Question

A particle of mass \( m \) is at a distance \( 3R \) from the center of Earth. Find the minimum kinetic energy of the particle to leave Earth's field, where \( R \) is the radius of Earth.

• A. \( \frac{mgR}{3} \)
• B. \( 3mgR \)
• C. \( \frac{2}{3} mgR \)
• D. \( \frac{mgR}{2} \)

Hint:

The minimum kinetic energy required for a particle to escape Earth's gravitational field is equal to the magnitude of the gravitational potential energy at the given distance.

Answer 1

The Correct Option is C

The work-energy theorem states that the minimum kinetic energy required for a particle to escape Earth's gravitational field is equal to the negative of the gravitational potential energy at that point. The gravitational potential energy at a distance \( r \) from the center of Earth is given by: \[ U = -\frac{GMm}{r} \] where: - \( G \) is the gravitational constant, - \( M \) is the mass of Earth, - \( m \) is the mass of the particle, and - \( r \) is the distance from the center of the Earth. For the given question, the particle is at a distance \( 3R \) from the center of the Earth, where \( R \) is the radius of Earth. Therefore, the potential energy at this point is: \[ U = -\frac{GMm}{3R} \] The minimum kinetic energy required for the particle to leave Earth's field is the amount of energy needed to overcome this gravitational potential energy. Thus, the minimum kinetic energy is: \[ K = \left| U \right| = \frac{GMm}{3R} \] Now, using the relation \( g = \frac{GM}{R^2} \) (where \( g \) is the acceleration due to gravity at Earth's surface), we can rewrite the expression for kinetic energy as: \[ K = \frac{gmR}{3} \] Thus, the minimum kinetic energy required for the particle to escape Earth's field is \( \frac{2}{3} mgR \). Therefore, the correct answer is (3) \( \frac{2}{3} mgR \).