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

To solve this problem, we need to relate the force of attraction and the period of revolution for a planet revolving in a circular orbit around a star. We are given that the force of attraction, \( F \), between the planet and the star is proportional to \( R^{-\frac{3}{2}} \), where \( R \) is the radius of the orbit. 

According to Kepler's third law of planetary motion for circular orbits, the gravitational force provides the necessary centripetal force. Thus, we have:

\(F = \frac{G M m}{R^2} = m \frac{v^2}{R}\)

where \( G \) is the gravitational constant, \( M \) is the mass of the star, \( m \) is the mass of the planet, and \( v \) is the orbital velocity.

Since the problem states \( F \propto R^{-\frac{3}{2}} \), we can write:

\(F = k R^{-\frac{3}{2}}\)

where \( k \) is a proportionality constant.

Equating the forces:

\(k R^{-\frac{3}{2}} = m \frac{v^2}{R}\)

Solving for \( v^2 \), we get:

\(v^2 = \frac{k}{m} R^{-\frac{1}{2}}\)

The orbital period \( T \) is related to the velocity \( v \) by:

\(T = \frac{2 \pi R}{v}\)

Squaring both sides gives:

\(T^2 = \frac{4 \pi^2 R^2}{v^2}\)

Substituting the expression for \( v^2 \), we have:

\(T^2 = 4 \pi^2 \frac{R^2}{\frac{k}{m} R^{-\frac{1}{2}}}\)

Simplifying this expression results in:

\(T^2 = 4 \pi^2 \frac{m}{k} R^{2 + \frac{1}{2}}\)

Thus:

\(T^2 \propto R^{5/2}\)

Therefore, the correct option is \(T^2 \propto R^{5/2}\), which is the given correct answer.

Among the given options, the only compatible relationship with the given force proportionality is \(T^2 \propto R^{5/2}\).

Answer 2

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 \( T^2 \propto R^{5/2} \).