Write $k=\mu(1-\beta)$. Taking the <cross product> of the equation of motion with $\mathbf r$ gives
$$
\frac d{dt}(\mathbf r\times\dot{\mathbf r})=0,
\qquad
\boxed{\mathbf h=\mathbf r\times\dot{\mathbf r}}.
$$
Since $\mathbf r\cdot\mathbf h=0$ at every time, the orbit lies in the fixed plane perpendicular to the conserved <specific angular momentum> $\mathbf h$. Taking the <inner product> with $\dot{\mathbf r}$ gives the conserved <specific orbital energy>
$$
\boxed{C=\frac12|\dot{\mathbf r}|^2-\frac{\mu(1-\beta)}r}.
$$
For $\beta>1$, the effective inverse-square force is repulsive and the potential term is positive.

Solution

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

 Solution

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