From: utzoo!decvax!harpo!floyd!vax135!ariel!houti!lime!houxe!hogpc!houxm!ihnp4!ihuxr!lew
Newsgroups: net.physics
Title: Explanation of pressure equation
Article-I.D.: ihuxr.392
Posted: Thu Apr 14 12:34:41 1983
Received: Mon Apr 18 22:44:09 1983

I posted an equation for the pressure of an isothermal atmosphere in
a linearly increasing gravitational field:

p(z) = p1 * exp( -(mg/kT) * 1/2 * z^2/R )

where p1 is the pressure at z=0 and g is the g-field value at z=R .
The formula applies to the pressure in a deep shaft, drilled from the
surface. It is obtained by integrating the differential equation for the
pressure:

dp = -q * g(z) * dz = -(p*m/kT) * (g*z/R) * dz

dp/p = -(mg/kT) * z/R * dz

log(p) = -(mg/kT) * 1/2 * z^2/R + constant


Here, q is the mass density and it goes to (p*m/kT) by the ideal gas law.
g(z)=g*z/R is an expression of the linear increase of gravity from 0
at the center (z=0) to g at the surface (z=R).

If you adjust p1 to give p(R)=p0, and then compare the values obtained
for z<R with the values obtained from a downward extrapolation of the
exponential atmosphere, you will find that the former are less than the
later. The values diverge slowly as z decreases from z=R.

The value at z=0 is the square root of the value obtained from a downward
extrapolation of the exponential atmosphere, in units of p0.

	Lew Mammel, Jr. ihuxr!lew