A double star is a system of two stars of masses `m` and `2m`, rotating about their centre of mass only under their mutual gravitational attraction. If `r` is the separation between these two stars then their time period of rotation about their centre of mass will be proportional to

Options

A. Heavier star revolves in orbit of radius `2r//3`.
B. Both the stars revolve with the same speed, period of which is equal to `(2pi//r^(3))(2GM^(2)//3)`
C. Kinetic energy of the heavier star is twice that of the other star.
D. None of the above

Answer

A, C

Explanation

The centre of mass of the double star system remains stationary and both the stars revolve around in circular orbits which are concentric with the centre of mass.
The distance of centre of mass from the heavier star is equal to `((Mr+2M.0)/(M+2M))=r/3`
Hence the heavier star revolves in a circle of radius `r//3` while the lighter star in a circle of radius `2r//3`. Hence, option a is incorrect
To calculate period of revolution of a double star system concept of reduced mass can be used.
The reduced mass of the system is equal to
`((M)(2M))/((M+2M))=(2M)/3`
Hence, the period of revolution will be equal to `(2pi)/sqrt(2(GM)/3)r^(3/2)`
where r is distance between two stars. Hence, option b is correct.
Kinetic energy of a star will be equal to `1/2mv^(2)` where `v` is speed of the star which is equal to (radius of its circular orbit)
`xxomega`
Hence `KE` of heavier stars is `E_(1)=1/2(2M)(r/3omega)^(2)` and that of lighter star `E_(2)=1/2M((2r)/3omega)^(2)`.
It means `KE` of the lighter star is twice that of the heavier star.
Hence, option c is incorrect.