A body starts from rest from a point distant `r_(0)` from the centre of the earth. It reaches the surface of the earth whose radius is `R`. The velocity acquired by the body is

Options

A. `2GMsqrt(1/R-1/(r_(0))`
B. `sqrt(2GM(1/R-1/(r_(0))))`
C. `GMsqrt(1/R-1/(r_(0))`
D. `sqrt(GM(1/R-1/(R_(0)))`

Explanation

To solve the problem of finding the velocity acquired by a body starting from rest at a distance \( r_0 \) from the center of the Earth and reaching the surface of the Earth (radius \( R \)), we can use the principle of conservation of mechanical energy. Here’s a step-by-step solution:

### Step 1: Understand the Energy Conservation Principle
The total mechanical energy (potential energy + kinetic energy) of the system remains constant if only conservative forces (like gravity) are acting on it.

### Step 2: Write the Initial and Final Energy Expressions
1. **Initial Energy (at distance \( r_0 \))**:
- Gravitational Potential Energy (U_i):
\[ U_i = -\frac{GMm}{r_0} \]
- Kinetic Energy (K_i):
\[ K_i = 0 \quad (\text{since it starts from rest}) \]
- Total Initial Energy (E_i):
\[ E_i = U_i + K_i = -\frac{GMm}{r_0} + 0 = -\frac{GMm}{r_0} \]

2. **Final Energy (at the surface of the Earth)**:
- Gravitational Potential Energy (U_f):
\[ U_f = -\frac{GMm}{R} \]
- Kinetic Energy (K_f):
\[ K_f = \frac{1}{2}mv^2 \]
- Total Final Energy (E_f):
\[ E_f = U_f + K_f = -\frac{GMm}{R} + \frac{1}{2}mv^2 \]

### Step 3: Set Initial Energy Equal to Final Energy
Using the conservation of energy:
\[ E_i = E_f \]
\[ -\frac{GMm}{r_0} = -\frac{GMm}{R} + \frac{1}{2}mv^2 \]

### Step 4: Simplify the Equation
1. Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ -\frac{GM}{r_0} = -\frac{GM}{R} + \frac{1}{2}v^2 \]
2. Rearranging gives:
\[ \frac{1}{2}v^2 = \frac{GM}{R} - \frac{GM}{r_0} \]
3. Factor out \( GM \):
\[ \frac{1}{2}v^2 = GM \left( \frac{1}{R} - \frac{1}{r_0} \right) \]

### Step 5: Solve for \( v^2 \)
Multiply both sides by 2:
\[ v^2 = 2GM \left( \frac{1}{R} - \frac{1}{r_0} \right) \]

### Step 6: Take the Square Root to Find \( v \)
\[ v = \sqrt{2GM \left( \frac{1}{R} - \frac{1}{r_0} \right)} \]

### Final Answer
The velocity acquired by the body when it reaches the surface of the Earth is:
\[ v = \sqrt{2GM \left( \frac{1}{R} - \frac{1}{r_0} \right)} \]