The <specific orbital energy> changes by $\Delta\varepsilon=\Delta(v^2)/2<0$, so a marginally bound orbit becomes more tightly bound. Neglecting its tiny initial binding energy,
$$
\boxed{a'\simeq\frac{GM_*}{B}\left(\frac\delta\mu\right)^2
=\frac{a_p}{4(7+5\sqrt2)}\left(\frac\delta\mu\right)^2}.
$$

To obtain the new periapsis, put $c=\sqrt2-1$ and $s=\mu/\delta$. Immediately after the encounter,
$$
v_r'\simeq-cv_K\theta,
\qquad
v_\theta'\simeq v_K\left(\sqrt2-\frac{c\theta^2}{2}\right).
$$
Using the post-encounter energy and angular momentum in the <Kepler orbit> relations and retaining $O(s^2)$ gives
$$
\boxed{q'\simeq q\left[1-(6+4\sqrt2)\left(\frac\mu\delta\right)^2\right]}.
$$
The inward radial kick therefore lowers the next periapsis as well as shrinking the semi-major axis.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 316 / 4 / v / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-316/4/v/solution