 Solution

In the center of mass frame, put the two positions at $(0,0,z(t))$ and $(0,0,-z(t))$, with $z>0$ during infall. Their separation is $2z$, so Newtonian gravity givesMultiplying by $\dot{z}$ and integrating givesThe negative square root is essential for the falling branch. All second mass moment tensor components except $I_{zz}=2m{z}^2$ vanish. Differentiating three times,Substitution yields ${\stackrel{...}{I}}_{zz}=-m^2\dot{z}/z^2$, hencewith every other component of ${\stackrel{...}{I}}_{ij}$ zero. The distance entering each acceleration is $2z$, which accounts for the factor one quarter in the equation of motion.