 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-66/3/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

For a regular, spherical star, hydrostatic equilibrium gives

d P / d r = - G m (r) ρ (r) / r^{2}

. Multiply by

4 π r^{3}

and integrate to the surface. Integration by parts gives

\int_{0}^{R} 4 π r^{3} \frac{d P}{d r} d r = 4 π R^{3} P (R) - 3 \int_{0}^{R} 4 π r^{2} P d r .

The central boundary term vanishes for finite central pressure. Since

d m = 4 π r^{2} ρ d r

, the right-hand side of the integrated hydrostatic equation is

- \int_{0}^{R} 4 π G m ρ r d r = - \int_{0}^{M} \frac{G m}{r ( m )} d m = Ω.

The last expression is the gravitational energy: adding a thin mass shell

d m

to the interior mass

m

contributes

- G m d m / r

, counting each interacting pair once. Hence

3 \int_{0}^{M} \frac{P}{ρ} d m + Ω = 4 π R^{3} P (R) .

With the specified zero surface pressure this proves the stellar virial theorem in the required form,

3 \int_{0}^{M} \frac{P}{ρ} d m + Ω = 0.

 New to topics? Read the docs here!