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

Table of contents
- Solution f

Solution

While resonance fixes

a_{d}

, substitute the phase relation from part (e) into the supplied eccentricity equation:

\overset{e}{˙}_{d} = \frac{βG M _{*}}{4 c a _{d}^{2} e _{d}} - \frac{5}{2} \frac{βG M _{*}}{c a _{d}^{2}} e_{d} = \frac{βG M _{*}}{4 c a _{d}^{2} e _{d}} (1 - 10 e_{d}^{2}) .

To the requested first order in eccentricity, discard the

e_{d}^{2}

correction. It follows that

\frac{d}{d t} (e_{d}^{2}) = \frac{βG M _{*}}{2 c a _{d}^{2}} .

Using

β_{m a x} = 4 c e_{d 0} a_{d}^{2} D / (G M_{*})

and integrating from

e_{d} (0) = e_{d 0}

gives

e_{d} (t) = e_{d 0}^{2} + 2 D e_{d 0} (\frac{β}{β _{m a x}}) t .

The neglected term eventually matters and prevents indefinite validity of this small-eccentricity growth law.

Solved by gpt-5.6-sol high.

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