Past exam of the physics 1 paper of the University of Oxford / 2015 resit / 9 / Solution

Orbital energy. Treat the star as stationary to leading order in $m/M$, with Newtonian gravitational parameter $\mu =GM$. For velocity components in plane polar coordinates, $v^2={\dot{r}}^2+r^2{\dot{\varphi }}^2$. The kinetic energy is $m{v}^2/2$. Integrating Newton's law of universal gravitation with zero gravitational potential energy at infinite separation gives $U(r)=-GMm/r$. Thuswhere $J=m{r}^2\dot{\varphi }$ is the satellite's angular momentum. The first term is radial kinetic energy; the second is transverse kinetic energy, also the centrifugal contribution to the radial effective potential; the last is the attractive gravitational potential energy. These are not three independent sources of energy: the centrifugal term is already part of the ordinary kinetic energy.

For an elliptic Kepler orbit of semi-major axis $a$ and orbital eccentricity $e$, the polar equation of a Kepler orbit $r=p/(1+e\cos \varphi )$ has semilatus rectum $p=J^2/(GM{m}^2)=a(1-e^2)$. Substituting $\dot{r}=(Je/(mp))\sin \varphi$ into the preceding expression reduces it toFor a circular Kepler orbit, $a=R$, $v_0^2=GM/R$, $K=GMm/(2R)$, and $U=-GMm/R$. Negative total energy describes a bound orbit.

Rapid, symmetric mass loss. Assume the mass is removed on a timescale short compared with the satellite's orbital period, without an impulsive kick or appreciable wind drag on the satellite. The post-loss gravitational parameter is ${\mu }^{\prime }=G(1-f)M$ with $f=0.01$. At the change, $r=R$, $\dot{r}=0$, and $v=v_0=\sqrt{GM/R}$ are unchanged. Spherical symmetry keeps the force central, so the specific angular momentum $j=J/m=\sqrt{GMR}$ is conserved. Orbital energy is not conserved across the changing gravitational field:The new semi-major axis is consequentlyThe original speed now exceeds the new circular speed at $R$, and the outward radial acceleration is $v_0^2/R-{\mu }^{\prime }/R^2=fGM/R^2>0$. Thus the release point is the new pericentre, not the apocentre. The impulsive orbital expansion under isotropic mass loss givesFor the stated one-percent loss,An independent angular momentum check is $j^2={\mu }^{\prime }{a}^{\prime }(1-e^{\prime 2})=GMR$. The impulsive orbit stays bound for $f<1/2$; loss of half the central mass is the zero-orbital energy limit.

What the wind timescale changes. The source specifies a spherically symmetric wind but does not give its duration. The preceding nearest and furthest distances require the rapid-loss assumption. In the opposite, slow limit, adiabatic orbital expansion under isotropic mass loss preserves an initially circular orbit to leading adiabatic order. Conserved specific angular momentum then gives $j^2=GM(t)r(t)$, hence $M(t)r(t)=MR$. After the same loss,For a finite intermediate timescale, the orbit depends on the mass-loss history; the percentage alone does not select a unique pair of turning distances. Neglecting wind recoil and wind drag is common to both idealizations.