Dome, Dome on the Range

Gerry Quinn · 15 Jan 2011, 17:50 · from 2011-January.txt.gz

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From: gerryq at indigo.ie (Gerry Quinn)
Date: Sat, 15 Jan 2011 17:50:51 -0000
Subject: (urth) Dome, Dome on the Range
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From: "Jeff Wilson" <jwilson at io.com>

> It occured to me at lunch today that a problem with the open air 
> atmosphere schemes, even ones that last thousands of years between 
> refills, is that they take half to 3/4 the mass of earth's atmosphere to 
> work. When this does escape from the moon, only small amount of that will 
> also escape from the earth, and  the reset will eventually come to fall on 
> the earth, raising its surface air pressure accordingly; this is a serious 
> problem

So suck it up and send it back!

Seriously, will this actually happen, anyway?  I couldn't find detailed 
references to a proper calculation, but here are my thoughts:

If an atom leaves the top of Lune's atmosphere at lunar escape velocity, the 
question of whether it is likely to end up in Urth's atmosphere on Urth 
depends on whether it exceeds Urth escape velocity at this point (unless 
it's headed directly for Urth, which is unlikely.

Escape velocity at a distance r from a body is sqrt(2GM/r) where M is the 
mass of the body and G is the gravitational constant.. Urth's mass is 81 
times that of Lune.  Let's call the distance from Urth to Lune D, and the 
height of the lunar atmosphere S, and calculate the escape velocity of an 
atom at the top of Lune's atmosphere:  K is the mass of Lune.
V(Lune) = sqrt( 2GK/S)
V(Urth) = sqrt( 2G(81K)/(D-S))

Setting both escape velocities to be equal:
sqrt( 2GK/S) = sqrt(2G(81K)/(D-S))
1/S = 81/(D-S)
81S = D-S

Since D is 250000 miles, S is about 3000 miles.  If Lune's atmosphere is 
less than this depth, atoms escaping from the top will exceed Urth's escape 
velocity.  (Strictly speaking I should consider escape velocity from the 
Urth-Lune system, but it gets more complicated, and the result will only be 
a little less.)

How deep is Lune's atmosphere?  To have a pressure at ground level equal to 
that of Urth, there must be 6 times as much air above any point, because 
lunar gravity surface is 1/6 that of Urth.  If we ignore the relative drop 
off in gravity above the surface of the two bodies, the atmosphere should be 
proportionaly the same as Urth's, except 6 times deeper.

For our purposes, the mesopause (the point of minimal atmospheric 
temperature), which is at about 100km on Earth, seems to be the relevant 
feature.  That would be the equivalent of 600km on Lune.  Unfortunately, 
this is enough to bring into question our assumption that the drop-off in 
gravitational force with height can be ignored - at this height, Lune's 
gravity is only around 55% of that at the surface.  That would mean the 
mesopause on Lune would actually be higher than 600 km.

However, I don't think that's too much of a problem: without going into 
details, I have convinced myself that the calculations do not 'run away' 
sufficiently to bring the lunar mesopause to 3000 miles.  Also, its height 
could be reduced by increasing the concentration of heavy inert constituents 
in the linar atmosphere (argon, perhaps, or some synethethic heavy 
molecule).

Anyway, if you have a link that says otherwise I would be interested to see 
it!  If not, as I said, the solution would be to gather up any excess air 
that lands on Urth and bring it back to Lune.

- Gerry Quinn

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