Derive the equation

$$\frac{d^{2} u}{d \theta^{2}}+u=\frac{f(u)}{m h^{2} u^{2}}$$

for the orbit $r^{-1}=u(\theta)$ of a particle of mass $m$ and angular momentum $h m$ moving under a central force $f(u)$ directed towards a fixed point $O$. Give an interpretation of $h$ in terms of the area swept out by a radius vector.

If the orbits are found to be circles passing through $O$, then deduce that the force varies inversely as the fifth power of the distance, $f=c u^{5}$, where $c$ is a constant. Is the force attractive or repulsive?

Show that, for fixed mass, the radius $R$ of the circle varies inversely as the angular momentum of the particle, and hence that the time taken to traverse a complete circle is proportional to $R^{3}$.

[You may assume, if you wish, the expressions for radial and transverse acceleration in the forms $\ddot{r}-r \dot{\theta}^{2}, 2 \dot{r} \dot{\theta}+r \ddot{\theta}$.]