Re: Finding a formula for the period of orbital motion in terms of the orbital

The question: "Three identical stars, at the vertices of an equilateral triangle, orbit about their common center of mass. Find the period of this orbital motion in terms of the orbital radius, R, and the mass of each star, M.

         vector v  
         <------- M
               *  @  *
          *      / \      *
                /   \     
               /     \
       *      /       \       *
             /         \
            /     @.    \
      *    /     CM  .R  \     *   /
          /            .  \       /vector v
     M @   -----------------     /
      \*                     *@M
       \
vector v\ *               *
              *      *

(v denotes velocity, CM denotes center of mass, M = mass of each star, R denotes orbital radius) [Difficulty] The answer given by the book is 2pi*sqrt[(sqrt(3)*R^3)/(GM)]. [Thoughts] The centripetal acceleration is v^2/R, and v = 2pi/T, where T is the period or time for one revolution, 2pi. F = ma = M*(v^2/R) = M* [(2pi/T)^2]/R. Do I have to use Newton's Law of Universal Gravitation F = G * m_1 * m_2/r^2, and set it equal to my above equation (I don't think so, since CM, or center of mass, is not a point mass and thus there is no attraction between each star and the CM)? If setting it equal to my above equation is the correct way, then m_1 = M and M_2 = CM, and the term M is eliminated, which I don't think is the correct way of solving this problem. Please help me, please! Any hint will be appreciated. Thank you in advance! (I think the sqrt(3)in the book's answer has to do with the 3 stars, since there are 3 of them.) Thank you so much. By the way, I am only 14 and am studying physics and calculus on my own. "

You are on the right track, but you made a mistake early. Your second equation, "v=2pi/T", should be "v=2piR/T".

From there, you can work out that
T = 2piR/sqrt(a_cR)
where a_c is the centripetal acceleration.

You had the right idea when you were talking about Newton's Law of Universal Gravitation, but the details are important! With a three-body problem you have to work out the gravitational force between each pair of bodies unless you know something about the symmetry which can simplify the task. In this case there is some symmetry: If you consider one star at a time, there are two forces acting on it, one from each of the other two stars. But neither of those forces points toward the center of the orbit, so we must work out the vector components of each force. Are you familiar with vector components? If not, look at this page at Wikipedia, for example. If you decompose the force (due to one of the other stars) into two components, one directed toward the center of the system and the other at 90 degrees from that, the 2nd component will be exactly in the opposite direction as the 2nd component due to the other star (the third star of the system), so the sum of those two components will be zero. Therefore the symmetry allows you to consider only the components directed toward the center of the system.

It turns out that the sqrt(3) factor is due to the 30-degree angle between each of the two forces and the direction of the center of the system, because the cosine of a 30-degree angle is sqrt(3)/2. Hopefully you know enough trigonometry to be familiar with sines and cosines.

The answer provided for you by your book is correct, so I am not going to work out all the details for you. But I will provide some hints:
1. What is the distance between each pair of stars?
2. What is the angle between each side of the triangle and the center of the system?
3. Use Newton's Law of Universal Gravitation to work out the force between each pair of stars.
4. How many star pairs do you need to consider when working out the total force on one of the stars?
5. Once you know the force, how do you figure the acceleration on each star? What is this acceleration equivalent to?

John Link, MadSci Physicist