Past exam of the astrophysics course of the University of Oxford / 2015 / 1 / b

Use constant component masses and a circular Keplerian orbit. Then $J=\mu \sqrt{GMA}$ gives $\dot{J}/J=\dot{A}/(2A)$, while $P=2\pi A^{3/2}/\sqrt{GM}$ gives $\dot{P}/P=3\dot{A}/(2A)=3\dot{J}/J$. Dividing the supplied loss law by $J$ therefore yieldsThe positive instantaneous orbital decay timescale is ${\tau }_P=P/|\dot{P}|$, since $\dot{P}<0$. For $t_G=10^{10}$ yr, the requested prescription givesIt is important to distinguish this prescription from the integrated gravitational-wave inspiral time. Integrating $\dot{A}=-C/A^3$ givesFinite contact radii supply a small lower-limit correction when the initial separation is much larger than the compact objects. Thus the periods below satisfy the printed instantaneous-timescale criterion and have a formal merger time $3.75$ Gyr. If instead the requirement is literally $t_{\mathrm{merge}}=10$ Gyr, every period limit is larger by $(8/3{)}^{3/8}=1.445$. Eccentric systems require a different decay calculation.

The binary of neutron stars has $M=2.8M_{\odot }$ and $\mu =0.7M_{\odot }$. Inserting these into the circular orbital decay timescale givesUsing the integrated ten-billion-year merger criterion instead would give $15.2$ h.

For the mixed pair, $M=11.4M_{\odot }$ and $\mu =(14/11.4)M_{\odot }=1.228M_{\odot }$. The prescribed instantaneous orbital decay timescale givesThe integrated ten-billion-year merger criterion instead gives $26.6$ h. Both are pre-disruption circular-orbit estimates; the final contact/capture process differs from that for two neutron stars.

For two black holes of equal mass, $M=20M_{\odot }$ and $\mu =5M_{\odot }$. The prescribed instantaneous orbital decay timescale givesThe integrated ten-billion-year criterion gives $51.9$ h. These large initial periods compared with a neutron-star pair reflect the stronger radiation reaction of the more massive system.