A body is released from a point of distance `R'` from the centre of earth. Its velocity at the time of striking the earth will be `(R’gtR_(e))`

Options

A. `sqrt(2gR_(e))`
B. `sqrt(R_(e)g)`
C. `sqrt(2g(R'-R_(e)))`
D. `sqrt(2gR_(e)(1-(R_(e))/(R^(r))))`

Explanation

To find the velocity of a body released from a distance \( R' \) from the center of the Earth when it strikes the Earth, we can use the principle of conservation of mechanical energy. Here’s a step-by-step solution:

### Step 1: Understand the Initial and Final States
- **Initial State:** The body is at a distance \( R' \) from the center of the Earth. The initial kinetic energy (KE) is 0 because it is released from rest.
- **Final State:** The body strikes the surface of the Earth. At this point, it has some kinetic energy and potential energy.

### Step 2: Write the Conservation of Mechanical Energy Equation
The conservation of mechanical energy states that the total mechanical energy (potential energy + kinetic energy) at the initial position is equal to the total mechanical energy at the final position.

\[ U_i + K_i = U_f + K_f \]

Where:
- \( U_i \) = Initial potential energy
- \( K_i \) = Initial kinetic energy
- \( U_f \) = Final potential energy
- \( K_f \) = Final kinetic energy

### Step 3: Calculate Initial and Final Energies
- **Initial Potential Energy \( U_i \):**
\[ U_i = -\frac{GMm}{R'} \]
Where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( m \) is the mass of the body.

- **Initial Kinetic Energy \( K_i \):**
\[ K_i = 0 \]

- **Final Potential Energy \( U_f \):**
At the surface of the Earth, the potential energy is:
\[ U_f = -\frac{GMm}{R_e} \]
Where \( R_e \) is the radius of the Earth.

- **Final Kinetic Energy \( K_f \):**
\[ K_f = \frac{1}{2}mv^2 \]
Where \( v \) is the velocity of the body just before it strikes the Earth.

### Step 4: Set Up the Equation
Substituting the values into the conservation of energy equation:
\[ -\frac{GMm}{R'} + 0 = -\frac{GMm}{R_e} + \frac{1}{2}mv^2 \]

### Step 5: Simplify the Equation
Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ -\frac{GM}{R'} = -\frac{GM}{R_e} + \frac{1}{2}v^2 \]

Rearranging gives:
\[ \frac{1}{2}v^2 = \frac{GM}{R_e} - \frac{GM}{R'} \]

### Step 6: Factor Out \( GM \)
\[ \frac{1}{2}v^2 = GM \left(\frac{1}{R_e} - \frac{1}{R'}\right) \]

### Step 7: Express in Terms of \( g \)
We know that \( g = \frac{GM}{R_e^2} \). Thus, we can express \( GM \) as \( g R_e^2 \):
\[ \frac{1}{2}v^2 = g R_e^2 \left(\frac{1}{R_e} - \frac{1}{R'}\right) \]
\[ \frac{1}{2}v^2 = g R_e \left(1 - \frac{R_e}{R'}\right) \]

### Step 8: Solve for \( v \)
Multiplying both sides by 2:
\[ v^2 = 2g R_e \left(1 - \frac{R_e}{R'}\right) \]
Taking the square root gives:
\[ v = \sqrt{2g R_e \left(1 - \frac{R_e}{R'}\right)} \]

### Final Answer
Thus, the velocity of the body at the time of striking the Earth is:
\[ v = \sqrt{2g R_e \left(1 - \frac{R_e}{R'}\right)} \]