 # 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 ?

11 years ago

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

11 years ago

see the eq 5 ,  the ratio instead of 2(R2/R1) ,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 ....