= Solution
For a family of spherical profiles with a fixed dimensionless shape, write the <gravitational potential energy> as $\Omega=-\alpha GM^2/R$, where $\alpha>0$ depends on that shape. The <virial theorem> in <2/b> and the sphere volume $V=4\pi R^3/3$ give
$$
\boxed{\overline P=\frac{\alpha}{4\pi}\frac{GM^2}{R^4}\propto\frac{GM^2}{R^4}.}
$$
For a uniform <density>, $\alpha=3/5$; formally integrating $-\int_0^R GM(r)\,dM(r)/r$ for $M(r)=Mr/R$ gives $\alpha=1$. Neither the <virial theorem> nor the initial profile guarantees a specific final shape, so the proportionality requires the stated homologous-profile assumption. Surface <pressure> must be handled consistently with <2/b> before using either coefficient as an equilibrium model.
Back to article page
