If `upsilon_(e)` is the escape velocity for earth when a projectile is fired from the surface of earth. Then, the escape velocity if the same projectile is fired from its centre is

Options

A. `sqrt((3)/(2)) upsilon_(e)`
B. `(3)/(2) upsilon_(e)`
C. `sqrt((2)/(3)) upsilon_(e)`
D. `(2)/(3) upsilon_(e)`

Explanation

To find the escape velocity of a projectile fired from the center of the Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Escape Velocity**:
The escape velocity \( V_e \) from the surface of the Earth is given by the formula:
\[ V_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Consider the Energy at the Center**:
When a projectile is fired from the center of the Earth, we need to consider the gravitational potential energy and kinetic energy at that point. The gravitational potential energy \( U \) at a distance \( r \) from the center of a sphere of uniform density is given by:
\[ U = -\frac{GMm}{r} \]
However, at the center, the effective gravitational force acting on the projectile is zero because the mass of the Earth is symmetrically distributed around it.

3. **Initial Energy Calculation**:
At the center of the Earth, the initial potential energy \( U_i \) is:
\[ U_i = -\frac{3}{2} \frac{GMm}{R} \]
The initial kinetic energy \( K_i \) is:
\[ K_i = \frac{1}{2} mv^2 \]
Therefore, the total initial energy \( E_i \) is:
\[ E_i = K_i + U_i = \frac{1}{2} mv^2 - \frac{3}{2} \frac{GMm}{R} \]

4. **Final Energy Calculation**:
At infinity, both potential energy \( U_f \) and kinetic energy \( K_f \) become zero:
\[ E_f = 0 \]

5. **Apply Conservation of Energy**:
According to the conservation of energy:
\[ E_i = E_f \]
Thus,
\[ \frac{1}{2} mv^2 - \frac{3}{2} \frac{GMm}{R} = 0 \]

6. **Solve for \( v \)**:
Rearranging the equation gives:
\[ \frac{1}{2} mv^2 = \frac{3}{2} \frac{GMm}{R} \]
Dividing both sides by \( m \) and multiplying by 2:
\[ v^2 = \frac{3GM}{R} \]
Taking the square root:
\[ v = \sqrt{\frac{3GM}{R}} \]

7. **Relate to Escape Velocity**:
We know that the escape velocity from the surface of the Earth is \( V_e = \sqrt{\frac{2GM}{R}} \). We can express our result in terms of \( V_e \):
\[ v = \sqrt{3} \cdot \sqrt{\frac{GM}{R}} = \sqrt{3} \cdot \frac{V_e}{\sqrt{2}} = \frac{\sqrt{3}}{\sqrt{2}} V_e \]
Thus,
\[ v = \frac{\sqrt{3}}{2} V_e \]

### Final Answer:
The escape velocity if the same projectile is fired from the center of the Earth is:
\[ v = \frac{\sqrt{3}}{2} V_e \]