 Solution

ID: past-exam-of-the-astrophysics-course-of-the-university-of-oxford/2018/6/c/solution

For a kick parallel or antiparallel to star 1's original velocity, the immediate post-explosion relative speed is

∣ v_{0} \pm v^{\pm} ∣

. Thus

E^{'} = \frac{1}{2} μ^{'} v_{0}^{2} [(1 \pm \frac{v ^{\pm}}{v _{0}})^{2} - 2 \frac{M ^{'}}{M _{0}}] .

A positive-energy, hyperbolically escaping pair therefore satisfies precisely

1 \pm \frac{v ^{\pm}}{v _{1} + v _{2}} > 2 \frac{M _{NS} + M _{2}}{M _{1} + M _{2}} .

The absolute value is essential when a sufficiently large opposite kick reverses the relative tangential velocity.

Marginal case. Strictly speaking, a pair is gravitationally bound only for

E^{'} < 0

. If unbound includes marginal escape, replace the printed strict inequality by

\geq

. For a concrete counterexample to an unrestricted “if and only if” with

>

, take no kick and

Δ M = M_{0} /2

. Both sides equal one and the energy is zero: this is not a bound ellipse, although the printed strict inequality is false.

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