= 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 gives
$$
\boxed{\frac12v^2-\frac{GM}{r}=-\frac{GM}{2a}}.
$$
Put $r=a(1-\cos2\eta)=2a\sin^2\eta$. Substitution into the energy equation and integration from the Big Bang gives the cycloidal orbit
$$
\boxed{2\eta-\sin2\eta=\left(\frac{GM}{a^3}\right)^{1/2}t}.
$$
Since $r^3=8a^3\sin^6\eta$,
$$
\boxed{\Omega t
=\sqrt{\frac{GM}{r^3}}\,t
=\frac{\eta-\sin\eta\cos\eta}{\sqrt2\,\sin^3\eta}}.
$$
Differentiating the parametric solution gives
$$
\boxed{\omega t=\frac vr\,t
=\frac{(\eta-\sin\eta\cos\eta)\cos\eta}{\sin^3\eta}}.
$$
Eliminating $\eta$ numerically gives the stated accurate fit
$$
\boxed{\Omega t+0.85\,\omega t=2^{-3/2}\pi}.
$$
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 requires
$$
\boxed{\Omega_i=\sqrt{\frac{GM}{r_i^3}}},
$$
rather 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 $\Omega_0t=C$. For the dwarf, $\Omega_it=C-0.85\,\omega_it$. Since $\Omega\propto r^{-3/2}$,
$$
\left(\frac{r_i}{r_0}\right)^{3/2}
=\frac{C}{C-0.85\,\omega_it}.
$$
Therefore
$$
\boxed{r_0=r_i\left[1-\frac{0.85}{2^{-3/2}\pi}\omega_it\right]^{2/3}}.
$$
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