= Solution
At rest with all separations initially infinite, both initial kinetic and gravitational energies vanish, so the conserved total energy is zero. During an isolated collapse, $T+W=0$, rather than the equilibrium relation $2T+W=0$. Hence at the stated instantaneous <gravitational potential energy>,
$$
\boxed{\langle v_{\rm coll}^2\rangle
=\frac{-2W}{Nm}=\frac{2GNm}{R_g}
=2\langle v_{\rm eq}^2\rangle.}
$$
The rms collapse speed is $\sqrt2$ times the rms equilibrium speed for the same instantaneous $W$ and mass. For a perfectly radial cold infall this mean-square motion includes coherent inward streaming; it need not equal a local random <velocity dispersion> after subtracting that streaming. The quoted comparison follows the question's global mean-square convention.
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