Question:
A mass \( m \) is taken from Earth's surface to infinity. Work done is:
A mass \( m \) is taken from Earth's surface to infinity. Work done is:
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To move a mass away from a massive body, an external agent must do positive work to overcome the attractive gravitational force. Thus, the work done in taking a mass to infinity from a surface will always be positive.
Updated On: Jun 3, 2026
- \( +\frac{GMm}{R} \)
- \( -\frac{GMm}{R} \)
- \( 0 \)
- \( \frac{GMm}{2R} \)
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The Correct Option is A
Solution and Explanation
Concept:
Work done by an external agent in moving a mass in a gravitational field is equal to the change in the gravitational potential energy of the system. The gravitational potential energy \( U \) of a mass \( m \) at a distance \( r \) from the center of the Earth (mass \( M \)) is given by:
\[ U = -\frac{GMm}{r} \]
The work done (\( W \)) is calculated as \( W = U_{\text{final}} - U_{\text{initial}} \).
Step 1: Identify the initial and final states
Initial position: On the Earth's surface, so \( r_1 = R \). Final position: At infinity, so \( r_2 = \infty \).
Step 2: Calculate the initial and final potential energy.
Initial potential energy, \( U_i = -\frac{GMm}{R} \).
Final potential energy, \( U_f = -\frac{GMm}{\infty} = 0 \).
Step 3: Calculate the work done by the external agent.
Work done \( W = U_f - U_i \) \[ W = 0 - \left( -\frac{GMm}{R} \right) \] \[ W = +\frac{GMm}{R} \]
Step 1: Identify the initial and final states
Initial position: On the Earth's surface, so \( r_1 = R \). Final position: At infinity, so \( r_2 = \infty \).
Step 2: Calculate the initial and final potential energy.
Initial potential energy, \( U_i = -\frac{GMm}{R} \).
Final potential energy, \( U_f = -\frac{GMm}{\infty} = 0 \).
Step 3: Calculate the work done by the external agent.
Work done \( W = U_f - U_i \) \[ W = 0 - \left( -\frac{GMm}{R} \right) \] \[ W = +\frac{GMm}{R} \]
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