= 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 gives
$$
\frac{dP}{dr}=-\frac{GM(r)\rho}{r^2}=-\frac{4\pi G\rho^2}{3}r.
$$
Integrating inward from negligible surface pressure at $r=R_p$ yields the <uniform-density planetary pressure profile>:
$$
\boxed{P(r)=\frac{2\pi G\rho^2}{3}(R_p^2-r^2)=P_c\left(1-\frac{r^2}{R_p^2}\right),\qquad P_c=\frac{2\pi G\rho^2R_p^2}{3}.}
$$
The surface gravity is $g_s=GM/R_p^2=4\pi G\rho R_p/3$. Eliminate $\rho R_p$ to obtain
$$
\boxed{P_c=\frac{3g_s^2}{8\pi G}.}
$$
The 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.
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