Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 4 / 9C / ii

The autonomous Hamiltonian is constant and positive on every trajectory, since $q\ne 0$. Hamilton's equations give $\dot{q}=2p$, $\dot{p}=2/q^3$, and henceThus $u=pq-2Ht$ is conserved. Put $s=u+2Ht$. Multiplying the energy equation by $q^2$ gives $H{q}^2=s^2+1$. The full classical path is thereforeThe sign $\eta$ labels the two disconnected half-lines and is fixed by initial data. Differentiating verifies both Hamilton equations. The inverse-square potential prevents the path from crossing zero.