<p>You can't without assuming something about the overall velocity.</p>

<p>The radial velocity is one component of a velocity vector; you are missing the other two components, which could in principle be anything. However, you could <em>assume</em> that as <strong>open</strong> clusters are mostly quite young and members of the Galactic disc population, that they are moving in the disc on roughly circular orbits. NB. This cannot possibly work for <strong>globular</strong> clusters, which are an entirely different population, many with non-circular orbits and orbits out of the galactic plane.</p>

<p>Having made that assumption, your radial velocity is the orbital velocity resolved through an angle that is formed between your line of sight to the cluster and a tangent to the Galactic centre at the cluster's Galactic position.</p>

<p>This method will not work well for clusters with a Galactic latitude much different to zero, or clusters with small Galactic longitude (which should have zero radial velocity in this model).</p>

<p>For all this to work you need the radial velocity, the Galactic latitude and longitude of the cluster, the distance to the cluster and an assumed distance to the Galactic centre.</p>

<p>I've had a go at sketching the situation. The cluster is on a circular orbit around the Galactic centre with a true orbital velocity $v$ (shown as a vector). We measure the component of this in our line of sight as the radial velocity, where 
$$v_r = v \cos \alpha.$$
The trick is to find $\alpha$ using the triangle formed by: us, the Galactic centre and the cluster. If the distance to the galactic centre is $d_{gc}$ and the distance to the cluster is $D$ and the galactic longitude of the cluster (angle measured from the direction of the Galactic centre) is $l$, then I think that the distance of the cluster from the Galactic centre $d_{Cl}$ is given by the cosine rule as
$$ d_{Cl} = \left( d_{gc}^{2} + D^2 - 2d_{gc}D \cos l \right)^{1/2}$$</p>

<p>Then the angle $\beta$ at the apex defined by the cluster id given by
$$\cos \beta = \frac{d_{Cl}^2 + D^2 - d_{gc}^2}{2d_{Cl}D}$$<br>
which leads to
$$ \cos \beta = \frac{D - d_{gc} \cos l}{d_{Cl}}$$</p>

<p>You then look at the diagram and note that $\beta = 90 - \alpha$, so that
$\cos \beta = \sin \alpha$. Hence you can calculate $\alpha$ and hence calculate $v$.</p>

<p><a href="https://i.sstatic.net/MO5dL.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/MO5dL.png" alt="Orbital geometry"></a></p>

CC BY-SA 4.0
2018-05-08T06:37:24.880Z
2018-06-30T20:19:27.313Z
2018-06-30T20:19:27.313Z
ProfRob
<p>I'm not really a physicist so I don't know if it's possible to get the real orbital velocity of a star from its radial velocity. I'm working on a paper for school and I want to get the mean velocity of selected open clusters to graph the galaxy's rotational curve, but in the databases I only encounter radial velocities. How do I get the real orbital velocity from those parameters? Thanks!</p>

CC BY-SA 4.0
2018-05-08T15:24:37.943Z
an object in a perfectly circular orbit has zero radial velocity. Then what? :-(
Carl Witthoft
CC BY-SA 4.0
2018-05-08T16:05:07.780Z
@CarlWithtoft  Radial velocity means line of sight velocity as seen from here.
ProfRob
CC BY-SA 4.0
2018-05-08T18:52:07.467Z
If the database has both radial velocities and [proper motions](https://en.wikipedia.org/wiki/Proper_motion), then you can compute 3D velocities.
Mike G
<astrophysics><orbital-mechanics><orbital-elements><velocity><star-cluster>
Calculating obital velocity from radial velocity
Camila Pazmino
# Calculating obital velocity from radial velocity

(Asked by: username_0 on 2018-05-08)

I'm not really a physicist so I don't know if it's possible to get the real orbital velocity of a star from its radial velocity. I'm working on a paper for school and I want to get the mean velocity of selected open clusters to graph the galaxy's rotational curve, but in the databases I only encounter radial velocities. How do I get the real orbital velocity from those parameters? Thanks!

**Tags:** astrophysics, orbital-mechanics, orbital-elements, velocity, star-cluster

> **Comments:**
> **username_1**: an object in a perfectly circular orbit has zero radial velocity. Then what? :-(  
> **username_2**: @CarlWithtoft  Radial velocity means line of sight velocity as seen from here.  
> **username_3**: If the database has both radial velocities and [proper motions](https://en.wikipedia.org/wiki/Proper_motion), then you can compute 3D velocities.  

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## 1 answers

### Answer by username_2 on 2018-05-08. Score: 5 | Accepted answer

You can't without assuming something about the overall velocity.

The radial velocity is one component of a velocity vector; you are missing the other two components, which could in principle be anything. However, you could *assume* that as **open** clusters are mostly quite young and members of the Galactic disc population, that they are moving in the disc on roughly circular orbits. NB. This cannot possibly work for **globular** clusters, which are an entirely different population, many with non-circular orbits and orbits out of the galactic plane.

Having made that assumption, your radial velocity is the orbital velocity resolved through an angle that is formed between your line of sight to the cluster and a tangent to the Galactic centre at the cluster's Galactic position.

This method will not work well for clusters with a Galactic latitude much different to zero, or clusters with small Galactic longitude (which should have zero radial velocity in this model).

For all this to work you need the radial velocity, the Galactic latitude and longitude of the cluster, the distance to the cluster and an assumed distance to the Galactic centre.

I've had a go at sketching the situation. The cluster is on a circular orbit around the Galactic centre with a true orbital velocity $v$ (shown as a vector). We measure the component of this in our line of sight as the radial velocity, where
$$v\_r = v \cos \alpha.$$
The trick is to find $\alpha$ using the triangle formed by: us, the Galactic centre and the cluster. If the distance to the galactic centre is $d\_{gc}$ and the distance to the cluster is $D$ and the galactic longitude of the cluster (angle measured from the direction of the Galactic centre) is $l$, then I think that the distance of the cluster from the Galactic centre $d\_{Cl}$ is given by the cosine rule as
$$ d\_{Cl} = \left( d\_{gc}^{2} + D^2 - 2d\_{gc}D \cos l \right)^{1/2}$$

Then the angle $\beta$ at the apex defined by the cluster id given by
$$\cos \beta = \frac{d\_{Cl}^2 + D^2 - d\_{gc}^2}{2d\_{Cl}D}$$  
which leads to
$$ \cos \beta = \frac{D - d\_{gc} \cos l}{d\_{Cl}}$$

You then look at the diagram and note that $\beta = 90 - \alpha$, so that
$\cos \beta = \sin \alpha$. Hence you can calculate $\alpha$ and hence calculate $v$.

[![Orbital geometry](https://i.sstatic.net/MO5dL.png)](https://i.sstatic.net/MO5dL.png)


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