Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 316 (Planetary System Dynamics)

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

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

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

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

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).

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

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 $\sqrt{G{M}_{\ast }{a}_p}$.

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 $a\to \infty$, $e\to 1$.

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.

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

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 $e=0$.

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.

Write the power-law size distribution as $n(D)=k{D}^{-\alpha }$. The total geometric cross-section of the optically thin debris disk isBecause blackbody grains at radius $r$ intercept the fraction ${\sigma }_{\mathrm{tot}}/(4\pi r^2)$ of the stellar luminosity, the fractional luminosity of a debris disk givesAnother integral then yieldsfor $D_{\min }\ll D\ll D_{\max }$.

For equal material densities, a projectile of diameter $D_{\mathrm{cc}}$ catastrophically disperses a target of diameter $D$ whenAssume $D_{\min }\ll D_{\mathrm{cc}}\ll D$, neglect gravitational focusing, and approximate the collision cross-section by $\pi D^2/4$. The collision rate per target isSubstitution giveswhere

For size-independent $Q_D^{\ast }$, the total rate of destructive events whose target exceeds $D$ isSince the integrand is proportional to $D^{\prime 3-2\alpha }$ and $\alpha >3$, the lower limit dominates. With $n(D)=k{D}^{-\alpha }$,The mean interval between such events is therefore

If $Q_D^{\ast }\propto D^b$, part (b) givesThe mass in one logarithmic size interval scales as $D^3[Dn(D)]\propto D^{4-\alpha }$. A steady collisional cascade requires the mass destroyed per unit time, and hence the mass flux through every logarithmic interval, to be independent of $D$. Its exponent is thereforeSolving gives

The total destructive-event rate above $D$ scales asThe steady-state relation from part (d) impliesso the exponent is exactly $-3$. ConsequentlyThe power $3$ is independent of the disruption-law index $b$: changing $b$ changes the equilibrium size-distribution index $\alpha$ in precisely the compensating way.

Let the shattering threshold be $Q_s=Q_a{D}^{-a}$ and the full catastrophic disruption threshold be $Q_D^{\ast }$. From part (b), the rate of impacts exceeding an energy threshold $Q$ is proportional to $Q^{(1-\alpha )/3}$. The expected number of shattering impacts during one catastrophic-disruption waiting time is thereforeExcluding the final catastrophic event, the number of rubblising collisions is approximatelyFor the expression printed in the paper, if the gravity-regime term $Q_b{D}^{-b}$ dominates, this becomesThis estimate assumes independent impacts drawn from the same projectile distribution and ignores structural evolution after each rubblising collision.

The radiation-pressure coefficient reduces the dust grain's effective stellar gravitational parameter to $(1-\beta )G{M}_{\ast }$. At the exterior 5:4 mean-motion resonance, $n_d/n_p=4/5$, so mean motion givesThis is a first-order mean-motion resonance. Its leading disturbing-function term is therefore linear in the small orbital eccentricity:Thus its dimensionless strength is of order $(M_p/M_{\ast })e_d$, up to the Laplace-coefficient combination $f_{5:4}$, and its resonant argument varies slowly near commensurability.

Let $\psi ={\lambda }_d-{\lambda }_p$ be azimuth in the frame rotating with the planet, with the planet fixed at $\psi =0$. Since $n_d/n_p=4/5$, one synodic circuit contains four radial oscillations. To first order in $e_d$, choosing conjunction at apoapsis givesThe requested sketch is therefore a four-lobed closed curve centred on the star. For $e_d=0.2$, its radii range from $0.8a_d$ to $1.2a_d$; an outer lobe points along the star-planet line, and the planet lies at $(a_p,0)$. At that radial alignment $\varphi =\pi$, because ${\lambda }_d={\lambda }_p$ and ${\lambda }_d={\varpi }_d+\pi$ at apoapsis.

At exact conjunction the planet's force is radial and has no instantaneous torque, but the approach and departure do not cancel when conjunction is displaced from apoapsis. Write the conjunction longitude as ${\lambda }_c={\varpi }_d+\pi +\delta$. At conjunction,The resonant part of the given evolution law has ${\dot{a}}_d=-8e_d{a}_{d}D\sin \varphi$. Thus a conjunction after apoapsis, $\delta >0$, gives $\sin \varphi <0$ and transfers angular momentum to the dust, increasing $a_d$. A conjunction before apoapsis removes angular momentum, while one exactly at apoapsis has zero secular transfer by symmetry.

Poynting–Robertson drag removes angular momentum, so a trapped grain requires positive resonant torque from the planet. Part (c) shows that conjunction must occur after the dust has passed apoapsis. In the planet's rotating frame, the four-lobed pattern is therefore rotated so that the relevant outer lobe lags the fixed planet in the direction of orbital motion. Equivalently,with the stable phase displaced beyond $\varphi =\pi$ rather than sitting at the torque-free value $\varphi =\pi$.

During resonant trapping of dust, the mean semi-major axis is stationary. Setting the supplied ${\dot{a}}_d$ to zero givesA real resonant phase exists only when $|\sin \varphi |\le 1$. Hencewhere $e_{d0}$ is the grain's orbital eccentricity when trapping begins. The inequality also has the expected sign $\sin \varphi <0$, allowing the planetary torque to oppose Poynting–Robertson drag.

While resonance fixes $a_d$, substitute the phase relation from part (e) into the supplied eccentricity equation:To the requested first order in eccentricity, discard the $e_d^2$ correction. It follows thatUsing ${\beta }_{\max }=4c{e}_{d0}{a}_d^{2}D/(G{M}_{\ast })$ and integrating from $e_d(0)=e_{d0}$ givesThe neglected term eventually matters and prevents indefinite validity of this small-eccentricity growth law.