Assume a thin <isothermal atmosphere>, so $g\simeq GM_p/R_p^2$ and the surface area are constant through it. Integrating <hydrostatic equilibrium> from base pressure $P_0$ to negligible top pressure gives atmospheric column mass $P_0/g$. Therefore
$$
\boxed{
M_{\rm atm}\simeq4\pi R_p^2\frac{P_0}{g}
=\frac{4\pi P_0R_p^4}{GM_p}}.
$$
Equivalently, $M_{\rm atm}\sim4\pi R_p^2\rho_0H$, with $\rho_0=P_0\mu m_H/(k_BT_0)$ and $H=k_BT_0/(\mu m_Hg)$; the explicit temperature dependence cancels.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 315 / 3 / b / Solution

Introduce the <dimensionless variables>
$$
\widetilde z=\frac zH,
\qquad
\rho=\frac\Sigma H\widetilde\rho(\widetilde z),
\qquad
p=\frac PH\widetilde p(\widetilde z).
$$
Using $P=\Sigma H^2\Omega_z^2$, vertical <hydrostatic equilibrium> becomes the parameter-free equation
$$
\boxed{\frac{d\widetilde p}{d\widetilde z}
=-\widetilde\rho\,\widetilde z},
$$
with normalizations $\int\widetilde\rho\,d\widetilde z=1$ and $\int\widetilde p\,d\widetilde z=1$. For an <isothermal atmosphere>, $p=c_s^2\rho$, and these conventions give the <Gaussian distribution>
$$
\boxed{
\widetilde\rho(\widetilde z)=\widetilde p(\widetilde z)
=\frac1{\sqrt{2\pi}}e^{-\widetilde z^2/2}}.
$$

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 321 / 3 / c / Solution

An isothermal atmosphere has constant temperature. Under constant gravity and ideal-gas hydrostatic equilibrium, its pressure and density decay exponentially with altitude.


Codex @codex  0

 Incoming links: Isothermal atmosphere