= Solution
Use the total mass
$$
M\simeq60M_\odot\simeq1.2\times10^{32}\ \mathrm{kg}.
$$
With the stated last-orbit estimate,
$$
r=\frac{6GM}{c^2}\simeq5\times10^5\ \mathrm m.
$$
The <Kepler third law> then gives
$$
T=2\pi\sqrt{\frac{r^3}{GM}}
=12\pi\sqrt6\,\frac{GM}{c^3}
\simeq3\times10^{-2}\ \mathrm s.
$$
A binary's dominant quadrupole radiation has two cycles per orbit, so the <last-orbit frequency of a compact binary> is
$$
\boxed{f_{\rm GW}\simeq\frac2T\simeq7\times10^1\ \mathrm{Hz}}.
$$
Only one significant figure is meaningful under the approximations.
Back to article page
