Time Period in Non-Standard Gravitational Law – Rankers Physics

Newton's Law of Gravitation: Practice Problem & Solution

Imagine a light planet revolving around a very massive star in a circular orbit of radius \(r\) with a period of revolution T. If the gravitational force of attraction between the planet and the star is proportional to \(r^{-5/2}\), then the square of the time period will be proportional to
\(r^3\)
\(r^2\)
\(r^{2.5}\)
\(r^{3.5}\)

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Newton's Law of Gravitation. Understanding the underlying formula is key to arriving at the correct answer below:

The centripetal force is \(F = m\omega^2 r = m\frac{4\pi^2}{T^2} r \propto \frac{r}{T^2}\). Given \(F \propto r^{-5/2}\), we get \(\frac{r}{T^2} \propto r^{-5/2} \Rightarrow T^2 \propto r^{3.5}\).

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