Problem 37 - A Complete Resource Book in Physics for JEE Main

Question

Three planets of same density have radii \(R_{1}, R_{2}\) and \(R_{3}\) such that \(R_{1}=2 R_{2}=3 R_{3}\). The gravitational field at their respective surfaces are \(g_{1}, g_{2}\) and \(g_{3}\) and escape velocities from their surfaces are \(v_{1}, v_{2}\) and \(v_{3}\) respectively, then (A) \(g_{1} / g_{2}=2\) (B) \(g_{1} / g_{3}=3\) (C) \(v_{1} / v_{2}=1 / 4\) (D) \(v_{1} / v_{3}=3\)

Step-by-Step Solution

Answer

The correct relationships for the given problem are: (A) \(g_{1}/g_{2} = 2\) (B) \(g_{1}/g_{3} = 3\) (D) \(v_{1}/v_{3} = 3\)

Step 1: Express the masses of the planets in terms of their volumes and densities

Since the planets have the same density, we can express their masses as: \(M_{1} = \rho V_{1}\) \(M_{2} = \rho V_{2}\) \(M_{3} = \rho V_{3}\) The volumes of the planets can be written as: \(V_{1} = \frac{4}{3}\pi R_{1}^3 \) \(V_{2} = \frac{4}{3}\pi R_{2}^3 \) \(V_{3} = \frac{4}{3}\pi R_{3}^3 \) Substitute the volumes in the mass equations: \(M_{1} = \rho \frac{4}{3}\pi R_{1}^3 \) \(M_{2} = \rho \frac{4}{3}\pi R_{2}^3 \) \(M_{3} = \rho \frac{4}{3}\pi R_{3}^3 \)

Step 2: Calculate the gravitational fields

Now, let's calculate the gravitational fields at their respective surfaces for planets 1, 2, and 3 using the formula g = (GM/R^2): \(g_{1} = \frac{GM_{1}}{R_{1}^2}\) \(g_{2} = \frac{GM_{2}}{R_{2}^2}\) \(g_{3} = \frac{GM_{3}}{R_{3}^2}\) Substitute the masses from Step 1 into the above expressions and simplify: \(g_{1}= \frac{G\rho \frac{4}{3}\pi R_{1}^3}{R_{1}^2} = \frac{4G\rho\pi R_{1}}{3} \) \(g_{2}= \frac{G\rho \frac{4}{3}\pi R_{2}^3}{R_{2}^2} = \frac{4G\rho\pi R_{2}}{3}\) \(g_{3}= \frac{G\rho \frac{4}{3}\pi R_{3}^3}{R_{3}^2} = \frac{4G\rho\pi R_{3}}{3}\)

Step 3: Find the relationships between gravitational fields

We have g1, g2, and g3 in terms of radii. Use the given relationships (R1 = 2 * R2 = 3 * R3) to find the relationships between them: \(\frac{g_{1}}{g_{2}} = \frac{\frac{4G\rho\pi R_{1}}{3}}{\frac{4G\rho\pi R_{2}}{3}} = \frac{R_{1}}{R_{2}} = 2 \) (Answer: A) \(\frac{g_{1}}{g_{3}} = \frac{\frac{4G\rho\pi R_{1}}{3}}{\frac{4G\rho\pi R_{3}}{3}} = \frac{R_{1}}{R_{3}} = 3 \) (Answer: B)

Step 4: Calculate the escape velocities

Now, let's calculate the escape velocities from the surface of the planets using the formula v = \(\sqrt\frac{2GM}{R}\) for planets 1, 2, and 3: \(v_{1} = \sqrt{\frac{2GM_{1}}{R_{1}}} \) \(v_{2} = \sqrt{\frac{2GM_{2}}{R_{2}}} \) \(v_{3} = \sqrt{\frac{2GM_{3}}{R_{3}}} \) Substitute the masses from Step 1 into the above expressions and simplify: \(v_{1} = \sqrt{\frac{2G\rho \frac{4}{3}\pi R_{1}^3}{R_{1}}} = \sqrt{8G\rho\pi R_{1}^2} \) \(v_{2} = \sqrt{\frac{2G\rho \frac{4}{3}\pi R_{2}^3}{R_{2}}} = \sqrt{8G\rho\pi R_{2}^2} \) \(v_{3} = \sqrt{\frac{2G\rho \frac{4}{3}\pi R_{3}^3}{R_{3}}} = \sqrt{8G\rho\pi R_{3}^2} \)

Step 5: Find the relationships between escape velocities

We have v1, v2, and v3 in terms of radii. Use the given relationships (R1 = 2 * R2 = 3 * R3) to find the relationships between the escape velocities: \(\frac{v_{1}}{v_{2}} = \frac{\sqrt{8G\rho\pi R_{1}^2}}{\sqrt{8G\rho\pi R_{2}^2}} = \frac{R_{1}}{R_{2}} = 2 \) (This result contradicts Answer C: \(v_{1} / v_{2}=1 / 4\), so C is not correct) \(\frac{v_{1}}{v_{3}} = \frac{\sqrt{8G\rho\pi R_{1}^2}}{\sqrt{8G\rho\pi R_{3}^2}} = \frac{R_{1}}{R_{3}} = 3 \) (Answer: D) In conclusion, the correct relationships are: (A) \(g_{1}/g_{2} = 2\) (B) \(g_{1}/g_{3} = 3\) (D) \(v_{1}/v_{3} = 3\)

Key Concepts

Universal Law of Gravitation

At the heart of understanding celestial mechanics and the motion of planets is the Universal Law of Gravitation. This law was formulated by Sir Isaac Newton in the 17th century and it revolutionized the way we understand the forces of attraction between masses. Simply put, it states that every two objects in the universe exert a gravitational force on each other that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

This law can be expressed mathematically by the equation:
\[\begin{equation} F = G \frac{m_1 m_2}{r^2} \end{equation}\]where \( F \) is the gravitational force between the masses, \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses of the two objects, and \( r \) is the distance between the centers of the two masses.

This fundamental law explains not only the falling of an apple, but also the orbiting of the moon around Earth, planets around the sun, and the vast motion of galaxies. It has been essential in understanding the exercise scenario of comparing gravitational fields among planets of different sizes but identical densities.

Gravitational Escape Velocity

When dealing with planetary bodies, an important concept is gravitational escape velocity. This is the speed needed for an object to break free from the gravitational influence of a celestial body without further propulsion. In practical terms, it's the speed a rocket needs to reach to leave a planet and move into space without falling back. The escape velocity depends on the mass and radius of the planet and can be calculated using the formula:
\[\begin{equation} v = \sqrt{\frac{2GM}{R}} \end{equation}\]Here, \( v \) is the escape velocity, \( G \) is the gravitational constant, \( M \) is the mass of the celestial body, and \( R \) is the radius of the body.

Implications on Exercise

The exercise demonstrates a crucial point: While the escape velocity does depend on the mass, because mass is a function of both density and volume, any change in volume (and thus radius) due to uniform density will alter the escape velocity notably. The relationship found in the exercise contradicts one of the answers, indicating an error. Understanding this formula helps comprehend the larger escape velocities of big planets like Jupiter compared to smaller ones like Earth.

Density and Radius Relationship

The density of a planet is its mass per unit volume, and this characteristic greatly affects gravitational attraction. When comparing planets or celestial objects with the same density, we can make direct comparisons of their gravitational fields and escape velocities based on their radii alone. This is evident in the given exercise, where the density is constant across three different planets.

Since volume is proportional to the cube of the radius (for spherical objects), the mass of the planet will increase with its volume assuming the density is fixed. Therefore, knowing the radius allows us to infer volumes and masses:
\[\begin{equation} V = \frac{4}{3}\pi R^3 \end{equation}\]and
\[\begin{equation} M = \rho V \end{equation}\]Here, \( \rho \) denotes the density, \( V \) the volume, and \( R \) the radius of the planet. This simple yet pivotal relationship plays a key role in the understanding of gravitational fields and escape velocities as it connects size with gravitational influence. Correctly using this relationship, we can predict how changes in a planet's size will affect its gravity and the escape velocity needed to break free from its gravitational pull, leading to accurate conclusions in the textbook exercise.