Problem 4 - Master Resource Book in JEE Main Physics

Question

Imagine a light planet revolving around a very massive star in a circular orbit of radius \(r\) with a period of revolution \(T\). If the gravitational force of attraction between the planet and the star is proportional to \(R^{-3 / 2}\), then \(T_{2}\) is proportional to (a) \(R^{3}\) (b) \(R^{5 / 2}\) (c) \(R^{3 / 2}\) (d) \(R^{7 / 2}\)

Step-by-Step Solution

Answer

Option (c) \(R^{3/2}\).

Step 1: Identify the given relationship

We are told that the gravitational force is proportional to \( R^{-3/2} \). This means that \( F \propto R^{-3/2} \).

Step 2: Relate force to centripetal motion

For a planet in a circular orbit, the gravitational force also acts as the centripetal force keeping the planet in orbit. Hence, we have \( F = \frac{mv^2}{R} \), where \( m \) is the mass of the planet and \( v \) is its orbital velocity.

Step 3: Use velocity and period relation

The orbital velocity \( v \) can be related to the period \( T \) using the equation \( v = \frac{2\pi R}{T} \). This gives us \( v^2 = \left(\frac{2\pi R}{T}\right)^2 \).

Step 4: Substitute and solve for period

Replace \( v^2 \) in the centripetal force equation with \( \frac{4\pi^2 R^2}{T^2} \). Thus, the equation becomes: \( F = \frac{m \cdot 4\pi^2 R^2}{T^2R} = \frac{4\pi^2 m R}{T^2} \).

Step 5: Express force in terms of radius

We know \( F \propto R^{-3/2} \). Equating the two expressions for \( F \), we have: \( \frac{4\pi^2 m R}{T^2} \propto R^{-3/2} \).

Step 6: Solve for the proportionality

Set up the proportionality: \( \frac{4\pi^2 m R}{T^2} = kR^{-3/2} \), where \( k \) is a proportionality constant. Simplifying gives \( T^2 \propto R^{3/2} \).

Step 7: Determine the relation for \( T \)

Taking the square root on both sides, we get \( T \propto R^{3/4} \), implying \( T^2 \propto R^{3/2} \).

Key Concepts

Kepler's Third Law

Kepler's Third Law of Planetary Motion is a profound concept that describes how the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit. This law is represented mathematically as \( T^2 \propto R^3 \) where:

• \( T \) is the orbital period.
• \( R \) is the semi-major axis, which can be understood as the average distance from the planet to the star around which it revolves.

However, in this particular exercise, what makes it interesting is the proportionality of the gravitational force to \( R^{-3/2} \), which alters the usual scenario described by Kepler. This implies a different relationship, hence allowing us to explore how the period \( T \) would vary with respect to \( R \). This deviation leads us through the calculation to understand that the specific conditions dictate \( T^2 \propto R^{3/2} \), showcasing a nuanced application of orbital dynamics.

Centripetal Force

Centripetal force is essential in maintaining an object in circular motion. It acts towards the center of the circle, ensuring that the object doesn't fly off tangentially but stays in its orbital path. In the context of celestial mechanics, the gravitational force between a star and a planet acting as this centripetal force is crucial.

• Mathematically, centripetal force \( F \) can be expressed as \( F = \frac{mv^2}{R} \), where:
• \( m \) is the mass of the object (in this case, the planet).
• \( v \) is the orbital velocity.
• \( R \) is the radius of the orbit.

Understanding that the same force can be provided by gravitational attraction allows us to equate these expressions, as shown in the exercise. This connection between gravitational force and centripetal needs helps us derive relationships and deepen our understanding of orbital velocities and periods.

Orbital Velocity

Orbital velocity is the speed needed for a planet or a satellite to maintain its orbit around a celestial body. It's calculated using the relation between the circumference of the orbit and the time period of revolution. In formulaic terms, the velocity \( v \) is expressed as \( v = \frac{2\pi R}{T} \), allowing us to understand the dependency of velocity on orbital radius and period.

• \( v \) represents the orbital velocity.
• \( 2\pi R \) is the circumference of the circular orbit.
• \( T \) is the time taken to complete one orbit.

This equation highlights how increasing the orbital radius without altering the period could increase the circumferential path, necessitating a higher velocity. These dynamics further illustrate how gravitational force balances out with these velocities to create stable orbits, contributing fundamentally to our understanding of celestial mechanics and the stability of planetary systems.