Question

A body which is initially at rest at a height \( R \) above the surface of the Earth of radius \( R \), falls freely towards the Earth. The velocity on reaching the surface of the Earth is:

• A. \( \sqrt{2gR} \)
• B. \( \sqrt{gR} \)
• C. \( \sqrt{\frac{3}{2} gR} \)
• D. \( \sqrt{4gR} \)

Hint:

For free fall from height \( h \), the velocity on reaching the surface can be found using energy conservation: \[ v = \sqrt{2 g h} \] or using gravitational potential.

Answer 1

The Correct Option is B

Step 1: {Apply conservation of energy}
The total energy remains constant: \[ {Increase in kinetic energy} = {Decrease in potential energy} \] Step 2: {Using gravitational potential energy}
\[ \frac{1}{2} mv^2 = mgR \left( \frac{1}{1 + \frac{h}{R}} \right) \] For \( h = R \): \[ mv^2 = mgR \] \[ v = \sqrt{gR} \] Thus, the correct answer is \( \sqrt{gR} \).

Answer 2

Step 1: Given:
- Initial height from Earth’s surface = \( R \)
- Radius of Earth = \( R \)
- So, initial distance from Earth’s center = \( 2R \)
- Final distance from Earth’s center (on the surface) = \( R \)
- Initial velocity = 0 (body starts from rest)

Step 2: Use conservation of mechanical energy:
Initial total energy = Final total energy
\( \Rightarrow \frac{-GMm}{2R} + 0 = \frac{-GMm}{R} + \frac{1}{2}mv^2 \)

Step 3: Rearranging terms:
\( \frac{-GMm}{2R} = \frac{-GMm}{R} + \frac{1}{2}mv^2 \)
Bring all potential terms to one side:
\( \frac{-GMm}{2R} + \frac{GMm}{R} = \frac{1}{2}mv^2 \)
\( \frac{GMm}{2R} = \frac{1}{2}mv^2 \)

Step 4: Cancel \( m \) and simplify:
\( \frac{GM}{2R} = \frac{1}{2}v^2 \Rightarrow v^2 = \frac{GM}{R} \)

Step 5: Use \( g = \frac{GM}{R^2} \Rightarrow GM = gR^2 \)
Substitute into the equation:
\( v^2 = \frac{gR^2}{R} = gR \)
So, \( v = \sqrt{gR} \)

Final Answer: \( \sqrt{gR} \)