# Orbital speeds of Components of Binary System

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?

• You can use the vis-viva equation. We have several questions about vis-viva here. Commented May 23, 2022 at 14:39
• Actually, the equation given there isn't quite correct. You need the combined mass, as mentioned here: en.wikipedia.org/wiki/Specific_orbital_energy I have a little more info here: physics.stackexchange.com/a/675868/123208 Commented May 23, 2022 at 14:45
• You mean how to find their speeds at any given time/phase in the orbit? How to measure the speeds? Commented May 23, 2022 at 16:45

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 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 $$G \approx 887.125 \frac{\text{AU}}{M_\odot}(\text{km}/\text{s})^2$$

• FWIW, Ryan S. Park et al 2021 AJ 161 105 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). Commented May 23, 2022 at 18:42
• @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. Commented May 23, 2022 at 19:25
• :) 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. Commented May 23, 2022 at 19:33
• 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. Commented May 25, 2022 at 7:48
• @MikeStone Good question. I added a note explaining. Commented May 25, 2022 at 15:09