= Solution
For an affine parameter $\lambda$, the <geodesic Lagrangian> is
$$
\mathcal L=\frac12\left[-(1+r^2)\dot t^2
+\frac{\dot r^2}{1+r^2}
+r^2\dot\theta^2+r^2\sin^2\theta\,\dot\phi^2\right].
$$
Varying $S=\int\mathcal L\,d\lambda$ gives the <geodesic equations>
$$
\ddot t+\frac{2r}{1+r^2}\dot r\dot t=0,
$$
$$
\ddot r+r(1+r^2)\dot t^2
-\frac{r}{1+r^2}\dot r^2
-r(1+r^2)(\dot\theta^2+\sin^2\theta\,\dot\phi^2)=0,
$$
$$
\ddot\theta+\frac{2\dot r}{r}\dot\theta
-\sin\theta\cos\theta\,\dot\phi^2=0,
\qquad
\ddot\phi+\frac{2\dot r}{r}\dot\phi
+2\cot\theta\,\dot\theta\dot\phi=0.
$$
The rotational <Killing vector fields> of the <spherical symmetry> conserve the angular-momentum vector. Its direction is fixed, so every nonradial orbit lies in the plane through the symmetry centre orthogonal to that vector. A spatial rotation can make this the equatorial plane. Equivalently, the initial conditions $\theta=\pi/2$ and $\dot\theta=0$ solve the $\theta$ equation uniquely. This is the <planarity of geodesics in spherical symmetry>; a radial geodesic has zero <angular momentum> and may be assigned any plane.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-309-equatorial-plane.png]
{title=An equatorial <geodesic> in a spherically symmetric spacetime}
On $\theta=\pi/2$, the cyclic coordinates $t$ and $\phi$ produce <geodesic conserved quantities from Killing vectors>,
$$
\boxed{E=(1+r^2)\dot t,
\qquad L=r^2\dot\phi}.
$$
For a timelike geodesic parametrized by <proper time>, the <proper-time normalization> is $g_{ab}\dot x^a\dot x^b=-1$. Substitution of $E$ and $L$ gives the first-order radial equation
$$
\boxed{\dot r^2=E^2-(1+r^2)\left(1+\frac{L^2}{r^2}\right)}.
$$
This is the <timelike geodesic effective potential> for static <Anti-de Sitter spacetime> with curvature radius one.
A geodesic passing through $r=0$ must have $L=0$. If it moves away from the origin, $E>1$, and
$$
\dot r^2=(E^2-1)-r^2.
$$
Taking proper time $s=0$ at departure gives
$$
r(s)=\sqrt{E^2-1}\sin s
\qquad(0\leq s\leq\pi).
$$
It turns around at $s=\pi/2$ and first returns to the origin at
$$
\boxed{\Delta s=\pi}.
$$
This energy-independent refocusing is the characteristic <radial timelike geodesic in anti-de Sitter spacetime> result; for curvature radius $a$, the answer is $\pi a$.
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