= Solution
For particle positions $\mathbf r_1,\mathbf r_2$, the two-particle <Schrodinger equation> is
$$
\boxed{
i\hbar\frac{\partial\Psi}{\partial t}
=\left[-\frac{\hbar^2}{2m_1}\nabla_1^2
-\frac{\hbar^2}{2m_2}\nabla_2^2
-\frac{Gm_1m_2}{|\mathbf r_1-\mathbf r_2|}\right]\Psi}.
$$
Write $|a\rangle_i$ for the localized <wave packet> $\psi_{ai}$ and $d_{ab}=|x_{a1}-x_{b2}|$. Neglecting packet spreading and branch overlap, the initial <product state> evolves branchwise as
$$
|\Psi(t)\rangle\simeq\frac12\sum_{a,b=0}^1
\exp\left(\frac{iGm_1m_2t}{\hbar d_{ab}}\right)
|a\rangle_1|b\rangle_2,
$$
up to phases generated independently on the two particles. These branch-dependent phases generally cannot be separated into one phase depending only on $a$ and one depending only on $b$, so the <Newtonian gravitational potential energy> creates <gravitationally induced entanglement>.
If $d_{10}=|x_{11}-x_{02}|=d$ is much smaller than the other separations, remove their nearly common phase and retain only
$$
\phi=\frac{Gm_1m_2t}{\hbar d}.
$$
The state is approximately
$$
|\Psi(t)\rangle
=\frac12\left(|00\rangle+|01\rangle
+e^{i\phi}|10\rangle+|11\rangle\right).
$$
Its <concurrence> is $|\sin(\phi/2)|$, so it becomes <maximally entangled state>[maximally entangled] first at $\phi=\pi$. For $m_1=m_2=m$,
$$
\boxed{t_{\rm ent}
=\frac{\pi\hbar d}{Gm^2}
=\frac{hd}{2Gm^2}}
=\frac{(6.6\times10^{-34})(2\times10^{-4})}
{2(6.7\times10^{-11})(10^{-14})^2}
\simeq9.9\ \mathrm{s}.
$$
Thus the near-maximal entanglement time is about $\boxed{10\ \mathrm{s}}$ within the stated approximation.
A single prescribed <Newtonian gravitational potential>[classical gravitational potential] gives a Hamiltonian of the form $H_1[\Phi]\otimes I+I\otimes H_2[\Phi]$. Its evolution factorizes as $U_1\otimes U_2$ and preserves every initial <product state>, so it cannot generate this entanglement. A semiclassical mean field sourced only by expectation values likewise gives each particle a local one-body potential and does not provide a quantum mediator carrying branch correlations.
An <entanglement witness> is a <Hermitian operator> $W$ whose expectation is nonnegative on every <separable quantum state>[separable state] but negative on at least one <entangled state>. At $\phi=\pi$, define
$$
|\Psi_*\rangle
=\frac12(|00\rangle+|01\rangle-|10\rangle+|11\rangle),
\qquad
W=\frac12I-|\Psi_*\rangle\langle\Psi_*|.
$$
The largest <Schmidt coefficient> of $|\Psi_*\rangle$ is $1/\sqrt2$, so every product state $|u\rangle|v\rangle$ in the four-dimensional branch subspace satisfies $|\langle\Psi_*|u,v\rangle|^2\leq1/2$. By closure under <convex combinations>, $\operatorname{Tr}(W\rho_{\rm sep})\geq0$ for every separable mixture, whereas
$$
\langle\Psi_*|W|\Psi_*\rangle=-\frac12.
$$
A negative measured value therefore certifies entanglement. Under the assumptions that the masses began unentangled and interacted only through gravity, such certification would show that the mediator can transmit quantum coherence; it would be evidence against a purely classical gravitational channel and for the quantum nature of gravity.
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