Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 54 / 3

The conventional complete-flyby impulse approximation uses a straight incoming path $x=x_0$, $y=-S{x}_{0}t$, with $S>0$. Integrate the satellite's transverse acceleration over the whole encounter. For $x_0>0$,The hint's integral is one over a half encounter, and two over the full encounter. The leading longitudinal impulse vanishes by oddness. In weak elastic scattering, conservation of the relative speed $V=S|x_0|$ gives the next-order change along the original velocity as $-\Delta v_x^2/(2V)$. With the original outer flow along negative $y$, this meansThe same signed expression holds for inner particles and has the opposite sign there. This is the complete-flyby gravitational impulse. Its weak-deflection requirement is $G{M}_s/(S^2|x_0{|}^3)\ll 1$. Rotation and tidal dynamics retained throughout the encounter give a more detailed response; this impulse model does not claim to solve the full Hill scattering problem exactly.

The printed coefficient $1/2$ is four times smaller than this complete-flyby value. It is obtained if the single-sided transverse impulse $G{M}_s/(S{x}_0^2)$ is inserted into the same quadratic longitudinal estimate, leaving out the other half. Thus the scaling and sign agree, but that numerical coefficient is not derived by the usual full-encounter prescription. To keep the subsequent requested formulas unambiguous, write the satellite impulse normalizationThe following parts are derived for general $\chi$ and specialize to the supplied value. The complete-flyby one-sided torque coefficient $8/27$ is also the normalization used in the primary coplanar impulse calculation summarized by Chametla and collaborators.

A strip $dx$ at $x>0$ passes the satellite at relative speed $Sx$, so its encounter mass flux is $Sx\Sigma (x)dx$. Each unit mass gains specific angular momentum $r_0\Delta v_y$. Therefore the one-sided impulse torque on a disk isPutting $\chi =1/2$ gives exactly the requested expression. Inner strips have the opposite signed impulse but the same positive encounter flux. The net disc torque is consequentlyAn even surface density makes the integrand vanish pointwise, so the total torque is zero although each one-sided torque can be nonzero. The cutoff must keep the encounters in the weak-deflection regime; an appropriate physical value is at least of order the thickness or the relevant strong-scattering radius. The infinite-sheet integral also assumes convergence or a specified outer truncation.

The odd mass density difference is $2{\Sigma }_0\beta x/r_0$. Integrating $x^{-3}$ from $H$ to infinity givesFor Keplerian shear, $S=3\Omega /2$ and $G{M}_s=q{\Omega }^2{r}_0^3$. Thus the density-slope satellite torque isUsing the complete-flyby normalization multiplies this supplied-model result by four. The linear mass density law is a local Taylor approximation, not a nonnegative mass density on the entire infinite real line. With an outer local cutoff $L\gg H$, the same calculation replaces $H^{-2}$ by $H^{-2}-L^{-2}$; the extension to infinity is the leading local result. Small $|\beta |H/r_0$ keeps the mass density perturbation small in the dominant encounter region.

Outer particles gain angular momentum and inner particles lose it. The satellite transmits angular momentum from inner to outer material. If $\beta >0$, the denser outer side receives the larger torque: the disc gains net angular momentum and the satellite loses it. If $\beta <0$, the net transfer reverses. At zero slope these exchanges cancel in the satellite's total torque.

For a fixed satellite mass on an adiabatically changing circular orbit, $L_s=M_s\sqrt{G{M}_{\ast }{r}_0}=M_s{r}_0^2\Omega$. The satellite torque is minus the disc torque. HenceThe circular-orbit migration rate from disk torque, in the normalization supplied by the question, isIt is inward for a positive mass density slope and outward for a negative one. The complete-flyby normalization makes its magnitude four times larger, with the same direction. This is a local torque estimate; the hypothesis of a nearly circular slowly migrating orbit is needed when using the derivative of the circular angular momentum.