A pair of stars rotates about a common centre of mass. One of the stars has a mass M and the other has mass m such that =2m. The distance between the centres of the stars is d ( d being large compare to the size of eithe star).
The period of rotation of the stars about their common centre of mass ( in terms of d,m,G) is

Answer

`m/M`

Explanation

The centre of mass of the system divides the distance between the starts in the inverse ratio of their masses. If `d_(1)` and `d_(2)` are the distance force for their circular motion.
`d_(1)=d/(M+m)xxm`
`d_(2)=d/(M+m)xxMimplies(d_(1))/(d_(2))=m/M`
the stars will rotate in circles of radii `d_(1)` and `d_(2)` about their centre of mass. The same force of attraction provies the necessary centripetal force for their circular motion.
`:. (GMm)/d^2=Momega_(1)^(2)d_(1)=momega_(2)^(2)d_(2)`
`=(G(M+m))/(d^(3))`
and `omega_(2)^(2)=(GM)/(d^(2)d_(2))=(GM)/(d^(2))xx(M+m)/(dxxM)=(G(M+m))/(d^(3))`
`omega_(1)=omega_(2)=sqrt((G(M+m))/(d^(3)))`
From the fact that the moment of momentum is also the angular momentum,
`(L_(M))/(L_(m))=((Mv_(1))d_(1))/((mv_(2))d_(2))=M/mxx(d_(1)^(2))/(d_(2)^(2))=(L_(M))/(L_(m)) =m/M`
`implies (K_(M))/(K_(m))=(1/2mv_(1)^(2))/(1/2mv_(1)^(2))=(M/m) (omega_(1)^(2)d_(1)^(2))/(omega_(2)^(2)d_(2)^(2))=M/m(d_(1)^(2))/(d_(2)^(2))` (`:'omega_(1)=omega_(2))`
Therfore `(K_(m))/(K_(m))=M/m(m/M)^(2)=m/M`