An isolated triple star systerm consists of two identical stars, each of mass `m` and f fixed star of mass `M`. They revolve around the central star in the same circular orbit to radius `r`. The orbiting stars are always at opposite ends of a diameter of the orbit. The time period of revolution of each star around the fixed is equal to:

Options

A. `(4pir^(3//2))/(sqrt(G(4M+m)))`
B. `(2pir^(3//2))/(sqrt(GM))`
C. `(2pir^(3//2))/(sqrt(G(M+m)))`
D. `(4pir^(3//2))/(sqrt(G(M+m)))`

Answer

A

Explanation

Net gravitational force on any orbitin star provides necessary centripetal force.
So, `(GMm)/(r^(2))+(Gm^(2))/(4r^(2))=(mV^(2))/r`
`impliesV=sqrt((G(4M+m))/(4r))`
`T=(2pir)/V=(4pir^(3//2))/(sqrt(G(4M+m)))`