= Solution
For a spherical <mass distribution> with constant circular <speed> $v_c$,
$$
M(<r)=\frac{v_c^2r}{G},\qquad
\boxed{\rho(r)=\frac{1}{4\pi r^2}\frac{dM}{dr}=\frac{v_c^2}{4\pi Gr^2}.}
$$
For $v_c=220\,\mathrm{km\,s^{-1}}$, this is about $0.0140(r/8\,\mathrm{kpc})^{-2}M_\odot\,\mathrm{pc^{-3}}$. It is the spherical model's total density; it is singular if extrapolated to the very centre and is intended only over the flat-rotation region.
For a sharp outer truncation at $r_*$, the exterior mass is fixed at $M_*=v_c^2r_*/G$. Integrating the gravitational work from $r$ to the boundary and then to infinity gives
$$
\frac12v_{\rm esc}^2=\int_r^{r_*}\frac{v_c^2}{s}\,ds+\int_{r_*}^\infty\frac{GM_*}{s^2}\,ds
=v_c^2\left[\log\frac{r_*}{r}+1\right].
$$
Thus
$$
\boxed{v_{\rm esc}(r)=\sqrt{2}v_c\sqrt{1+\log(r_*/r)}.}
$$
The exterior contribution cannot be dropped: at $r=r_*$ it correctly gives the point-mass escape speed $\sqrt2v_c$.
If the fastest measured radial speed is assumed to approximate both the total speed and the local escape speed of a bound star, then at $r=8\,\mathrm{kpc}$,
$$
\boxed{r_*\simeq r\exp\left(\frac{430^2}{2(220)^2}-1\right)\simeq19.9\,\mathrm{kpc}.}
$$
Without those extra assumptions, the measurement gives only a bound: a bound star's total speed is at least its radial component, and the sample maximum need not reach escape, so $r_*\gtrsim19.9\,\mathrm{kpc}$ within this model. An unbound interloper would invalidate even that bound. This is not a determination of the real Milky Way halo edge.
Back to article page
