Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 320 (Modern Stellar Dynamics)

For a bound Kepler orbit about mass $M$, the specific orbital energy and specific angular momentum areThe apsidal radii areequivalently $a=(r_{-}+r_{+})/2$ and $e=(r_{+}-r_{-})/(r_{+}+r_{-})$. Combining the formulas also gives

At fixed energy, the circular orbit has $L_c^2=GMa$, whileAn isotropic galactic distribution function has conditional angular-momentum density $n(L)dL\propto LdL$. ThereforeNormalization on $0\le e<1$ gives the thermal eccentricity distribution

The sudden potential change leaves the star's position and velocity unchanged. Its new energy is thereforeThus $E_1>0$ precisely when $r<a$. Parametrize the orbit by eccentric anomaly $\eta$:A phase-mixed orbit is uniform in the mean anomaly ${\theta }_r$. For $e>0$, the condition $r<a$ is $-\pi /2<\eta <\pi /2$ modulo one period. The corresponding mean-anomaly interval has lengthHenceThe energy increase $G(M_0-M_1)/r$ is largest near periapsis, so stars are more readily unbound there than near apoapsis. A perfectly circular orbit is the measure-zero marginal case $E_1=0$ at every phase.

Average the phase probability over the thermal eccentricity distribution:Therefore

For slow mass loss, the adiabatic invariance of an orbital action applies. Spherical symmetry conserves $L$ exactly, and adiabatic evolution conserves the radial action. For a Kepler orbit,Consequently $Ma$ and $e$ remain constant. When the central mass halves, every semimajor axis doubles while each eccentricity is unchanged:No orbit becomes unbound as long as the final mass remains positive, soBecause the action mapping changes no eccentricity and introduces no orientation preference, the initially isotropic distribution remains isotropic rather than becoming radially or tangentially anisotropic.

SetThe spherical Poisson equation givesThe numerator identitysplits this into two scale-free density-potential components:Thus

At small radius, $D\sim 2b$ and the first term dominates:The cusp has finite enclosed mass because $r^2\rho \sim r^{1/2}$. Moreover,so the circular speed tends to zero as $r^{1/4}$. At large radius, $D\sim r$ andSince $r^2\rho \sim r^{-3/2}$ is integrable at infinity and $\Phi \sim -GM/r$, the total mass is finite and equals $M$.

Use the relative energy $\mathcal{E}=-E=\Psi -v^2/2$. At fixed $r$, write velocity-space spherical coordinates with polar angle $\alpha$ from the radial direction, so $L=rv\sin \alpha$. ThenThe beta function integrals givewhereThus $\gamma =2\beta$ and $p=n+3/2-\beta$, with convergence for $\beta <1$ and $n>-1$.

Comparison with part a givesHence the model has the constant-anisotropy distribution functionwhere matching the two density coefficients gives

For one component with density ${\rho }_k=A_k{r}^{-2{\beta }_k}{\Psi }^{p_k}$, direct velocity integration, or the Spherical Jeans equation, givesEach component has velocity-anisotropy parameter ${\beta }_k$. Therefore the combined coefficient is the radial-pressure-weighted meanHeresoSince $q\sim \sqrt{r/b}/2$ at the origin and $q\sim \sqrt{b/r}$ at infinity,The second, less radial component reduces the anisotropy only at intermediate radii.

The enclosed masses of the $r^{-1}$ host and truncated satellite areFor a circular orbit, the Jacobi tidal radius isSince $d\log M_h/d\log r_o=2$, substitution givesand henceThe assumption $A_s\ll A_h$ makes $r_t/r_o\ll 1$, precisely the scale separation required by the local tidal approximation.

PutPart a giveswhile the host circular speed is $v_c=\sqrt{C{r}_o}$. The Chandrasekhar dynamical friction acceleration reduces to the radius-independent valueAssume stripped material leaves with the satellite's instantaneous specific angular momentum. The remaining orbit then obeysSince $r_o{v}_c=\sqrt{C},r_o^{3/2}$,For initial radius $r_{o0}$,andIn this ideal cusp, radius and remaining mass both reach zero at the finite time $t_d$.

If $M_s$ is constant, the frictional acceleration instead scales as $r_o^{-2}$. The angular-momentum equation givesThereforeThe orbit again reaches the centre in finite time, but its inward speed accelerates without the strong tidal-mass suppression present in part b.

For a singular isothermal sphere host, ${\rho }_h=A_h{r}_o^{-2}$,Now $d\log M_h/d\log r_o=1$, so the circular tidal formula has a factor two:Using $M_s=2\pi A_s{r}_t^2$ givesThe frictional acceleration is now $a_{\mathrm{df}}=-G{\mu }_i\mathcal{L}{r}_o^2$. Since the specific angular momentum is $V{r}_o$,ThusNeither radius nor mass reaches zero at finite time. Compared with the $r^{-1}$ host, the steeper isothermal cusp shrinks the tidal radius as $r_o^2$, so stripping suppresses the friction rapidly enough to produce dynamical-friction stalling by tidal stripping.