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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2025

Paper 1

A circular orbit at $r=r_0$ requires $V'(r_0)=0$, hence
$$
\frac{r_0}{a^2}-\frac{h^2}{r_0^3}=0,
\qquad
h^2=\frac{r_0^4}{a^2}.
$$
This can be satisfied for every $r_0>0$. Moreover,
$$
V''(r_0)=\frac1{a^2}+\frac{3h^2}{r_0^4}=\frac4{a^2}>0,
$$
so every such circular orbit is stable. The radial energy equation gives
$$
E^2=\left(1+\frac{r_0^2}{a^2}\right)^2,
$$
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 $\dot\phi=\pm1/a$.

Solution

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

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

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

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