Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 316 / 2

Here $a_p$ is the planet's semi-major axis, while $a,e,I$ are the particle's semi-major axis, orbital eccentricity, and orbital inclination relative to the planet's plane. The formula assumes the circular restricted three-body problem: the planet-to-star mass ratio is small, the particle has negligible mass, and its motion is approximately heliocentric and Keplerian away from brief encounters. The planet's own orbit is circular.

The Jacobi constant is exactly conserved in that ideal rotating problem. Expressing it in heliocentric orbital elements away from the planet gives the approximately conserved Tisserand parameterThe first term measures the particle's normalized binding energy. The second is twice the component of its specific angular momentum normal to the planet's plane, normalized by $\sqrt{G{M}_{\ast }{a}_p}$.

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Set $x=a/a_p$ and $c_I=\cos I$. At fixed $T_p$ and $I$,For a prograde orbit, the unsquared equation also requires $x\ge 1/T_p$. Differentiation shows that the only stationary values occur atThe first is an endpoint maximum with $e=1$. The second is the interior minimumIt lies on the physical locus when $T_p^3\le 27{\cos }^{2}I$. Equality makes the minimum circular. If $T_p^3>27{\cos }^{2}I$, a forbidden interval surrounds $3/T_p$ and the allowed locus splits into branches ending at $e=0$. As $a\to \infty$, $e\to 1$.

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For $T_p=5/2$ and $I=0$, the fixed-Tisserand parameter curve begins atIt falls smoothly toand then rises asymptotically back toward $e=1$ as $a/a_p\to \infty$. Since $T_p<3$, this coplanar locus never reaches a circular Kepler orbit.

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For $e=0$, regardIts minimum occurs at $x=(\cos I{)}^{-2/3}$ and equals $3(\cos I{)}^{2/3}$. A circular orbit is therefore possible for $T_p=5/2$ only ifThe smallest required orbital inclination is consequently

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A non-coplanar encounter may change $I$ while preserving $T_p$, so the projection of the accessible phase space onto the $(a,e)$ plane is a region rather than one curve. A pair $(x,e)$ is accessible for some inclination exactly whenFor $T_p=5/2$, the boundary curves are obtained by taking $\cos I=1$ or $-1$; intermediate inclinations fill the region between them. In particular, inclinations at least as large as the value found in part (d) allow the region to meet $e=0$.

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At a planet-crossing encounter, let $U$ be the planetocentric relative speed far from the planet and $v_p=\sqrt{G{M}_{\ast }/a_p}$. Combining the particle's heliocentric energy and normal angular momentum givesThus the Tisserand parameter fixes the encounter-speed scale and confines every post-encounter orbit to the same allowed region. For impact parameter $b$, a two-body estimate givesAn encounter with smaller $U$ is bent through a larger angle, whereas one with larger $U$ is less strongly focused. The actual displacement on the $(a,e)$ plot therefore depends on encounter geometry as well as $T_p$; very small $U$ limits the velocity vector available to redirect, and very large $U$ gives weak deflection, with the largest typical kicks between those limits.

Solved by gpt-5.6-sol high.