In the planetocentric frame, the incoming relative velocity $\mathbf w_i$ is prograde and the encounter outside the planet bends it inward through $\theta$, without changing $w=|\mathbf w|$. In the inertial frame $\mathbf v=\mathbf v_p+\mathbf w$, so
$$
\Delta(v^2)=2\mathbf v_p\cdot(\mathbf w_f-\mathbf w_i)
=2v_Kv_\infty(\cos\theta-1)
\simeq-v_Kv_\infty\theta^2.
$$
Substitution of the previous result yields the negative <gravitational assist> energy kick
$$
\boxed{\Delta(v^2)=-B\left(\frac\mu\delta\right)^2,
\qquad
B=\frac{4}{(\sqrt2-1)^3}\frac{GM_*}{a_p}
=4(7+5\sqrt2)\frac{GM_*}{a_p}}.
$$

Solution

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

 Solution

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