= Solution
A geodesic passing through $r=0$ must have $h=0$. Its radial equation is then
$$
\dot r^2+\frac{r^2}{a^2}=E^2-1.
$$
On the outgoing and returning portion it is the harmonic solution
$$
r(\tau)=a\sqrt{E^2-1}\sin(\tau/a),\qquad 0\leq\tau\leq\pi a.
$$
It reaches
$$
\boxed{r_{\max}=a\sqrt{E^2-1}}
$$
when $\tau=\pi a/2$ and returns to the origin at
$$
\boxed{\Delta\tau=\pi a.}
$$
The return time is independent of the launch energy, a characteristic focusing property of global anti-de Sitter spacetime.
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