In a double star, two stars (one of mass `m` and the other of mass `2m`) distance `d` apart rotate about their common centre of mass., Deduce an expressioin for the period of revolution. Show that te ratio of their angular momenta about the centre of mass is the same as the ratio of their kinetic energies.

Explanation

The centre of mass `C` will be at distances `d//3` and `2d//3` from the masses `2m` and m respectively Both the stars rotate round `C` in their respective orbits with the same angular velocity `omega` the gravitational force acting on each star due to the other supplies the necessary centripetal force The gravitational force on either star is `(G(2m)m)/(d^(2))` If we consider the rotation of the smaller star, the centripetal force `(m r omega^(2))is [m((2d)/(3))omega^(2)]` and for bigger star `[(2mdomega^(2))/(3)]` i.e same
`:. (G (2m)m)/(d^(2)) = m ((2d)/(3))omega^(2)` or `omega = sqrt((3gm)/(d^(3))`
Therefore, the period of revolution is given by `T = (2pi)/(omega) = 2pi sqrt(d^(3)/(3Gm))`
The ratio of the angular momentum is `(Iomega)_(big)/((iomega)_(small)) = (I_(big))/(I_(small)) = ((2m)((d)/(3))^(2))/(m((2d)/(3))^(2))=(1)/(2)`
since `omega` is same for both The ratio of their kinetic energygies is `((1)/(2)Iomega^(2))_(big)/(((1)/(2)Iomega^(2))_(small))=I_(big)/(I_(small))=(1)/(2)`
which is the same as the ratio of their angular momentum .
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