Question

Three identical stars of mass M form an equilateral triangle that rotates around the triangle’s center as the stars move in a common circle about that center. The triangle has an edge length of L. What is the speed of the stars?

Answer 1

Answer:

$v=\sqrt{\dfrac{2GM}{L}}$

Explanation:

M = Mass of planets

R = Radius of circle

v = Velocity

$\theta$ = Angle

The circle is inside the triangle

$cos\theta=\dfrac{\dfrac{L}{2}}{R}\\\Rightarrow R=\dfrac{L}{2cos\theta}$

The centripetal acceleration

$\dfrac{Mv^2}{R}=2\dfrac{GM^2}{L^2}cos\theta\\\Rightarrow \dfrac{Mv^2}{\dfrac{L}{2cos\theta}}=2\dfrac{GM^2}{L^2}cos\theta\\\Rightarrow \dfrac{Mv^22cos\theta}{L}=2\dfrac{GM^2}{L^2}cos\theta\\\Rightarrow v^2=\dfrac{2GM}{L}\\\Rightarrow v=\sqrt{\dfrac{2GM}{L}}$

The speed of the stars is $v=\sqrt{\dfrac{2GM}{L}}$