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

Step 1: Understanding the Concept:
For a circular orbit, the centripetal force is provided by the gravitational force.
We relate the period \( T \) to the radius \( R \) by balancing these forces.
: Key Formula or Approach:
1. Centripetal Force \( F_c = m \omega^2 R = m \left( \frac{2\pi}{T} \right)^2 R \).
2. Given \( F_g \propto R^{-5/2} \).
Step 2: Detailed Explanation:
Set the forces equal (ignoring constants for proportionality):
\[ F_c \propto F_g \]
\[ \frac{m R}{T^2} \propto R^{-5/2} \]
Divide by \( R \) on both sides:
\[ \frac{1}{T^2} \propto R^{-5/2} \cdot R^{-1} \]
\[ \frac{1}{T^2} \propto R^{-7/2} \]
Inverting the proportionality:
\[ T^2 \propto R^{7/2} \]
Step 3: Final Answer:
\( T^2 \) is proportional to \( R^{7/2} \).