Hydrostatic equilibrium

Remarks

Part 1 is the Euler equation with $v = 0$. For part 2, integrate Poisson's equation over the ball of radius $r$ and apply the divergence theorem. This gives $4 \pi r^2 \Phi'(r) = 4 \pi G m(r)$, so $\Phi'(r) = G m(r)/r^2$, which is Newton's law of gravitation for the mass inside $r$. Example: in a ball of uniform density $\rho_0$ and radius $R$ with $P(R) = 0$, integrating gives the central pressure $P(0) = \frac{2}{3} \pi G \rho_0^2 R^2$. Example: in a thin layer near the ground with constant $g = \abs{\nabla \Phi}$, part 1 gives $dP/dz = -\rho g$, the barometric law of a column of air or water. The hypothesis $v = 0$ is needed: in a rotating star the centrifugal term adds to $\nabla \Phi$ and the surfaces of equal pressure are no longer spheres.

Going deeper

For a student

In the equations of stellar structure, hydrostatic equilibrium is the equation for $dP/dr$. Combined with a relation between pressure and density, such as a polytropic law $P = K \rho^{1 + 1/n}$, it reduces to a single second-order equation, the Lane–Emden equation. Multiplying by $4 \pi r^3$ and integrating over the star gives the virial theorem for a self-gravitating gas: $3 \int P \, dV$ equals minus the gravitational energy. In general relativity the equation is replaced by the Tolman–Oppenheimer–Volkoff equation, which leads to an upper bound on the mass of neutron stars.