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(A) For two stars of equal mass $M$ orbiting each other in a circular path of radius $R$,the distance between them is $d = 2R$.The gravitational force

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$R^{\frac{3}{2}}$

Vedclass · Answer

(A) For two stars of equal mass $M$ orbiting each other in a circular path of radius $R$,the distance between them is $d = 2R$.The gravitational force between them provides the necessary centripetal force for circular motion.The gravitational force is $F_g = \frac{G M^2}{(2R)^2} = \frac{G M^2}{4R^2}$.The centripetal force required for a star of mass $M$ to move in a circle of radius $R$ with angular velocity $\omega$ is $F_c = M \omega^2 R$.Equating the forces: $M \omega^2 R = \frac{G M^2}{4R^2}$.Simplifying,$\omega^2 = \frac{G M}{4R^3}$.Since the time period $T = \frac{2\pi}{\omega}$,we have $T^2 = \frac{4\pi^2}{\omega^2} = \frac{4\pi^2 (4R^3)}{GM} = \frac{16\pi^2 R^3}{GM}$.Thus,$T^2 \propto R^3$,which implies $T \propto R^{\frac{3}{2}}$.

Two stars of equal masses $M$ are orbiting in a circle of radius $R$. Their orbital time period is proportional to

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