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For a star the proper motion, $\mu$, is usually measured in arcseconds per year. The diagram below illustrates how the quantities of radial, tangential and true velocity are related. The radial and tangential velocities are clearly components of the true velocity that have been resolved usefully relative to an observer on the Sun.

Diagram to show True Velocity and Proper Motion

Given a known value of $\mu$, this link states that the tangential velocity, $V_\theta$, is related to the proper motion by the equation

$$V_\theta = 4.7\mu d$$

where $d$ is the distance, and $V_\theta$ is the tangential velocity. I am unsure of where the factor of $4.7$ comes from. However, other sources give it as $V_\theta = \mu d$, although they do not define any units.

Could anyone confirm the units of the above quantities, and which equation is correct, as well as the origin of the 4.7?

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  • $\begingroup$ Note that in the eqn with the factor 4.7 the units of $\mu$ are arcsec/yr. That eqn is only correct for the inconsistent units given - ie km/s $\ne$ arcsec/yr x parsec. If you convert to consistent units you'll get the 2nd equation which is correct for any set of consistent units, eg km/s = rad/s x km. The equation is just $v=\omega r$. $\endgroup$ – sammy gerbil Oct 28 '17 at 21:16
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Both equations are correct. The factor of $4.7$ comes from the unit conversion:

$$\frac{({\rm arcsec}\,{\rm yr}^{-1})({\rm pc})}{{\rm km}\,{\rm s}^{-1}} = \frac{(4.84\times10^{-6}\,{\rm rad})\,(3.154\times10^{7}\,{\rm s})^{-1}(3.086\times10^{13}\,{\rm km})}{{\rm km}\,{\rm s}^{-1}} = 4.7$$

Note that I've used the "small angle approximation" $\sin(x)\approx x$ for $x\approx0$, so I can treat ${\rm rad}$ as dimensionless.

If you use simple $V_{\theta}=\mu d$, you will get the same answer, but when you input the values you'll need to do any applicable unit conversion yourself.

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