Three identical bodies of mass M are located at the vertices of an equilateral triangle with side L. At what speed must they move if they all revolve under the influence of one another's gravity in a circular orbit circumscribing the triangle while still preserving the equilateral triangle
Answers
The distance between a vertex and the center of an equilateral triange is 2/3 of the altitude, and the altitude is √3/2 times the side. Each star will be moving in a radius of L√3/3 around the center.
Two stars exerts a gravitational force of Fg = GM^2/L^2 on the third.
The net force is toward the middle of the circle. Part of the force, Fg cos 60, is in opposing directions and cancels out, but the other component, Fg sin 60, is added togther √3/2 Fg + √3/2 Fg = √3 Fg is the total force on each star. This means that each star experiences an acceleration of
√3Fg/M = (√3GM^2)/(ML^2) = √3GM/L^2
We have an acceleration and a radius, we can use the formula for centripetal acceleration a = v^2/r to find the velocity
√3GM/L^2 = v^2/(L√3/3)
GM/L = v^2
√(GM/L) = v
There's definitely at least a 1/1000 chance I did this all correctly. Hope it's clear enough for you to follow.