How fast is fuel escaping a rocket for it to reach the escape velocity 11 km/s?
I was having a discussion with a person and we were talking about whether or no the fuel had to be pushing out of a rocket at 11 km/s in order for the rocket to reach 11 km/s.
I said it has to be because of the fact that there has to be an equal and opposite reaction to force the rocket to that speed. Neither of us know enough about physics to find the actual answer.
escape-velocity
AdamAdam
$\begingroup$ Consider the question, "11km relative to what?" $\endgroup$ – Russell Borogove Oct 13 '20 at 2:08
$\begingroup$ Just to be clear, are you asking if the speed the fuel is pumped creates thrust? $\endgroup$ – GdD Oct 13 '20 at 7:11
$\begingroup$ As a rule of thumb, trying to make a single stage of a rocket have a delta V (change in velocity) that is more than three times exhaust velocity is pushing the rocket equation a bit too hard. That would require, at a minimum, a rocket whose initial mass is 95% propellant. A more realistic value is in the neighborhood of 90%. $\endgroup$ – David Hammen Oct 13 '20 at 8:07
$\begingroup$ You see, 11km/s is the instantaneous speed with which a non-accelerating projectile needs to escape Earth's gravity. Our rockets, on the other hand, are accelerating bodies, there is no need to exceed 11km/s, in fact, they don't. The highest speed of rockets within the Earth's atmosphere is about 7km/s. $\endgroup$ – varun Oct 13 '20 at 20:04
$\begingroup$ @varun Every spacecraft humanity has sent into interplanetary space has exceeded 11 km/s relative to Earth in the Low Earth Orbit region, with New Horizons as the reigning champion at 16.7 km/s. Even the Apollo missions went up to over 10.4km/s on chemical rockets. A spacecraft that wants to get anywhere further than LEO efficiently must exceed 7km/s.. $\endgroup$ – notovny Oct 14 '20 at 4:14
The velocity of a rocket can exceed its exhaust velocity.
It is possible for the velocity of a rocket to be greater than the exhaust velocity of the gases it ejects. ...The thrust of the rocket does not depend on the relative speeds of the gases and rocket, it simply depends on conservation of momentum.
Source https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/9-7-rocket-propulsion/
See also on the physics stack https://physics.stackexchange.com/q/73692
Also from Rocket and Spacecraft Propulsion: Principles, Practice and New Developments
It can be seen in Figure 1.6 that the rocket can travel faster than the speed of its exhaust. This seems counter-intuitive when thinking in terms of the exhaust pushing against something. In fact, the exhaust is not pushing against anything at all, and once it has left the nozzle of the rocket engine it has no further effect on the rocket. All the action takes place inside the rocket, where a constant accelerating force is being exerted on the inner walls of the combustion chamber and the inside of the nozzle. So while the speed of the rocket depends on the magnitude of the exhaust velocity, as shown in Figure 1.6, it can itself be much greater. A stationary observer sees the rocket and its exhaust passing by, both moving in the same direction, although the rocket is moving faster than the exhaust.
original screenshot
Note that based on this graph, a rocket with an exhaust velocity of 4 km/s can attain 11 km/s.
Organic MarbleOrganic Marble
$\begingroup$ Your choice of quotes is excellent! "All the action takes place inside the rocket..." is the clearest most concise sentence I've read all year. This is an "Aha!" answer. $\endgroup$ – uhoh Oct 13 '20 at 11:32
$\begingroup$ @uhoh I'm not sure it's related to the kind of action that you linked. $\endgroup$ – user253751 Oct 13 '20 at 11:50
$\begingroup$ @user253751 I'm way ahead of you What can "action" mean in "All the action take place inside the rocket."? $\endgroup$ – uhoh Oct 13 '20 at 11:51
$\begingroup$ @uhoh thanks for the edit/transcription! $\endgroup$ – Organic Marble Oct 13 '20 at 12:04
$\begingroup$ @CarlWitthoft I'd linked to Euler's "action" but I was thinking of Newton's $\endgroup$ – uhoh Oct 13 '20 at 19:00
An intuitive way to think about it: You have a big rocket, composed of two parts of equal mass: the payload part and the fuel part. You launch the exhaust (fuel) backwards at 1 km/s (for simplicity: all at once so you don't accelerate any of it forward first), and so clearly the payload part is now moving at 1 km/s forward. Open the fairing and reveal the payload: a rocket just like the first, only half as big, but same construction: half of it fuel, half payload! Launch its fuel backwards at 1 km/s, and the front part will accelerate by 1 km/s forward - but it was already moving at 1 km/s so now it moves at 2... and your newly ejected fuel is at stop relative to the launch speed. Now reveal the payload... which is again same type of rocket except half the size again! Once you launch its fuel backwards, the rocket will move at 3 km/s forward while the fuel - will move at 1 km/s forward!
Of course halving the size every time puts a serious damper on how far you can push it, and if you look at how much of the launch mass was fuel and how much payload, you can see how it goes - why rockets are ~95% fuel by weight, and why you need such an enormous rocket to launch a pretty tiny spacecraft. It also shows where specific impulse comes into play: what speed would you reach if instead of pushing the fuel away at 1 km/s every time you could push it at 3 km/s?
SF.SF.
$\begingroup$ I get it now, you said "half of the mass" +1 $\endgroup$ – Organic Marble Oct 13 '20 at 3:51
$\begingroup$ @OrganicMarble uh, you out of all people? $\endgroup$ – SF. Oct 13 '20 at 5:27
$\begingroup$ @OrganicMarble No, I mean you could run in circles around me where it comes to that stuff - the one weakness of your answers is that you know that stuff so clearly and in-depth you sometimes don't even realize someone might not know something "this fundamental" and skim a simpler part as something too obvious to explain. So how'd you not get my answer when it's essentially your answer dumbed down for general public? $\endgroup$ – SF. Oct 13 '20 at 13:47
$\begingroup$ I was tired last night and missed the "half of the mass" part when I first read it. It's an excellent explanation. $\endgroup$ – Organic Marble Oct 13 '20 at 13:48
$\begingroup$ @EricDuminil I rephrased it a bit - I guess the original phrasing didn't get that part properly in focus. I think at this point the reader will realize the original rocket was in fact 87.5% fuel, and get the general intuition of how the discretized example expands into continuous spectrum. Throwing logarithms at people fresh to nderstanding conservation of momentum is a quick way to confuse them worse. We kinda-sorta understand them intuitively, but only as long as we don't dive too deep. $\endgroup$ – SF. Oct 14 '20 at 8:12
For a rocket that is not subject to external forces, conservation of momentum dictates that $$m(t)\,\dot v(t) + v_e(t)\,\dot m(t) = 0$$ where
$m(t)$ is the mass of the rocket, including propellant, at time $t$,
$v(t)$ is the rocket's velocity at time $t$, relative to some inertial observer,
$\dot v(t)$ is the rocket's acceleration at time $t$,
$v_e(t)$ is the velocity at which the rocket expels exhaust, relative to the rocket, and
$\dot m(t)$ is the rate at which the rocket is losing mass.
Assuming a constant exhaust velocity, integrating this with respect to time results in the ideal rocket equation for a single stage of a rocket: $$\Delta v = v_e \ln\left(\frac{m_0}{m_f}\right)$$ where
$\Delta v$ is the rocket's change in velocity,
$v_e$ is the effective velocity at which exhaust leaves the rocket,
$m_0$ is the rocket's initial mass (payload, structure, and propellant),
$m_f$ is the rocket's final mass (payload and structure), and
$\ln(x)$ is the natural logarithm function.
This means that having a rocket's change in velocity exceed the rocket's exhaust velocity is eminently achievable. It requires that the rocket's initial mass be at least 63% propellant. Rockets that propel things into space typically have an initial mass that is about 90% propellant.
As a rule of thumb, trying to make a single stage of a rocket have a delta V that is more than three times exhaust velocity is pushing the rocket equation a bit too hard. That would require, at a minimum, a rocket whose initial mass is 95% propellant. A more realistic value is a rocket whose initial mass is in the neighborhood of 90% propellant. This results in a rocket that ideally has a delta V that is 2.3 times exhaust velocity.
This, by the way, is why the concept of a single stage to orbit rocket is appealing and yet seemingly just out of grasp. The delta V needed to put a payload into low Earth orbit ranges from over 9 km/s to a bit over 10 km/s, depending on the rocket. A rocket whose initial mass is 90% propellant and whose exhaust velocity is 4 km/s (both of which are feasible) can ideally achieve a delta V of 9.2 km/s. This is just in range of what is feasible. The problem is the word "ideally". That a single stage to orbit rocket is just on the cusp of what is feasible means that organizations that want to put things into orbit or beyond inevitably take on the significant added complexities associated with multistage rockets.
David HammenDavid Hammen
Here is what I hope is an intuitive argument that it does not need to be the case that the exhaust velocity is greater than whatever velocity the rocket is to achieve.
First of all, think about a 'rocket' which is powered by someone sitting in the back throwing pebbles out of it. The pebbles weigh $0.1\,\mathrm{kg}$ and the person can throw them at $10\,\mathrm{m/s}$ relative to the rocket.
At some point the person has thrown all but one pebble. The remaining mass of the rocket and the person is $100\,\mathrm{kg}$, so the total mass of the thing, including the pebble, is $100.1\,\mathrm{kg}$.
So what happens when the person throws this last pebble? Well, we can use conservation of momentum to tell us: If the rocket is travelling at $v$ just before the pebble is thrown, then its initial momentum is $100.1v$. Afterwards, the pebble is going at $v - 10$, so the final momentum is $100 (v + \Delta v) + 0.1(v - 10)$, where $\Delta v$ is the change in speed. So we know these are the same, so
$$100.1v = 100(v + \Delta v) + 0.1(v - 10)$$
And from this we work out that $\Delta v = 1/100\,\mathrm{m/s}$: it does not depend on $v$ at all.
Well, OK. So let's imagine now that, just before the last pebble is thrown, the rocket was travelling at $11\,\mathrm{km/s} - 0.005\,\mathrm{m/s}$. Well, just after the pebble is thrown is it now travelling at $11\,\mathrm{km/s} + 0.005\,\mathrm{m/s}$: it's now going faster than $11\,\mathrm{km/s}$.
But the pebble was thrown at far, far less than $11\,\mathrm{km/s}$
And obviously this is true for any velocity: if I start at $0\,\mathrm{km/s}$ and I have enough pebbles to throw, I can get to any velocity I like.
I will need a lot of pebbles though.
tfbtfb
$\begingroup$ I like this answer as It is much simpler and involve minimum equations, and convey clearly the important parts ("it doesn't depend on v at all"), and all the exemples clarify things. $\endgroup$ – Olivier Dulac Oct 13 '20 at 22:57
$\begingroup$ There's a bit of a sign problem on the momentum of the pebble, isn't there? Otherwise a very good answer. $\endgroup$ – jcaron Oct 15 '20 at 21:19
$\begingroup$ @jcaron: yes, thank you, I've fixed it now I hope! $\endgroup$ – tfb Oct 16 '20 at 8:08
Consider a rocket floating motionless in space. You'd like to start moving, so you throw some mass out the back in the form of exhaust, which accelerates your craft in the opposite direction. The speed of the mass exiting your ship doesn't matter at all - so long as it's moving, your ship will have equal momentum in the opposite direction.
Now consider the situation in a different reference frame. Your rocket isn't motionless at all, it's actually coasting through space at a constant speed (note: this scenario is fundamentally identical to being at rest). That speed can be anything, depending on your reference frame. But even in this new reference frame, throwing mass out the back of the rocket will still accelerate it. It doesn't matter what your speed is in your chosen reference frame, or how fast you throw the mass - it will always result in the rocket moving faster. From this, you can see that even a rocket moving very quickly can accelerate by ejecting mass an an arbitrarily low speed.
Any mass ejected from the back of a rocket will increase the rocket's forward velocity. Therefore, a rocket can move faster than the speed of its exhaust (and in fact can reach speeds arbitrarily close to c, so long as it has mass to eject). If this was not the case, it would violate the conservation of momentum - if you change the momentum of the exhaust but not the rocket, the overall momentum of the system is no longer constant. A change in exhaust momentum will always result in a change in rocket momentum, regardless of their relative speeds.
Nuclear HoagieNuclear Hoagie
$\begingroup$ Does this mean that a rocket's top speed is essentially infinite (well, the speed of light as far as we know)? Of course it would still be limited by the amount of propellant you can burn. $\endgroup$ – maaw Oct 13 '20 at 19:37
$\begingroup$ @maaw Right, the only limiting factor is how much fuel you have. There are practical limits due to the tyranny of the rocket equation, since for any amount of fuel you bring with you, you need more fuel to accelerate it, so there are diminishing returns. But if we had a supply ship that could deliver fuel when needed, there's nothing at all about a rocket's design that would limit its top speed (except c, of course). It can always go a little faster, so long as you have something to throw out the back, and it doesn't matter how hard you throw it. $\endgroup$ – Nuclear Hoagie Oct 13 '20 at 19:45
$\begingroup$ @maaw, It's actually explained by the Tsiolkovsky eqn (given in the answers of SF and Organic Marble). the limiting factor is not directly the fuel, but more the ratio of fuel mass to "rocket+payload" mass. For rocket+payload approaching 0 mass, the terminal velocity will approach infinity (well in non-relativistic mechanics anyway). $\endgroup$ – Carl Berger Oct 14 '20 at 20:46
There is no lower limit; however, as you lower the exhaust velocity, more mass must be exhausted.
Rocket propulsion works by the conservation of momentum. The change in momentum of the exhaust (its mass times its velocity) is equal but opposite in sign of the momentum of the rocket. I've illustrated this below.
Notice how the mass of the exhaust (drawn with a large m) is bigger than the mass of the rocket (drawn with a small m). For example, the Saturn 3rd stage carried 109,000 kg of propellant which propelled 64,000 kg of the empty third stage and Apollo payload.
In order for the products of mass times velocity to be equal in amount, the exhaust's large mass must be multiplied by a small velocity (small v above), and the rocket's small mass is multiplied by a large velocity (large v above).
As you increase the ratio between the masses, you likewise increase the ratio between the velocities. You could hypothetically get 11 km/s with a slow 1 m/s exhaust, if the exhaust was 11,000 times the mass of the remaining rocket. In practice, the mass ratios are not that extreme.
The minus sign in the above equation means that the velocities are in opposite directions. The exhaust moves backward, and the rocket moves forward.
A similar phenomenon happens with a gun and a bullet. The gun is like the exhaust: high mass but low velocity (the "recoil"). The bullet is like the rocket: small mass but high velocity.
DrSheldonDrSheldon
Suppose you throw something out of the back of your spaceship. The total momentum doesn't change, so the velocity of what you threw times its mass will be equal to the (negative of) the velocity of the remaining ship times its mass (in the frame of reference where the original velocity of the spaceship is zero). If what you threw out was more massive than the rest of your spaceship, then the velocity of your spaceship will be greater than that of the exhaust.
It's more complicated with actual rockets, since the exhaust is coming out over time rather than in a single instance, and rockets are generally used in gravitational fields, but the basic principle holds.
AcccumulationAcccumulation
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