Question

A mass \( m \) is taken from Earth's surface to infinity. Work done is:

• A. \( +\frac{GMm}{R} \)
• B. \( -\frac{GMm}{R} \)
• C. \( 0 \)
• D. \( \frac{GMm}{2R} \)

Hint:

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.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Work done to move a body in a gravitational field is equal to the change in its gravitational potential energy.
Because we are moving the mass away from the Earth (against gravity), we must perform positive work.
Step 2: Key Formula or Approach:
Work Done ($W$) = Potential Energy at final position ($U_f$) - Potential Energy at initial position ($U_i$).
The formula for gravitational potential energy is:
\[ U = -\frac{GMm}{r} \] Detailed Explanation:
1. Initial position is on the Earth's surface, so $r = R$.
\[ U_i = -\frac{GMm}{R} \] 2. Final position is at infinity, so $r = \infty$.
\[ U_f = -\frac{GMm}{\infty} = 0 \] 3. Work done by an external agent is the difference:
\[ W = U_f - U_i \] \[ W = 0 - \left( -\frac{GMm}{R} \right) \] \[ W = +\frac{GMm}{R} \] This is the amount of energy that must be supplied to move the mass entirely out of the Earth's gravitational influence.
Step 3: Final Answer:
The work done is $+\frac{GMm}{R}$.