Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 59 / 3 / c / Solution

For an incompressible planetary interior of density $\rho$, the enclosed mass and local gravity are $M(r)=4\pi \rho r^3/3$ and $g(r)=4\pi G\rho r/3$. Hydrostatic equilibrium therefore givesIntegrating inward from negligible surface pressure at $r=R_p$ yields the uniform-density planetary pressure profile:The surface gravity is $g_s=GM/R_p^2=4\pi G\rho R_p/3$. Eliminate $\rho R_p$ to obtainThe units of $g_s^2/G$ are pressure. For a nonzero imposed surface pressure, add $P_s$ throughout and interpret the boxed central value as $P_c-P_s$. The constant-density approximation is crucial; real centrally concentrated rocky planets need a different profile and generally a larger central pressure at the same mass and radius.