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

Answer: (A)

$T^2 \propto R^{5/2}$

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 $T^2 \propto R^{5/2}$.