A small planet is revolving around a very massive star in a circular orbit of radius ' $R$ ' with a period…

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 ' $\mathrm{R}{ }^{-5 / 2}$, then ' T ', would be proportional to
  1. $\mathrm{R}^{3 / 2}$
  2. $R^{3 / 5}$
  3. $\mathrm{R}^{7 / 2}$
  4. $\mathrm{R}^{7 / 4}$

Solution

For the planet to orbit around the star, the centripetal force must be provided by gravitational force. Hence, $\mathrm{F}_{\mathrm{G}}=\mathrm{F}_{\mathrm{a}}$ $\mathrm{F}_{\mathrm{a}} \propto-\mathrm{R}^{-5 / 2}$ ....(Given) Here, -ve sign indicates force is towards the centre of orbit. $\begin{array}{ll} & \Rightarrow \mathrm{a} \propto-\mathrm{R}^{-5 / 2} \\ \therefore \quad & -\omega^2 \mathrm{R} \propto-\mathrm{R}^{-5 / 2} \\ \therefore \quad & \omega^2 \propto \mathrm{R}^{-(5+2) / 2} \\ \therefore & \frac{4 \pi^2}{\mathrm{~T}^2} \propto \mathrm{R}^{-7 / 2} \\ \therefore \quad & \mathrm{~T}^2 \propto \mathrm{R}^{7 / 2} \\ & \Rightarrow \mathrm{~T} \propto \mathrm{R}^{7 / 4} \end{array}$

Asked in: MHT CET 2024 (10 May Shift 1)

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