= 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> gives
$$
\boxed{\ddot z=-\frac{m}{4z^2},\qquad z(0)=z_0,\quad\dot z(0)=0.}
$$
Multiplying by $\dot z$ and integrating gives
$$
\frac12\dot z^2=\frac m4\left(\frac1z-\frac1{z_0}\right),\qquad
\boxed{\dot z=-\sqrt{\frac m2}\sqrt{\frac1z-\frac1{z_0}}.}
$$
The negative square root is essential for the falling branch. All <second mass moment tensor> components except $I_{zz}=2mz^2$ vanish. Differentiating three times,
$$
\dddot I_{zz}=4m(3\dot z\ddot z+z\dddot z),\qquad
\dddot z=\frac{m\dot z}{2z^3}.
$$
Substitution yields $\dddot I_{zz}=-m^2\dot z/z^2$, hence
$$
\boxed{\dddot I_{zz}=\frac{m^2}{z^2}\sqrt{\frac m2}\sqrt{\frac1z-\frac1{z_0}},}
$$
with every other component of $\dddot I_{ij}$ zero. \b[The distance entering each acceleration is $2z$], which accounts for the factor one quarter in the equation of motion.
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