The kick changes the tangential velocity by $\gamma v_K\cos(\phi-f)$. With $s=\sqrt{1-e^2}$ and $r/a=s^2/(1+e\cos f)$, the new <specific angular momentum> is
$$
\frac{h'}{\sqrt{\mu a}}
=s+\gamma\frac{s^2\cos(\phi-f)}{1+e\cos f}.
$$
Every <Kepler orbit> satisfies $h'^2=\mu a'(1-e'^2)$. Combining this identity with the value of $a/a'$ found in part (d) gives
$$
\boxed{
e'^2=1-\frac a{a'}
\left[
\sqrt{1-e^2}
+\gamma\frac{(1-e^2)\cos(\phi-f)}{1+e\cos f}
\right]^2},
$$
where $a/a'$ is the explicit function of $e,f,\gamma,\phi$ in part (d).

Solved by gpt-5.6-sol high.

Solution

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

 Solution

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