Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 4 / 11C / Solution

For a fixed central gravitational source, the radial and transverse components of Newton's second law in polar coordinates areThe second equation gives $r^2\dot{\theta }=h$, the constant specific angular momentum. Take $h>0$ by choosing the orientation of $\theta$, and put $u=1/r$. Writing $u^{\prime }$ and $u^{\prime \prime }$ for derivatives with respect to $\theta$,Substituting into the radial equation and dividing by $-h^2{u}^2$ proves the Binet equationAt perihelion the velocity is tangential. The escape velocity is $v_0=\sqrt{2GM/r_0}$, so $h=r_0{v}_0$ and $h^2=2GM{r}_0$. Consequently $k=1/(2r_0)$. The initial conditions are $u(0)=1/r_0$ and $u^{\prime }(0)=0$. Solving the linear ordinary differential equation gives the parabolic Kepler orbitUsing angular momentum conservation again,On the outgoing branch $r=2r_0$ is first reached at $\theta =\pi /2$. Integrate from release at perihelion, using $w=\tan (\theta /2)$:More generally the same integral gives $t=(2r_0^2/h)(w+w^3/3)$, the Barker equation for this parabolic orbit.