 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ii/paper-2/34d/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

V^{'} = 0

. Differentiating the Schwarzschild effective potential at fixed

h

gives

V^{'} = \frac{r _{s}}{2 r ^{2}} - \frac{h ^{2}}{r ^{3}} + \frac{3 r _{s} h ^{2}}{2 r ^{4}}, h^{2} = \frac{r ^{2} r _{s}}{2 r - 3 r _{s}} .

A timelike circular orbit requires

r > 3 r_{s} /2

. Evaluating the second derivative after this substitution gives

V^{''} = \frac{r _{s} ( r - 3 r _{s} )}{r ^{3} ( 2 r - 3 r _{s} )} .

A stable nondegenerate minimum therefore requires $r > 3 r_{s}$ . The boundary

r = 3 r_{s}

is the marginal innermost stable circular orbit, not a strict stable minimum.

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