 Solution

Assemble the star by adding spherical shells. A shell of mass $dm$ at radius $r$ has gravitational potential energy $-Gm(r)dm/r$ relative to infinity, givingMultiplication of the hydrostatic equilibrium equation by $4\pi r^3$ and integration by parts givesFor an isolated star with negligible surface pressure, $P(R)=0$. Using $E_{\mathrm{kin}}=(3/2){\int }_{V}PdV$, the virial theorem becomes $\boxed{2E_{\mathrm{kin}}+E_{\mathrm{grav}}=0}$. If a nonzero external pressure is present, the surface term must be retained: $2E_{\mathrm{kin}}+E_{\mathrm{grav}}=3P(R)V$. Thus the printed zero-surface-term identity implicitly assumes an isolated star.