What is the safe distance to a supernova explosion?
In other words, what stars near the Sun may have an impact on the Solar system equilibrium or the Earth life if they become supernova ? Is SN 1987 A too far ?
astrophysics atmospheric-science plasma-physics solar-system supernova
honeste_vivere
$\begingroup$ When I asked this one , I was checking this assumption : "more the question is stupid, better it is upvoted". $\endgroup$ – user46925 Oct 12 '16 at 22:53
I worked this out a little while back in order to check something said on one of these Nova or other science show specials. I wanted to know how much energy would be required to remove the entire atmosphere of the Earth and whether a supernova (or other astronomical event) could possibly do this.
Let's assume the following quantities:
$M_{E}$ = mass of Earth $\sim 5.9742 \times 10^{24} \ kg$
$R_{E}$ = mean equatorial radius of Earth $\sim 6.378140 \times 10^{6} \ m$
$h_{E}$ = mean scale height of Earth's atmosphere $\sim 10 \ km$
$AU$ = astronomical unit $\sim 1.49598 \times 10^{11} \ m$ or $\sim 2.0626 \times 10^{5} \ parsecs$
Let's assume Earth's atmosphere has the following concentrations by volume:
$N_{2} \sim 78.08$%
$O_{2} \sim 20.95$%
$Ar \sim 0.93$%
$C O_{2} \sim 0.039$%
To start, we find the total volume of Earth's atmosphere given by: $$ V_{atm} = 4 \pi \int_{a}^{b} \ dr r^{2} = \frac{4 \pi}{3} r^{3} \vert_{a}^{b} $$ where we assume $a = R_{E}$ and $b = R_{E} + h_{E}$, which gives us a volume of $V_{atm} \sim 5.120 \times 10^{18} \ m^{3}$. Thus, we can estimate fractional volumes of each constituent gas to be:
$N_{2} \sim 3.998 \times 10^{18} \ m^{3}$
$O_{2} \sim 1.073 \times 10^{18} \ m^{3}$
$Ar \sim 4.762 \times 10^{16} \ m^{3}$
$C O_{2} \sim 1.997 \times 10^{15} \ m^{3}$
This allows us to estimate to total number of particles for each constituent gas using: $$ N_{j} = V_{j} \times \frac{ 1 }{ V_{atm} } \times N_{A} $$ where $N_{A}$ is the Avogadro constant and $V_{j}$ is the fractional volume of species $j$. This gives us the following values for $N_{j}$:
$N_{2} \sim 1.074 \times 10^{44} \ molecules$
$O_{2} \sim 2.882 \times 10^{43} \ molecules$
$Ar \sim 1.279 \times 10^{42} \ molecules$
$C O_{2} \sim 5.365 \times 10^{40} \ molecules$
Now we estimate the total number of moles of each constituent gas using: $$ M_{j} = \frac{ N_{j} }{ N_{A} } $$ which gives us:
$N_{2} \sim 1.784 \times 10^{20} \ moles$
$O_{2} \sim 4.786 \times 10^{19} \ moles$
$Ar \sim 2.124 \times 10^{18} \ moles$
$C O_{2} \sim 8.910 \times 10^{16} \ moles$
Ionizing Earth's Atmosphere
As a first approximation, we can assume that if the atmosphere were ionized, it may be easier to lose it (e.g., see answer that discusses this). Thus, let's see how much energy is needed to ionize the atmosphere.
We can look up the ionization energy for argon and the dissociation energy for each of the other molecules, given to be:
$E_{i,Ar} \sim 1520.6 \ kJ \ mole^{-1}$
$E_{d,N2} \sim 945 \ kJ \ mole^{-1}$
$E_{d,O2} \sim 497 \ kJ \ mole^{-1}$
$E_{d,CO} \sim 360 \ kJ \ mole^{-1}$
$\rightarrow E_{d,CO2} \sim 720 \ kJ \ mole^{-1}$
Using these values and the number of moles of each species, we can estimate the total energy needed to ionize all the argon and dissociate all the other constituent gases, which gives us:
$N_{2} \sim 1.686 \times 10^{26} \ J$
$O_{2} \sim 2.378 \times 10^{25} \ J$
$C O_{2} \sim 6.414 \times 10^{22} \ J$
$Ar \sim 3.230 \times 10^{24} \ J$
A typical supernova (i.e., Type Ia) releases something like $\sim 10^{44} \ J$ of total energy (Note that hypernovae can release more and other stellar events can produce more energy, but more on that later.). If we assume all of that energy is directly injected to ionize the atmosphere and that it radiates from the source in a spherically symmetric manner, then the intensity will decrease as $\sim r^{-2}$, where $r$ is the distance from the source emitter (i.e., supernova) to the absorber (i.e., Earth's atmosphere). Ignoring angle of incidence issues, the absorbing area of the Earth is just $4 \ \pi R_{E}^{2} \sim 5.099 \times 10^{8} \ km^{2}$ or $\sim 5.099 \times 10^{14} \ m^{2}$.
We can estimate the minimum safe distance by comparing the energies and ignore any losses by the absorber, which gives us a zeroth approximation: $$ A_{source} \ E_{abs} = A_{abs} \ E_{source} \\ r_{source}^2 = r_{abs}^2 \frac{ E_{source} }{ E_{abs} } $$ where $source$ is the energy source (i.e., supernova here) and $abs$ is the absorber (i.e., Earth's atmosphere). If we solve for $r_{source}$ as our minimum safe distance for each constituent gas individually, we have:
$r_{source}$ for $N_{2} \sim 4.906 \times 10^{15} \ m$ or $\sim 33,000 \ AU$ or $\sim 0.16 \ parsecs$
$r_{source}$ for $O_{2} \sim 1.307 \times 10^{16} \ m$ or $\sim 87,000 \ AU$ or $\sim 0.42 \ parsecs$
$r_{source}$ for $C O_{2} \sim 2.515 \times 10^{17} \ m$ or $\sim 1,680,000 \ AU$ or $\sim 8.15 \ parsecs$
$r_{source}$ for $Ar \sim 3.544 \times 10^{16} \ m$ or $\sim 237,000 \ AU$ or $\sim 1.15 \ parsecs$
So in the grand scheme of things, these distances are small enough to suggest that most stars are far enough away that they will not completely ionize our atmosphere.
Energizing Earth's Atmosphere
What if we tried to determine how much energy would be necessary to increase the particles mean kinetic energy such that their most probable speeds exceeded the escape speed of Earth's gravity?
At STP the constituent gases considered have thermal speeds (i.e., most probable speeds) of:
$N_{2} \sim 417.15 \ m/s$
$O_{2} \sim 390.31 \ m/s$
$C O_{2} \sim 332.82 \ m/s$
$Ar \sim 349.33 \ m/s$
The difference in kinetic energy between their STP energy and escape speed energy is given by: $$ \Delta K_{j} = \frac{ 1 }{ 2 } m_{j} \left( V_{esc}^{2} - V_{Tj}^{2} \right) $$ which is, for each constituent gas, given by:
$N_{2} \sim 2.904 \times 10^{-18} \ J$
$O_{2} \sim 3.318 \times 10^{-18} \ J$
$C O_{2} \sim 4.565 \times 10^{-18} \ J$
$Ar \sim 4.143 \times 10^{-18} \ J$
If we multiply these values by the total number of particles we estimated previously, $N_{j}$, then we can estimate the total energy needed to effectively evaporate the atmosphere of each constituent gas. The energies needed are:
which corresponds to a total energy of $\sim 4.131 \times 10^{26} \ J$. Using a similar approach as for the ionization above, we get minimum safe distances of:
$r_{source}$ for $C O_{2} \sim 1.287 \times 10^{17} \ m$ or $\sim 860,000 \ AU$ or $\sim 4.17 \ parsecs$
So again, these distances are small enough to suggest that most stars are far enough away that they will not completely evaporate our atmosphere.
The above estimates are for absolute devistation and are only valid given the assumptions. Note that an extinction level event probably would not require the total ionization or evaporation of Earth's atmosphere. Rather, only a fraction of the atmosphere would need to be ionized or evaporated to cause problems, as the two links provided by @BowlOfRed suggest.
In my original post I eluded to more energetic events like hypernova but forgot to discuss them. Typically, hypernova are not much more than ~50 times as energetic as supernova, would not alter the above distances much. Gamma-ray bursts, again have comparable total energy releases, but here the energy is focused into a relatively narrow beam rather than spherical. Even so, the beam would need to be focused directly on Earth and the source relatively close to evaporate and/or ionize the Earth's atmosphere.
I should also point out that a significant fraction (in some cases, nearly all the energy) of the energy in a supernova goes to neutrinos, which do effectively nothing to our atmosphere. Thus, the above values are grossly underestimated. Meaning, a supernova (or other huge energy release) would need to be significantly closer to cause the same effects.
What I did not mention is that the entire atmosphere need not be ionized or evaporated for there to be significant problems. Simply ionizing a significant fraction of the $N_{2}$ could produce damaging levels of $NO_{x}$'s that lead to acid rain and other polluting effects.
Further, a significant enhancement in the level of ionizing radiation could damage enough of the ozone layer to lead to large scale crop failures. Though an atmospheric chemist/physicist would be better suited to estimate the minimum safe distance for these effects.
honeste_viverehoneste_vivere
$\begingroup$ TY for this great answer $\endgroup$ – user46925 Feb 1 '16 at 20:58
$\begingroup$ While this is a great answer, is removing the atmosphere the mechanism by which a supernova would kill? E.G. raising the temperature of the atmosphere by 50 C would likely do it. Or raising the temperature of the ground. These would take much less energy than baking away the atmosphere. There may be other mechanisms. $\endgroup$ – mmesser314 Feb 28 '16 at 13:35
$\begingroup$ Just want to point out that the errors you would incur by treating the atmosphere as entirely nitrogen are completely irrelevant when making order of magnitude estimates like this. $\endgroup$ – DanielSank Mar 3 '16 at 23:24
$\begingroup$ No mistake. I'm saying tha tthe whole analysis where you include the various different atmospheric gasses is not needed. $\endgroup$ – DanielSank Mar 4 '16 at 16:08
$\begingroup$ It bears noting that Proxima Centuri, the closest star to the solar system, is about 1.3 parsecs away, and that the inert gas Argon is not ecologically very important to life on Earth. There are at least 56 star systems, however, within 5 parsecs, so the CO2 number is the one that matters. en.wikipedia.org/wiki/List_of_nearest_stars_and_brown_dwarfs A recent evaluation of the risk of a extinction causing gamma ray burst risk can be found at arxiv.org/abs/1609.09355 $\endgroup$ – ohwilleke Oct 14 '16 at 5:32
According to Phil Plait and others, anything over 100 light years (and probably a fair bit closer) should be safe. There aren't any known supernova candidates that close.
http://earthsky.org/space/supernove-distance https://twitter.com/BadAstronomer/status/201708339904778240
SN 1987A isn't even in our galaxy. It's over 150,000 light years distant.
BowlOfRedBowlOfRed
$\begingroup$ Twitter content as a reference? Come on. The other linked article gives no calculations or citations to literature. $\endgroup$ – Rob Jeffries Jan 28 '16 at 8:16
protected by Qmechanic♦ Feb 1 '16 at 21:42
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