Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 4 / 10B / i / Solution

Multiplying the tangential equation of motion by $r$ givesTherefore $h=r^2\dot{\theta }$ is constant. It is the signed specific angular momentum; reverse the angular orientation if necessary to take $h>0$ for a nonradial orbit. Set $u=1/r$ and denote differentiation with respect to $\theta$ by a prime. ThenSubstitution into the radial equation of motion and division by $-h^2{u}^2$ give the Binet equationIts general solution is $u=GM/h^2+A\cos \theta +B\sin \theta$. Write $A=(GM/h^2)e\cos {\theta }_0$ and $B=(GM/h^2)e\sin {\theta }_0$, where $e\ge 0$. This produces the Kepler orbitHere $e$ is the orbital eccentricity, and ${\theta }_0$ points toward periapsis when $e>0$. Only portions with $u>0$ describe positive radial distances. The expression assumes $h\ne 0$; a zero-angular momentum trajectory is radial and must be treated separately.