Fluids at Rest — Question 9

Question 9

A compressible atmosphere is not a constant-density column

An idealized static atmosphere satisfies the supplied relation $\rho=p/c^2$ with constant $c$, and constant downward gravity $g$. At altitude $z=0$, pressure is $p_0$; $z$ increases upward.

1. Derive $p(z)$ and the altitude where it is half of $p_0$.

2. Use $c^2=8.0\times10^4\,\mathrm{m^2/s^2}$ and $g=10\,\mathrm{m/s^2}$. At one scale height, compare the exact pressure with the constant-density approximation based on ground density.

Question 9 - Solution

A compressible atmosphere is not a constant-density column

(a) Derivation and calculation

$$\frac{dp}{dz}=-\rho g=-\frac{g}{c^2}p\Rightarrow\int_{p_0}^{p(z)}\frac{dp}{p}=-\frac g{c^2}\int_0^z dz.$$$$\boxed{p(z)=p_0e^{-z/H_s}},\qquad\boxed{H_s=c^2/g}.$$$$p=p_0/2\Rightarrow\boxed{z_{1/2}=H_s\ln2}.$$

(b) Result and interpretation

$$H_s=8.0\times10^4/10=8000\,\mathrm m,\quad\boxed{z_{1/2}=5545.18\,\mathrm m}.$$$$p(H_s)=\boxed{p_0/e=.367879p_0}.$$$$p_{\mathrm{linear}}(H_s)=p_0-\rho_0gH_s=p_0-\frac{p_0}{c^2}g\frac{c^2}{g}=0.$$The linear approximation fails over a scale height because density changes substantially.

Physical check

For $z/H_s\ll1$, expansion of the exponential reproduces $p_0-\rho_0gz$.