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
Sarial
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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~
 

FAQ: Three body, equal mass star system

What is a three body, equal mass star system?

A three body, equal mass star system is a system of three stars that have an equal mass. These stars are gravitationally bound to each other and orbit around their common center of mass.

How common are three body, equal mass star systems?

Three body, equal mass star systems are relatively rare, as most star systems consist of two stars. However, they have been observed in some globular clusters and are thought to be more common in the early universe.

How does a three body, equal mass star system form?

Three body, equal mass star systems are thought to form from the collapse of a dense cloud of gas and dust. As the cloud collapses, it can fragment into multiple equal mass clumps, which then go on to become the stars in the system.

What are the consequences of having three equal mass stars in a system?

Having three equal mass stars in a system can lead to complex gravitational interactions, making the system unstable. This can result in the ejection of one or more stars from the system, or even a collision between two of the stars.

Can life exist in a three body, equal mass star system?

It is unlikely that life could exist on planets in a three body, equal mass star system. The gravitational forces and radiation from three stars would create a highly unstable and chaotic environment, making it difficult for planets to maintain a stable orbit and for life to evolve.

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