stability of hypothetical lunar atmosphere

Question

Assume that by some means, the moon could be given an atmosphere, of the same density and pressure at the surface as the earth's. Obviously in a stable atmosphere there are temperature variations from pole to equator, day to night, and a lot of interesting dynamics follow from that, but for argument's sake assume an average of something like 0 to 5 C.

How long would it last?

Clearly there's no air there now (unless you count traces of potassium, or whatever it has), so such at artificial atmosphere would not be permanent, but presumably the moon could hang onto some air, for some time. With the moon's lower gravity, the density or pressure vs altitude curve will be different, and I'd guess that a significantly larger fraction of the mass would be at higher altitudes than on earth, and there's no protection from the solar wind. Someone with the right equations could probably work out how long it would take the pressure at the surface to drop by half.

Answer 1

Not sure about the solar wind but when it comes to Jeans escape: The escape velocity from the Moon is 2.4 km/s:

http://www.wolframalpha.com/input/?i=escape+velocity+from+moon

This is not much higher than the velocity of the molecules. The mass of Nitrogen molecule, the prevailing one, is 28 u,

http://www.wolframalpha.com/input/?i=mass+of+nitrogen+molecule

which is $4.65 \times 10^{-26}$ kg. In the $\exp(-mv^2/2kT)$ Maxwell factor for $T=275$ Kelvin, we may see that $$\langle v^2\rangle = kT/m = 1.38 \times 10^{-23} \times 275 / 4.65 \times 10^{-26} =81,600$$ squared meters over squared seconds. We need $v^2=2,400^2 = 5,760,000$ meters per second which is 70 times more than $81,600$, so the relevant exponential is $\exp(-35)$ (don't forget the factor of $1/2$) which is $6\times 10^{-16}$ or so. Only this fraction of molecules going up escape. It takes about 100 seconds for a molecule to try another value of its velocity, after a collision and a return from the upper atmosphere, so one needs something like $10^{16}\times 100 = 10^{18}$ seconds for the Nitrogen molecules to reach the escape velocity. That's something like 30 billion years.

Nitrogen in the lunar atmosphere (and heavier molecules) would be safe against Jeans escape, I think. Of course, it would be much easier for Hydrogen to escape. 28 drops to 2, 35 drops to something like 2 as well, so much of the Hydrogen escape during its 10th attempt or so, within minutes. Modifying the prefactors in the exponent makes a huge impact on the result.

Answer 2

The solar wind tends to disassociate molecules. So without the magnetic field, you should test nitrogen as single atoms with atomic mass of 14 instead of the molecular mass of 28.

1.38×10−23×275/4.65×10−26 = 81600 = ~1/70
1.38×10−23×275/2.325×10−26 = 163200 = ~1/35

1/EXP(70/2*-1)100=158601345231343072.81296446257747 = ~5.026 billion years
1/EXP(35/2-1)*100=3982478439.7576225021870676349852 = ~126 years

Answer 3

Calculating Jeans escape time for temperature of 275 would be correct if you were talking about temperature near the surface, on the average, but (1) the escape time is dominated by the hottest part of the atmosphere, which for the moon would be more like 400K, and (2) the surface temperature isn't what's important for Jeans escape; it is the exosphere temperature, which due to ultraviolet heating is much hotter.