Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 316 / 1

Expand the secular Poynting–Robertson drag rates using $(1-e^2{)}^{-3/2}=1+\frac{3}{2}{e}^2+O(e^4)$ and $(1-e^2{)}^{-1/2}=1+O(e^2)$. Through second order in orbital eccentricity,The second formula follows from $d(e^2)/dt=2e\dot{e}$. The cubic correction in $\dot{e}$ contributes only at fourth order in $d(e^2)/dt$.

In resonant trapping of dust, the resonant torque must compensate the inward Poynting–Robertson drag. At the nearly fixed resonant semi-major axis, ${\dot{a}}_{\mathrm{res}}+{\dot{a}}_{\mathrm{pr}}=0$ requiresSince $|\sin \varphi |\le 1$, a necessary torque-balance condition is $|C|e^q\ge A(1+3e^2)/[a^2(p+q)]$. When the limiting orbital eccentricity is small, this givesEquality allows only the extreme phase $|\sin \varphi |=1$, with no margin for restoring torque; a robust trapped state needs $e>e_{\min }$. This is a necessary condition for resonance capture, not a sufficient probability-of-capture criterion for a migrating particle.

Write $u=e^2$ and use the torque balance from part (ii) to eliminate the resonant argument. The resonant contribution isAdding the Poynting–Robertson drag contribution gives the dissipative resonant eccentricity equilibrium equationwhereFor $5p>q$, integrating with $u(0)=e_0^2$ givesHere $a$ is the fixed resonant semi-major axis and the expression describes averaged evolution, neglecting the small resonant-argument libration. The physical bound-orbit interpretation also requires $B<1$; a quantitatively controlled second-order calculation needs $B\ll 1$.

At $e_{\max }=\sqrt{B}$, torque balance givesThe magnitude of the right-hand side must not exceed one. This fixes the dissipative offset of the resonant-argument libration centre.

For $C>0$, put $S=5A/[C{a}^2(5p-q)B^{q/2}]$. The two torque-balanced phases are ${\varphi }_0=-\arcsin S$ and ${\varphi }_0=\pi +\arcsin S$, modulo $2\pi$. In the usual weak-perturbation pendulum approximation, $\delta \dot{\varphi }\simeq -3(p+q)n\delta a/(2a)$ and $\delta \dot{a}\simeq -2(p+q)Ca{B}^{q/2}\cos {\varphi }_0\delta \varphi$. Restoring libration thus selects $\cos {\varphi }_0<0$, givingIt approaches $\pi$ when drag is weak compared with resonant torque. For $C<0$, the corresponding restoring branch has $\cos {\varphi }_0>0$ and approaches zero. The sine equation is the direct result independent of this leading-order stability approximation.

At periapsis, the particle's mean longitude equals its longitude of periapsis: $\lambda =\varpi$. Hence the pericentre geometry of an exterior resonance givesThe resonant argument therefore fixes the angular separation from the planet whenever the particle reaches periapsis. With $r_{\mathrm{peri}}=a(1-e)$ and planetary radius $a_p$, the actual separation isThus a restricted libration phase can prevent close encounters even when the radial ranges of the two orbits overlap: this is resonance protection.

The specific angular momentum is $h=\sqrt{G{M}_{\star }a(1-e^2)}$, so the periapsis angular speed isAt the exterior mean-motion resonance, neglecting radiation pressure, $n/n_p=p/(p+q)$. Thus at the limiting orbital eccentricity $e=\sqrt{2q/(5p-q)}$,The condition for overtaking the planet in angle isFor the formal second-order equilibrium, it holds for every physical bound value $0<q<5p/3$. To see this, rewrite $q/p=5e^2/(2+e^2)$, so $p/(p+q)=(2+e^2)/(2+6e^2)$. For $0<e<1$, $\sqrt{1+e}/(1-e{)}^{3/2}>1+2e$, whileThe instantaneous angular motion is therefore prograde relative to the planet near periapsis, although its mean relative motion is retrograde. The range $q\ll p$ is the regime where the eccentricity expansion is most reliable.

For the $5:3$ exterior mean-motion resonance, $p=3$, $q=2$. The truncated dissipative resonant eccentricity equilibrium givesFor weak drag and $C>0$, take the resonant-argument libration centre ${\varphi }_0\simeq \pi$. The pericentre geometry of an exterior resonance then places the three pericentres at angles $\pi /3$, $\pi$, and $5\pi /3$ relative to the planet. The trajectory closes after three particle orbits and five planet orbits. It moves retrograde in the rotating frame on average but develops a prograde loop near each periapsis, because the periapsis angular speed exceeds the planet's angular speed.

For a finite drag offset ${\varphi }_0=\pi +\arcsin S$, rotate the whole particle pattern through $\arcsin S/3$. The sketch uses the eccentricity requested by the second-order model; $e\simeq 0.555$ is not small, so the curve is an illustration of that approximation rather than a quantitatively accurate integration of the full dissipative resonance.