Parameters of dark halo

Given the previous failure of our MW model, let's assume there to be a dark halo. In addition, let us assume the MW to be spherical, with a constant circular speed $V_c$ out to some radius $R_\star$. Since $V_c^2=GM/R={\rm constant}$, we have that

$$M(R) = {V_c^2 R\over G} \hbox{\rm when $R<R_\star$}$$

$$= {V_c^2 R_\star\over G} \hbox{\rm when $R\ge R_\star$} .$$ (7.2)

Now the equation for the potential $\Phi$ is

$${d\Phi\over dR} = {GM\over R^2} = {V_c^2\over R}\,,$$ (7.3)

for $R\le R_\star$. Integration gives $\Phi={\rm constant}-V_c^2\ln(R_\star/R)$. We can determine the value of the constant at $R=R_\star$, since then $\Phi=-G M/R_\star=-V_c^2$. Hence

$$\Phi(R) = -V_c^2\,\left[1+\ln(R_\star/R)\right]\,.$$ (7.4)

Using Eq.(7.1) for the escape speed, we obtain

$$v_e^2 = 2\,V_c^2\left[1+\ln(R_\star/R)\right]\,.$$ (7.5)

At the position of the Sun, we have $R\approx 8.5{\hbox{\rm kpc}}$, $V_c=220{\hbox{\rm km s$^{-1}$}}$ and $v_e\ge 500{\hbox{\rm km s$^{-1}$}}$. This implies that $R_\star\ge 41{\hbox{\rm kpc}}$, and hence

$$M(R=R\star)\ge 4.6\times 10^{11}\hbox{$M_\odot$} .$$ (7.6)

Since the total luminosity of the MW is of order $1.4\times 10^{10}\hbox{$L_\odot$}$ (in the V -band), it means^7.4 that the mass-to-light ratio is at least 30. Given that the mass-to-light ratio of the MW stars is around 3 or so, it again suggest that a dominant fraction of the MW mass is not in stars.

Another estimate of the MW mass comes from the motion of neighbouring galaxies in the Local Group.