# Binary stars period help

Hello.
Two binary stars (m1 and m2) orbit each other around their center of mass. They orbit in circles of radii r1 and r2. I am to show that the period is given by T2=4pi2(r1+r2)3/(G(m1+m2))

The mutual force between m1 and m2 is F=Gm1m2/(r1+r2)2
Considering m1, the acceleration of m1 is given by a1=Gm2/(r1+r2)2=v12/r1
Solving for v and using T1=2(pi)r1/v1=T2, I get T2=4pi2r1(r1+r2)2/(Gm2)

What did I do wrong?

I would reconsider the definition of "center of mass."

What precisely do you mean by "reconsider?" If I take the CM to be my reference frame (which I do), then m1r1=m2r2. I don't know how this helps. However, I noticed that if I take m2 to be my reference frame, then I get T2=4pi2(r1+r2)3/(Gm2). This is unexpected to me. I thought that omega was independent of reference frame so long as these reference frames have only translational motion wrt each other. Since T is indirectly proportional to omega, T should also be independent of reference frame. I'm clearly making some mistake here, but I can't figure out what it is.

Stephen, your original approach was exactly on target. You simply didn't finish it.

The mutual force between m1 and m2 is F=Gm1m2/(r1+r2)2 Correct.
Considering m1, the acceleration of m1 is given by a1=Gm2/(r1+r2)2=v12/r1 Correct.
Solving for v and using T1=2(pi)r1/v1=T2, I get T2=4(pi)2r1(r1+r2)2/(Gm2) Correct.

Now, remembering that m1r1 = m2r2
divide numerator by r1 and denominator by m2r2 /m1 giving you
T2=4(pi)2(r1+r2)2/(Gm1/r2)
Now multiply numerator and denominator by (r1+r2) giving
T2=4(pi)2(r1+r2)3/(G(m1r1 /r2+m1)

But m1r1 /r2 = m2 which is exactly what you wanted.

But I'm puzzled by your later statement which seemed to suggest that it would be OK to use m2 as a reference frame on the assumption that these frames have only translational motion wrt each other. How can you make this assertion when these frames are rotating about each other?

The frames would not rotate with respect to each other. If their axes were originally parallel, they would always be parallel. In that way, omega would be the same in either frame.

Edit:
I learned it yesterday actually in my physics book. Perhaps my inexperience with the new concept has caused me to use it incorrectly. In the book, it says that if an object rotates in a plane, then it rotates about any axis perpendicular to that plane. And no matter which axis is chosen, the angular velocity omega is the same (however, as they are noninertial frames, linear velocity is affected).

Edit: