Question

Assume a planet with orbiting radius R and period of revolution T around the sun experiences a gravitational force which follows inverse cube law instead of inverse square law. In this case, the period of revolution is proportional to:

• A. \(\frac{1}{R}\)
• B. \(R^2\)
• C. \(R^{3/2}\)
• D. \(R^4\)

Hint:

For non-Newtonian gravitational laws, use \(F = m v^2/R\) to relate velocity and radius, then find period via \(T = 2 \pi R/v\).

Answer 1

The Correct Option is B

Step 1: Use centripetal force relation.
For circular motion: \(F = \frac{m v^2}{R}\). Here \(F \propto \frac{1}{R^3}\).

Step 2: Express velocity.
\[ \frac{m v^2}{R} = \frac{k}{R^3} \Rightarrow v^2 = \frac{k}{m R^2} \Rightarrow v = \sqrt{\frac{k}{m}} \frac{1}{R} \]

Step 3: Relate period T to velocity.
\[ T = \frac{2 \pi R}{v} = \frac{2 \pi R}{k^{1/2}/\sqrt{m} \cdot 1/R} = 2 \pi \sqrt{\frac{m}{k}} R^2 \]

Step 4: Conclusion.
Hence, \(T \propto R^2\).