# Weird gravitation problem

1. Nov 20, 2006

### chaose

A triple star system consists of two stars each of mass m revoling in the same circular orbit of radius r around a central star of mass M. The two orbiting stars are always at opposite ends of a diameter of the orbit. Derive an expression for the period of the revolution of the stars.

I don't understand this problem that well. Any hints?

I thought that if the two stars are at opposite ends of the orbit, won't the gravitational forces from one cancel out the forces from the other star?
Then won't the period just be Kepler's Third law?

T^2 = (r^3)*(4pi^2)/(GM), G = universal gravitational constant, M = mass of central star, T = period of revolution, r = radius of the orbit, pi = 3.14159...

2. Nov 20, 2006

### OlderDan

The force on each orbiting star is the sum of the forces from the other two stars. The net force on the central star will be zero, so it is stationary at the center of the motion of the other two stars. You can do this as a basic centripetal force problem if you get the forces right.

3. Nov 21, 2006

### chaose

thanks, I believe I got it now.

My end result was
T^2 = (16pi^2 * r^3)/(G*(4M+m))