Recall the geodesic equation in polar coordinates: \begin{align} \ddot{r}-r\dot\phi^2 &= 0 \\ r\ddot\phi+2\dot{r}\dot\phi &= 0 \end{align} Divide the second equation by $r\dot\phi$ to obtain \begin{equation} \frac{\ddot\phi}{\dot\phi} + \frac{2\dot{r}}{r} = 0 \end{equation} which can be integrated to yield \begin{equation} \dot\phi = \frac{A}{r^2} \label{geosolp} \end{equation} where $A$ is a constant. Inserting this into the first equation and multiplying by $\dot{r}$ now yields \begin{equation} \dot{r}\ddot{r} - \frac{A^2\dot{r}}{r^3} = 0 \end{equation} which can be integrated to yield \begin{equation} \hbox{constant} = \dot{r}^2 + \frac{A^2}{r^2} = \dot{r}^2 + r^2 \dot\phi^2 = |\vv|^2 \end{equation} Thus, geodesics must be traversed at constant speed, and can be classified by the constant $A$, which corresponds to angular momentum about the origin. It remains, of course, to solve ($\ref{geosolp}$), which we do in Geodesics in Polar Coordinates.

Solving the geodesic equation on the sphere is similar. We now start with \begin{align} \ddot\theta - \sin\theta\cos\theta\,\dot\phi^2 &= 0 \\ \ddot\phi + 2\cot\theta\,\dot\theta\,\dot\phi &= 0 \end{align} Divide the second equation by $\dot\phi$ to obtain \begin{equation} \frac{\ddot\phi}{\dot\phi} + 2\cot\theta\,\dot\theta = 0 \end{equation} which can be integrated to yield \begin{equation} \dot\phi = \frac{B}{r^2\sin^2\theta} \label{geosols} \end{equation} where $B$ is a constant (and the constant factor of $r^2$ is added for later convenience). Inserting this in the first equation and multiplying by $r^2\dot\theta$ yields \begin{equation} r^2\ddot\theta\,\dot\theta - \frac{B^2\cos\theta}{r^2\sin^3\theta} \>\dot\theta = 0 \end{equation} which can be integrated to yield \begin{equation} \hbox{constant} = r^2 \dot\theta^2 + \frac{B^2}{r^2\sin^2\theta} = r^2 \dot\theta^2 + r^2 \sin^2\theta \,\dot\phi^2 = |\vv|^2 \end{equation} and again we see that geodesics must be traversed at constant speed. Thus, geodesics on the sphere can be classified by the constant $B$, which again corresponds to (the $z$-component of) angular momentum. As before, it remains to solve ($\ref{geosols}$), which we do in Geodesics on the Sphere.