Put $u=1/r$. Since $h=r^2\dot\theta$, the radial equation becomes the <Binet equation>
$$
u''+u=\frac\mu{h^2},
$$
where primes denote derivatives with respect to the polar angle. Choosing the angular origin at <periapsis> gives
$$
u=\frac\mu{h^2}(1+e\cos f).
$$
Writing the semi-latus rectum as $h^2/\mu=a(1-e^2)$ yields the <Kepler orbit>
$$
\boxed{r=\frac{a(1-e^2)}{1+e\cos f}}.
$$
Hence the <specific angular momentum> and <specific orbital energy> are
$$
\boxed{h=\sqrt{\mu a(1-e^2)}},
\qquad
\boxed{C=-\frac\mu{2a}}.
$$

Solved by gpt-5.6-sol high.

Solution

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

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.

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

An orbit is unbound precisely when its <specific orbital energy> is nonnegative, equivalently $a/a'\leq0$. At <periapsis>, $f=0$, and for $e=1/2$ the condition from part (d) is
$$
1-\gamma^2-2\sqrt3\,\gamma\cos\phi\leq0,
\qquad
\cos\phi\geq\frac{1-\gamma^2}{2\sqrt3\,\gamma}.
$$
The directions are sampled from the <uniform distribution> on a circle. The fraction satisfying $\cos\phi\geq x$ is $\arccos x/\pi$; setting it equal to $1/6$ gives $x=\sqrt3/2$. Therefore
$$
\frac{1-\gamma^2}{2\sqrt3\,\gamma}=\frac{\sqrt3}{2},
\qquad
\gamma^2+3\gamma-1=0.
$$
The positive root is
$$
\boxed{\gamma=\frac{\sqrt{13}-3}{2}}.
$$

Solved by gpt-5.6-sol high.

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

{title2=$\varepsilon$}
{wiki}

The specific orbital energy is the mechanical energy per unit mass,
$$
\varepsilon=\frac{v^2}{2}-\frac{\mu}{r}=-\frac{\mu}{2a}.
$$
Its sign distinguishes bound elliptic, marginal parabolic, and unbound hyperbolic Kepler orbits.


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