# Calculating orbital velocities of stars in disk galaxies

In Barbara Ryden's Intro to Cosmology, chapter 8.2, she derives and equation for the radial velocities of stars around the center of their disk galaxy.

First, she states that when we look at disk galaxy, we see it with an inclination angle $\theta$, hence it appears elliptic with axis ratio $\frac{b}{a} = \cos\theta$.

Then, she states that by measuring the redshift of absorption or emission lines from the disk, we can find the radial velocity $v_r(R) = cz(R)$, where $R$ is the distance from the center of the galaxy. She states that this radial velocity is along the apparent long axis of the galaxy. Two questions about this:

1. Cosmological redshift is not generally a Doppler shift. Doppler shift occurs due to peculiar velocities. What kind of redshift does she mean here? If it is the Doppler shift, how can one distinguish between the redshift caused by the expansion of the universe and the redshift caused by the peculiar velocity?
2. This 'radial velocity along the apparent long axis of the galaxy' is not clear to me. Does she talk about the radial velocity towards the observer? Or maybe the radial velocity around the center of the galaxy? If the latter, how is the redshift related to it?

She then states that 'Since the redshift contains only the component of the stars' orbital velocity that lies along the light of sight, thre radial velocity we measure will be $$v_r(R) = v_{gal} + v(R)\sin{\theta}$$ where $v_{gal}$ is the radial velocity of the galaxy as a whole, resulting from the expansion of the universe.' This last derivation was really unclear.

The measured radial velocity here is the projected radial velocity relative to the average velocity of the galaxy as a whole - termed $v_{\rm gal}$ here. This average velocity could well be large due to cosmological redshift, but these are differential measurements with respect to this.
If a star is in an orbit with an inclination $\theta$, then the usual convention is that $\theta =\pi/2$ is edge on. Thus resolving $v(R)$ into the line of sight involves multiplication by $\sin \theta$.