Question

A light planet is revolving around a massive star in a circular orbit of radius $R$ with a period of revolution $T$. If the force of attraction between the planet and the star is proportional to $R^{-\frac{3}{2}}$, then choose the correct option:

• A. $T^2 \propto R^{5/2}$
• B. $T^2 \propto R^{7/2}$
• C. $T^2 \propto R^{3/2}$
• D. $T^2 \propto R^3$

Answer 1

The Correct Option is A

Given: - The force of attraction between the planet and the star is proportional to \( R^{-3/2} \).

Step 1: Expressing the Force of Attraction

Let the force of attraction between the planet and the star be given by:

\[ F \propto R^{-3/2} \]

We can write:

\[ F = \frac{k}{R^{3/2}} \]

where \( k \) is a proportionality constant.

Step 2: Applying Centripetal Force Condition

For a planet revolving in a circular orbit, the centripetal force is provided by the gravitational force:

\[ F = m \cdot \frac{v^2}{R} \]

where \( m \) is the mass of the planet and \( v \) is its orbital velocity.

Equating the two expressions for \( F \):

\[ \frac{k}{R^{3/2}} = m \cdot \frac{v^2}{R} \]

Rearranging terms:

\[ v^2 = \frac{k}{m} \cdot R^{-1/2} \]

Taking the square root:

\[ v \propto R^{-1/4} \]

Step 3: Relating Orbital Velocity to Period of Revolution

The orbital velocity is also given by:

\[ v = \frac{2 \pi R}{T} \]

Substituting \( v \propto R^{-1/4} \):

\[ \frac{2 \pi R}{T} \propto R^{-1/4} \]

Rearranging to find \( T \):

\[ T \propto R^{5/4} \]

Squaring both sides:

\[ T^2 \propto R^{5/2} \]

Conclusion:

The correct relationship is \( T^2 \propto R^{5/2} \).