Three identical bodies of mass M are located at the vertices of an equilateral triangle of side L. They revolve under the effect mutual gravitational force in a circular orbit , circumscribing the triangle while preserving the equilateral triangle . Their orbital velocity is

Options

A. `sqrt((GM)/(L))`
B. `sqrt((3GM)/(2L))`
C. `sqrt((3GM)/(L))`
D. `sqrt((2GM)/(3L))`

Answer

A

Explanation

Given `F_(1)=F_(2)=Fandtheta=60^(@)`
Resultant force `=sqrt(3)F`

Therefore Force om mass at A due to mass at B and C
`=sqrt(3)((GM^(2))/(L^(2)))`
Centripetal force for circumscribing the triangle in a circular orbit is provided by mutual gravitational interaction .
i.e ,`(Mv^(2))/((L//sqrt(3)))=sqrt(3)((GM^(2))/(L^(2)))or v=sqrt((GM)/(L))`