Escape speed of a truncated flat-rotation halo ($v_{\mathrm{esc}}^2=2v_c^2[1+\log (r_{\ast }/r)]$)

A spherical halo with flat circular speed $v_c$ has $M(r)=v_c^{2}r/G$ and $\rho =v_c^2/(4\pi G{r}^2)$. If it is sharply truncated at $r_{\ast }$, the work per unit mass from $r<r_{\ast }$ to the boundary is $v_c^2\log (r_{\ast }/r)$ and from the boundary to infinity is $v_c^2$. Adding them gives the displayed escape velocity. A bound star's observed radial speed is only a lower bound on its total speed and on the escape speed, so one maximum radial measurement generally gives a lower bound on $r_{\ast }$ rather than an exact halo edge.