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

The force is central force, so its torque about the star vanishes. The specific angular momentum is therefore constant:Taking the scalar product of $\ddot{\mathbf{r}}=-\mu \mathbf{r}/r^3$ with $\dot{\mathbf{r}}$ givesThus conservation of energy gives the second constant

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Put $u=1/r$. Since $h=r^2\dot{\theta }$, the radial equation becomes the Binet equationwhere primes denote derivatives with respect to the polar angle. Choosing the angular origin at periapsis givesWriting the semi-latus rectum as $h^2/\mu =a(1-e^2)$ yields the Kepler orbitHence the specific angular momentum and specific orbital energy are

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In axes directed toward periapsis and along the motion there, the position is $\mathbf{r}=r(\cos f,\sin f)$. Differentiating it and using $\dot{f}=h/r^2$ givesDifferentiating the orbit equation gives $\dot{r}=(\mu e/h)\sin f$, and substitution simplifies the components to

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Let $v_K=\sqrt{\mu /a}$ be the speed of the circular Kepler orbit at radius $a$. The release velocity relative to the planetesimal isThe particle starts at the same position as its parent, so the change in specific orbital energy isUsing the velocity components from part (c),Since $C=-\mu /(2a)$ and $C^{\prime }=-\mu /(2a^{\prime })$, rearrangement gives

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The kick changes the tangential velocity by $\gamma v_K\cos (\varphi -f)$. With $s=\sqrt{1-e^2}$ and $r/a=s^2/(1+e\cos f)$, the new specific angular momentum isEvery Kepler orbit satisfies $h^{\prime 2}=\mu a^{\prime }(1-e^{\prime 2})$. Combining this identity with the value of $a/a^{\prime }$ found in part (d) giveswhere $a/a^{\prime }$ is the explicit function of $e,f,\gamma ,\varphi$ in part (d).

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An orbit is unbound precisely when its specific orbital energy is nonnegative, equivalently $a/a^{\prime }\le 0$. At periapsis, $f=0$, and for $e=1/2$ the condition from part (d) isThe directions are sampled from the uniform distribution on a circle. The fraction satisfying $\cos \varphi \ge x$ is $\arccos x/\pi$; setting it equal to $1/6$ gives $x=\sqrt{3}/2$. ThereforeThe positive root is

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