Monday, 13 April 2020

astrophysics - stability of hypothetical lunar atmosphere


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



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.


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...