In a Double star system one of mass m1 qnd another of mass m2 with a separation d rotate about their common centre of mass. Then rate of sweeps of area of star of mass m1 to star of mass m2 about their common centre of mass is

(A) m1/m2

(B)m2/m1

(C) m1^2/ m2^2

(D) m2^2/ m1^2

here in the question ,we have two masses which are moving under the influence of their mutual gravitational force...
both of them are moving about their center of mass in a circular orbits of different radii..
if R1 & R2 are radius of lighter and heavior particle respectively then
 R1 +R2 = R ................1
Fc =mW12 R1       ..................2              (gravitional force will provide necassery centripital force)
Fc=2mW22R2     .......................3
dividing 1 & 2
 (W1/W2)2 =2(R2/R1) .............4
ratio of kinetic energy is (KE)2/(KE)1=2(W2/W1)2 (R2/R1)2
from eq 4 substituting (W2/W1)
    ratio =2(R2/R1) ...................5
now we have R1=2mR/3m=2R/3   &  R2=mR/3m=R/3
after substituting these we get
 ratio = 1:1

 

Regards

Arun