= Virial mean escape speed
{title2=$\langle v_{\rm esc}^2\rangle=4\langle v^2\rangle$}
For an isolated equilibrium of equal-mass points, the <virial theorem> gives $\langle v^2\rangle=-W/M$. A particle's instantaneous <escape velocity> satisfies $v_{\rm esc}^2=-2\Phi$ with zero potential at infinity. Since $W=\tfrac12\sum_i m\Phi_i$, averaging gives $\langle v_{\rm esc}^2\rangle=-4W/M$. These are three-dimensional mean-square speeds, not one-component <velocity dispersions>.
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