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. `2GM(1/R-1/R_0)`
B. `sqrt(2GM(1/R-1/R_0))`
C. `GM(1/R-1/R_0)`
D. `2GMsqrt((1/R-1/R_0))`

Explanation

To find the velocity acquired by a body that starts from rest at a distance \( r_0 \) from the center of the Earth and reaches the surface of the Earth (radius \( R \)), we can use the principle of conservation of energy.

### Step-by-Step Solution:

1. **Identify Initial and Final States**:
- The body starts from rest at a distance \( r_0 \) from the center of the Earth. Therefore, its initial kinetic energy (KE) is 0.
- The potential energy (PE) at this point is given by:
\[ PE_i = -\frac{GMm}{r_0} \]
- When the body reaches the surface of the Earth (radius \( R \)), its potential energy is:
\[ PE_f = -\frac{GMm}{R} \]
- At this point, it has acquired some kinetic energy, which we denote as \( KE_f \).

2. **Apply Conservation of Energy**:
- According to the conservation of mechanical energy:
\[ PE_i + KE_i = PE_f + KE_f \]
- Since the body starts from rest, \( KE_i = 0 \):
\[ -\frac{GMm}{r_0} + 0 = -\frac{GMm}{R} + KE_f \]

3. **Rearranging the Equation**:
- Rearranging the equation gives:
\[ KE_f = -\frac{GMm}{r_0} + \frac{GMm}{R} \]
- This simplifies to:
\[ KE_f = GMm \left( \frac{1}{R} - \frac{1}{r_0} \right) \]

4. **Express Kinetic Energy in Terms of Velocity**:
- The kinetic energy can also be expressed as:
\[ KE_f = \frac{1}{2} mv^2 \]
- Setting the two expressions for kinetic energy equal gives:
\[ \frac{1}{2} mv^2 = GMm \left( \frac{1}{R} - \frac{1}{r_0} \right) \]

5. **Solving for Velocity**:
- We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v^2 = GM \left( \frac{1}{R} - \frac{1}{r_0} \right) \]
- Multiplying both sides by 2 gives:
\[ v^2 = 2GM \left( \frac{1}{R} - \frac{1}{r_0} \right) \]
- Finally, taking 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:
\[ v = \sqrt{2GM \left( \frac{1}{R} - \frac{1}{r_0} \right)} \]