Let $v_K=\sqrt{\mu/a}$ be the speed of the <circular Kepler orbit> at radius $a$. The release velocity relative to the <planetesimal> is
$$
\Delta\mathbf v=\gamma v_K(-\sin\phi,\cos\phi).
$$
The particle starts at the same position as its parent, so the change in <specific orbital energy> is
$$
C'-C=\mathbf v\mathbin\cdot\Delta\mathbf v+\frac12|\Delta\mathbf v|^2.
$$
Using the velocity components from part (c),
$$
\mathbf v\mathbin\cdot\Delta\mathbf v
=\frac{\gamma\mu}{a\sqrt{1-e^2}}
\bigl[\cos(\phi-f)+e\cos\phi\bigr].
$$
Since $C=-\mu/(2a)$ and $C'=-\mu/(2a')$, rearrangement gives
$$
\boxed{\frac a{a'}=1-\gamma^2-
\frac{2\gamma}{\sqrt{1-e^2}}
\bigl[\cos(\phi-f)+e\cos\phi\bigr]}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 316 / 1 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-316/1/d/solution