Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 316 / 4 / v / Solution

To first order, the initial-phase correction in $M_0$ can be omitted inside a term already multiplied by $e$. ThusFor $|a-a_g|/a\ll 1$ and $|\delta \lambda |\ll 1$, expand the exact rotating reference frame coordinates. Products of the small radial and angular displacements are second order. This givesAt $t=0$ 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 $|(n-n_g)t|\ll 1$, so small frequency mismatch alone does not justify it for arbitrarily large times.