A small planet is revolving around a very massive star in a circular orbit of radius ' R ' with a period of revolution ' T '. If the gravitational force between the planet and the

Question

A small planet is revolving around a very massive star in a circular orbit of radius ' R ' with a period of revolution ' T '. If the gravitational force between the planet and the star were proportional to ' R − 5/2 , then ' T ', would be proportional to

TestHub · Accepted Answer

The gravitational force between the planet and the star provides the necessary centripetal force for circular motion.

$F = m ω^{2} R = m (\frac{2 π}{T})^{2} R$

Given that the gravitational force is proportional to $R^{- 5/2}$ :

$F \propto R^{- 5/2}$

Equating the two expressions for force:

$m (\frac{2 π}{T})^{2} R \propto R^{- 5/2}$

$\frac{1}{T ^{2}} \propto R^{- 5/2 - 1}$

$\frac{1}{T ^{2}} \propto R^{- 7/2}$

$T^{2} \propto R^{7/2}$

$T \propto R^{7/4}$

Answer: $R^{7/4}$

Physics - Gravitation Question with Solution | TestHub

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