Hydrostatic balance makes the atmospheric column mass $(P_0-P_{\rm top})/g$. Multiplying by the surface area gives the <mass of a thin hydrostatic atmosphere>
$$
\boxed{M_{\rm atm}\simeq
\frac{4\pi R_p^2(P_0-P_{\rm top})}{g}
\simeq\frac{4\pi R_p^4P_0}{GM_p}},
$$
where the second expression neglects the top pressure and atmospheric self-gravity.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 315 / 2 / e / Solution

The column mass between pressures $P_0$ and $P_{\rm top}$ is $(P_0-P_{\rm top})/g$. A thin atmosphere over a spherical planet therefore has
$$
M_{\rm atm}\simeq\frac{4\pi R_p^2(P_0-P_{\rm top})}{g}
\simeq\frac{4\pi R_p^4P_0}{GM_p}
$$
when its top pressure is negligible.


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