1. Nov 24, 2009

### theneedtoknow

I just have a quick question about a binary star system consisting of 2 starts of unequal masses. Is the period of rotation of the 2 masses about their centre of mass equal? Do they both take the same amount of time to complete 1 revolution about the centre of mass?

2. Nov 24, 2009

### ideasrule

Yup. The two-body system's center of mass cannot move because there are no external forces, so the line between the two bodies must always pass through the center of mass. Imagine a bar connecting the two bodies; the bar's rotating, expanding, and contracting, but not bending or moving. You can imagine that the two bodies will have the same orbital period.

3. Nov 24, 2009

### theneedtoknow

Thanks! :) That makes sense.
So how do I determine this period of orbit of the 2 bodies around the centre of mass? What if, for example, I was considering a planet orbiting around a star. Would I use the general version of Kepler's 3rd law? [ P^2 = (4pi^2*a^3)/ (G(m1 + m2)) ] with a being the average distance between the 2 bodies (since, as you said, this distance will contract and expand as they orbit) (I'll assume circular orbits for simplicity)? Or would I calculate the period in some other way?

4. Nov 25, 2009

### Ich

Have a look at http://en.wikipedia.org/wiki/Reduced_mass" [Broken]

Last edited by a moderator: May 4, 2017
5. Nov 25, 2009

### theneedtoknow

So , the radial acceleration the 2 bodies would experience towards each other (and therefore toward the centre of mass) is F/mreduced = G*m1*m1/(distance between objects)^2 /mreduced ?
and the radial acceleration (assuming circular orbit) is equal to v^2 / (distance of body to centre of mass), so I can calculate the orbital velocities of the 2 bodies as:

velocity = square root [ (G*m1*m1/(distance between objects)^2 ) * (distance of body to centre of mass)/(mreduced) ]

is this correct?