Question

The amount of work done to raise a mass ‘m’ from the surface of the Earth to a height equal to the radius of the Earth ‘R’, will be: ____.

• A. mgR
• B. 2mgR
• C. mgR/4
• D. mgR/2

Hint:

A useful shortcut for work done to lift a mass to height $h$ is $W = \frac{mgh}{1 + h/R}$. Here $h=R$, so $W = \frac{mgR}{1 + R/R} = \frac{mgR}{2}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Topic:
This question is from "Gravitation." It addresses the change in gravitational potential energy when an object is moved to a height comparable to the Earth's radius. At such distances, the gravitational field is not uniform, so the simple formula $W = mgh$ is no longer accurate.
Step 2: Key Formulas and Approach:

Potential Energy $U = -GMm / r$.
Work Done $W = \Delta U = U_{final} - U_{initial}$.
Relation at surface: $g = GM / R^2 \implies GM = gR^2$.

Step 3: Detailed Explanation:

Initial State: At the Earth's surface, the distance from the center is $r_i = R$. \[ U_i = -\frac{GMm}{R} \]
Final State: At a height '$R$' above the surface, the distance from the center is $r_f = R + R = 2R$. \[ U_f = -\frac{GMm}{2R} \]
Calculate Work Done: \[ W = U_f - U_i = \left( -\frac{GMm}{2R} \right) - \left( -\frac{GMm}{R} \right) \] \[ W = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R} \]
Convert to 'g': Using $GM = gR^2$: \[ W = \frac{(gR^2)m}{2R} = \frac{mgR}{2} \]
If we had used $W = mgh$ where $h=R$, we would get $mgR$, which is exactly double the correct answer. This shows why the energy change method is necessary for large heights.
Step 4: Final Answer:
The work done is mgR/2.