 Solution

Treat the Milky Way and M31 as a radial Kepler orbit of total mass $M$. At maximum separation $r=2a$ the radial speed vanishes, so conservation of energy givesPut $r=a(1-\cos 2\eta )=2a{\sin }^2\eta$. Substitution into the energy equation and integration from the Big Bang gives the cycloidal orbitSince $r^3=8a^3{\sin }^6\eta$,Differentiating the parametric solution givesEliminating $\eta$ numerically gives the stated accurate fit

A remote dwarf sees the Local Group primarily through its total monopole mass, so the same radial timing relation applies with its own $r_i,v_i$. Dimensional consistency requiresrather than the cube-root exponent printed in the converted statement. Let $C=2^{-3/2}\pi$. At the zero-velocity radius $r_0$, ${\omega }_0=0$ and ${\Omega }_{0}t=C$. For the dwarf, ${\Omega }_{i}t=C-0.85{\omega }_{i}t$. Since $\Omega \propto r^{-3/2}$,Therefore