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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2025

Paper 316

Resolving the conserved energy into radial and azimuthal parts gives
$$
C=\frac12\left(\dot r^2+\frac{h^2}{r^2}\right)-\frac{\mu(1-\beta)}r.
$$
Using the release values of $C$ and $h$, and choosing the outward root,
$$
\boxed{v_r(r)=\dot r
=\sqrt{\frac{\mu}{a_p}\left[(2\beta-1)
-2(\beta-1)\frac{a_p}{r}-\left(\frac{a_p}{r}\right)^2\right]}}.
$$
Equivalently, with $x=r/a_p$,
$$
v_r=\sqrt{\frac{\mu}{a_p}}
\frac{\sqrt{(x-1)[(2\beta-1)x+1]}}x.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 316 / 2 / v / Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 316 / 2 / v / Solution

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Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 316 / 2 / v / Solution

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