Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

It's not going to last forever - Jeans Escape is going to eventually act on the atmosphere after trillions of trillions of years: see http://faculty.washington.edu/dcatling/Catling2009_SciAm.pdf

share|improve this question
add comment

2 Answers

up vote 3 down vote accepted

This is a quick answer, but the last paragraph of the article you linked says that the process will be important within about 1 billion years. I'm not sure if this is distinct from the fact that the Sun will be significantly hotter anyway because it is gradually expanding during its main-sequence life (i.e. before it even starts to become a giant) which will heat the Earth and thus increase the Jeans escape rate.

I think what the question is really asking is: how rapidly is the atmosphere escaping now, and how long would it take the atmosphere to escape? I tried briefly to dig around for an appropriate equation. The relevant Wiki section is mostly tagged "citation needed", which didn't really help, but claims that $O_2$ is sufficiently bound that it would take more than a trillion years to escape.

For what it's worth, one can compute the fraction of gas particles that has a vertical velocity greater than the local escape velocity from the Maxwell distribution, which is basically a normal distribution.

$\int_{v_\text{esc}}^\infty f_vdv=\int_{v_\text{esc}}^\infty\sqrt{\frac{m}{2kT}}\exp\left(-\frac{mv^2}{2kT}\right)dv$

$=\int_{v_\text{esc}}^\infty\sqrt{\frac{1}{2c_s^2}}\exp\left(-\frac{v^2}{2c_s^2}\right)$

$=\frac{\sqrt{\pi}}{2}\text{erfc}\left(\frac{v_\text{esc}}{\sqrt{2}c_s}\right)$

At the surface, you can use $v_\text{esc}\approx11.2$km.s$^{-1}$ and $c_\text{s}\approx331$m.s$^{-1}$ which, according to my computer, gives $5.02\times10^{-251}$. This is a very crude approximation. Ideally you'd average over the total mass of the atmosphere with the appropriate distributions. To get a rate you'd need the acceleration, since this just gives an unbound fraction rather than how rapidly that fraction is lost. But I hope this gives some solidity to the fact that it'd take a very long time.

share|improve this answer
    
Wow - good answer! By acceleration, do you mean that the escape rate would accelerate as more atmosphere is lost? –  InquilineKea Aug 11 '11 at 9:19
    
No, I mean that just because the gas is unbound at a certain point in time, it doesn't mean it will escape immediately. Basically, the little calculation I made doesn't involve time at all, but just shows that the gas is almost entirely bound to the Earth. –  Warrick Aug 11 '11 at 14:39
    
You took the escape velocity for escaping from the earth's surface. But the atmosphere has its thickness. For a particle at 100 km altitude a much smaller velocity is needed. –  Anixx Nov 30 '13 at 3:41
add comment

What other assumptions are you going to make? Once you make one non-physical assumption, every further extrapolation is self-inconsistent. Comets are at least speculated if not accepted to make a decent contribution to atmospheric volatiles. Does comet bombardment continue at the rate it would if the Sun stayed happily main-sequence? At the presumably higher rate as if the Sun did go red giant and perturbed every orbit in the Solar System? Life continues to survive and maintains near-present-day atmospheric composition? Life gets wiped out some five billion years from now for no reason?

I don't think this question is answerable, unless you just do

logbase2(present atmospheric pressure/criterion pressure for calling the atmosphere "gone")*(Jeans escape characteristic time scale for N2), which I believe will be the shortest of the critical components of the atmosphere.

share|improve this answer
    
Well - the fact is - all models have non-physical assumptions. It's necessary when we want to study one particular process in detail. Atmospheric science is full of non-physical assumptions (slab-ocean model, aquaplanet, Lovelock's daisies, etc) because we have to get a physical interpretation of the dynamics –  InquilineKea Aug 10 '11 at 22:24
    
There's a difference between an approximation and a non-physical assumption. The approximation has a well-defined limit of applicability. –  Andrew Aug 11 '11 at 10:44
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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