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Past exam of the mathematics course of the University of Cambridge

2026

Paper 316

While resonance fixes $a_d$, substitute the phase relation from part (e) into the supplied eccentricity equation:
$$
\dot e_d
=\frac{\beta GM_*}{4ca_d^2e_d}
-\frac52\frac{\beta GM_*}{ca_d^2}e_d
=\frac{\beta GM_*}{4ca_d^2e_d}(1-10e_d^2).
$$
To the requested first order in eccentricity, discard the $e_d^2$ correction. It follows that
$$
\frac d{dt}(e_d^2)=\frac{\beta GM_*}{2ca_d^2}.
$$
Using $\beta_{\max}=4ce_{d0}a_d^2D/(GM_*)$ and integrating from $e_d(0)=e_{d0}$ gives
$$
\boxed{e_d(t)=\sqrt{e_{d0}^2+2De_{d0}
\left(\frac\beta{\beta_{\max}}\right)t}}.
$$
The neglected term eventually matters and prevents indefinite validity of this small-eccentricity growth law.

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 316 / 4 / f / Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 316 / 4 / f / Solution

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Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 316 / 4 / f / Solution

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