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. Find out its velocity on reaching the surface of earth. Take `g=` acceleration due to gravity on the surface of the Earth.

Options

A. `sqrt((2gR))`
B. `sqrt((gR))`
C. `sqrt((3)/(2)gR)`
D. `sqrt((4gR))`

Explanation

To find the velocity of a body falling freely from a height \( R \) above the surface of the Earth to the surface itself, we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Understand the Initial and Final States
- The body is initially at rest at a height \( R \) above the Earth's surface.
- The radius of the Earth is also \( R \).
- Therefore, the initial height \( h \) from the center of the Earth is \( R + R = 2R \).

### Step 2: Calculate Initial Potential Energy
- The gravitational potential energy (U) at a height \( h \) from the center of the Earth is given by:
\[ U = -\frac{GMm}{r} \]
where \( G \) is the universal gravitational constant, \( M \) is the mass of the Earth, and \( m \) is the mass of the body.
- At height \( 2R \):
\[ U_i = -\frac{GMm}{2R} \]

### Step 3: Calculate Final Potential Energy
- When the body reaches the surface of the Earth (at height \( R \)):
\[ U_f = -\frac{GMm}{R} \]

### Step 4: Apply Conservation of Energy
- The change in potential energy (ΔU) is equal to the change in kinetic energy (ΔK):
\[ \Delta U = U_f - U_i = \Delta K \]
- The kinetic energy at the surface when the body reaches the ground is given by:
\[ K = \frac{1}{2}mv^2 \]
- Therefore, we have:
\[ \Delta K = \frac{1}{2}mv^2 \]

### Step 5: Set Up the Equation
- The change in potential energy can be expressed as:
\[ \Delta U = -\frac{GMm}{R} + \frac{GMm}{2R} = \frac{GMm}{2R} \]
- Setting the change in potential energy equal to the change in kinetic energy:
\[ \frac{GMm}{2R} = \frac{1}{2}mv^2 \]

### Step 6: Simplify and Solve for Velocity
- Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{GM}{2R} = \frac{1}{2}v^2 \]
- Multiply both sides by 2:
\[ \frac{GM}{R} = v^2 \]
- Taking the square root gives:
\[ v = \sqrt{\frac{GM}{R}} \]

### Step 7: Relate \( g \) to \( GM/R \)
- We know that \( g = \frac{GM}{R^2} \), thus \( GM = gR \).
- Substitute \( GM \) into the velocity equation:
\[ v = \sqrt{\frac{gR}{R}} = \sqrt{g} \]

### Final Answer
The velocity of the body when it reaches the surface of the Earth is:
\[ v = \sqrt{gR} \]