In polar coordinates in the orbital plane, $h=r^2\dot\theta$. Setting $u=1/r$, the <Binet equation> for the effective inverse-square force is
$$
u''+u=\frac{\mu(1-\beta)}{h^2}.
$$
Its general solution is
$$
u=\frac{\mu(1-\beta)}{h^2}
\left[1+e\cos(\theta-\varpi)\right],
$$
and hence
$$
\boxed{r=\frac{h^2/[\mu(1-\beta)]}
{1+e\cos(\theta-\varpi)}}.
$$
This is a branch of a <hyperbolic Kepler orbit>; because $1-\beta<0$, its physical branch has a negative denominator.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 316 / 2 / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-316/2/ii/solution