= Solution
Take the <gravitational potential> to vanish at infinity. At the position of star $i$, excluding its own point-mass field, the instantaneous test-particle <escape velocity> obeys $v_{{\rm esc},i}^2=-2\Phi_i$. The factor of one half in the total <gravitational potential energy> is important:
$$
W=\frac12\sum_i m\Phi_i.
$$
Therefore the equal-mass average is
$$
\boxed{\langle v_{\rm esc}^2\rangle
=-\frac2N\sum_i\Phi_i=-\frac{4W}{Nm}
=\frac{4GNm}{R_g}=4\langle v_{\rm eq}^2\rangle.}
$$
Thus the rms escape speed is twice the equilibrium rms speed. This is the <virial mean escape speed> relation, using the same three-dimensional averaging convention as the preceding part. It concerns escape in the instantaneous system potential, rather than a subsequent collective rearrangement of the cluster.
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