 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-317/4/i/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

For

P = K ρ^{2}

2 K \nabla ρ = - \nablaΦ

\nabla^{2} ρ + k^{2} ρ = 0, k^{2} = \frac{2 π G}{K} .

ρ (r) = ρ_{c} \frac{sin ( k r )}{k r} .

The singular

cos (k r) / r

solution is excluded by finite central mass density. Taking the first zero as the stellar surface yields

R = \frac{π}{k} = \frac{π K}{2 G} .

M = 4 π \int_{0}^{R} ρ r^{2} d r = \frac{4 π ρ _{c}}{k ^{3}} \int_{0}^{π} u sin u d u = \frac{4 π ^{2} ρ _{c}}{k ^{3}} .

Dividing by the volume

4 π R^{3} /3

gives

\frac{ρ ˉ}{ρ _{c}} = \frac{3}{π ^{2}} .

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