 Solution

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.