It's a well known problem that Earth is slowly losing its supply of helium due to helium's ability to "bubble off" the atmosphere. All the gas giants have significant percentages of helium in their atmosphere so they are big enough. What's the minimum planet mass to hold onto atmospheric helium?

  • $\begingroup$ Since helium is not obtained from the atmosphere, I'm not sure it is an 'industrial' problem (well, until we exhaust the natural gas fields with reasonable helium concentrations). $\endgroup$ – Jon Custer Jun 16 '15 at 13:37

You have to compare the most probable speed of the atoms of He with the escape velocity from the planet.

So you are working out whether $$ \left(\frac{2kT}{m} \right)^{1/2} > \alpha\ \left(\frac{2GM}{R}\right)^{1/2},$$ where $T$ is the local temperature, $m$ is the He atom mass, $M$ is the planet mass and $R$ is the radius at which you are considering the escape speed.

The parameter $\alpha$ is there because even if the average thermal velocity is well below the escape speed, there will still be some fraction of those atoms in a Maxwell-Boltzmann distribution that will have enough kinetic energy to escape. But the escape also needs to occur in a region of the atmosphere that is sufficiently sparse that an energetic atom can escape before interacting with something else. The value of $\alpha$ is not an exactly determined quantity, the rate of "Jeans escape" (as it is known) will increase with decreasing $\alpha$. Often a number like $\alpha \sim 0.2$ is used if the atmosphere is to escape within hundreds of millions of years.

Anyway, the bottom line to this is that although planet mass features in this criterion, it is not the only parameter. The radius of the atmosphere from which an escape is hypothesised and its temperature are equally influential.

  • $\begingroup$ Thank you for the equations. Those are incredibly helpful. $\endgroup$ – Green Jun 16 '15 at 14:28
  • $\begingroup$ Shouldn't the thermal velocity scale be $\sqrt{3kT/m}$ rather than $\sqrt{2kT/m}$? Not that it matters when $\alpha$ isn't precisely known, but still. $\endgroup$ – Michael Seifert Jun 16 '15 at 14:49
  • $\begingroup$ @MichaelSeifert Why would it be 3 instead of 2? $\endgroup$ – Green Jun 16 '15 at 15:08
  • $\begingroup$ Your formula uses the most probable speed from the Maxwell-Boltzmann distribution, but I would think that the RMS velocity is more relevant in this case. You're really doing an energy calculation in comparing the particle's thermal energy to the kinetic energy required to escape the surface; and the average value of the kinetic energy is given by $\frac{1}{2} m v_\text{RMS}^2$. That said, the difference could be easily folded into $\alpha$, and it's possible that the form you wrote it in is the more conventional one (I'm not an expert in this field.) $\endgroup$ – Michael Seifert Jun 16 '15 at 15:18
  • $\begingroup$ @MichaelSeifert "That said, the difference could be easily folded into α". Correct. If you wish. $\endgroup$ – ProfRob Jun 16 '15 at 15:32

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