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Question

In a solar system, the time-period of revolution of a planet tracing a circular orbit of radius $R$ is proportional to :

This question was previously asked in
NEET UG Re-Exam 2026 Question Paper (21-Jun-2026)
The correct answer is
$R^{3/2}$

Physics: Planetary Motion in Solar System

Understanding Planetary Time Period

The question asks for the relationship between the time period of revolution ($T$) of a planet and the radius ($R$) of its circular orbit within a solar system. This relationship is governed by Kepler's Laws of Planetary Motion.

Applying Kepler's Third Law

Kepler's Third Law states that the square of the time period of revolution ($T^2$) is directly proportional to the cube of the semi-major axis of the orbit ($a^3$). For a circular orbit, the semi-major axis is equal to the radius ($R$). Therefore, the law can be expressed as:

$ T^2 \propto R^3 $

Derivation for Circular Orbit

1. Force Balance: In a circular orbit, the gravitational force exerted by the central star (mass $M$) on the planet (mass $m$) provides the necessary centripetal force.

$ \frac{G M m}{R^2} = \frac{m v^2}{R} $

Where $G$ is the gravitational constant and $v$ is the orbital velocity. 2. Orbital Velocity: Simplifying the force balance gives $ v^2 = \frac{G M}{R} $. 3. Time Period Formula: The time period $T$ is the circumference ($2\pi R$) divided by the velocity ($v$).

$ T = \frac{2\pi R}{v} $

Squaring both sides:

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

4. Substituting $v^2$: Substitute $ v^2 = \frac{G M}{R} $ into the equation for $T^2$:

$ T^2 = \frac{4\pi^2 R^2}{(G M / R)} = \frac{4\pi^2 R^3}{G M} $

5. Proportionality: Since $ \frac{4\pi^2}{G M} $ is constant for a given solar system, we get:

$ T^2 \propto R^3 $

Final Relationship

Taking the square root of both sides of the proportionality $ T^2 \propto R^3 $, we find the relationship for the time period $T$:

$ T \propto \sqrt{R^3} $

$ T \propto R^{3/2} $

Therefore, the time period of revolution of a planet is proportional to $ R^{3/2} $.
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