Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 316 / 3

Let $\mathbf{r}={\mathbf{r}}_2-{\mathbf{r}}_1$ and $r=|\mathbf{r}|$. Newton's second law for the two planetesimals givesSubtracting yields the forced relative two-body problemThe internal gravitational forces cancel from the centre of mass equation but add in the relative equation; external forces affect the internal orbit through their difference in acceleration.

Put $m=m_1+m_2$, $\mu =Gm$, $n=\sqrt{\mu /a^3}$, $v=na$, and $\theta =nt$. Work instantaneously in the inertial frame in which the centre of mass is at rest. The unperturbed circular relative orbit hasThe individual velocities are ${\mathbf{v}}_1=-(m_2/m)\mathbf{v}$ and ${\mathbf{v}}_2=(m_1/m)\mathbf{v}$. Linear gas drag gives ${\mathbf{F}}_i/m_i=-({\mathbf{v}}_i-v_g\hat{\mathbf{y}})/t_i$, where $t_i=t_{si}$ are the aerodynamic stopping times. DefineThen the gas drag on a planetesimal binary isEquivalently, $\mathbf{f}=-\kappa \mathbf{v}+\Delta v_g\hat{\mathbf{y}}$. If the centre of mass moves, $v_g\hat{\mathbf{y}}$ here is its relative gas velocity, held approximately fixed during one short binary orbit.

With $\hat{\mathbf{r}}=(\cos \theta ,\sin \theta )$ and $\hat{\mathbf{\theta }}=(-\sin \theta ,\cos \theta )$, project the relative gas drag to obtainThe specific orbital energy is $\epsilon =v^2/2-\mu /r=-\mu /(2a)$. Differentiation cancels the conservative gravitational work and gives $\dot{\epsilon }=\mathbf{v}\cdot \mathbf{f}$. For the circular orbit,Using $v^2=\mu /a$, the orbit-averaged drag work relation therefore gives the instantaneous leading-order changeThe differential headwind term can change sign during the orbit; the damping term always removes orbital energy.

For weak gas drag, evaluate the work on the unperturbed circular orbit and average over one orbital period. The gas velocity changes negligibly during that period and $⟨\cos \theta ⟩=0$. Hence gas-driven contraction of a planetesimal binary obeysWith constant aerodynamic stopping times,This is the secular, orbit-averaged separation. It assumes the perturbations are small enough for the orbit to remain nearly circular; a large differential headwind can excite eccentricity or disrupt a weakly bound pair instead.

Let $\mathbf{V}$ be the centre of mass velocity and $\mathbf{v}$ the binary relative velocity. Summing the two linear gas drag forces givesThe last term averages to zero over the internal circular orbit. On the nearly circular stellar orbit, $\mathbf{V}\simeq v_K\hat{\mathbf{\theta }}$ and ${\mathbf{v}}_g=(1-\eta )v_K\hat{\mathbf{\theta }}$, so the averaged tangential acceleration is $-\eta {\kappa }_b{v}_K$.

With specific orbital energy $-G{M}_{\star }/(2a_b)$, the orbit-averaged drag work is $-\eta {\kappa }_b{v}_K^2$, where $v_K^2=G{M}_{\star }/a_b$. Thus gas-driven migration of a binary centre of mass givesThis uses the weak-drag, nearly Keplerian approximation implicit in applying the circular-orbit work relation. Strongly gas-coupled orbits require a coupled radial-azimuthal drift solution.

Put $ϵ=m_2/m_1\ll 1$ and $t_{s2}/t_{s1}={ϵ}^k$, with $0<k<1$. The gas-driven contraction of a planetesimal binary and gas-driven migration of a binary centre of mass times obeyThereforeThe low-mass secondary has the shorter aerodynamic stopping time and chiefly controls internal damping, whereas the primary chiefly controls centre of mass drag. The pair can contract substantially, potentially merging, before its stellar orbital radius changes appreciably. This conclusion is restricted to a pair that remains bound and within the adopted linear-drag regime.