 Solution

Treat the cylinder as infinitely long, with a vacuum exterior and no surface tension. Let ${\Phi }_0$ be its gravitational potential. Integrating the gravitational Poisson equation over a coaxial cylinder of radius $r<R$ and length $L$, then applying the divergence theorem, givesThus the inward gravitational acceleration is $-2\pi G\rho r\hat{\mathbf{r}}$. Hydrostatic equilibrium requiresImpose zero pressure on the free surface, $p_0(R)=0$, and integrate inward:This is the hydrostatic self-gravitating incompressible cylinder. Its central pressure is $\pi G{\rho }^2{R}^2$, and the units of the expression are $\mathrm{Pa}$. A nonzero external pressure would add that constant to $p_0(r)$.