The escape velocity for a planet is `v_(e)`. A tunnel is dug along a diameter of the planet and a small body is dropped into it at the surface. When the body reaches the centre of the planet, its speed will be

Options

A. `v_(e)`
B. `(v_(e))/(sqrt(2))`
C. `(v_(e))/2`
D. zero

Explanation

To solve the problem, we need to determine the speed of a small body when it reaches the center of a planet after being dropped from the surface through a tunnel dug along the diameter of the planet.

### Step-by-Step Solution:

1. **Understanding Escape Velocity**:
The escape velocity \( v_e \) for a planet is given by the formula:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

2. **Potential Energy at the Surface**:
The gravitational potential energy \( V_s \) at the surface of the planet is given by:
\[ V_s = -\frac{GMm}{R} \]
where \( m \) is the mass of the small body.

3. **Potential Energy at the Center**:
The gravitational potential energy \( V_0 \) at the center of the planet is given by:
\[ V_0 = -\frac{3GMm}{2R} \]

4. **Change in Potential Energy**:
The change in potential energy as the body moves from the surface to the center is:
\[ \Delta V = V_0 - V_s = \left(-\frac{3GMm}{2R}\right) - \left(-\frac{GMm}{R}\right) \]
Simplifying this gives:
\[ \Delta V = -\frac{3GMm}{2R} + \frac{GMm}{R} = -\frac{3GMm}{2R} + \frac{2GMm}{2R} = -\frac{GMm}{2R} \]

5. **Kinetic Energy at the Center**:
According to the work-energy principle, the loss in potential energy equals the gain in kinetic energy. The kinetic energy \( K \) at the center is given by:
\[ K = \frac{1}{2}mv^2 \]
Setting the change in potential energy equal to the kinetic energy gives:
\[ -\Delta V = K \implies \frac{GMm}{2R} = \frac{1}{2}mv^2 \]

6. **Solving for Speed \( v \)**:
Canceling \( m \) from both sides and solving for \( v \):
\[ \frac{GM}{2R} = \frac{1}{2}v^2 \implies v^2 = \frac{GM}{R} \implies v = \sqrt{\frac{GM}{R}} \]

7. **Relating to Escape Velocity**:
Since we know that \( v_e^2 = \frac{2GM}{R} \), we can express \( v \) in terms of \( v_e \):
\[ v = \sqrt{\frac{GM}{R}} = \sqrt{\frac{1}{2} \cdot \frac{2GM}{R}} = \frac{v_e}{\sqrt{2}} \]

### Final Answer:
The speed of the body when it reaches the center of the planet is:
\[ v = \frac{v_e}{\sqrt{2}} \]