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> All else being equal, the terminal velocity is inversely proportional to the
> square root of air density. Air density drops off pretty quickly, and I 
> really should be doing something other than digging up the math for that. I
> think it involves calculus to integrate the amount of mass as the column of
> the atmosphere trails off.

Chemistry types have a method for
dealing with this question without
dragging in the calculus:

Suppose an atmosphere to be mainly
affected by gravity, resulting in
the potential energy for a mass M
to be linear in height h: Mgh.

What relative concentrations will
we have when two different packets
of air are in equilibrium?

If they are at the same height, we
will have half the mass in one, and
half the mass in the other, and the
amount flowing from one to the other
balances the amount flowing in the
opposite direction.[0]

If they are at differing heights,
then a greater percentage of the
higher air tends to descend than
that percentage of the lower air
which ascends.  In order for the
two flows to balance, the higher
packet must contain less air than
the lower, and the mass balance
of the flows corresponds thusly:

    high percentage of thin air
    ---------------------------
    low percentage of dense air

Now, rates are exponential in
energy differences[1], so that
theoretically we should expect
an exponential decay in height,
to compensate.  How does it go
in practice?

-Dave

[0] How well does it balance?
    Chemical equilibria seem 
    stable, as they deal with
    very large numbers over a
    very long time.  Economic
    equilibria are viewed from
    the mayfly standpoint of
    individual people, and so,
    at best, the shot noise is
    very visible.

[1] That is to say, rates will
    be exponential in the free
    energy differences between
    endpoints and a transition
    state.  We can ignore that
    complication in this model.


