a double star consists of two stars having masses m and 2m respsctively separated by a distance R.find the ratio of the kinetic energy of the bigger star to the smaller star ?

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

F_c =mW_{1² R}1       _{..................2              (gravitional force will provide necassery centripital force)}

F_c=2mW₂²R₂     .......................3

dividing 1 & 2

 (W1/W2)² =2(R₂/R_{1) .............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

see the eq 5 ,  the ratio instead of 2(R₂/R_{1) ,it is}  (R₂/R₁),

now u will will get ratio as 1:2 not 1:1 ...

sometimes calculations mistake may be there in the sol  ,so see the concept and do it urself ....