7
$\begingroup$

Given the following graph: Escape velocities of gases at various temperatures on top of planets and celestial bodies,

Relations exist among the escape velocities, molar masses of gases and the masses of natural planets and satellites. Is there a general math formula where inserting a molar mass of a gas, mass of the celestial body and generating the boolean whether the gas can stay inbound of the body?

For what it worth, the graph shows a xenon atmosphere would have retained on top of Lunar. Even if radioactive Xenon poison from nuclear fission reactor gives away their neutrons to become stable isotope and is released at the moon or an intrinsic Xenon atmosphere would be possible, humans would have mined such precious Xenon resource.

How about Krypton?

$\endgroup$
4
  • 1
    $\begingroup$ "can stay inbound of the body" Its a statistical thing. all gasses escape, eventually, for all planets. It's just the timescale involved. The image supplied is for retaining atmosphere "long enough", being a billion years or so. $\endgroup$
    – PcMan
    Apr 28 at 2:25
  • $\begingroup$ Page 24: tcd.ie/Physics/people/Peter.Gallagher/lectures/PY4A03/pdfs/… $\endgroup$ Apr 28 at 4:12
  • $\begingroup$ @SurpriseDog Please do not answer questions in comments. $\endgroup$
    – Frostfyre
    Apr 28 at 12:09
  • $\begingroup$ @SurpriseDog Page19: If m>3kTR/2GM, the gas is retained? $\endgroup$
    – Kav
    Apr 29 at 14:24
11
$\begingroup$

In general, you want to compare some characteristic thermal velocity $v_{\mathrm{th}}$ of a gas molecule with the escape velocity of the planet. There are different choices you can make, all within a factor of 2 or so of each other. I tend to take $$v_{\mathrm{th}}=\sqrt{\frac{3k_BT}{m}}$$ where $k_B$ is Boltzmann's constant, $T$ is the atmospheric temperature$^{\dagger}$ and $m$ is the mass of a single molecule. If $v_{\mathrm{th}}\gg v_e$, where $v_e$ is the escape velocity, then most of that particular gas will be lost to space. If $v_{\mathrm{th}}\ll v_e$, then most of that gas will be retained. In the regime in between, where $v_{\mathrm{th}}\approx v_e$, you'll get a result somewhere in the middle; I believe a rule of thumb is that if $v_{\mathrm{tm}}\gtrsim v_e/6$, most of that gas will be lost over several billion years.

As gases follow Maxwell-Boltzmann distributions, obviously not all particles move with $v=v_{\mathrm{th}}$, so you'd expect some amounts of each gas to escape and some amounts to retain. It's impossible to say that all molecules of a particular gas will escape, but you can determine whether the majority of the gas will escape or be retained.

I should note that the above assumes the planet loses gases due to non-thermal escape mechanisms, such as the pathways by which Earth loses $\mathrm{O}^+$ near the poles. However, thermal escape is usually the dominant mechanism, and as a rule of thumb, you'll want to follow the above recipe and just compare $$v_{\mathrm{th}}=\sqrt{\frac{3k_BT}{m}},\quad v_e=\sqrt{\frac{2GM}{R}}$$ where $M$ and $R$ are the mass and radius of the planet and $G$ is the gravitational constant.


$^{\dagger}$Stricly speaking, I believe that this should be the temperature in the upper atmosphere, given that the temperature varies dramatically with altitude.

$\endgroup$
5
  • $\begingroup$ You’ll notice that, on larger timescales, the graph tends to shift upwards. Mars, for instance, per the graph should have an oxygen atmosphere— it has a CO2 one instead. The equation provides a “mean” velocity, and shifting the graph up provides a factor of safety for faster-moving gas to still be unable to escape the gravity cup $\endgroup$ Apr 28 at 10:22
  • $\begingroup$ You’ll also notice that Earth should be, according to the charts, holding on to a rich methane atmosphere. It’s believed that sizeable near-impacts can shave lighter gasses off the top. I don’t think there’s any equation for these two effects, but knowing they exist can help you build the world you want. $\endgroup$ Apr 28 at 10:25
  • $\begingroup$ I assume G is the gravitation constant, M is the mass of the celestial body (mass of Mars, the moon, Venus et cetera) and R is the radius of the celestial body? $\endgroup$
    – Kav
    Apr 29 at 7:21
  • $\begingroup$ @Kav Yes, you're right - sorry, I maybe should have said that explicitly. $\endgroup$
    – HDE 226868
    Apr 29 at 13:57
  • $\begingroup$ @JamesMcLellan To have an oxygen atmosphere you need either a source of oxygen (plants) or to have already oxidized everything around that can be oxidized. The latter isn't likely to happen, thus in practice oxygen atmospheres only happen with life. $\endgroup$ Oct 12 at 0:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.