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

`v=\sqrt{(2GM)/(L)}`

Step-by-step explanation:

M = Mass of planets

R = Radius of circle

v = Velocity

`\theta` = Angle

The circle is inside the triangle

`cos\theta=((L)/(2))/(R)\\\Rightarrow R=(L)/(2cos\theta)`

The centripetal acceleration

`(Mv^2)/(R)=2(GM^2)/(L^2)cos\theta\\\Rightarrow (Mv^2)/((L)/(2cos\theta))=2(GM^2)/(L^2)cos\theta\\\Rightarrow (Mv^22cos\theta)/(L)=2(GM^2)/(L^2)cos\theta\\\Rightarrow v^2=(2GM)/(L)\\\Rightarrow v=\sqrt{(2GM)/(L)}`

The speed of the stars is
`v=\sqrt{(2GM)/(L)}`