Question

If an object of mass 1 kg is taken to a height which is equal to three times the radius of the earth, then the change in its potential energy is
(Radius of the earth \( = 6400 \, \text{km} \), acceleration due to gravity \( g = 10 \, \text{m/s}^2 \))

• A. \( 48 \times 10^6 \, \text{J} \)
• B. \( 24 \times 10^6 \, \text{J} \)
• C. \( 36 \times 10^6 \, \text{J} \)
• D. \( 12 \times 10^6 \, \text{J} \)

Hint:

For potential energy at large heights, use \( U = -\frac{GMm}{r} \), not \( mgh \). Use energy difference between surface and height.

Answer 1

The Correct Option is A

Step 1: Gravitational potential energy at a distance \( r \): \[ U = -\frac{GMm}{r} \] Step 2: Potential energy difference: \[ \Delta U = GMm \left( \frac{1}{R} - \frac{1}{4R} \right) = GMm \cdot \frac{3}{4R} \] Step 3: Use \( g = \frac{GM}{R^2} \Rightarrow GM = gR^2 \) \[ \Delta U = \frac{3}{4R} \cdot gR^2 \cdot m = \frac{3}{4} g R m = \frac{3}{4} \cdot 10 \cdot 6.4 \times 10^6 \cdot 1 = 48 \times 10^6 \, \text{J} \]