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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 38E / c / Solution

A circular orbit at

r = r_{0}

requires

V^{'} (r_{0}) = 0

, hence

\frac{r _{0}}{a ^{2}} - \frac{h ^{2}}{r _{0}^{3}} = 0, h^{2} = \frac{r _{0}^{4}}{a ^{2}} .

This can be satisfied for every

r_{0} > 0

. Moreover,

V^{''} (r_{0}) = \frac{1}{a ^{2}} + \frac{3 h ^{2}}{r _{0}^{4}} = \frac{4}{a ^{2}} > 0,

so every such circular orbit is stable. The radial energy equation gives

E^{2} = (1 + \frac{r _{0}^{2}}{a ^{2}})^{2},

and future direction selects

E = 1 + r_{0}^{2} / a^{2} = f (r_{0})

. Therefore

\dot{t} = \frac{E}{f ( r _{0} )} = 1,

so proper time and coordinate time differ only by an additive constant on the orbit. Also

\dot{ϕ} = \pm 1/ a

.

Solved by gpt-5.6-sol high.

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