<p>Use the vis-viva equation for 2 bodies:</p>
<p><span class="math-container">$$v^2=G(M_1+M_2)\left(\frac{2}{r}-\frac{1}{a}\right)$$</span></p>
<p>where <span class="math-container">$v$</span> is the relative velocity, <span class="math-container">$G \approx 887.125 \frac{\text{AU}}{M_\odot}(\text{km}/\text{s})^2$</span> is the gravitational constant, <span class="math-container">$M_1 \approx 1.1 M_\odot$</span> and <span class="math-container">$M_2 \approx 0.907 M_\odot$</span> are the masses of the stars, <span class="math-container">$r $</span> is the distance between them, and <span class="math-container">$a \approx 23 \text{AU}$</span> is the semi-major axis of the orbit.</p>
<p>Solving for <span class="math-container">$v$</span>, I get 15.5 km/s at closest approach and 4.49 km/s at furthest distance.  As a comparison, Earth's orbital speed is around 30 km/s and Saturn's is around 9.7 km/s</p>
<p><strong>Note:</strong> The Gravitational constant is given <a href="https://en.wikipedia.org/wiki/Gravitational_constant" rel="nofollow noreferrer">here</a> as <span class="math-container">$G \approx 4.3009 \times 10^{-3} \frac{\text{pc}}{M_\odot}(\text{km}/\text{s})^2$</span>, but we want it in AUs instead of parsecs for our units to work out, so we can change units by multiplying by <span class="math-container">$\frac{206265 \text{AU}}{1 \text{pc}}$</span> to get <span class="math-container">$G \approx 887.125 \frac{\text{AU}}{M_\odot}(\text{km}/\text{s})^2$</span></p>

CC BY-SA 4.0
2022-05-23T18:42:14.850Z
FWIW, [Ryan S. Park et al 2021 AJ 161 105](https://doi.org/10.3847/1538-3881/abd414) lists $GM_\odot=132712440041.279419\,\rm{km^3/s^2}$, estimated from DE440. In your units, that comes to ~887.1278676647596. (Of course, all those digits are irrelevant when our other data only has 3 significant figures).
PM 2Ring
CC BY-SA 4.0
2022-05-23T19:25:01.397Z
@PM2Ring I am perhaps not as careful as I should be with my values in some of these answers.  I converted $G$ from a wikipedia value with the same number of sig. figs, and used google's parsec to AU conversion value.  I also used the first google reference I could find for the stellar masses and periapsis/apoapsis.  Hopefully I can be forgiven for my numerical sloppiness in this case since the question asks specifically about *how* to calculate the speeds.
Connor Garcia
CC BY-SA 4.0
2022-05-23T19:33:04.420Z
:) Your answer's fine. I just thought you might like a better value for that constant, for future reference. I used 149597870700 m for the AU; it was defined to be an exact number in 2012. I often use Google's built-in calculator, since it knows lots of units, but it's not perfect. Eg, it uses the tropical year in its definition of the light-year, but the IAU definition uses the Julian year.
PM 2Ring
CC BY-SA 4.0
2022-05-25T07:48:20.893Z
Thanks for the formula but, just to show my ignorance yet again, could you clarify how the 887,125 is arrived at? I've googled it in conjunction with the word "gravitational" but so far have had no joy. Cheers.
Mike Stone
CC BY-SA 4.0
2022-05-25T15:09:09.467Z
@MikeStone Good question.  I added a note explaining.
Connor Garcia
CC BY-SA 4.0
2022-05-23T17:03:45.493Z
2023-03-18T23:54:55.480Z
2023-03-18T23:54:55.480Z
Connor Garcia
<p>How does one calculate the orbital speed s of the members of a binary system?</p>
<p>Iirc, the average separation of Alpha Centauri A and B is 23 AU, but the orbit is highly elliptical, so that the separation varies from 11.2 AU to 35.6 AU. Iirc their combined mass is 1.93 times the Sun.</p>
<p>From other things I've read, this means that at maximum separation the components will be moving at a bit less than one-third the speed at closest separation, but despite browsing I'm far from clear how to find what the actual speeds <em>are</em>. I'm sure it will be something obvious <em>after</em> it has been explained but right now I'm stumped.</p>
<p>Can anyone help?</p>

CC BY-SA 4.0
2022-05-23T14:39:15.890Z
You can use the [vis-viva equation](https://en.wikipedia.org/wiki/Vis-viva_equation). We have several questions about vis-viva here.
PM 2Ring
CC BY-SA 4.0
2022-05-23T14:45:05.110Z
Actually, the equation given there isn't quite correct. You need the combined mass, as mentioned here: https://en.wikipedia.org/wiki/Specific_orbital_energy I have a little more info here: https://physics.stackexchange.com/a/675868/123208
PM 2Ring
CC BY-SA 4.0
2022-05-23T16:45:57.510Z
You mean how to find their speeds at any given time/phase in the orbit? How to measure the speeds?
ProfRob
<binary-star><orbital-elements><eclipsing-star-systems>
Orbital speeds of Components of Binary System
Mike Stone
# Orbital speeds of Components of Binary System

(Asked by: username_0 on 2022-05-23)

How does one calculate the orbital speed s of the members of a binary system?

Iirc, the average separation of Alpha Centauri A and B is 23 AU, but the orbit is highly elliptical, so that the separation varies from 11.2 AU to 35.6 AU. Iirc their combined mass is 1.93 times the Sun.

From other things I've read, this means that at maximum separation the components will be moving at a bit less than one-third the speed at closest separation, but despite browsing I'm far from clear how to find what the actual speeds *are*. I'm sure it will be something obvious *after* it has been explained but right now I'm stumped.

Can anyone help?

**Tags:** binary-star, orbital-elements, eclipsing-star-systems
> **Comments:**
> **username_1**: You can use the [vis-viva equation](https://en.wikipedia.org/wiki/Vis-viva_equation). We have several questions about vis-viva here.  
> **username_1**: Actually, the equation given there isn't quite correct. You need the combined mass, as mentioned here: https://en.wikipedia.org/wiki/Specific_orbital_energy I have a little more info here: https://physics.stackexchange.com/a/675868/123208  
> **username_2**: You mean how to find their speeds at any given time/phase in the orbit? How to measure the speeds?  

---

## 1 answers

### Answer by username_3 on 2022-05-23. Score: 5 | Accepted answer

Use the vis-viva equation for 2 bodies:

$$v^2=G(M\_1+M\_2)\left(\frac{2}{r}-\frac{1}{a}\right)$$

where $v$ is the relative velocity, $G \approx 887.125 \frac{\text{AU}}{M\_\odot}(\text{km}/\text{s})^2$ is the gravitational constant, $M\_1 \approx 1.1 M\_\odot$ and $M\_2 \approx 0.907 M\_\odot$ are the masses of the stars, $r $ is the distance between them, and $a \approx 23 \text{AU}$ is the semi-major axis of the orbit.

Solving for $v$, I get 15.5 km/s at closest approach and 4.49 km/s at furthest distance. As a comparison, Earth's orbital speed is around 30 km/s and Saturn's is around 9.7 km/s

**Note:** The Gravitational constant is given [here](https://en.wikipedia.org/wiki/Gravitational_constant) as $G \approx 4.3009 \times 10^{-3} \frac{\text{pc}}{M\_\odot}(\text{km}/\text{s})^2$, but we want it in AUs instead of parsecs for our units to work out, so we can change units by multiplying by $\frac{206265 \text{AU}}{1 \text{pc}}$ to get $G \approx 887.125 \frac{\text{AU}}{M\_\odot}(\text{km}/\text{s})^2$

> **Comments:**
> **username_1**: FWIW, [Ryan S. Park et al 2021 AJ 161 105](https://doi.org/10.3847/1538-3881/abd414) lists $GM_\odot=132712440041.279419\,\rm{km^3/s^2}$, estimated from DE440. In your units, that comes to ~887.1278676647596. (Of course, all those digits are irrelevant when our other data only has 3 significant figures).  
> **username_3**: @PM2Ring I am perhaps not as careful as I should be with my values in some of these answers.  I converted $G$ from a wikipedia value with the same number of sig. figs, and used google's parsec to AU conversion value.  I also used the first google reference I could find for the stellar masses and periapsis/apoapsis.  Hopefully I can be forgiven for my numerical sloppiness in this case since the question asks specifically about *how* to calculate the speeds.  
> **username_1**: :) Your answer's fine. I just thought you might like a better value for that constant, for future reference. I used 149597870700 m for the AU; it was defined to be an exact number in 2012. I often use Google's built-in calculator, since it knows lots of units, but it's not perfect. Eg, it uses the tropical year in its definition of the light-year, but the IAU definition uses the Julian year.  
> **username_0**: Thanks for the formula but, just to show my ignorance yet again, could you clarify how the 887,125 is arrived at? I've googled it in conjunction with the word "gravitational" but so far have had no joy. Cheers.  
> **username_3**: @MikeStone Good question.  I added a note explaining.  

---

