Problem 22 - Master Resource Book in JEE Main Physics

Question

The masses and radii of the earth and moon are \(M_{1}, R_{1}\) and \(M_{2}, R_{2}\) respectively. Then centres are distance \(d\) apart. The minimum velocity with which a particle of mass \(M\) should be projected from a point midway between their centres so that it escapes to infinity is (a) \(2 \sqrt{\frac{G}{d}\left(M_{1}+M_{2}\right)}\) (b) \(2 \sqrt{\frac{2 G}{d}\left(M_{1}+M_{2}\right)}\) (c) \(2 \sqrt{\frac{G M}{d}\left(M_{1}+M_{2}\right)}\) (d) \(2 \sqrt{\frac{G M\left(M_{1}+M_{2}\right)}{d\left(R_{1}+R_{2}\right)}}\)

Step-by-Step Solution

Answer

The correct answer is option (b), \(2 \sqrt{\frac{2 G}{d}(M_{1}+M_{2})}\).

Step 1: Understanding the Problem

We are asked to find the minimum velocity to project a particle from a point midway between Earth and Moon that escapes their combined gravitational influence. This is a classic escape velocity problem considering both celestial bodies.

Step 2: Diagram and Distances

Since the particle is projected from the midpoint of the line connecting the centers of Earth and Moon, each is a distance of \(\frac{d}{2}\) from the particle. Therefore, we should apply the formula for the escape velocity considering potential energies due to both celestial bodies.

Step 3: Gravitational Potentials

The gravitational potential at the midpoint due to Earth is \(-\frac{G M_1}{d/2} = -\frac{2G M_1}{d}\) and due to the Moon is \(-\frac{G M_2}{d/2} = -\frac{2G M_2}{d}\). The total potential energy experienced by the particle is the sum of potentials due to Earth and Moon.

Step 4: Total Gravitational Potential Energy

Combine the potential energies to get the total potential energy at the midpoint: \[ U = -\frac{2G M_1}{d} - \frac{2G M_2}{d} = -\frac{2G}{d}(M_1 + M_2) \]

Step 5: Escape Velocity Formula Application

The escape velocity \( v \) from a potential \( U \) is given by the formula \( v = \sqrt{-2U} \). Plugging in our total potential energy, we get:\[ v = \sqrt{-2 \left( -\frac{2G}{d}(M_1 + M_2) \right)} = 2 \sqrt{\frac{2G}{d}(M_1 + M_2)} \]

Step 6: Select the Correct Option

Based on the derived formula for escape velocity, the correct answer is option (b): \(2 \sqrt{\frac{2 G}{d}(M_1 + M_2)}\).

Key Concepts

Gravitational Potential Energy

Gravitational potential energy is a key concept in understanding how energy is stored in an object due to its position relative to a massive body like Earth or the Moon. It represents the energy an object possesses because of the gravitational forces acting upon it. This energy is potential because it has the possibility of being converted to kinetic energy, particularly when an object moves in the gravitational field. For a two-mass system, such as a particle influenced by both Earth and the Moon, the gravitational potential energy from each body contributes to the total energy. The formula used for calculating this potential energy for a mass at a distance from another mass is given by:\[U = - \frac{GM}{r}\]Where:

• \(U\) is the gravitational potential energy,
• \(G\) is the gravitational constant,
• \(M\) is the mass of the celestial body, and
• \(r\) is the distance between the centers of the masses.

In the exercise, this principle is applied to determine the escape velocity of a particle situated between both the Earth and the Moon.

Celestial Mechanics

Celestial mechanics is the branch of astronomy that deals with the motions and gravitational interactions of celestial bodies. This field explains how objects in our solar system, like planets, moons, and comets, move and interact based on the gravitational forces they exert on one another. By using Newton's laws of motion and universal gravitation, celestial mechanics allows us to predict the trajectories of these bodies. In the context of escape velocity, celestial mechanics explores how a body, such as a rocket, can overcome the gravitational hold of larger celestial objects. The underlying science ensures safe space missions and satellite deployments by calculating precise velocities for launching spacecraft. A fascinating aspect of celestial mechanics in the problem at hand is the combined influence of both Earth and the Moon's gravitation, which dictates the minimum velocity required for a particle to escape their gravitational fields. This complex interaction highlights the intricacy of gravitational influence in a multiple body system.

Two-Body Problem

The two-body problem is a fundamental problem in physics and celestial mechanics, describing the motion of two massive objects interacting through their mutual gravitational attraction. It is well-known for having a precise and exact solution, which can help predict the behavior of planets and satellites. Solving a two-body problem involves calculating factors like velocities and distances where both bodies affect the motion of any masses they encounter. In the featured problem, the two masses are Earth and the Moon, with a particle positioned midway. Here, the two-body problem helps us find the impact these two celestial bodies have on the particle's potential to escape. The challenge is understanding how their individual and combined gravitational affects the particle. To solve this, the forces on the particle are calculated based on its positions relative to both the celestial bodies. This calculation eventually aids in determining the minimum velocity required for the particle to break free from these combined gravitational forces. Thus, although this problem is abstract, understanding the two-body problem is crucial for many real-world applications in space travel and other astrophysical simulations.