Question

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 \(T^2\) is proportional to :

• A. \(R^3\)
• B. \(R^{7/2}\)
• C. \(R^{3/2}\)
• D. \(R^{3.75}\)

Hint:

For circular motion, combine centripetal force with given force law to derive time-period relations.

Answer 1

The Correct Option is B

Concept: Centripetal force: \[ \frac{mv^2}{R} = F \] Also: \[ v = \frac{2\pi R}{T} \]

Step 1: Force relation.
\[ F \propto R^{-5/2} \]

Step 2: Substitute velocity.
\[ \frac{m(2\pi R)^2}{T^2 R} \propto R^{-5/2} \] \[ \frac{R^2}{T^2 R} \propto R^{-5/2} \Rightarrow \frac{R}{T^2} \propto R^{-5/2} \]

Step 3: Solve relation.
\[ T^2 \propto R^{1 + 5/2} = R^{7/2} \]