Head-on quadrupole radiation from equal masses ($P=m^5(1/z-1/z_0)/(15z^4)$)

For two equal point masses at $\pm z$ falling from rest at $\pm z_0$ under Newtonian gravity, in geometrized units one has $\ddot{z}=-m/(4z^2)$ and $\dot{z}=-\sqrt{m(1/z-1/z_0)/2}$. The only nonzero component of the second mass moment tensor is $I_{zz}=2m{z}^2$, and its third derivative is $-m^2\dot{z}/z^2$. The trace-free mass quadrupole moment and quadrupole formula give the displayed power. From infinity, integrating down to $z=z_c$ gives $E=2\sqrt{2}{m}^{9/2}/(105z_c^{7/2})$. Relative to the total initial mass $2m$ the fraction is $\sqrt{2}(m/z_c{)}^{7/2}/105$. The weak-field slow-motion approximation must be distinguished from its extrapolation to a compact endpoint.