early models were too imprecise and flawed to work appropriately, and the ptolomaic model still was the model with the better prediction of the positions of the heavenly bodies. The idea that the Earth rotates was much less revolutionary than heliocentrism, but was already more-or-less accepted with reluctancy in the middle ages . This happens because if the stars rotated around Earth, they would need do so at an astonishing velocity, dragging the Sun, the Moon and the planets with it, so it would be easier if Earth itself rotated. People were uncomfortable with this idea, but they still accepted it, and this became easier to be accepted after the Earth sphericity was an established concept. In the first years of the 1500's, while the Portuguese and Spanish were sailing around the globe, a polish and very skilled matemathical and astronomer called Nikolaus Kopernikus took some years thinking about the mechanics of the heavenly bodies. After some years making calculations and observations, he created a model of circular orbits of the planets around the Sun and perceived that his model were much more simpler than the ptolomaic geocentric model and was at least as precise. His model also features a rotating Earth and fixed stars. Further, his model implied that the Sun was much larger than the Earth, something that was already strongly suspected at the time due to calculations and measurements and also implied that Jupiter and Saturn were several times larger than Earth, so Earth would definitively be a planet just like the other five then-known planets were. This could be seen as the borning of the model today know as Solar System. Fearing persecution and harsh criticism, he avoided to publish many of his works, sending manuscripts to only his closest acquaintances, however his works eventually leaked out and he was convinced to allow its full publication anyway. Legend says that he was presented to his finally fully published work in the very day that he died in 1543, so he could die in peace. There was a heated debate between supporters and oppositors of Copernic's heliocentric theory in the middle 1500's. One argument for the opposition was that star parallaxes could not be observed, which implied that either the heliocentric model was wrong or that the stars were very very far and many of them would be even larger than the Sun, which seemed to be a crazy idea at the time. Tycho Brache , which did not accepted heliocentrism, in the latest years of the 1500's tried to save geocentrism with an hybrid geo-heliocentric model that featured the five heavenly planets orbiting the Sun while the Sun and the Moon orbited Earth. However, he also published a theory which better predicted the position of the Moon. Also, by this time, the observation of some supernovas showed that the celestial sphere of the stars were not exactly immutable. In 1600, the astronomer William Gilbert provided strong argument for the rotation of Earth, by studing magnets and compasses, he could demonstrate that the Earth was magnetic, which could be explained by the presence of enourmous quantities of iron in its core. 7. With telescopes All of what I wrote above happened without telescopes, only by using naked eye observations and measurements around the globe. Now, add even some small telescopes and things change quickly. The earliest telescopes were invented in 1608 . In 1609, the astronomer Galieu Galilei heard about that, and constructed his own telescope. In January of 1610, Galieu Galilei , using a small telescope, observed four small bodies orbiting Jupiter at different distances, figuring out that they were Jupiter's "moons", he also could predict and calculate its positions along their orbits. Some months later, he also observed that Venus had phases as seen from the Earth. He also observed Saturn's rings, but his telescope was not powerful enough to resolve them as rings, and he tought that they were two moons. These observations were incompatible with the geocentric model. A contemporary of Galilei, Johannes Kepler , working on Copernicus' heliocentric model and making a lot of calculations, in order to explain the differing orbital velocities, created an heliocentric model where the planets orbits the Sun in elliptic orbits with one of the ellipse's focus in the Sun. His works were published in 1609 and 1619. He also suggested that tides were caused by the motion of the Moon, though Galilei was skeptical to that. His laws predicted a transit of Mercury in 1631 and of Venus in 1639, and such transit were in fact observed. However, a predicted transit of Venus in 1631 could not be seen due to imprecision in calculations and the fact that it was not visible in much of the Europe. In 1650 the first double star were observed. Further in the 1600's, the Saturn rings were resolved by the use of better telescopes by Robert Hooke , who also observed a double star in 1664 and developed microscopes to observe cellular structures. From them on, many stars were discovered to be double. In 1655, Titan were discovered orbiting Saturn, putting more confidence on the heliocentric model. More four Saturnian moons were discovered between 1671 and 1684. 8. Gravitation Heliocentrism was reasonably well-accepted in the middle 1600's, but people was not confortable with it. Why the planets orbits the Sun afterall? Why the Moon orbits Earth? Why Jupiter and Saturn had moons? Although Keplerian mechanics could predict their moviment, it was still unclear what was the reason that makes them move that way. In 1687, Isaac Newton who was one of the most brilliant physic and mathematic that ever lived (although he was also an implacable persecutor of his opponents), provided the gravitational theory (based on prior work by Robert Hooke). Ideas for the gravitation theory and the inverse square law already were developed in the 1670's, but he could publish a very simple and clear theory for gravitation, very well-fundamented in physics and mathematics and it explained the motions of the celestial bodies with a great precision, including comets. It also explained why the planets, the Moon and the Sun are spherical, explained tides and it also served to explain why things falls to the ground. This made heliocentrism to be definitely widely accepted. Also, Newton gravitational law predicted that Earth rotation would make it not exactly spherical, but a bit ellipsoidal by a factor of 1:230. Something that agreed with measures done using pendulums in 1673. 9. What are the stars and the Solar System afterall? In the early 1700's, Edmund Halley , already knowing about newtonian laws (he was a contemporary of Newton) perceived that comets who passed near Earth would eventually return, and he found that there was a particular instance of sightings every 76 years, so he could note that those comets in reality were all the same comet, which is called after him. The only remaining problem with the heliocentric model was the lack of observation of parallax to the stars. And nobody knew for sure what the stars were. However, if they in fact are very distant bodies, most of them would be much larger than the Sun. At the first half of the 1700's, trying to observe parallax, James Bradley perceived phenomena like the aberration of light and the Earth's nutation, and those phenomena also provides a way to calculate the speed of light. But the observation of parallax remained a challenge during the 1700's. In 1781, Uranus were discovered orbiting the Sun beyond Saturn. Although barely visible to the naked eye in the darkest skies, it was so dim that it escaped observation from astronomers until then, and so were discovered with a telescope. The first asteroids were also discovered in the early 1800's. Investigation on pertubations on Uranus' orbit due to the predicted newtonian and keplerian movement eventually leaded to the discovery of Neptune in 1846. In 1838, the astronomer Friedrich Wilhelm Bessel who measured the position of more than 50000 stars with the greatest precision as possible, could finally measure the parallax of the star 61 Cygni successfully, which proved that stars were in fact very distant bodies and that many of them were in fact larger than the Sun. This also demonstrates that the Sun is a star. Vega and Alpha Centauri also had their parallaxes measured successfully in 1838. Further, those measurements permitted to estimate the distance between those stars and the Solar System to be on the order of many trillions of kilometers, or several light-years. | {
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11,673 | Black Holes are regions of space where things get weird [Citation Needed] . Past the event horizon of a black hole, any moving particle instantaneously experiences a gravitational acceleration towards the black hole that will cancel out it's current velocity, even light. That means that the gravity well of the black hole must be able to accelerate from -C¹ to 0 instantly². Given that fact, we can assume the gravitational acceleration of black holes is C/instant³. Given this, it stands to reason that in successive instants, the particle will be moving at speeds greater than C, because it is experiencing greater gravitational forces and continuous gravitational acceleration. Does this actually make sense? Is there something I'm missing here? By this logic, it seems like anything inside of the event horizon of a black hole could and should move faster than C due to gravitational acceleration. Edit: I showed this question to a friend and he questioned if the hypothetical particles that were radiating from the singularity (The photon traveling exactly away from the black hole) might be hawking radiation; that is, the gravitation acceleration of a black hole is only strong enough to curve the path of light around a non-zero radius (thus not actually stopping it, but altering it's course), and not powerful enough to decelerate light. Is this actually what hawking radiation is, or is he as confused as I am? ¹ Where movement towards the singularity would be considered a positive value, movement away from the singularity is a negative value, that is, anything moving at the speed of light away from the singularity would be moving with a velocity of -C relative to the singularity. ² If it couldn't accelerate from -C to 0 instantly, any photon traveling exactly away from the black hole would be able to escape the event horizon. ³ An instant is an arbitrary amount of time, it could be a fraction of a second, a second, a minute.... | I think your initial question is a good one, but the text gets a bit more jumbled and covers a few different points. Can things move faster than light inside the event horizon of a black
hole? Nice question. Black Holes are regions of space where things get weird. I'm 100% OK with this statement. I think it's a true enough summary and I'm sure I've heard physicists say this too. Even if "Weird" isn't a clearly defined scientific term, I'm 100% fine with this (even without a citation). Past the event horizon of a black hole, any moving particle
instantaneously experiences a gravitational acceleration towards the
black hole that will cancel out it's current velocity, even light. That means that the gravity well of the black hole must be able to
accelerate from -C* to 0 instantly ✝. Are you quoting somebody here? Anyway, this isn't quite true. Black holes don't accelerate things from -c (which I'm guessing would be a light beam trying to fly away from the singularity but inside the event horizon), to 0 "instantaneously". Perhaps a better way to look at it is to consider curvature of space, and inside a black hole, space curves so much that all directions point to the singularity. It's the "all roads go to Rome" scenario, even if you do a complete 180, you're still on a road that leads to the singularity. I understand the temptation to look at that as deceleration, but I think that's a bad way to think about it. Light doesn't decelerate, it follows the curvature of space. Given that fact, we can assume the gravitational acceleration of black
holes is C/instant**. Given this, it stands to reason that in
successive instants, the particle will be moving at speeds greater
than C, because it is experiencing greater gravitational forces and
continuous gravitational acceleration.
Does this actually make sense? Is there something I'm missing here? By
this logic, it seems like anything inside of the event horizon of a
black hole could and should move faster than C due to gravitational
acceleration. outside of a black hole, continuous acceleration would never lead to a speed greater than C. You can accelerate for billions and trillions of years and all you'd do is just add more 9s to the right of the decimal point. You seem to be assuming that inside a black hole this can happen, but I'm not sure why you'd assume that. "continuous gravitational acceleration" - no matter how strong, is no guarantee for faster than light travel. That's logically inconsistent with the laws of relativity. Edit: I showed this question to a friend and he questioned if the
hypothetical particles that were radiating from the singularity (The
photon traveling exactly away from the black hole) might be hawking
radiation; that is, the gravitation acceleration of a black hole is
only strong enough to curve the path of light around a non-zero radius
(thus not actually stopping it, but altering it's course), and not
powerful enough to decelerate light. Is this actually what hawking
radiation is, or is he as confused as I am? I think, a more correct way to look at hawking radiation is to see it as something that forms just outside of the black hole, a particle/anti particle pair and one escapes and the other falls inside, and that's probably not 100% correct either, but the singularity itself doesn't send out particles. Hawking radiation has to do with quantum properties of space. It's not a property of black holes. The black hole just happens to be unique in that it can capture one half of a virtual particle pair and the other half can escape. This also is a pretty different topic than your original question. *Where movement towards the singularity would be considered a positive value, movement away from the singularity is a negative value, that
is, anything moving at the speed of light away from the singularity
would be moving with a velocity of -C relative to the singularity.
✝If it couldn't accelerate from -C to 0 instantly, any photon
traveling exactly away from the black hole would be able to escape the
event horizon. **An instant is an arbitrary amount of time, it could be a fraction of a second, a second, a minute.... I think it's a good idea to differentiate mass-less pure energy particles and particles with mass. You seem to be saying that a ray of light can be traveling away from a black hole at the speed of light, get caught in the gravity, slow down and then fall back into the black hole like a ball that's tossed straight up into the air from the surface of the Earth. That's probably not what happens. The ray of light follows the path of space time ahead of it, which happens to be curved so much that it points into the black hole, even if, in the classical sense, the light begins by pointing away. All space curves into the singularity once you're inside the event horizon, so there is no "away from" anymore. At least, that's how I think it works. | {
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11,960 | Could dark matter compress and form black holes? Since dark matter is even more abundant than normal matter, a dark matter black hole should not be rare...right? | The problem with trying to form a black hole with dark matter is that dark matter can only weakly interact (if at all) with normal matter and itself, other than by gravity. This poses a problem. To get dark matter concentrated enough to form a black hole requires it to increase its (negative) gravitational binding energy without at the same time increasing its internal kinetic energy by the same amount. This requires some sort of dissipative interaction between dark matter and normal matter (or itself). The following scenario should make this clear. Suppose we have a lump of dark matter that gravitationally attracts another lump of dark matter. As the two approach each other, they accelerate and gain kinetic energy. The kinetic energy gained will be exactly enough to then separate them to a similar degree to which they started, unless some dissipative process takes place. An example is to suppose that dark matter is weakly interacting massive particles (WIMPs). WIMPs are gravitationally drawn towards the centres of stars. If the weak interactions happen sufficiently frequently then it might be possible for them to accumulate in stars, rather than shoot through and out the other side. It has been hypothesised that black holes could be made like this near the centre of a Galaxy, seeded by dense neutron stars. The density of neutron star matter, combined with the enhanced density of dark matter near galaxy centres could result in dark matter accumulation in the neutron stars, leading to the formation of black holes. Once a black hole is formed then any dark matter that enters the event horizon cannot emerge regardless of what kinetic energy it gains in the process. However, there is still a problem. Material in orbit around a black hole has less angular momentum the closer it orbits. To pass inside the event horizon requires the dark matter to lose angular momentum. Normal matter does this via an accretion disc that can transport angular momentum outwards by viscous torques, allowing matter to accrete. Dark matter has almost zero viscosity so this can't happen. So building a supermassive black hole from a smaller seed would be difficult, but forming small black holes out of neutron stars might be easier. It has been proposed that a relative lack of pulsars observed towards our own Galactic centre could be due to this process. | {
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12,777 | I have heard a lot of buzz about distant planets that could potentially be second homes for human existence, but what is that approximate number? | There is currently only one planet known to be capable of supporting human life, and you're on it. Several planets have been found in the region in which we expect water to be liquid on much of the planet. Of these, only one fits the criteria of being Earth-sized and well placed in the habitable zone: Kepler 186-f However we know nothing about it's atmosphere (or lack of one). The star is a red dwarf, so it could be subject to dramatic solar flares. The planet is rather colder than earth, so could be in a perpetual "snowball world" state, depending on the composition of the atmosphere and the strength of the greenhouse effect. The atmosphere would be very unlikely to be even close to breathable, and it is nearly 500 light-years from Earth, so could not be reached in a reasonable amount of time, even with much more advanced propulsion. At the moment we can't usually detect most Earth-like planets in the habitable zone of brighter stars like the sun, though the probably do exist and may be common. | {
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12,781 | I want to buy some meteorite and put stones made from it into gold or platinum ring. Or maybe I can use some metal meteorite for making ring? What kinds of meteorites can I use? It must be strong, non radioactive, it will be good if it pretty look. What minerals do fit? | There is currently only one planet known to be capable of supporting human life, and you're on it. Several planets have been found in the region in which we expect water to be liquid on much of the planet. Of these, only one fits the criteria of being Earth-sized and well placed in the habitable zone: Kepler 186-f However we know nothing about it's atmosphere (or lack of one). The star is a red dwarf, so it could be subject to dramatic solar flares. The planet is rather colder than earth, so could be in a perpetual "snowball world" state, depending on the composition of the atmosphere and the strength of the greenhouse effect. The atmosphere would be very unlikely to be even close to breathable, and it is nearly 500 light-years from Earth, so could not be reached in a reasonable amount of time, even with much more advanced propulsion. At the moment we can't usually detect most Earth-like planets in the habitable zone of brighter stars like the sun, though the probably do exist and may be common. | {
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12,854 | Sound can't travel through outer space. But if it could, how loud would the Sun be? Would the sound be dangerous to life on Earth, or would we barely hear it from this distance? | The Sun is immensely loud. The surface generates thousands to tens of thousands of watts of sound power for every square meter. That's something like 10x to 100x the power flux through the speakers at a rock concert, or out the front of a police siren. Except the "speaker surface" in this case is the entire surface of the Sun, some 10,000 times larger than the surface area of Earth. Despite what "user10094" said, we do in fact know what the Sun "sounds" like -- instruments like SDO's HMI or SOHO's MDI or the ground-based GONG observatory measure the Doppler shift everywhere on the visible surface of the Sun, and we can actually see sound waves (well, infrasound waves) resonating in the Sun as a whole! Pretty cool, eh? Since the Sun is large, the sound waves resonate at very deep frequencies -- typical resonant modes have 5 minute periods, and there are about a million of them going all at once. The resonant modes in the Sun are excited by something. That something is the tremendous broadband rushing of convective turbulence. Heat gets brought to the surface of the Sun by convection -- hot material rises through the outer layers, reaches the surface, cools off (by radiating sunlight), and sinks. The "typical" convection cell is about the size of Texas, and is called a "granule" because they look like little grains when viewed through a telescope. Each one (the size of Texas, remember) rises, disperses its light, and sinks in five minutes. That produces a heck of a racket. There are something like 10 million of those all over the surface of the Sun at any one time. Most of that sound energy just gets reflected right back down into the Sun, but some of it gets out into the solar chromosphere and corona. No one can be sure, yet, just how much of that sound energy gets out, but it's most likely between about 30 and about 300 watts per square meter of surface, on average. The uncertainty comes because the surface dynamics of the Sun are tricky. In the deep interior, we can pretend the solar magnetic field doesn't affect the physics much and use hydrodynamics, and in the exterior (corona) we can pretend the gas itself doesn't affect the physics much. At the boundary layers above the visible surface, neither approximation applies and the physics gets too tricky to be tractable (yet). In terms of dBA, if all that leaked sound could somehow propagate to Earth, well let's see... Sunlight at Earth is attenuated about 10,000 times by distance (i.e. it's 10,000 times brighter at the surface of the Sun), so if 200 W/m2 of sound at the Sun could somehow propagate out to Earth it would yield a sound intensity of about 20 mW/m2. 0dB is about 1pW/m2 , so that's about 100dB. At Earth, some 150,000,000 kilometers from the sound source. Good thing sound doesn't travel through space, eh? The good folks at the SOHO/MDI project created some sound files of resonant solar oscillations by speeding up the data from their instrument by 43,000 times. You can hear those here, at the Solar Center website . Someone else did the same thing with the SDO/HMI instrument, and superposed the sounds on first-light videos from SDO . Both of those sounds, which sound sort of like rubber bands twanging, are heavily filtered from the data -- a particular resonant spatial mode (shape of a resonant sound) is being extracted from the data, and so you hear mainly that particular resonant mode. The actual unfiltered sound is far more cacophonous, and to the ear would sound less like a resonant sound and more like noise. | {
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13,270 | I've seen a number of news reports indicating there is likely a 9th planet in our Solar System , something with an orbital period of between 10k-20k years, that is 10 times Earth's mass. I haven't seen any real indication of where this object might be. If I had access to a sufficient telescope, would I be able to find this planet, and what way would I point a telescope to find it? How far is it likely to be, or is that not well known? | It's too dim to be seen during a normal survey during the majority of its
orbit. Update: Scientists at the University of Bern have modeled a hypothetical 10 Earth mass planet in the proposed orbit to estimate its detectability with more precision than my attempt below. The takeaway is that NASAs WISE mission would have probably spotted a planet of at least 50 Earth masses in the proposed orbit and that none of our current surveys would have had a chance to find one below 20 earth masses in most of its orbit. They put the planets temperature at 47K due to residual heat from formation; which would make is 1000x brighter in infrared than it is in visible light reflected from the sun. It should however be within reach of the LSST once it is completed (first light 2019, normal operations beginning 2022); so the question should be resolved within a few more years even if its far enough from Batygin and Brown's proposed orbit that their search with the Subaru telescope comes out empty. My original attempt to handwave an estimate of detectability is below. The paper gives potential orbital parameters of $400-1500~\textrm{AU}$ for the semi major axis, and $200-300~\textrm{AU}$ for perihelion. Since the paper doesn't give a most-likely case for orbital parameters, I'm going to go with the extreme case that makes it most difficult to find. Taking the most eccentric possible values from that gives an orbit with a $1500~\textrm{AU}$ semi-major axis and a $200~\textrm{AU}$ perihelion has a $2800~\textrm{AU}$ aphelion. To calculate the brightness of an object shining with reflected light, the proper scaling factor is not a $1/r^2$ falloff as could be naively assumed. That is correct for an object radiating its own light; but not for one shining by reflected light; for that case the same $1/r^4$ scaling as in a radar return is appropriate. That this is the correct scaling factor to use can be sanity checked based on the fact that despite being similar in size, Neptune is $\sim 6x$ dimmer than Uranus despite being only $50\%$ farther away: $1/r^4$ scaling gives a $5x$ dimmer factor vs $2.25$ for $1/r^2$ . Using that gives a dimming of 2400x at $210~\textrm{AU}\;.$ That puts us down $8.5$ magnitudes down from Neptune at perihelion or $16.5$ magnitude. $500~\textrm{AU}$ gets us to $20$ th magnitude, while a $2800~\textrm{AU}$ aphelion dims reflected light down by nearly $20$ magnitudes to $28$ magnitude. That's equivalent to the faintest stars visible from an 8 meter telescope ; making its non-discovery much less surprising. This is something of a fuzzy boundary in both directions. Residual energy from formation/radioactive material in its core will be giving it some innate luminosity; at extreme distances this might be brighter than reflected light. I don't know how to estimate this. It's also possible that the extreme cold of the Oort Cloud may have frozen its atmosphere out. If that happened, its diameter would be much smaller and the reduction in reflecting surface could dim it another order of magnitude or two. Not knowing what sort of adjustment to make here, I'm going to assume the two factors cancel out completely and leave the original assumptions that it reflects as much light as Neptune and reflective light is the dominant source of illumination for the remainder of my calculations. For reference, data from NASA's WISE experiment has ruled out a Saturn-sized body within $10,000~\textrm{AU}$ of the sun. It's also likely too faint to have been detected via proper motion; although if we can pin its orbit down tightly Hubble could confirm its motion. Orbital eccentricity can be calculated as: $$e = \frac{r_\textrm{max} - r_\textrm{min}}{2a}$$ Plugging in the numbers gives: $$e = \frac{2800~\textrm{AU} - 200~\textrm{AU}}{2\cdot 1500~\textrm{AU}} = 0.867$$ Plugging $200~\textrm{AU}$ and $e = 0.867$ into a cometary orbit calculator gives a $58,000$ year orbit. While that gives an average proper motion of $ 22~\textrm{arc-seconds/year}\;,$ because the orbit is highly eccentric its actual proper motion varies greatly, but it spends a majority of its time far from the sun where its values are at a minimum. Kepler's laws tell us that the velocity at aphelion is given by: $$v_a^2 = \frac{ 8.871 \times 10^8 }{ a } \frac{ 1 - e }{ 1 + e }$$ where $v_a$ is the aphelion velocity in $\mathrm{m/s}\;,$ $a$ is the semi-major axis in $\mathrm{AU},$ and $e$ is orbital eccentricity. $$v_a = \sqrt{\frac{ 8.871 \times 10^8 }{ 1500 } \cdot \frac{ 1 - 0.867 }{ 1 + 0.867 }} = 205~\mathrm{m/s}\;.$$ To calculate the proper motion we first need to convert the velocity into units of $\textrm{AU/year}:$ $$205 \mathrm{\frac{m}{s}}\; \cdot \mathrm{\frac{3600 s}{1 h}} \cdot \mathrm{\frac{24 h}{1 d}} \cdot \mathrm{\frac{365 d}{1 y}} \cdot \mathrm{\frac{1\; AU}{1.5 \times 10^{11}m}} = 0.043~\mathrm{\frac{AU}{year}}$$ To get proper motion from this, create a triangle with a hypotenuse of $2800~\textrm{AU}$ and a short side of $0.043~\textrm{AU}$ and then use trigonometry to get the narrow angle. $$\sin \theta = \frac{0.044}{2800}\\ \implies \theta = {8.799×10^{-4}}^\circ = 3.17~\textrm{arc seconds}\;.$$ This is well within Hubble's angular resolution of $0.05~\textrm{arc seconds};$ so if we knew exactly where to look we could confirm its orbit even if its near its maximum distance from the sun. However its extreme faintness in most of its orbit means that its unlikely to have been found in any survey. If we're lucky and it's within $\sim 500~\textrm{AU},$ it would be bright enough to be seen by the ESA's GAIA spacecraft in which case we'll located it within the next few years. Unfortunately, it's more likely that all the GAIA data will do is to constrain its minimum distance slightly. Its parallax movement would be much larger ; however the challenge of actually seeing it in the first place would remain. | {
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13,282 | Theoretical Planet Nine , proposed by Caltech's Mike Brown and Konstantin Batygin , is said to have a 15-20 thousand year orbit. Approximately how bright would the sun appear from Planet Nine aphelion and perihelion? | Between $1/40,000$ and $1/8,000,000$ of the brightness as seen from Earth, depending on what the actual orbit would turn out to be, and where the planet is in its $15,000$ year orbit period. Brightness drops as $\dfrac{1}{r^2}$ with distance from the light source. Earth is at $1~\textrm{AU}.$ The theoretical planet is at $200~\textrm{AU}$ when it's closest to the sun, and up to $2800~\textrm{AU}$ at the point furthest from the sun at the upper end of the estimated orbit. So e.g. $1/200^2 = 1/40,000$ of the brightness (luminance) as seen from Earth. For how a human would experience it, we can convert to exposure value as used in photography: The difference in exposure value (photographic 'stops') is $\log 2$ of the luminance ratio, so we would have $15$ to $23$ stops less light than on Earth.
Sunny noon on Earth is $15~\textrm{EV}.$ So the brightness at noon on the planet surface would be: At $200~\textrm{AU},$ planet orbit is closest to the sun: About $0~\textrm{EV},$ roughly the same as a dimly lit interior $400~\textrm{AU},$ lower bound on semi-major axis: $-2 ~\textrm{EV},$ similar to a landscape lit by the full moon $1500~\textrm{AU},$ upper bound on semi-major axis: $-6~\textrm{EV},$ similar to landscape lit by a quarter moon $2800~\textrm{AU},$ upper bound on aphelion (the point on orbit most distant from the sun): $-8~\textrm{EV}.$ This would be dark, but you would probably still see enough to avoid running into things. | {
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13,330 | In the comments to this question , there was considerable doubt placed as to the subject of if the so called "9th planet" really exists. That wasn't really the intent of the previous question, so I ask this. Why hasn't this "9th planet" been detected before now, if it even exists? | Brown and Batygin, the authors of the paper on the possible planet, have a webpage addressing this. A few reasons not already covered: It moves quite slowly - the authors estimate 0.2-0.6 arc seconds per hour - so standard surveys may not notice the movement and fail to recognize it as a solar system object. Eris, which is the most distant confirmed object still known in the
solar system, moves at a speed of 1.5 arcseconds per hour, which is so
slow that it was missed the first time around. Most surveys of the
outer solar system would not be able to find Planet Nine, even if it
were quite bright, as they would just think it is a stationary star. If the planet is near aphelion, it might be an order of magnitude further away than any major or minor planet we've found so far (excluding exoplanets, which are found by methods that don't apply in this case). The authors suggest an aphelion between 500 and 1200 AU. For comparison, Pluto is at 30-50 AU, while Eris at around 100 AU wasn't discovered until 2005. The potential 9th planet would be far larger than Eris, but is also likely to be much further away, and thereby fainter. The WISE survey eliminated Saturn-sized planets within 10,000 AU , and Jupiter-sized planets within 26,000 AU. But the potential 9th planet is far smaller than those. WISE has also done a more sensitive search, which would pick up Neptune-sized objects, but that search has so far covered only a limited part of the sky. The planet will be far harder to spot if it has the Milky Way in the background - there are too many stars potentially drowning out a faint object. Here's the authors' summary: Estimated orbit for the putative 9th planet. The horizontal axis is the right ascension. The colored segments are regions where it should have been found by existing surveys. Illustration by Brown and Batygin, assuming fair use applies. The biggest unexplored territory is where, statistically, it is most
likely to be: near aphelion. Sadly, aphelion is also very close to the
Milky Way galaxy. Ugh. So where is it? Probably distant. 500 AU+. Probably fainter than 22nd
magnitude. Very possibly in the middle of the Milky Way galaxy. Now go find planet nine. More details on the authors' webpage: http://www.findplanetnine.com/p/blog-page.html Finally, the gravitational dominance of the Sun reaches halfway to the nearest star. There's still plenty of unexplored territory for planets smaller than Saturn to hide in. Note the log axis. We have a good map for the inner 50 AU, and are starting to find objects around 100 AU, but solar system objects might exist all the way to the outer edges of the Oort cloud. Illustration from wikipedia . | {
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13,552 | I often read that the CMB was released from everywhere in the Universe, in every direction. If that statement is true, can someone elaborate what "everywhere in the universe, in every direction" means ? If not, why can we observe it in every direction ? | Until the Universe was 380,000 years old, it was filled with a gas of protons an electrons. There was also radiation, in thermal equilibrium with the matter, and because it was so hot, the protons and electrons couldn't form neutral hydrogen, since every time it "tried", an energetic photon would knock off the electron. This gas was everywhere. And photons traveled and scattered in all directions: Photons ( purple ) scatter on free electrons ( green ), and both are mixed with protons ( red ). 380,000 years after the Big Bang, the temperature had fallen sufficiently that neutral atoms could form (this is called recombination ). The radiation, which until now had scattered continously on free electrons, could now stream freely between the atoms (this is called decoupling ). So they did. Still in all directions: This free streaming is still taking place. Photons travel in all directions, and are everywhere. The photons that you are able to see, are the ones that started out at a particular distance from you, and in a particular direction, but other photons started out at smaller and larger distances, and in other directions. You just don't see them, because you happen to be right here . But a person in another place of the Universe would see the same as you. The photons that we observe as the CMB come from a region we call the surface of last scattering , because it corresponds to the surface of a shell centered at us. But there is nothing special about this "surface", except that is consists of all points in the Universe that are so far from us, that it takes a photon roughly 13.8 billion years to travel. And because of the expansion, these points are now roughly 47 billion lightyears away from us. In the figure below, the arrows show CMB photons. All have the same length; they start where they were emitted and end where they are today. What we observe as the CMB are all arrows that end at the Milky Way (in the center). Other arrows may be observed by other observers in other galaxies that have their own surface of last scattering around them. | {
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13,688 | Of course everyone knows by now of the detection of Gravitational waves But, since General Relativity and Quantum Mechanics don't get along , can we say now that this detection proves that Quantum Mechanics doesn't actually apply and that General Relativity did prevail? Another question: how can we identify the ripple's origin (let's say whether it's a result of the big bang or another big event)? EDIT 16-2-2016 I was reading an article today and I thought I'd share it here; It's basically saying that without a third detector we can't triangulate the signal.
Some scientists tried ways to observe the light of the event directly after the observations of the wave but they couldn't detect the merger simply because it's too far away or too faint to be observed with our current technology. | No more than the observation of light waves disproves quantum mechanics. Light has properties of both a particle and a wave. At low energies, the particle nature of light is hard to detect: radio waves are made of photons, but individual radio wave photons are pretty hard to detect. I'm not sure that we have directly detected individual photons with energies below the infrared band. Gravitational waves (probably) also have both a wave and a particle nature. The gravitational field is probably quantised. But at the frequencies and sensitivity at which LIGO operates, individual quanta cannot be measured. So this detection does not prove the ascendency of GR over QM. If anything, understanding extreme events like black hole mergers might lead to a theoretical understanding of the quantum nature of gravity. | {
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13,782 | I am trying to find an original video/image of what LIGO actually saw, but all I can find is artist renditions of gravitational waves. | The actual image isn't much. I was able to find it from Science , and this is all it is: It's just a ripple, seen at slightly different times from two different observatories. The shift fits perfectly by shifting it by the speed of light difference in their locations. Thus is the proof of gravity waves. It should be noted that the reason there are two instruments is to provide a cross check against other vibration sources. Each observatory works by detecting vibrations on a 4 km scale, down to a very small order of magnitude (1/10,000 the width of a proton). When the two are compared, then one can assume the signal must have come from a non-local source, which only Gravity Waves fit that definition. | {
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14,082 | Stars twinkle because their light has to squeeze through several different layers of the Earth's atmosphere. So why doesn't the moon twinkle as well? | The first handful of hits on Google actually return incomplete and even wrong answers (e.g. "Because the Moon is much brighter" which is plain wrong, and "Because the Moon is closer" which is incomplete [see below]). So here's the answer: As you mention, when light enters our atmosphere, it goes through several parcels of gas with varying density, temperature, pressure, and humidity. These differences make the refractive index of the parcels different, and since they move around (the scientific term for air moving around is "wind"), the light rays take slightly different paths through the atmosphere. Stars are point sources Stars are immensely far away, effectively making them point sources. When you look at a point source through the atmosphere, the different paths taken from one moment to another makes it "jump around" — i.e. it twinkles (or scintillates ). The region in which the point source jumps around spans an angle of the order of an arcsecond. If you take a picture of a star, then during the exposure time, the star has jumped around everywhere inside this region, and thus it's no longer a point, but a "disk". …the Moon is not The same is true for the Moon, but since the Moon (as seen from Earth) is much larger (roughly 2000 times larger, to be specific) than this "seeing disk" as it's called, you simply don't notice it. However, if you are observing details on the Moon through a telescope, then the seeing puts a limit on how fine details you can see. The same is even true for planets. The planets you can see with the naked eye span from several arcsec up to almost an arcmin. Although they look like point sources (because the resolution of the human eye is roughly 1 arcmin), they aren't, and you will notice that they don't twinkle (unless they're near the horizon where their light goes through a thicker layer of atmosphere). The image below may help understanding why you see the twinkling of a star, but not of the Moon (greatly exaggerated): EDIT: Due to the comments below, I added the following paragraph: Neither absolute size, nor distance is important in itself. Only the ratio is. As described above, what makes a light source twinkle depends on its apparent size compared to the seeing $s$, i.e. its angular diameter $\delta$ defined by the ratio between its absolute diameter $d$ and its distance $D$ from Earth:
$$
\delta = 2 \arctan \left( \frac{d}{2D} \right)
\simeq \frac{d}{D}\,\,\,\mathrm{for\,small\,angles}
$$ If $\delta \lesssim s$, the object twinkles. If it's larger, it doesn't. Hence, saying that the Moon doesn't twinkle because it's close is an incomplete answer, since for instance a powerful laser 400 km from Earth — i.e. 1000 times closer than the Moon — would still twinkle because it's small. Or vice versa, the Moon would twinkle even at the distance it is, if it were just 2000 times smaller. Finally, to achieve good images with a telescope you not only want to put it at a remote site (to avoid light pollution), but also — to minimize the seeing — at high altitudes (to have less air) and at particularly dry regions (to have less humidity). Alternatively you can just put it in space. | {
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14,102 | Does turning the colour of the sky blue need more luminous light? Does it depend on luminosity or some other factors are also responsible for this phenomenon? Why can't the moon light turn the sky blue even a little bit (at least the area near the disc). Thanks | The simple answer is that it does, but it's not bright enough to be visible to the naked eye. Earth's atmosphere scatters the moon light just like sunlight. The full moon (like the sun) fills about 1/2 of 1 degree of the sky, the entire sky being 180 degrees, give or take, so the full moon fills less than 1 part in 100,000 of the night sky, so there simply isn't enough blue light to be visible over the brighter stars even with the brightest full moon. Our eyes are very good at seeing variations in brightness, but not that good. . . . and, for what it's worth, the night sky has always appeared to have a dark bluish tint to me, but that might just be my brain playing tricks on me because logically I know it's there. I'm not sure whether it's actually visible. With a good sized telescope, moonlight scattering acts as a form of light pollution. Telescope users know that you get better visuals when there's no moon. Source . | {
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14,319 | A couple days ago, I zoomed in with my 30x optical zoom camera, and after some exposure adjustments, a bright star in the night sky turned into this: Are those other planets or other stars? Or is that a lens effect? EDIT: The bright object in question was ~60 degrees above the horizon, and ESE of me (East-south-east). I took the picture on 3-25-16 from Madison, Wisconsin. EDIT: Question answered, more clear picture added FYI. | You don't say what time you were looking. Here is a screenshot from Stellarium at 10pm Wisconsin time on 25th March 2016. Jupiter is in the ESE, but the altitude is a bit lower than 60 degrees. Seems fairly conclusive. You were seeing Ganymede and a Europa/Io combination. | {
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14,733 | I'm curious to know what the sky would look like without any other galaxies out there. How much do other galaxies factor into the stars we see? Does the Milky Way account for most of them? Would the night sky look normal? Or would it be very empty? | I feel sure this is a repeat, but couldn't immediately find it. The only things in the night sky we can see (with the naked eye) that are not part of our own Galaxy are (on a good night) the Andromeda galaxy and the Large and small Magellanic clouds. Every individual star brighter than $V=6$ and visible to the naked eye is in the Milky Way. So it would hardly look any different. | {
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14,740 | The Chandrasekhar limit is the maximum mass of a stable white dwarf star. Beyond this, a carbon-oxygen white dwarf will typically explode in a type 1a supernova, due to the nuclear reactions at those temperatures. I've heard that oxygen-neon-magnesium white dwarfs, on the other hand, will not ignite. Rather, electron capture becomes energetically favorable and they become neutron stars. If I understand correctly, this also happens for carbon-oxygen white dwarfs in binaries, if the white dwarf has most of the mass in the system. Why is this? Can anyone walk me through, step by step, what happens if a white dwarf (of both compositions) exceeds the Chandrasekhar limit, and how each differs? Source: https://arxiv.org/abs/astro-ph/9701225 | I feel sure this is a repeat, but couldn't immediately find it. The only things in the night sky we can see (with the naked eye) that are not part of our own Galaxy are (on a good night) the Andromeda galaxy and the Large and small Magellanic clouds. Every individual star brighter than $V=6$ and visible to the naked eye is in the Milky Way. So it would hardly look any different. | {
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14,889 | The density of the Sun is $1410~\frac{\text{kg}}{\text{m}^{3}}$ and Mercury's is $5430~\frac{\text{kg}}{\text{m}^{3}}$, but shouldn't the Sun be denser? Because when the Solar System was forming, there was a big disk of debris, and depending on the density of the debris it went closer or further from the centre, which then formed the planets, but the Sun is in the centre, and it's less dense than Mercury, why? | The sun isn't the same density all the way through. According to MSFC's solar interior page , the core density at the centre of the sun is a whopping 150,000 kg/m$^3$. Surrounding it the radiative zone is around 20,000 - 200 kg/m$^3$ (already less dense than water). Eventually at the edge is the convective zone - the density at the part that we see is much less dense than our own air... So although the Sun's average density isn't very remarkable, the core is the densest place in the solar system. (Sun cross section from Wikipedia.org ) | {
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14,921 | Wolfram Alpha's query: Sun Earth distance in AU gives: 1.012 au (astronomical units) They definitely use this value in other places too - e.g. Earth-Sun $L_1$ distances from Earth and from Sun don't add up to 1 AU - they are 0.01009 AU and 1.001 AU respectively. Contrarily, Google gives Earth Sun distance in AU = 1.000 AU. The top search results for Sun Earth distance 1.012 AU yield a page with a big heading of "Paranormal UFO Aliens Wow!". Is the 1.012 AU value an error in Wolfram Alpha, or is it some little-known adjustment, redefinition or other quirk of the unit or the Solar System? | As of this moment (2016 May 18, 13:15 UTC), the Earth is 1.0116 astronomical units from the Sun. WA is smart enough to know that "Sun Earth distance in AU" is a time-dependent question. Update: The stricken text that follows from my original answer is incorrect. I am leaving it present (but stricken) for the sake of humility. It interpreted your query to mean Sun Earth distance today in AU , and because today is 2016 May 18 and because you didn't specify a time of day, it picked noon (UTC), it in turn interpreted your query to mean Sun Earth distance on 2016 May 18 at noon UTC in AU . The correct answer : I happened to ask WA the very question posed in the OP a couple of minutes apart and got two different answers (1.014 and 1.015 AU), and this did not occur across a day boundary. WA apparently interpreted your query to mean the current distance between the Sun and the Earth. You have to be very careful of what you ask WA. A better query is to ask WA What is the mean distance between the earth and the sun in AU? That query will give you an answer of 1.0000010178 au. An even better question is What is the semimajor axis of Earth's orbit? That will give you an answer of 1.00000011 au. Note the extra zero, but also note that for some reason, the precision is reduced. | {
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15,086 | The vastness of space brings me a sense of chilliness even though I have never experienced it, although I wish to. Just how cold is interstellar space (on average)? How is this even measured? I mean you can't just stick a thermometer in space, right? | You can stick a thermometer in space, and if it is a super-high-tech one, it might show you the temperature of the gas. But since the interstellar medium (ISM) is so dilute, a normal thermometer will radiate energy away faster than it can absorb it, and thus it won't reach thermal equilibrium with the gas.
It won't cool all the way to 0 K, though, since the cosmic microwave background radiation won't allow it to cool further than 2.7 K, as described by David Hammen. The term "temperature" is a measure of the average energy of the particles of a gas (other definitions exist e.g. for a radiation field). If the gas is very thin, but particles move at the same average speed as, say, at the surface of Earth, the gas is still said to have a temperature of, say, 27º C, or $ 300\,\mathrm{K}$ . The ISM consists of several different phases, each with their own physical characteristics and origins. Arguably, the three most important phases are (see e.g. Ferrière 2001 ): Molecular clouds Stars are born in dense molecular clouds with temperatures of just 10-20 K. In order for a star to form, the gas must be able to collapse gravitationally, which is impossible if the atoms move too fast. The warm neutral medium The molecular clouds themselves form from gas that is neutral, i.e. not ionized. Since most of the gas is hydrogen, this means that it has a temperature of roughly $10^4\,\mathrm{K}$ , above which hydrogen tends to get ionized. The hot ionized medium Gas that accretes onto the galaxy in its early phases tend to have much larger temperature, of roughly $10^6\,\mathrm{K}$ . Additionally, the radiative feedback from the hot stars (O and B), and the kinetic and radiative energy injected by supernova explosions ionize and heat gas bubbles that expand. This gas comprises the hot ionized medium. Cooling The reason that the ISM is so sharply divided into phases, as opposed to just being a smooth mixture of particles of all sorts of energies, is that gas cools by various physical processes that have a rather temperature-specific efficiency.
"Cooling" means converting the kinetic energy of particles into radiation that is able to leave the system. Hot gas Very hot gas is fully collisionally ionized and thus cools mainly through free electron emitting Bremsstrahlung. This mechanism becomes inefficent below $\sim10^6\,\mathrm{K}$ . Warm gas Between $10^4\,\mathrm{K}$ and $10^6\,\mathrm{K}$ , recombinations (i.e. electrons being caught by ions) and collisonal excitation and subsequent de-excitation lead to emission, removing energy from the system.
Here the metallicity $^\dagger$ of the gas is important, since various elements have different energy levels. Cool gas At lower temperatures, the gas is almost fully neutral, so recombinations cease to have any influence. Collisions between hydrogen atom become too weak to excite the atoms, but if molecules or metals are present, it is possible through fine/hyperfine lines, and rotational/vibrational lines, respectively. The total cooling is the sum of all these processes, but will be dominated by one or a few processes at a given temperature. The figures below from Sutherland & Dopita (1993) shows the main cooling processes (left) and the main cooling elements ( right ), as a function of temperature: The thick line show the total cooling rate. The figure below, from the same paper, shows the total cooling rate for different metallicities. The metallicity is a logarithmic scale, so [Fe/H] = 0 means Solar metallicity, and [Fe/H] = –1 means 0.1 times Solar metallicity, while "nil" is zero metallicity. Since these processes don't cover equally the full temperature range, the gas will tend to reach certain "plateaus" in temperatures, i.e. it will tend to occupy certain specific temperatures. When gas cools, it contracts. From the ideal gas law, we know that the pressure $P$ is proportional to the product of the density $n$ and the temperature $T$ . If there's pressure equilibrium in the ISM (which there isn't always, but in many cases is a good assumption), then $nT$ is constant, and thus if a parcel of hot ionized gas cools from $10^7\,\mathrm{K}$ to $10^4\,\mathrm{K}$ , it must contract to increases its density by a factor $10^3$ . Thus, cooler clouds are smaller and denser, and in this way the ISM is divided up in its various phases. So, to conclude, interstellar space is not as cold as you may think. However, being extremely dilute, it is difficult to transfer heat, so if you leave your spaceship, you will radiate away energy faster than you can absorb it from the gas. $^\dagger$ In astronomy, the term "metal", refers to all elements that are not hydrogen or helium, and "metallicity" is the fraction of gas that consists of metals. | {
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15,092 | I am new to using SDSS image files and have noticed that some of the values in the FITS file are negative. Can somebody explain to me what these values actually mean? | You can stick a thermometer in space, and if it is a super-high-tech one, it might show you the temperature of the gas. But since the interstellar medium (ISM) is so dilute, a normal thermometer will radiate energy away faster than it can absorb it, and thus it won't reach thermal equilibrium with the gas.
It won't cool all the way to 0 K, though, since the cosmic microwave background radiation won't allow it to cool further than 2.7 K, as described by David Hammen. The term "temperature" is a measure of the average energy of the particles of a gas (other definitions exist e.g. for a radiation field). If the gas is very thin, but particles move at the same average speed as, say, at the surface of Earth, the gas is still said to have a temperature of, say, 27º C, or $ 300\,\mathrm{K}$ . The ISM consists of several different phases, each with their own physical characteristics and origins. Arguably, the three most important phases are (see e.g. Ferrière 2001 ): Molecular clouds Stars are born in dense molecular clouds with temperatures of just 10-20 K. In order for a star to form, the gas must be able to collapse gravitationally, which is impossible if the atoms move too fast. The warm neutral medium The molecular clouds themselves form from gas that is neutral, i.e. not ionized. Since most of the gas is hydrogen, this means that it has a temperature of roughly $10^4\,\mathrm{K}$ , above which hydrogen tends to get ionized. The hot ionized medium Gas that accretes onto the galaxy in its early phases tend to have much larger temperature, of roughly $10^6\,\mathrm{K}$ . Additionally, the radiative feedback from the hot stars (O and B), and the kinetic and radiative energy injected by supernova explosions ionize and heat gas bubbles that expand. This gas comprises the hot ionized medium. Cooling The reason that the ISM is so sharply divided into phases, as opposed to just being a smooth mixture of particles of all sorts of energies, is that gas cools by various physical processes that have a rather temperature-specific efficiency.
"Cooling" means converting the kinetic energy of particles into radiation that is able to leave the system. Hot gas Very hot gas is fully collisionally ionized and thus cools mainly through free electron emitting Bremsstrahlung. This mechanism becomes inefficent below $\sim10^6\,\mathrm{K}$ . Warm gas Between $10^4\,\mathrm{K}$ and $10^6\,\mathrm{K}$ , recombinations (i.e. electrons being caught by ions) and collisonal excitation and subsequent de-excitation lead to emission, removing energy from the system.
Here the metallicity $^\dagger$ of the gas is important, since various elements have different energy levels. Cool gas At lower temperatures, the gas is almost fully neutral, so recombinations cease to have any influence. Collisions between hydrogen atom become too weak to excite the atoms, but if molecules or metals are present, it is possible through fine/hyperfine lines, and rotational/vibrational lines, respectively. The total cooling is the sum of all these processes, but will be dominated by one or a few processes at a given temperature. The figures below from Sutherland & Dopita (1993) shows the main cooling processes (left) and the main cooling elements ( right ), as a function of temperature: The thick line show the total cooling rate. The figure below, from the same paper, shows the total cooling rate for different metallicities. The metallicity is a logarithmic scale, so [Fe/H] = 0 means Solar metallicity, and [Fe/H] = –1 means 0.1 times Solar metallicity, while "nil" is zero metallicity. Since these processes don't cover equally the full temperature range, the gas will tend to reach certain "plateaus" in temperatures, i.e. it will tend to occupy certain specific temperatures. When gas cools, it contracts. From the ideal gas law, we know that the pressure $P$ is proportional to the product of the density $n$ and the temperature $T$ . If there's pressure equilibrium in the ISM (which there isn't always, but in many cases is a good assumption), then $nT$ is constant, and thus if a parcel of hot ionized gas cools from $10^7\,\mathrm{K}$ to $10^4\,\mathrm{K}$ , it must contract to increases its density by a factor $10^3$ . Thus, cooler clouds are smaller and denser, and in this way the ISM is divided up in its various phases. So, to conclude, interstellar space is not as cold as you may think. However, being extremely dilute, it is difficult to transfer heat, so if you leave your spaceship, you will radiate away energy faster than you can absorb it from the gas. $^\dagger$ In astronomy, the term "metal", refers to all elements that are not hydrogen or helium, and "metallicity" is the fraction of gas that consists of metals. | {
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16,379 | Theoretically, planets would have an approximately equal chance of going one way in their orbit or another but in reality, this is not the case (at least in our solar system). Why is this? | The same reason (almost) all of them rotate in the same direction: because of the conservation of angular momentum. Before a star and its planets exist, there’s just a cloud of disorganized gas and small molecules. The Solar System formed from such a cloud around 4.6 billion years ago. On that scale, there is some small amount of rotation within the cloud. It could be caused by the gravity of nearby stellar objects, local differences in mass as the cloud churns, or even the impact of a distant supernova. The point is, all molecular clouds have at least a little rotation. In a large system like a molecular cloud, each particle has some angular momentum, and it all adds together across a very wide area. That’s a lot of momentum, and it is conserved as the cloud continues to collapse under its own gravity. That angular momentum also flattens the cloud, which is the reason why the Solar System is near-planar. When the cloud finally collapses, it forms a star and shortly after planets. However, angular momentum is always conserved. That's why planets all follow the same orbit, and why almost all of them rotate in the same direction. There's nothing to turn them the other direction, so they will continue spinning in the same direction as the original gas cloud. There are a few exceptions, though. Whenever objects formed in such a way that sent them orbiting the opposite direction, they usually collided with objects going in the same direction as the original cloud. This destroyed any outlying objects or sent them in the same direction as the original cloud. Still, two huge exceptions are planets Venus and Uranus. Uranus spins on an axis of almost 90-degrees (on its side). Venus meanwhile spins the opposite direction as Earth and the other planets. In both cases there is strong evidence that these planets were struck by large objects at some point in the distant past. The impacts were large enough to overcome the angular momentum of the bodies, and give them a different spin. There are also a range of other theories; for example, some astronomers think that Venus may have been flipped upside-down. Point is, there were irregular events that happened to both of these planets. | {
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16,630 | To cite the German newpaper article Astronomen beobachten erwachendes Schwarzes Loch : Das Materie-Monster sitzt den Angaben zufolge im Herzen der 42 Millionen Lichtjahre entfernten Polarring-Galaxie NGC 660, deren Aktivität innerhalb weniger Monate Hunderte Male zugenommen hatte. Erst wenn die Massemonster große Mengen Materie verschlucken, werden sie aktiv. Bei diesem Prozess wird so viel Energie frei, dass die Materie hell aufleuchtet, bevor sie im Schwarzen Loch verschwindet und ein Teil von ihr in Form von Jets weit ins Weltall hinaus geschleudert wird. This translates roughly to: According to the data the matter-monster is in the middle of the 42 million light years far away polar ring galaxy NGC 660, whose activity has increased a lot in only a few months. Only when these Matter-monsters swallow big amounts of matter, they become active. This process releases so much energy, that it brightly enlightens the matter, before it disappears inside the black hole. A part of the matter is flung out in the universe in the form of jets. My physics teacher told me once, that a black hole is just a very small-sized and heavy object which has so much gravity that nothing at all, not even light can't escape of its gravity. This explanation is also supported by this SE.astronomy - If nothing travels at the speed of light, except light, how can a black hole also pull light into itself? question. If a "normal" (not supermassive) black hole can already prevent light to escape, how can matter which is pulled into the black hole can produce energy/light which can't escape the gravity of the black hole? How can a supermassive black hole be able to pull matter, but not the light photons of the energy? Additionally: Why is part of the matter which is pulled into the black hole flung (i.e. accelerated ) into the universe? I understand why this matter is maybe divided - the acceleration increases as far as i know quadratic, i.e. the differences in terms of the actual acceleration, in dependence of the location may be so huge that the matter can't be held together. But i don't understand why a part of the matter is accelerated in the exact opposite direction, as a force bigger as the gravity of the supermassive black hole would be needed. Therefore: Why is a part of the matter accelerated in the opposite direction (i.e. out of the black hole's gravity) than the other part? Note: My physics education is rather limited. I know a bit about the Newtonian gravity and a bit about the theory of Conservation of energy. But that's all I know about physics. | It is quite correct that a black hole has so much mass that light cannot escape from a region around the black hole. The edge of this region is called the event horizon. If you cross an event horizon you are never coming back. That applies equally to light, and matter. Around the black hole there may be matter in orbit. Since the Black hole has such strong gravity, the speed of the orbiting matter will be very fast. In fact it will be close to the speed of light. This high speed gives it lots of energy. The matter will form a disc, called an accretion disc, around the black hole, and collisions in this disc will cause the matter to heat up, to millions of degrees. At these temperatures, the disc will glow with X-rays. On the part of the disc closest to the black hole, matter will be falling in from the disc, but before it reaches the black hole it can get enough energy to be ejected out, at very high speed, close to the speed of light. It gets ejected at right angles to the disc, at the poles of the black hole. These are the "jets". Intense radiation is produced along these jets. Blazars are distant supermassive black holes with jets that are pointed right at us. So the black hole itself is "black", but the matter orbiting around it may be very bright. | {
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16,704 | Shouldn't the Moon appear as bright as a full Moon seen at midnight from Earth? The photo was taken by DSCOVR at Lagrange point 1. In the picture, The Moon appears dark gray. Of course the Earth appears bright, reflecting sunlight from clouds and water. The Moon's surface is gray and should reflect less light than the Earth. It should be irrelevant that we see the far side, since the reflectivity of the Moon's surface should be the same on the far side as the side that faces the Earth. The midnight full Moon appears much, much brighter as seen from Earth than it does in this picture, despite the fact that the amount of sunlight reflecting from the surface of the Moon is the same in both instances. I understand the photo was taken with 3 separate exposures of red, blue and green, but this should not affect the brightness. So why does it appear so dull? | That's what it really would look like if you were there with DSCOVR. The albedo of the Moon is only about 0.136 , about half of the Earth's average albedo . Of course the part with clouds is higher. I was shocked too, but it was explained in written copy that accompanied the release of the original image. Shouldn't the Moon appear as bright as a full Moon seen at midnight from Earth? It does. If the moon were a diffuse, white ball, a full moon would be about seven times brighter! If you watch the image or GIF, the Moon is roughly the same brightness as central Australia or the Sahara region. Phil Plait explains well in Bad Astronomy . There's a lot to read here . EDIT: I just ran across these images of astronauts on the surface while reading this answer . Their suits are not 100% white to begin with, but the Lunar soil - at least in these locations - is significantly darker. It is close to the same color as the (presumably) nearly-black radiator fins for the heat sink of the RTG unit (2nd photo) at the astronaut's foot. above: "Buzz Aldrin carries the EASEP." from here above: "Astronaut Alan L. Bean from Apollo 12, put the Plutonium 238Pu Fuel from the Lunar Module into the SNAP 27 RTG" from here . above: Image from NASA/NOAA from Bad Astronomy | {
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17,875 | If nothing else wipes out human existence prior to this, at what point will the Sun make Earth uninhabitable for humans? | The Sun is gradually getting larger and brighter. In fact, as called2voyage pointed out, its brightness is increasing by 1% every 100 million years. You can see how the Sun will change in the future from this graph: ( Source ) According to this paper , within 1 billion (short scale) years from now, the ever-increasing luminosity will have made Earth nearly uninhabitable. The average temperature will have reached 47°C, compared to its current 15°C. Essentially no water will be left either, except at the poles. This may allow for simple life to survive for a while. By 3.5 billion years, Earth will no longer resemble its current self. Its oceans, magnetic field and ozone layer and plate tectonics will be no more. Its surface temperature will skyrocket to roughly 1,330°C, hot enough to melt surface rock. No longer will our planet resemble a pale blue dot, and it will be more like Venus. Our planet is officially dead, along with all life on it. ( Source ) ~4.5 billion years from now, the Sun will become a red giant and possibly consume Earth. However, according to this paper , it may heat up potentially habitable bodies like Triton, to the point where they would support life. Unfortunately, the Sun won't remain in this stage for long enough — life usually takes billions of years to develop. | {
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17,909 | On a casual star walk, (for comparison of identifying different objects) I witnessed some meteors, quite a few satellites gliding on their smooth and straight trajectories, and some airplanes with their much brighter (sometimes blinking) lights, lower altitude and the characteristic sense of moving towards and away from me. Suddenly one of the satellites seemed as if its motion had been a bit jittery, and its path started to curve inwards in an asymmetrical, gentle and large C-shape. The airplanes made noise, this one did not. Visually perfectly identical with all of the normal satellites. I had never seen anything like that before so I wondered if it could've been some debris colliding into things, but probably not. The best explanation I have for now is that it apparently might have been a satellite on a Tundra or Molniya orbit, as their ground tracks aren't the usual straight or sine wave shaped trajectories that we're used to. Most satellite tracking websites either didn't seem to have tracking options for those special types at all, or you could only see their locations for now, not a custom time. I'm in the capital city area of Finland, and when I saw the object, the time was 23:10 on the UTC+3 time zone, 12th of August 2016. Of what I found, the only strange looking ground track that comes even close would be Molniya 1-62, drawing a loop above Finland. What I can't find is a website that would allow me to rewind time to check where the satellite was at 23:10. But whether or not I should keep searching for that info. I first have this question: If I look at the satellite tracks going across a map, how do I determine how far a satellite would be visible when viewed from the ground? If I looked at similar tracks for airplanes, it's much easier to understand that the ground area from which an airplane can be spotted doesn't span vast distances, but how this compares to a satellite is difficult to imagine as we don't perceive them in a three dimensional space, but spots gliding across the flat surface of a dome. | The Sun is gradually getting larger and brighter. In fact, as called2voyage pointed out, its brightness is increasing by 1% every 100 million years. You can see how the Sun will change in the future from this graph: ( Source ) According to this paper , within 1 billion (short scale) years from now, the ever-increasing luminosity will have made Earth nearly uninhabitable. The average temperature will have reached 47°C, compared to its current 15°C. Essentially no water will be left either, except at the poles. This may allow for simple life to survive for a while. By 3.5 billion years, Earth will no longer resemble its current self. Its oceans, magnetic field and ozone layer and plate tectonics will be no more. Its surface temperature will skyrocket to roughly 1,330°C, hot enough to melt surface rock. No longer will our planet resemble a pale blue dot, and it will be more like Venus. Our planet is officially dead, along with all life on it. ( Source ) ~4.5 billion years from now, the Sun will become a red giant and possibly consume Earth. However, according to this paper , it may heat up potentially habitable bodies like Triton, to the point where they would support life. Unfortunately, the Sun won't remain in this stage for long enough — life usually takes billions of years to develop. | {
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17,939 | I was bopping around YouTube and observed this enjoyably produced video . In it, when describing the behavior of a black hole with the mass of a US nickel, the narrator says, "Its 5 grams of mass will be converted to 450 terajoules of energy, which will lead to an explosion roughly three times bigger than the bombs dropped on Hiroshima and Nagasaki combined." Of all the fun things illustrated there, that was the one whose pretense I didn't understand. Do black holes explode after they've radiated away all of their mass? Or would the "explosion" just come from the rapid pace at which the black hole would consume nearby matter? The Googling I've done thus far has provided no firm answer. The closest I've come is from the Wikipedia page on Hawking Radiation , which states, "For a black hole of one solar mass, we get an evaporation time of 2.098 × 10^67 years—much longer than the current age of the universe at 13.799 ± 0.021 x 10^9 years. But for a black hole of 10^11 kg, the evaporation time is 2.667 billion years. This is why some astronomers are searching for signs of exploding primordial black holes." Some other websites refer to the last "explosion" of the Milky Way's supermassive black hole being some 2 million years ago, but is that the same mechanic mentioned on the Wikipedia page? Or the YouTube video, for that matter? | What this video is talking about is Hawking Radiation, as you've linked. Hawking Radiation is a proposed hypothetical (by no means verified or proven) way for a black hole to radiate its energy into space. The basic idea is that a black hole is nothing but mass/energy compressed to an infinitesimal point, which is radiating its energy into space over time. For large black holes (such as solar mass or bigger), this radiation process is tiny and the time taken to leak all the black hole's energy into space (and thus for the black hole to "evaporate") is exceedingly long. For tiny black holes however, the time to radiate all the black hole's energy is exceedingly short. You can calculate how long it will take for a black hole of mass $m$ to evaporate (and release all its mass/energy) with the equation $$t_{\text{ev}} = \frac{5120\pi G^2 m^3}{\hbar c^4} = (8.41\times 10^{-17}\:\mathrm{s}\:\mathrm{kg}^{-3})\:m^3$$ For $m = 5\:\mathrm{g}=0.005\:\mathrm{kg}$, you get $t_{\text{ev}} \simeq 4\times10^{-19}\:\mathrm{s}$. Now that means in this tiny amount of time, the black hole will radiate all of its mass/energy away and completely evaporate. But the output of all the energy in 5 g of mass is a huge output. Putting out 450 Terajoules of energy in $10^{-19}\:\mathrm{s}$ is basically just an explosion. You can determine the total energy output from the famous equation $$E=mc^2$$ Just plug in $m = 0.005\:\mathrm{kg}$ and $c = 3\times 10^8\:\mathrm{m/s}$ and you'll get $E=4.5\times 10^{14}\:\mathrm{J} = 450\:\mathrm{Terajoules}$. So in short, hypothetical calculations (not even theory at this point) suggest that a tiny black hole with the mass of a nickel would immediately explode out in a huge amount of energy. Whether such a black hole can form, or if such an evaporation would/could occur is still highly debated and ultimately unknown at this point. | {
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17,942 | I'm looking for a function that given a time (for example Unix time ) gives back the current distance between Mars and Earth in light-seconds. It can be a mathematical function or programming function (preferably in Python or pseudocode), but must not rely on an Internet connection. This question is similar to a question I asked erlier , but again I need to get the distance without an Internet connection. Thanks in advance! Sidenote: I'm planning on learning orbital mechanics, but I need this function earlier. :) | What this video is talking about is Hawking Radiation, as you've linked. Hawking Radiation is a proposed hypothetical (by no means verified or proven) way for a black hole to radiate its energy into space. The basic idea is that a black hole is nothing but mass/energy compressed to an infinitesimal point, which is radiating its energy into space over time. For large black holes (such as solar mass or bigger), this radiation process is tiny and the time taken to leak all the black hole's energy into space (and thus for the black hole to "evaporate") is exceedingly long. For tiny black holes however, the time to radiate all the black hole's energy is exceedingly short. You can calculate how long it will take for a black hole of mass $m$ to evaporate (and release all its mass/energy) with the equation $$t_{\text{ev}} = \frac{5120\pi G^2 m^3}{\hbar c^4} = (8.41\times 10^{-17}\:\mathrm{s}\:\mathrm{kg}^{-3})\:m^3$$ For $m = 5\:\mathrm{g}=0.005\:\mathrm{kg}$, you get $t_{\text{ev}} \simeq 4\times10^{-19}\:\mathrm{s}$. Now that means in this tiny amount of time, the black hole will radiate all of its mass/energy away and completely evaporate. But the output of all the energy in 5 g of mass is a huge output. Putting out 450 Terajoules of energy in $10^{-19}\:\mathrm{s}$ is basically just an explosion. You can determine the total energy output from the famous equation $$E=mc^2$$ Just plug in $m = 0.005\:\mathrm{kg}$ and $c = 3\times 10^8\:\mathrm{m/s}$ and you'll get $E=4.5\times 10^{14}\:\mathrm{J} = 450\:\mathrm{Terajoules}$. So in short, hypothetical calculations (not even theory at this point) suggest that a tiny black hole with the mass of a nickel would immediately explode out in a huge amount of energy. Whether such a black hole can form, or if such an evaporation would/could occur is still highly debated and ultimately unknown at this point. | {
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17,990 | I assume that about 99.9% of the sun-rays that do not fall on any planet or any other celestial body keep on traveling farther and farther unto infinity. Apparently such rays get lost. Keeping in mind the colossal energy Sun has produced since 4.5 billion years I am somehow reluctant to reconcile myself to the idea that Nature would have allowed wastage of so much energy produced by the Sun. Nonetheless I want to get enlightened whether it got really lost or got utilized. If it got utilized I want to know how it may have got utilized at all and whether any sustainable evidence is available in support of any such finding? | The light from the Sun spreads, at least initially, in a roughly isotropic fashion into the universe. As it gets further from the Sun, some of that light will interact with the interstellar medium (ISM) and therefore some of the energy emitted by the Sun will be used to excite atoms and molecules or even ionise some atoms. This will be the fate of almost all the light which is emitted from the Sun in the direction of the plane of our Galaxy, which contains sufficient molecular gas and dust to block starlight travelling through it for any distance. We know this happens because we can "see" dark clouds in the Milky Way, that can be penetrated by longer wavelength radiation to reveal all the billions of Sun-like stars that lie behind them. Roughly speaking, about half the visible light from the Sun will be absorbed every 1000 light years when travelling in directions within $\pm 5$ degrees of the Galactic plane, so it is essentially all absorbed within a few thousand light years. But most of the Sun's light is not travelling in the direction of the Galactic plane, and interstellar and intergalactic space has a very low density of gas and dust. The equivalent extinction number for the intergalactic medium is that light travels many billions of light years with almost no chance of being absorbed (see Zu et al. 2010 ). This means that most of the light from the Sun will travel to cosmological distances (billions of light years) over the course of the next billions of years. Indeed, light emitted from the Sun shortly after its birth has already travelled 4.5 billion light years. That this happens is demonstrated by the fact that we can observe galaxies (the light from which is nothing more than the summation of light from many stars like the Sun) that are 4.5 billion (and more) light years away. As the sunlight travels towards cosmological distances, its wavelength is "stretched" by the expansion of the universe, becoming redder and redder. We know this happens because distant galaxies have redshifted spectra. If the universe keeps expanding, then its matter density will continue to decrease and there is little to stop the radiation from the Sun travelling on forever, with a wavelength that scales as the scale factor, $a$ , of the universe. We could consider a co-moving and co-expanding cube containing the Sun's radiation as the universe expands. The total radiative energy inside that cube diminishes as $a^{-1}$ - that is, the energy content of the universe in the form of radiation from stars (and other sources) becomes energetically less important as the universe expands and appears to be being superseded by the energy contained in the vacuum itself (a.k.a. dark energy). In conclusion, most of the energy emitted by the Sun is not "used" for anything; it propagates into space, becoming more and more dilute and less energetically important as cosmic time marches on. | {
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18,039 | I know that the light coming from Proxima Centauri is not bright enough to make it naked-eye visible from the Earth. Is the Sun naked-eye visible from Proxima Centauri? | Well, there's two things we'll need for this: apparent magnitude (the brightness that an object appears to have) and absolute magnitude (the actual brightness an object has). Both of these scales are logarithmic, with brighter objects being lower and dimmer objects being higher. Astronomers have determined that the Sun's absolute magnitude is 4.83. Knowing this, we can find the apparent magnitude of the Sun from Proxima Centauri's location. Apparent and absolute magnitudes are related by the equation: $$M = m - 5 (\log_{10}{d}-1)$$ Where $M$ is the absolute magnitude, $m$ is the apparent magnitude, and $d$ is the distance, in parsecs. Astronomers have determined that Proxima Centauri is 1.3 parsecs from us. So the apparent magnitude can be determined as: $$m = 4.83 + 5(\log_{10}{1.3}-1) ≈ +0.4$$ As made clear in this paper , most humans can see objects with apparent magnitudes as dim as $5$ without using tools. So yes, it is certainly visible, and would be quite bright. It is between Procyon and Achernar, the 9th and 10th brightest star on Earth's night sky. For a comparison, Proxima Centauri has an apparent magnitude of $+11.13$ from Earth's perspective. If we want to compare the two, we could use this formula: $$v_b = 10^{0.4 x} = 10^{0.4×(11.13-0.4)} \approx 19588$$ So the Sun would appear almost 20,000 times brighter from Proxima Centuari than PC appears from the Sun. | {
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18,112 | What is the exact measurement of a light year? I searched google for the answer in meters and came up with $9.461\cdot 10^{15}$ meters. When I calculated the answer considering $299\,792\,458 \;\text{m/s}$ as the speed of light, I came up with: $$
299\,792\,458 \times 365 \times 24 \times 3\,600 = 9\,454\,254\,955\,508\,926 \;\text{m}
$$ Why is there such a gap? Did I miss something to add in the equation or is it just wrong? | By convention, astronomy uses the Julian Year for the computation of a light year : Although there are several different kinds of year, the IAU regards a year as a Julian year of 365.25 days (31.5576 million seconds) unless otherwise specified. Wikipedia gives the length as 31 557 600 s ✕ 299 792 458 m/s = 9 460 730 472 580 800 m (exactly) The reason that it's exact (and not subject to experimental error) is that the metre itself is defined in terms of the speed of light, so the quantities are fixed by definition. | {
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18,120 | How do we know the laws of physics are the same throughout the universe? Intuitively I would say they would vary in two natural ways: the constants in the equations may vary or the math in the equations may vary. As a guess they could change over a long time. What is the farthest radius we can prove from Earth, with absolute certainty, that the laws of physics do not vary? I am aware this may not be a radius but a more complex shape that cannot be simply described by a radius. The nearest answer I can think of for a radius is a guess. And that guess is based on the farthest physics experiment we have done from earth, which I think is an experiment with mirrors on the moon. Therefore if we assume (I don't know if this assumption is 100% reasonable) all physics laws hold because this experiment works. Then the radius is the distance to the moon. This doesn't give a concrete answer for the radius, merely an educated guess. | Nothing can be proved "with absolute certainty"; that is not how science works. We adopt a working hypothesis that the constants of nature are exactly that; both constant in time and space. Then we conduct experiments that attempt to falsify that hypothesis or at least place limits on by how much things might vary. For reasons that are explained in answers to this Physics SE question (see also this question ), only the dimensionless parameters like the fine structure constant can be assessed for their variation - other constants like $G$, $c$ and $h$ are tied up in our system of (measuring) units so we are unable to say whether they are changing or not. Taking the example of the fine structure constant, observations of absorption lines towards distant quasars put strong limits on by how much this can have varied in space and time (the two are inseparable, since it takes finite time for information to travel to us). So you can find lots of different attempts to do this in the literature - I dug out a few. Albareti et al. (2015) say the variation is less than a couple of parts in a 100,000 out to a redshift of 1 (a lookback time of about 8 billion years or so. Similar constraints exist for experiments carried out in different parts of the solar system. On the other hand, some authors do claim variations of a few parts per million on similar lookback times or in different directions ( Murphy et al 2008 ; King et al. 2012 ), but these claims are disputed by many, if not most workers in the field. There is a massive review of this topic by Uzan (2011) , which you could read - this really is a broad question. My summary would be - at the moment there is no convincing evidence for any variation in space and time. | {
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18,128 | I don't know if this is the right place for asking this question, my apologies in advance. I would like to be updated on the latests news on space and I am looking for a website devoted to space news, or a blog. I was only able to find minor blogs which were not updated often. Could you suggest me a high quality blog/space related website that you know? Thanks! | Nothing can be proved "with absolute certainty"; that is not how science works. We adopt a working hypothesis that the constants of nature are exactly that; both constant in time and space. Then we conduct experiments that attempt to falsify that hypothesis or at least place limits on by how much things might vary. For reasons that are explained in answers to this Physics SE question (see also this question ), only the dimensionless parameters like the fine structure constant can be assessed for their variation - other constants like $G$, $c$ and $h$ are tied up in our system of (measuring) units so we are unable to say whether they are changing or not. Taking the example of the fine structure constant, observations of absorption lines towards distant quasars put strong limits on by how much this can have varied in space and time (the two are inseparable, since it takes finite time for information to travel to us). So you can find lots of different attempts to do this in the literature - I dug out a few. Albareti et al. (2015) say the variation is less than a couple of parts in a 100,000 out to a redshift of 1 (a lookback time of about 8 billion years or so. Similar constraints exist for experiments carried out in different parts of the solar system. On the other hand, some authors do claim variations of a few parts per million on similar lookback times or in different directions ( Murphy et al 2008 ; King et al. 2012 ), but these claims are disputed by many, if not most workers in the field. There is a massive review of this topic by Uzan (2011) , which you could read - this really is a broad question. My summary would be - at the moment there is no convincing evidence for any variation in space and time. | {
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18,260 | As like Earth, does the moon have a day (24 hours). If it has, how many hours are in a moon day?. | The answer depends on your definition of a day. If you define a day as we usually define it in the Earth (time between the Sun is at noon or average time between sunrises, a 24 hours day in Earth), the length of a day in the Moon is the synodic period of the Moon and it takes 29.530589 days (29 d 12 h 44 min 2.9 s) . Anyway, if you define a day as rotation on own axis, that is, time between a far away star being at noon, you have what in Earth is called sidereal day (23 hours, 56 minutes in Earth). In the moon the day defined this way would be 27.321661 days (27 d 7 h 43.19 min 11.5 s) . Please notice that since the Moon is tidally locked to the Earth, rotation periods equal orbital periods. | {
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18,423 | I am writing a science fiction novel, where a ship is stranded in a single star system (a red supergiant). One of the plot points is the star becoming supernova in several hours, so the characters have to fix their ship before that happens. I have basic knowledge of how it works: Iron generated from nuclear fusion gets accumulated in the core, until it reaches a point when iron fusion starts. As iron fusion is an endothermic reaction, the core is no longer able to generate enough energy to hold against its own gravity and external layers pressure, so it collapses, and explode. I have read that once the iron fusion starts inside the core, the collapse occurs within minutes, that the collapse itself lasts a few seconds (even less than a second), and that the shockwave takes several hours to reach the surface. Is all that correct? The thing is that I need the characters to bee able to predict the explosion in a short term. A few hours or even minutes. It would be great if they could be aware of the core collapse and start a countdown. So, are there any external cue of these events, like changes in luminosity or color? Does the star spectrum change when iron fusion starts, or when the core collapses? I know that the core collapse generates a huge amount of neutrinos. Is this amount so intense that it can be easily detectable? (that is, without a huge detector in an underground facility). Can the amount of iron in the core be estimated from tha star spectrum and size, so the aproximate time of the collapse could be predicted? | I think your best bet would be detecting neutrinos generated by nuclear burning inside the star (as we do for the Sun). Once the star hits the carbon-burning stage, it's actually putting out more energy in neutrinos than in photons. During the silicon-burning phase, which lasts for a few days and is what creates the degenerate iron core (that collapses once it is massive enough), the neutrino flux increases to about 10 47 erg/s a few seconds before core collapse. (The peak flux during core collapse is about 10 52 to 10 53 erg/s). This paper by Asakura et al. estimates that the Japanese KamLAND detector could detect the pre-supernova neutrino flux for stars at distances of several hundred parsecs , and provide advance warning of a core-collapse supernova several hours or even days in advance. Since your characters are in the same system as the star, they'd hardly need a large underground detector to pick up the neutrinos. This plot shows an example of neutrino luminosity (for anti-electron neutrinos) versus time for a pre-supernova star (from Asakura et al. 2016, based on Odrzywolek & Heger 2010 and Nakazato et al. 2013); core collapse begins at t = 0s. By measuring the spectrum of energies for different types of neutrinos and their time evolution, you could probably get a very good idea of how far along the star was, particularly as we can probably assume your characters have much better models for stellar evolution than we currently do. (They'd also want to get accurate measurements of the star's mass, rotation rate, maybe internal structure via astroseismology, etc., in order to fine-tune the stellar-evolution model; these are all things they could do pretty easily.) The core collapse itself would be signalled by the enormous increase in neutrino flux. This "What If" article by Randall Munroe estimates that the neutrino flux from a core-collapse supernova would be lethal to a human being at a distance of around 2 AU. Which, as he points out, could actually be inside of a supergiant star, so your characters would probably be a bit further away than that. But it does show that the neutrino flux would be easily detectable, and that your characters might well get radiation poisoning from it if they were closer than 10 AU. (Of course, you'd want to detect it more directly than just waiting around till you started to feel sick, since that might take longer than the shock wave takes to reach the surface of the star.) This is just to bring home the fact that they wouldn't have any problem detecting the neutrinos.... | {
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18,609 | Andromeda is about 2.5 million ly away. Actually, in this universe, at what "speed" (in km/h) are two objects separating cosmologically - I mean strictly due to the "expansion of the universe" - if they are 2.5 million ly apart? I do understand that local "ordinary" or "peculiar" motion completely swamps this effect. If I'm not mistaken, the "local" "ordinary" motion of Andromeda per our galaxy happens to be about 400,000 km/h towards us. Is the "speed" due to the "expansion of the universe" drastically smaller than this? I assumed that the expansion of the universe (or "of the spacetime metric") is even everywhere: it's well known that it only affects "the largest structures" but I still assumed that the expansion is the same in my room, my galaxy, my cosmological region. Perhaps this assumption is totally wrong? | The rate of expansion, measured in the customary units of (km/s)/Megaparsec
is not known with great accuracy. Recent measurements include 67.6 (SDSS-III), 73(HST) 67.8 (Plank) 69.3 (WMAP) [ wikipedia ] The Andromeda galaxy is 0.78 Mpc from us, so taking the Hubble constant to be about 70, gives a recession of about 55 km/s. This is not a very great speed: compare with the orbital velocity of the sun around the galaxy at over 200 km/s, or the escape velocity of the galaxy (over 500km/s) As you note, this is pretty much swamped by the proper relative motion of our galaxies. Its blueshift indicates that Andromeda is approaching us at over 100 km/s. For galaxies outside of the the Local group , the Hubble flow dominates. Now the value of 55km/s assumes that space is smooth and homogeneous. This is approximately true on a universal scale, but it is not true on the scale of a galaxy cluster, where local gravitational effects dominate the curvature of spacetime. The general expansion of spacetime has very little effect on the motion of galaxies in the local group, as discussed by Iorio's paper on the motion of a gravitationally bound binary system | {
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18,875 | We're (basically) in the middle of an arm of our galaxy. That is to say, we're sitting in the middle of a dense disc of stars. It would seem to me that. You should see: the thick line of the milky way all around you , i.e. on that plane, in all four directions. (Additionally - sure - in the particular direction of the galaxy center, you would additionally see the huge central bulge.) However: this does not seem to be the case : when you look at the milky way from the vicinity of our solar system, you basically see it "in one direction". What am I misunderstanding? How come the sky object "the milky way" is well-known as only a lump/strip in one direction, rather than, a lump/strip that goes right around us? -- Put it this way... Quite simply does anyone have any photography of the anti-galactic point? (Near "Auriga" right?) Does it show any "milky way band" going through it? If not why not? Looking outwards, we are still looking through ~30k lightyears of the dense disk we are sitting in. | I'll turn my comment into a full answer. To put it simply, we actually do see the Milky Way all around us, even in the diametrically opposite direction from the galactic core. You can see this in the image below which is a full sky image I took from APOD . If you look at the edges of the disk in that image, you're looking at what is actually the edge of our galaxy, opposite in direction from the core. In effect, this is precisely the image you requested because it does contain the part of the sky which contains the anti-core part of the galaxy. It is certainly not as bright, but there are still stars and dust out there. In fact, if you look very closely, you still see lots of dark splotches that mask distance stars and galaxies, indicating there is dust there. I think the issue you may be having is that you expect many more stars to be within the outer regions of the disk than there actually are. The stellar density profile for our disk is roughly exponential, meaning there are literally exponentially more stars near the core than at the edges. If this means anything to you, the scale length for the exponential radial density profile is ~4 kpc. To really get a good understanding of the stellar distribution, take a look at Jurić et al. (2008) . They looked at (~48 million) stars from the SDSS and analyzed the stellar distribution across our galaxy (that is visible to us). You should find figures 10 through 18 of particular interest, however I'll present part of figure 16 here. This image shows the (logarithmic) density of stars as a function radius from the galactic core. The varying shades of grey indicate varying heights above the galactic plane (numbered in parsecs). The dashed lines are various exponential decay models with differing scale heights. You can see that these star densities, even within the limited radial ranges covered by the SDSS fall off by an order of magnitude! Hopefully this helps you appreciate the significant difference between the core brightness/visibility and that of the galactic edge. Here's a silver of Andromeda spectacularly demonstrating the falloff: even though galaxy photography tends to suggest to the casual eye an evenly-dense plate: This synthetic image referenced by Andy , from the Tycho Catalog Skymap also shows the situation clearly. | {
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19,059 | Imagine two major surface observatories, perhaps the "northermost" and "southernmost" such (ideally on a similar longitude). For nearby stars, could they each take a photo at the same time, and achieve a distance measurement to that star, based on about the size of the Earth as the baseline? (I appreciate that either of the scopes need only wait a few days, for their viewpoint to have moved a much longer baseline distance!) Or is that distance far too short? What then is about the minimum baseline we could parallax-measure the nearest stars, with our currently best telescopes? 100,000km, a million? Far more? | In principle, it's not impossible. The Gaia spacecraft, designed primarily for measuring stellar positions, is able to measure parallaxes up to 10 kpc away with 20% uncertainty. Its baseline is 2 AU; $2.3\times10^{4}$ times larger than the diameter of Earth. Thus, placing two Gaias on each side of Earth would be able to measure parallaxes of stars up to a distance of $10\,\mathrm{kpc} \,/\, 2.3\times10^{4} \simeq 0.4\,\mathrm{pc}$, meaning that you would almost be able to measure the distance of our nearest neighboring star, $\alpha$ Centauri, which lies at 1.3 pc. So you would just need improve your Gaias a little bit. This ignores small complications such as the atmosphere, but if you're willing to put them outside the atmosphere, you could do it. Of course, it would sort of be a waste of time, since we already know the distances to the nearest stars, but hey, go ahead. | {
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19,614 | I've read various articles and books (like this one ) stating that we are not certain about the geometry of the universe, but there were experiments on-going or planned that would help us find out. Recently though, I've watched a lecture by cosmologist Lawrence Krauss where he seems to categorically assert that the universe has been proven to be flat by the BOOMERanG experiment . Here's the relevant portion of the talk . I've looked around and there are still articles stating that we still don't know the answer to this question, like this one . So, my question is two-fold: Am I mixing concepts and talking about different things? If not then is this evidence not widely accepted by some reason? What reason would that be? | I think the reason you're suffering from conflicting sources is that you're mixing both new and old, out-of-date pieces of information. First off, the book you cited was published in 2001 - 15 years ago - and the other article you cite was published in 1999 - 17 years ago. There's been a lot of work done in the past 15 years, often under the term "precision cosmology", in an attempt to really nail down the precise content, shape, size, etc. of our Universe. By the early 2000's we pretty much knew the science behind everything (we knew about dark matter, dark energy, had well-developed theories on the Big Bang, etc.) but what we didn't have, were good, solid, believable numbers to put into these theories, explaining why the flatness of the universe was still contested in your sources. I'll direct you to two incredibly important observatories which have been paramount in achieving our goal of having "good numbers". The first is the Wilkinson Microwave Anistropy Probe (WMAP) , launched in 2001, and the second is the Planck satellite , launched in 2009. Both missions were designed to stare intently at the Cosmic Microwave Background (CMB) radiation and try to sort out the treasure trove of information which can be gleaned from it. In this vein, you might also come upon the Cosmic Background Explorer (COBE) , launched in 1989. This satellite had a similar purpose as the other two, but was not nearly as precise as the later two missions as to provide us with good numbers and definitive statements by the early 2000's. For that reason I'll mostly focus on what WMAP and Planck have told us. WMAP was a hugely successful mission which stared at the CMB for 9 years and created the most detailed and comprehensive map of its day. With 9 years of data, scientists were really able to reduce the observational errors on various cosmological quantities, including the flatness of the universe. You can see a table of their final cosmological parameters here . For the flatness, what you want to do is add up $\Omega_b$ (the baryonic matter density), $\Omega_d$ (the dark matter density), and $\Omega_\Lambda$ (the dark energy density). This will give you the overall density parameter , $\Omega_0$, which tells you the flatness of our universe. As I'm sure you know from your sources, if $\Omega_0 < 1$ we have a hyperbolic universe, if $\Omega_0 = 1$ our universe is flat, and $\Omega_0 > 1$ implies a spherical universe. From the results of WMAP, we have that $\Omega_0 = 1.000 \pm 0.049$ (someone can check my math) which is very close to one, indicating a flat universe. As far as I know, WMAP was the first instrument to give a truly precise measurement of $\Omega_0$, allowing us to say definitively that our universe appears flat. As you say, the BOOMERanG experiment also provided good evidence for this, but I don't think the results were nearly as powerful as WMAP's was. The other important satellite here is Planck. Launched in 2009, this satellite has provided us with the best high precision measurements of the CMB to-date. I'll let you dig through their results in their paper , but the punchline is that they measure the flatness of our universe to be $\Omega_0 = 0.9986 \pm 0.0314$ (calculated from this result table ), again extremely close to one. In conclusion, recent results (within the past 15 years) allow us to definitively state that our Universe appears flat. I don't think, at this time, anyone contests that or believes it is still uncertain. As it usually goes with science, answering one question has only resulted in more questions. Now that we know $\Omega_0 \simeq 1$, we have to ask why is it one? Current theory suggests it shouldn't be - that it should be either enormously small or enormously large. This is known as the Flatness Problem . That in turn delves into the Anthropic Principle as an attempted answer, but then, I'm getting out of the scope of this question. | {
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19,966 | I was trying to describe how vast the largest known star is to someone and felt I wasn't quite able to relate the scale difference. I know it's roughly 1500 times larger than the sun. Anyone know of a good analogy? | After some playing around with wolfram alpha and google my best comparison has been The sun compared to VY Canis Majoris is like a donut compared to the London Eye. The London Eye is about 120m in diameter, this divided by 1500 is about 8cm which is roughly the diameter of a ring donut. | {
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20,127 | The TRAPPIST-1 system is around an ultra-cool dwarf star. I went looking for more information on that kind of star, and found very little. The Wikipedia article on it lengthened from a minimal stub to one paragraph in the 15 minutes from when I began the search, after the press conference on the discovery of 7 Earth-sized planets in orbit around TRAPPIST-1. What kind of star is this? | Breaking the phrase down: Dwarf star - a term I will never understand - is used to describe relatively small, dim stars. Unfortunately, this encompasses most main-sequence stars, which are indeed dwarfs compared to large giants and supergiants. Ultra-cool , as called2voyage already discussed, means that the star has an effective temperatures of less than 2,700 kelvin. This is about half of the Sun's surface temperature. Spectroscopically, it means that these stars are of class M7-L8 , i.e. really, really cool stars that just barely reach the threshold for nuclear fusion. Actually, some L dwarfs will never fuse hydrogen as normal stars do, and will become brown dwarfs. Therefore, "ultra-cool dwarf" doesn't necessarily refer to just main sequence stars. Other characteristics in some cases (from Cruz et al .): Lithium in some L-type objects, indicative of low temperatures, low masses, and young ages in those particular objects. Very low surface gravities in late M-type dwarfs (though likely not as much in an M7 dwarf like TRAPPIST-1). Very low metallicity in certain other L-type objects. TRAPPIST-1 actually appears to be a rather massive ultra-cool dwarf, then, in comparison to some of these other objects. Here's an annotated Hertzsprung-Russell diagram, with a box for ultra-cool dwarfs and a circle for where TRAPPIST-1 is, approximately: Image from Wikipedia courtesy of user Saibo, under the Creative Commons Attribution-Share Alike 2.5 Generic license . | {
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20,142 | According to the NASA announcement and the Wikipedia article on TRAPPIST-1 , the planets around the TRAPPIST-1 star are named TRAPPIST-1 b through h. What happened to "TRAPPIST-1 a"? | The convention for planetary naming is that the closest planet to the star (if multiple planets are found at the same time) is named "star"b, then "star"c and so on. As correctly pointed out by Zephyr, if the discoveries are more haphazard, the order of discovery takes precedence over distance from the star). So, there is no Trappist-1a. Or you can think of it being the star itself if you like, although that becomes confusing if the star is itself part of a multiple star system, where it might be known as "star"A or "star"B etc. (note, capital letters), depending on where it is in the hierarchy. | {
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20,164 | I was going through all the questions in this community related to TRAPPIST-1 in order to know how planets TRAPPIST-1b to TRAPPIST-1h were discovered, but there aren't any. How were they discovered? | The star at the center of TRAPPIST-1 is called 2MASS J23062928-0502285 . It was discovered by the Two Micron All-Sky Survey (2MASS), which imaged the whole sky in the infrared between 1997 and 2001. This resulted in a catalog of over 300 million objects. TRAPPIST-1 itself was cataloged in 1999. The name is actually its coordinates in right ascension and declination. The planets of the TRAPPIST-1 were discovered by the method of transit photometry . The way this works is a telescope watches a star for a period of time and records the amount of light coming from the star. They chart how much light is coming from the star as a function of time, creating a light curve . If they see periodic dips in the intensity from the star, there is a high likelihood that that star has a planet in orbit around it . The planet blocks the light from the star every time it passes between us and the star. This causes the dips in the light curve. One advantage to this method is that you can scan multiple stars in the same field of view, analyzing them all for planets. By measuring how long it takes planets to pass in front of the star, how much light it blocks, and how often they orbit, scientists can calculate the masses of these planets and how far away from the star they are by using Kepler's laws of motion . TRAPPIST-1 was initially determined to have planets orbiting it by the Transiting Planets and Planetesimals Small Telescope - South team. From their data they determined that it had at least 3 planets. One of these planets was in the habitable zone of the star. They published their results in the journal Nature in May 2016. Once TRAPPIST had determined that the system had planets around it, NASA trained the Spitzer Space Telescope on it. Ground-based observations of Trappist-1 are difficult because it is so dim. Spitzer, an infrared telescope, made more precise measurements of the light curves and determined that there were at least 7 planets in orbit around it, 3 of which were in the habitable zone. Additional observations were made by numerous other telescopes including the Very Large Telescope, UKIRT, the Liverpool Telescope, and the William Herschel Telescope. The results were also published in Nature . Here's an image showing the light curve of the TRAPPIST-1 system as measured by Spitzer . | {
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20,276 | I have just been watching a podcast called "Deep Astronomy" and the discussion was about a super fast spinning black hole discovered with the NuSTAR space observatory. This black hole was modeled with high confidence to be spinning at about 99% of the maximum spin rate. They stopped short of saying that the tangential velocity of this spin rate is "c" (and how can a singularity have a "tangential velocity"?) They did say that the event horizon at maximum spin of a stellar black hole is about 1-1/2 km. and that if a black hole were to spin faster then the result would be a "naked singularity" that would defy the laws of physics (GR). Also, shouldn't all black holes spin extremely fast (conservation of angular momentum) or would a retrograde accretion disk slow it down. Could somebody clarify this whole "black hole spin thing" without getting too complicated?? | Since I like math, let's throw some math into this. I'll try to keep it as simple as possible though. Kerr Black Holes A rotating black hole is known as a Kerr Black Hole (named after Roy Kerr who found the numerical solution to GR equations for rotating black holes). In the case of a rotating black hole, there are two important parameters used to describe the black hole. The first is of course the mass of the black hole $M$ . The second is the spin $a$ . Really $a$ isn't the spin itself $-$ it's defined by $a=J/M$ (see footnote) where $J$ is the angular momentum of the black hole $-$ but it's a good proxy for spin so often you'll see scientists get lazy and just call it the spin of the black hole. The mathematics will tell you that Kerr black holes have the limitation that $$0 \le a/M \le 1$$ Black Hole Event Horizon The important parameter we want to calculate is the black hole's radius. If you run through the math, you find that this radius is given by $$r_e = M+(M^2-a^2)^{1/2}$$ In the case when $a/M=0$ (and thus $a=0$ ), this reduces to just $r_e=2M$ , or in regular units (instead of geometrized units) $r_e=2GM/c^2$ . Hopefully you can see that this just reduces to the normal Schwarzchild radius for a non-rotating black hole and thus the equation above is a generalization to account for spin. Let's look at the other limit when $a/M=1$ (and thus $a=M$ ). In this case, you find that the radius is $r_e=M$ . When $a/M=1$ , you have a maximally rotating black hole, and your radius is half the normal Schwarzchild radius of a non-rotating black hole. This equation defines the radius of the Event Horizon, the point after which there is no returning from the black hole. Ergosphere As it turns out, when you define your equation to calculate the radius of the black hole, there are actually multiple solutions! The section above shows one such solution, but there's another important solution as well. This radius, sometimes called the static limit is given by the equation $$r_{s} = M+\left(M-a^2\cos^2(\theta)\right)^{1/2}$$ Notice that this is almost exactly the same as above, except for that extra $\cos^2(\theta)$ . This defines a different, slightly larger and somewhat "pumpkin-shaped" horizon which encompasses the inner Event Horizon defined above. The region between this outer horizon and the inner horizon is known as the Ergosphere . Without getting into the nitty gritty details, I'll just say that one important point about the Ergosphere is that anything inside it (that is, $r_e<r<r_s$ ) must rotate exactly with the black hole - it is physically impossible to stay still here! Answers They stopped short of saying that the tangential velocity of this spin rate is "c" (and how can a singularity have a "tangential velocity"?) When you talk about the tangential velocity, there are multiple components of this black hole you/they may be talking about. One such tangential velocity is the tangential velocity of the event horizon (defined by $r_e$ above). We can take a look at the case of a maximally rotating black hole and say that the angular momentum, based on the equations above, of such a black hole is given by $$J_{max} = a_{max}Mc = M^2c$$ Note that I've dropped the geometrized units just to be completely explicit. This has introduced an extra $c$ now. Remember that $a_{max}$ is achieved when $a/M=1$ . We can also define the angular momentum using the standard equation from physics 101, $J=rMv_\perp$ , where of course $r$ is the radius of your object, and $v_\perp$ is the perpendicular, or else tangential, velocity of your spinning object. Recall from above that for a maximally rotating black hole, $r_e=M$ so we also have that $$J_{max} = r_eMv_{\perp} = M^2v_\perp$$ You can see that these two equations for $J_{max}$ only equal each other if the tangential velocity $v_\perp$ is equal to the speed of light $c$ . So yes, you are correct to presume that at the fastest possible rotations, the black hole's event horizon is rotating at the speed of light! I said though that there are multiple components you could talk about when discussing rotating black holes. The other, as you allude to, is the rotating singularity. You correctly point out - "how can a singularity have a tangential velocity"? As it turns out, Kerr black holes don't have point singularities, they have ring singularities . These are "rings" of mass with zero width but some finite radius. Almost like a disk of no height. These rings of course can then have a tangential velocity. You were correct to be suspicious of a point singularity having tangential velocity though. That is not possible. They did say that the event horizon at maximum spin of a stellar black hole is about 1-1/2 km. and that if a black hole were to spin faster then the result would be a "naked black hole" that would defy the laws of physics (GR). We know the equation exactly, since I defined it above. The radius of a stellar black hole (that is a black hole with mass exactly equal to the mass of the Sun, $M_\odot$ ) is given by $$r=\frac{GM_{\odot}}{c}=1.48\:km$$ So yes, they were correct on their radius. They also state that spinning any faster results in a naked singularity. This is entirely true. To see this, go back to the equation for the event horizon. Remember that our upper spin limit is that $a=M$ . What happens to our event horizon radius when $a>M$ (and thus $a/M>1$ )? For arguments say, lets say $a=2M$ . Then our event horizon radius becomes $$r_e=M-(M^2-a^2)^{1/2}=M-(M^2-4M^2)^{1/2}=M-(-3M^2)^{1/2}=M-i\sqrt{3}M$$ Suddenly our radius is complex and has an imaginary component! That means it is not physical and thus cannot exist . Now that we don't have an event horizon, our singularity can't hide behind it and is "naked", exposed to the universe for anyone to see. GR tells us such an event shouldn't be allowed to happen because it results in all sorts of violations of physics. So somehow, something has to prevent black holes from spinning faster than a maximal black hole. Shouldn't all black holes spin extremely fast (conservation of angular momentum) or would a retrograde accretion disk slow it down. Yes, that is true in general. All black holes should spin extremely fast, simply because of conservation of angular momentum. In fact, I don't think I can come up with a case where a black hole was found to be not spinning. Shown below is a plot from this Nature paper which shows the measured spin of 19 supermassive black holes. They're all spinning pretty fast with some of them almost at the speed of light. None of them are even close to not spinning. Footnote: In GR, to make the math easier, often scientists adopt special units known as geometrized units . These are units chosen in just such a way that the gravitational constant, $G$ , and the speed of light, $c$ , are equal to one. There's infinitely many units which allow this. Essentially this means no GR equations have $G$ or $c$ in them, but they're implicitly there, they're just equal to one and so not shown. | {
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20,362 | I recently saw this image, but somebody told me that it is fake, possibly photoshopped. The moon and the sun appear to be on the same perpendicular from the horizon. Is there any way to tell if this shot was actually captured? | No, it's not real. Given that the moon appears to be nearly full, it would be near to the horizon opposite the sun, which is clearly not the case. Also the framing of the sun and moon between the trees means that the sun and moon are on the same side of the trees, which is not possible when the moon is full. | {
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20,383 | It seems to me that the Moon has appeared in many locations - apparent locations - in the 'dome of the sky'. I know that the Sun follows an arc from east to west and that the arc is smaller in winter than in summer. Therefore the Sun cannot appear in every location in the 'dome of the sky'. My question is: can the moon appear in any (every?) location in the 'dome of the sky'? | Like the sun, the moon, from our perspective on the surface of the Earth, rises in the East and sets in the West. However, it does not rise exactly due East and set exactly due West. If you were to track the position of the moonrise along the horizon over the course of a year, you would notice that is varies considerably. It cannot, however, be located everywhere in the night sky, even over the course of a year. To understand why, we need to understand some celestial geometry. The ecliptic plane is the plane which a hypothetical line between the Earth and the sun sweeps out over the course of a year. In the image above, you can see
how the Earth's orbit exists within this plane. Also notice that the line from the Earth's geographic south pole through its north pole (its rotational axis) is not perpendicular to this ecliptic plane. In fact it is 23.5 degrees off this perpendicular direction. Notice that the Earth's rotational axis points in the same direction regardless of which side of the Sun the Earth is currently on. This rotational axis always points towards the star Polaris. This means that on different parts of the year, the sun will rise at varying points along the horizon. The Earth's equatorial plane is a plane defined to extend out from the Earth's equator in all directions. This plane is inclined relative to the ecliptic plane by 23.5 degrees (the same amount as the rotational axis is tilted relative to the perpendicular direction of the ecliptic plane). Now consider the Earth moon system. In order to determine how far North or South the moon can appear to rise from, you need to consider the angle between the Earth's equatorial plane and the moon's orbital plane. Since the Earth's rotational axis is inclined 23.5 degrees relative to the ecliptic, and the Moon's orbit is inclined 5.14 degrees relative to the ecliptic, the highest the Moon's orbit can be relative to the equatorial plane of the Earth is 28.64 degrees. Thus, the Moon cannot be directly overhead of any point on the surface of the Earth if that point is 28.64 degrees above or (by an symmetric argument) below the equator. Thus, the Moon cannot appear in every location in the sky for a given location on Earth. There are parts of the sky that the geometry of the Earth moon system simply will not permit the Moon to exist. | {
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20,389 | We know that a planet is formed from the nebula of gas and dust. Gravity pulls the particles together in a core. So we all know that the gas cloud is a cloud of hydrogen. But how did water form? From where did oxygen come? How did it combine to make $\text{H}_2\text{O}$? | Like the sun, the moon, from our perspective on the surface of the Earth, rises in the East and sets in the West. However, it does not rise exactly due East and set exactly due West. If you were to track the position of the moonrise along the horizon over the course of a year, you would notice that is varies considerably. It cannot, however, be located everywhere in the night sky, even over the course of a year. To understand why, we need to understand some celestial geometry. The ecliptic plane is the plane which a hypothetical line between the Earth and the sun sweeps out over the course of a year. In the image above, you can see
how the Earth's orbit exists within this plane. Also notice that the line from the Earth's geographic south pole through its north pole (its rotational axis) is not perpendicular to this ecliptic plane. In fact it is 23.5 degrees off this perpendicular direction. Notice that the Earth's rotational axis points in the same direction regardless of which side of the Sun the Earth is currently on. This rotational axis always points towards the star Polaris. This means that on different parts of the year, the sun will rise at varying points along the horizon. The Earth's equatorial plane is a plane defined to extend out from the Earth's equator in all directions. This plane is inclined relative to the ecliptic plane by 23.5 degrees (the same amount as the rotational axis is tilted relative to the perpendicular direction of the ecliptic plane). Now consider the Earth moon system. In order to determine how far North or South the moon can appear to rise from, you need to consider the angle between the Earth's equatorial plane and the moon's orbital plane. Since the Earth's rotational axis is inclined 23.5 degrees relative to the ecliptic, and the Moon's orbit is inclined 5.14 degrees relative to the ecliptic, the highest the Moon's orbit can be relative to the equatorial plane of the Earth is 28.64 degrees. Thus, the Moon cannot be directly overhead of any point on the surface of the Earth if that point is 28.64 degrees above or (by an symmetric argument) below the equator. Thus, the Moon cannot appear in every location in the sky for a given location on Earth. There are parts of the sky that the geometry of the Earth moon system simply will not permit the Moon to exist. | {
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20,466 | In astronomy distances are generally expressed in non-metric units like: light-years, astronomical units (AU), parsecs, etc. Why don't they use meters (or multiples thereof) to measure distances, as these are the SI unit for distance? Since the meter is already used in particle physics to measure the size of atoms, why couldn't it be used in astrophysics to measure the large distances in the Universe? For example: The ISS orbits about 400 km above Earth. The diameter of the Sun is 1.39 Gm (gigameters). The distance to the Andromeda Galaxy is 23 Zm (zettameters). At its furthest point, Pluto is 5.83 Tm (terameters) from the Sun. Edit: some have answered that meters are too small and therefore not intuitive for measuring large distances, however there are plenty of situations where this is not a problem, for example: Bytes are used for measuring gigantic amounts of data, for example terabytes (1e+12) or petabytes (1e+15) The energy released by large explosions is usually expressed in megatons, which is based on grams (1e+12) The SI unit Hertz is often expressed in gigahertz (1e+9) or terahertz (1e+12) for measuring network frequencies or processor clock speeds. If the main reason for not using meters is historical, is it reasonable to expect that SI-unites will become the standard in astronomy, like most of the world switched from native to SI-units for everyday measurements? | In addition to the answer provided by @HDE226868, there are historical reasons. Before the advent of using radar ranging to find distances in the solar system, we had to use other clever methods for finding the distance from the Earth to the sun; for example, measuring the transit of Venus across the surface of the sun . These methods are not as super accurate as what is available today, so it makes sense to specify distances, that are all based on measuring parallaxes, in terms of the uncertain, but fixed, Earth-Sun distance. That way, if future measurements change the conversion value from AU to meters, you don't have to change as many papers and textbooks. Not to mention that such calibration uncertainties introduce correlated errors into an analysis that aren't defeatable using large sample sizes. I can't speak authoritatively on the actual history, but solar system measurements were all initially done in terms of the Earth/sun distance. For example, a little geometry shows that it's pretty straightforward to back out the size of Venus's and Mercury's orbit in AU from their maximum solar elongation. I don't know how they worked out the orbital radii of Mars, etc, but they were almost certainly done in AU long before the AU was known, and all of that before the MKS system existed, let alone became standardized. For stars, the base of what is known as the "cosmological distance ladder" (that is "all distance measures" in astronomy) rests on measuring the parallax angle: $$\tan \pi_{\mathrm{angle}} = \frac{1 AU}{D}.$$ To measure $D$ in 'parsecs' is to setup the equation so that the angle being measured in arcseconds fits the small angle approximation. That is: $$\frac{D}{1\, \mathrm{parsec}} = \frac{\frac{\pi}{180\times60\times60}}{\tan\left(\pi_{\mathrm{angle}} \frac{\pi\, \mathrm{radians}}{180\times60\times60 \, \mathrm{arcsec}}\right)}.$$ In other words, $1\operatorname{parsec} = \frac{180\times 3600}{\pi} \operatorname{AU}$ . Astronomers also have a marked preference for the close cousin of mks/SI units, known as cgs . As far as I can tell, this is due to the influence of spectroscopists who liked the "Gaussian units" part of it for electromagnetism because it set Coulomb's constant to 1, simplifying calculations. | {
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20,499 | Is there a stable geostationary orbit around the Moon? My feeling is, that the orbit would collide with Earth, because of the Moon's slow rotation. | First off, such an orbit wouldn't be a geostationary orbit since geo- refers to the Earth. A more appropriate name would be lunarstationary or selenostationary . I'm not sure if there is an officially accepted term since you rarely hear people talk about such an orbit. You can calculate the orbital distance of a selenostationary orbit using Kepler's law: $$a = \left(\frac{P^2GM_{\text{Moon}}}{4\pi^2}\right)^{1/3}$$ In this case, $a$ is your orbital distance of interest, $P$ is the orbital period (which we know to be 27.321 days or 2360534 seconds), $G$ is just the gravitational constant, and hopefully it is obvious that $M_{\text{Moon}}$ is the mass of the Moon. All we have to do is plug in numbers. I find that $$a = 88,417\:\mathrm{km}=0.23\:\mathrm{Earth\mathit{-}Moon\:Distance}$$ So I at least match your calculation pretty well. I think you were just relying on Wolfram Alpha a bit too much to get the units right. The units do work out right though. If you want to determine if this orbit can exist however, you need to do a bit more work. As a first step, calculate the Moon's Hill Sphere . This is the radius at which the Moon still maintains control over it's satellite, without the Earth causing problems. The equation for this radius is given by $$r \approx a_{\text{Moon}}(1-e_{\text{Moon}})\sqrt[3]{\frac{M_{\text{Moon}}}{3M_{\text{Earth}}}}$$ In this equation, $a_{\text{Moon}} = 348,399\:\mathrm{km}$ is the Moon's semi-major axis around the Earth and $e_{\text{Moon}} = 0.0549$ is the Moon's orbital eccentricity. I'm sure you can figure out that the $M$'s are the masses of the respective bodies. Just plug and chug and you get $$r \approx 52,700\:\mathrm{km}$$ A more careful calculation , including the effects of the Sun is slightly more optimistic and provides a Hill radius of $r = 58,050\:\mathrm{km}$. In either case though, hopefully you can see that the radius for a selenostationary orbit is much farther than the Hill radius, meaning that no stable orbit can be achieved as it would be too much perturbed by the Earth and/or the Sun. One final, semi-related point. It turns out almost no orbits around the Moon are stable, even if they're inside the Hill radius. This is primarily to do with mass concentrations (or mascons) in the Moon's crust and mantle which make the gravitational field non-uniform and act to degrade orbits. There are only a handful of "stable" orbits and these are only achieved by orbiting in such a way as to miss passing over these mascons. | {
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20,823 | I've heard that the more an object spins, the less of a true sphere it is.
Using this logic most of neutron stars would be far from spherical,in general what shape are most neutron stars? | I don't think you'll find a single agreed shape for a rotating neutron star, not least because we don't have an agreed single model for the equation of state of the material in a neutron star (which is more complex than the name suggests). I found one openly available paper (I'm sure there are more) which will give you a rough flavor for the complexity of modeling the shape of neutron stars. As you'll see the difficulty of no single model for an equation of state (EOS is the shorthand typically used) is just one issue. I think "ellipsoid" should be considered as an approximation, although it's not something I'd consider written in stone. Remember that to be useful a paper has to provide not just a model for what the shape might be, but also someone has to provide a way to measure this, which is challenging. I think one of the hopes for the new era of gravitational wave astronomy is to be able (eventually) to make more useful and measurements that help us investigate the interior of neutron stars. So this is an open question, I think. @Rob-Jeffries asked a question in comment about typical numbers for the deformation, and I answered in comment but comments can be removed by the system, so I'm adding that information as an edit : In the first section of the paper I linked to they do quote fractional deformations as being typically $10^{−5}$, perhaps $10^{−4}$ in special cases and in extreme cases up to $10^{−3}$. However another paper gives an analysis based on crustal rigidity and a very small deformation for a particular neutron star. The paper I initially liked to describes an upper limit based on gravitational wave considerations, I think, rather than a general analysis. | {
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20,918 | I always hear the narrator of documentaries say that a star explodes because it ran out of fuel. Usually things explode when they have too much fuel, not when they run out of fuel.
Please explain... | Short answer: A tiny fraction of the gravitational potential energy released by the very rapid collapse of the inert iron core gets transferred to the outer layers and this is sufficient to power the observed explosion. In more detail: Consider the energetics of an idealised model star. It has a "core" of mass $M$ and initial radius $R_0$ and an outer envelope of mass $m$ and radius $r$. Now suppose the core collapses to a much smaller radius $R \ll R_0$ on such a short timescale that it decouples from the envelope. The amount of gravitational potential energy released will be $\sim GM^2/R$. A fraction of this released energy can be transferred to the envelope in the form of outward moving shocks and radiation. If the transferred energy exceeds the gravitational binding energy of the envelope $\sim Gm^2/r$ then the envelope can be blown into space. In an exploding star (a type II core collapse supernovae) $R_0\sim 10^4$ km, $R\sim 10$ km and $r \sim 10^8$ km. The core mass is $M \sim 1.2M_{\odot}$ and the envelope mass is $m \sim 10M_{\odot}$. The dense core is mostly made of iron and supported by electron degeneracy pressure . The star is said to have "run out of fuel" because fusion reactions with iron nuclei do not release significant amounts of energy. The collapse is triggered because nuclear burning continues around the core and so the core mass is gradually increased and as it does so it gradually shrinks (a peculiarity of structures supported by degeneracy pressure), the density increases and then an instability is introduced either by electron capture reactions or photodisintegration of iron nuclei. Either way, electrons (which are what is providing the support for the core) are mopped up by protons to form neutrons and the core collapses on a free fall timescale of $\sim 1$ s! The collapse is halted by the strong nuclear force and neutron degeneracy pressure. The core bounces; a shock wave travels outwards; most of the gravitational energy is stored in neutrinos and a fraction of this is transferred to the shock before the neutrinos escape, driving away the outer envelope. An excellent descriptive account of this and the previous paragraph can be read in Woosley & Janka (2005) . Putting in some numbers.
$$GM^2/R = 4\times 10^{46}\ {\rm J}$$
$$Gm^2/r = 3\times 10^{44}\ {\rm J}$$ So one only needs to transfer of order 1% of the collapsing core's released potential energy to the envelope in order to drive the supernova explosion. This is actually not yet understood in detail, though somehow supernovae find a way to do it. A key point is that the rapid collapse takes place only in the core of the star. If the entire star collapsed as one, then most of the gravitational potential energy would escape as radiation and neutrinos and there would be insufficient energy even to reverse the collapse. In the core collapse model, most (90%+) of the released gravitational energy is lost as neutrinos, but what remains is still easily sufficient to unbind the uncollapsed envelope . The collapsed core remains bound and becomes either a neutron star or black hole. A second way to cause a star (a white dwarf) to explode is a thermonuclear reaction. If the carbon and oxygen can be ignited in nuclear fusion reactions then enough energy is released to exceed the gravitational binding energy of the white dwarf. These are type Ia supernovae. | {
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