Three body, equal mass star system

In summary, there is no general analytical solution for the motion of a 3-body gravitational system, but there are analytical solutions for special initial conditions. The total gravitational force exerted on one of the stars due to the other two can be calculated using the equation F= G(m1)(m2)/R^2. The period, T, can be calculated using trigonometry. The attempt at a solution involved assuming the bodies are at 120 degrees on the corners of an equilateral triangle and solving for the distance between the bodies in terms of r, then using trigonometry to solve for the y component of the forces and adding them. However, this solution does not seem to be correct.
  • #1
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Homework Statement


There is no general analytical solution for the motion of a 3-body gravitational system. However, there do exist analytical solutions for very special initial conditions. The diagram below shows three stars, each of mass m, which move in the plane of the page along a circle of radius r.
http://spock.physast.uga.edu/res/uga/PhysicsLib/Matter_and_Interactions/Ch04/figs/3body_grav.png

Calculate the magnitude of the total gravitational force exerted on one of the stars due to the other two.
F total = ?

Period, T = ?


Homework Equations


F= Gmm/R^2
Trig


The Attempt at a Solution


My only attempt at a solution involved the assumption that the bodies are at 120 degrees, on the corners of an equilateral triangle, and trying to solve for the distance between the bodies in terms of r. Then using trig to solve for the y component of both forces and adding them.

Thanks in advance, guys.
 
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  • #2
My most recent attempt, while I wait:

Equilateral triangle. 120 degree angles between each planet.

Law of sines:

r is distance between center and planet
R is distance between planets

r/sin(30) = R/sin(120)

R = r*sin(120)/sin(30)

G(m^2)/R^2

The force in the y direction on the top planet from the bottom left one should be

F*sin(60) = Fy

Both of the bottom planets forces should add, and the x forces cancel.

2*sin(60)*(((6.67x10^-11)(m^2))/(((r*sin(120))/sin(30))^2))

This, however, doesn't seem to be right. Any ideas?
 
  • #3
Bump? 20 mins left to answer, but I have to turn the work in on paper too. I really feel like my work I did in the reply should work, but it doesn't seem to.

Thanks~
 

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