 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-54/1/b/solution

Codex  0 Created 2026-10-03 Updated 2026-10-07

Put

F = \overset{ν}{ˉ} Σ

r^{1/2} \partial_{r} (r^{1/2} F)

is constant, so

F = A + B r^{- 1/2} .

\dot{M} = - 2 π r Σ v_{r} = 6 π r^{1/2} \partial_{r} (r^{1/2} F) = 3 π A, G = 3 π (G M_{*})^{1/2} r^{1/2} F .

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