The escape velocity for a planet is `v_e`. A particle starts from rest at a large distance from the planet, reaches the planet only under gravitational attraction, and passes through a smooth tunnel through its centre. Its speed at the centre of the planet will be

Options

A. `v_(e)`
B. `1.5 v_(e)`
C. `sqrt(1.5) v_(e)`
D. `2 v_(e)`

Explanation

To solve the problem, we need to determine the speed of the particle when it reaches the center of the planet after starting from rest at a large distance. We will use the concept of conservation of energy.

### Step-by-Step Solution:

1. **Understanding Escape Velocity**:
The escape velocity \( v_e \) is the minimum velocity an object must have to escape the gravitational pull of a planet without any further propulsion. It 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. **Initial Energy at Large Distance**:
When the particle is at a large distance from the planet, it starts from rest, so its initial kinetic energy \( KE_i \) is:
\[ KE_i = 0 \]
The gravitational potential energy \( PE_i \) at a large distance is taken to be zero:
\[ PE_i = 0 \]
Therefore, the total initial energy \( E_i \) is:
\[ E_i = KE_i + PE_i = 0 + 0 = 0 \]

3. **Energy at the Center of the Planet**:
When the particle reaches the center of the planet, its potential energy \( PE_f \) can be calculated using the formula for gravitational potential energy inside a uniform sphere:
\[ PE_f = -\frac{GMm}{R} \quad \text{(where \( m \) is the mass of the particle)} \]
The kinetic energy \( KE_f \) at the center will be:
\[ KE_f = \frac{1}{2} mv^2 \]
where \( v \) is the speed of the particle at the center.

4. **Applying Conservation of Energy**:
According to the conservation of mechanical energy, the total energy at the beginning must equal the total energy at the center:
\[ E_i = E_f \]
Therefore:
\[ 0 = KE_f + PE_f \]
Substituting the expressions for \( KE_f \) and \( PE_f \):
\[ 0 = \frac{1}{2} mv^2 - \frac{GMm}{R} \]

5. **Solving for Speed \( v \)**:
Rearranging the equation gives:
\[ \frac{1}{2} mv^2 = \frac{GMm}{R} \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v^2 = \frac{GM}{R} \]
Multiplying both sides by 2:
\[ v^2 = \frac{2GM}{R} \]
Taking the square root:
\[ v = \sqrt{\frac{2GM}{R}} \]

6. **Relating to Escape Velocity**:
We know that \( v_e = \sqrt{\frac{2GM}{R}} \). Thus:
\[ v = v_e \]

### Final Answer:
The speed of the particle at the center of the planet will be equal to the escape velocity \( v_e \).