If we put a solid sphere in space, it will lose some molecules which will form a sort of an atmosphere around it so that we have the required vapour pressure for solid-vapour equilibrium (Temp. of body is the average temp of space ~ 3K) Now, this gas has some velocity distribution profile. A fraction of the molecules will have velocity greater than the escape velocity of the sphere. This fraction will escape, and more vapour will be formed, and more molecules will escape. Basically the system will never attain equilibrium and the solid will finally become gas.

Is there a flaw in this analysis? If not, can we predict the time required for evaporation for our hypothetical planet given its total mass, initial radius, molecular mass, etc.?

  • 2
    $\begingroup$ Should you not factor in the infall from other evaporated planets? $\endgroup$
    – DJohnM
    Commented Apr 23, 2014 at 23:46
  • $\begingroup$ Conceptually, I'm guessing you're probably right. However, the vapor pressure of most high-molecular weight materials is vanishingly small, so I would guess that this is a very small effect for anything other than very light molecules and atoms. $\endgroup$ Commented Apr 24, 2014 at 0:07
  • $\begingroup$ some materials have no significant vapour pressure. Iron, for instance, only reaches a vapour pressure of 1Pa at around 1700K. If you had an iron planet sitting at around 3K, it's vapour pressure would be so low that the protons would probably decay before it boiled away $\endgroup$
    – Jim
    Commented Apr 24, 2014 at 0:54
  • $\begingroup$ @Jim: Haha, that's a long time $\endgroup$ Commented Apr 24, 2014 at 0:57

1 Answer 1


Is there a flaw in this analysis?

Yes, it doesn't point out that as the fastest moving molecules escape, the remaining molecules are at a lower temperature. In other words, the sphere is cooled by the evaporation process. There would be two types of cooling, one from sublimation of the solid, and the other from the fastest molecules escaping.

If you want to calculate the evaporation time, you would need to consider the rate of heating from the cosmic microwave background radiation versus the cooling from evaporation.


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