= Solution
The separation of the stars is $2R$. <Newton's law of universal gravitation> and circular acceleration give
$$
M\Omega^2R=\frac{M^2}{(2R)^2},\qquad
\boxed{\Omega^2=\frac{M}{4R^3}.}
$$
To leading nonrelativistic order, the <mass density> is the sum of the two translated point-mass <Dirac delta distributions>. Thus
$$
I_{xx}=MR^2(1+\cos2\Omega t),\qquad
I_{yy}=MR^2(1-\cos2\Omega t),\qquad
I_{xy}=I_{yx}=MR^2\sin2\Omega t,
$$
with all components involving $z$ zero. The trace is $2MR^2$, a constant. Define the trace-free <mass quadrupole moment> $Q_{ij}=I_{ij}-\delta_{ij}I_{kk}/3$; its third derivatives equal those of $I_{ij}$. If $K=8MR^2\Omega^3$, then
$$
\dddot Q_{xx}=K\sin2\Omega t,\qquad
\dddot Q_{yy}=-K\sin2\Omega t,\qquad
\dddot Q_{xy}=\dddot Q_{yx}=-K\cos2\Omega t.
$$
Both off-diagonal entries count in the contraction. Therefore $\dddot Q_{ij}\dddot Q_{ij}=2K^2=128M^2R^4\Omega^6$, independent of phase. The <quadrupole formula> gives the <equal-mass circular-binary quadrupole luminosity>
$$
\boxed{\langle P\rangle=\frac{128}{5}M^2R^4\Omega^6=\frac{2M^5}{5R^5}.}
$$
Restoring units, $\Omega^2=GM/(4R^3)$ and
$$
\boxed{\langle P\rangle=\frac{2G^4M^5}{5c^5R^5}.}
$$
Here $R$ is each star's radius about the centre of mass, not the separation. The rest-frame prescription supplies the leading <mass density>; a moving star does not still have zero momentum density and spatial stress. The calculation uses the assumed <quadrupole formula> and the Newtonian orbit rather than imposing those rest-frame zeros on the moving binary.
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