Rotation curves measured from HI 21-cm emission

The HI 21-cm emission is able to penetrate virtually the entire galaxy, and so can be used to measure the rotation curve $V_c(R)$. The biggest problem with this is how to determine $d$, the distance to the cloud that produces the absorption.

A solution is to recognise from Eqs.(5.2) that $V_r$ has a maximum

$$V_{r,{\rm max}}=V_c-V_{c,0}\sin(l)\,,$$ (5.9)

for any $l$ with $\vert l\vert<\pi/2$, which occurs for $\alpha=0$, at which point

$$R=R_0\sin(l)\,,$$ (5.10)

is a minimum, and ^5.4 $d=R_0\cos(l)$. Combining these, we find

$${dV_c(R)\over dR}={dV_{r,{\rm max}}(l)\over dl}\,/R_0\,\cos(l)+V_{c,0}/R_0\,.$$ (5.11)

Performing this analysis essentially confirmed Oort's measurements, i.e. $\vert dV_c/dR\vert$, the rate of change of the circular velocity with $R$, is far smaller than you'd expect for a Keplerian rotation curve. In fact, from measurements of Cepheids with $R>R_0$ it became apparent that $dV_c/dR\approx 0$, i.e. the Milky Way's rotation curve is nearly flat, or $V_c(R)\approx {\rm const}$, as opposed to $V_c\propto (M/R)^{1/2}$ for Keplerian fall-off - it is as if there is considerable mass outside the solar circle.