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

Assume a thin isothermal atmosphere, so $g\simeq G{M}_p/R_p^2$ and the surface area are constant through it. Integrating hydrostatic equilibrium from base pressure $P_0$ to negligible top pressure gives atmospheric column mass $P_0/g$. ThereforeEquivalently, $M_{\mathrm{atm}}\sim 4\pi R_p^2{\rho }_{0}H$, with ${\rho }_0=P_0\mu m_H/(k_B{T}_0)$ and $H=k_B{T}_0/(\mu m_{H}g)$; the explicit temperature dependence cancels.