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:
- $T^2 \propto R^{5/2}$
- $T^2 \propto R^{7/2}$
- $T^2 \propto R^{3/2}$
- $T^2 \propto R^3$
The Correct Option is A
Approach Solution - 1
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 \(T^2 \propto R^{5/2}\).
Approach Solution -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 \( T^2 \propto R^{5/2} \).
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