 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-316/4/v/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

To first order, the initial-phase correction in

M_{0}

can be omitted inside a term already multiplied by

e

. Thus

\frac{r}{a} = 1 - e cos (n t - ϖ) + O (e^{2}), δ λ = (n - n_{g}) t + 2 e sin (n t - ϖ) + 2 e sin ϖ + O (e^{2}) .

For

∣ a - a_{g} ∣/ a ≪ 1

and

∣ δ λ ∣ ≪ 1

, expand the exact rotating reference frame coordinates. Products of the small radial and angular displacements are second order. This gives

\frac{x}{a} ≃ \frac{a - a _{g}}{a} - e cos (n t - ϖ), \frac{y}{a} ≃ (n - n_{g}) t + 2 e sin (n t - ϖ) + 2 e sin ϖ .

t = 0

the two sine terms cancel, as required. Equal mean motion gives a closed local epicyclic motion with radial-to-tangential amplitude ratio

1 : 2

; unequal mean motion adds along-track drift. The local expansion also requires

∣ (n - n_{g}) t ∣ ≪ 1

, so small frequency mismatch alone does not justify it for arbitrarily large times.

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