Period of rotation for three stars in equilateral triangle orbit

[Vitani11]

Homework Statement

Three identical stars, each with mass m, form the verticies of an equilateral triangle with side length d and rotate in a circular orbit due to their mutual gravitation. What is the period τ of their rotation?

I set up the FBD for each star and am now trying to figure out what to do from there. I know how to get period into the equation from circular motion, but is that all there is to it? The FBD gives different equations for each mass so when I solve I get different periods for each star. What should I be doing here?

Force equations

Star at bottom left point:

x: Fcosθ+F = ma_x
y: Fsinθ = ma_y

Star at bottom right:

x: -(Fcosθ+F) = ma_x
y: Fsinθ = ma_y

Star at top:

y: -2F = ma_y (or should this be -2Fcos(θ/2)?)
x: forces cancel to 0

If you draw an equilateral triangle you will see what I'm talking about and then how I oriented my axes around the stars.

 

[Vitani11]

When I solve for period my units are seconds - so I know my method is correct. Anyway I would solve each of the force equations above for period and compare because they should be the same (or at least what I have been thinking) and they are off by a factor due to the cosθ or sinθ

 

[TSny]

Instead of working with x and y components of force, think about centripetal components of forces. Use what you know about circular motion.

 

[Vitani11]

I should neglect the force that each is exerting on each other in x or y and only focus on the forces going towards the center of the system (triangle)? I thought about that but I was afraid to try it.

 

[Vitani11]

Okay I think this might be right. Using the centripetal components I was able to solve for period using the star at the top which my answer was T = (2πd^3/2)/(√(Gm))(3^1/4). Units check out.

(2Gm²cos(θ/2))/d² = mv²/d this was my equation from F = ma.

 

[TSny]

Is the orbital radius equal to d?

 

[haruspex]

Okay I think this might be right. Using the centripetal components I was able to solve for period using the star at the top which my answer was T = (2πd^3/2)/(√(Gm))(3^1/4). Units check out.

(2Gm²cos(θ/2))/d² = mv²/d this was my equation from F = ma.

I get a different constant factor. Please post all your working.

 

[Vitani11]

dsinθ-dsinθ/2 is the distance? Here is the work. Sorry for the blur.

 

[Vitani11]

OH MY GOD I DID IT THANK YOU haruspex as usual and Tsny . No dsinθ -dsinθ/2 was definitely not it