Question

The gravitational potential energy of a rocket of mass 200 kg at a distance \( 10^6 \, \text{m} \) from the centre of the earth is \( 3 \times 10^8 \, \text{J} \). The weight of the rocket at a distance \( 10^7 \, \text{m} \) from the centre of the earth is:

• A. \( 1.5 \times 10^{-2} \, \text{N} \)
• B. \( 3 \times 10^{-2} \, \text{N} \)
• C. \( 6 \times 10^{-2} \, \text{N} \)
• D. \( 4.5 \times 10^{-2} \, \text{N} \)

Hint:

The weight of an object decreases with the square of the distance from the center of the earth according to the inverse square law of gravitation.

Answer 1

The Correct Option is B

Step 1: Gravitational Potential Energy Formula.
The gravitational potential energy is related to the distance from the centre of the earth by: \[ U = - \frac{GMm}{r} \] where \( G \) is the gravitational constant, \( M \) is the mass of the earth, \( m \) is the mass of the rocket, and \( r \) is the distance from the centre of the earth. The weight of the rocket is given by: \[ W = \frac{GMm}{r^2} \] At a distance \( 10^7 \, \text{m} \), the weight can be calculated based on the inverse square law: \[ W = \frac{GMm}{(10^7)^2} = \frac{3 \times 10^8}{(10^7)^2} = 3 \times 10^{-2} \, \text{N} \] Step 2: Final Answer.
Thus, the weight of the rocket is \( 3 \times 10^{-2} \, \text{N} \).