 Virial mean escape speed

For an isolated equilibrium of equal-mass points, the virial theorem gives $⟨v^2⟩=-W/M$. A particle's instantaneous escape velocity satisfies $v_{\mathrm{esc}}^2=-2\Phi$ with zero potential at infinity. Since $W=\frac{1}{2}{\sum }_{i}m{\Phi }_i$, averaging gives $⟨v_{\mathrm{esc}}^2⟩=-4W/M$. These are three-dimensional mean-square speeds, not one-component velocity dispersions.