If V is the gravitational potential on the surface of the earth, then what is its value at the centre of the earth ?

Options

A. 2 V
B. 3 V
C. `(3)/(2)V`
D. `(2)/(3)V`

Explanation

To solve the problem, we need to understand the gravitational potential at different points relative to the Earth.

### Step-by-Step Solution:

1. **Understand Gravitational Potential (V):**
The gravitational potential \( V \) at a distance \( r \) from the center of a mass \( M \) is given by the formula:
\[ V = -\frac{GM}{r} \]
where \( G \) is the universal gravitational constant.

2. **Gravitational Potential at the Surface of the Earth:**
At the surface of the Earth, the gravitational potential \( V \) can be expressed as:
\[ V_{\text{surface}} = -\frac{GM}{R} \]
where \( R \) is the radius of the Earth.

3. **Gravitational Potential at the Center of the Earth:**
The gravitational potential at the center of the Earth can be derived from the formula for gravitational potential inside a uniform sphere. The potential at the center \( V_c \) is given by:
\[ V_c = -\frac{3GM}{2R} \]

4. **Relate the Two Potentials:**
Now, we can express \( V_c \) in terms of \( V_{\text{surface}} \):
\[ V_c = -\frac{3GM}{2R} = \frac{3}{2} \left(-\frac{GM}{R}\right) = \frac{3}{2} V_{\text{surface}} \]

5. **Final Answer:**
Therefore, the gravitational potential at the center of the Earth is:
\[ V_c = \frac{3}{2} V_{\text{surface}} \]

### Conclusion:
If \( V \) is the gravitational potential on the surface of the Earth, then the value at the center of the Earth is:
\[ V_c = \frac{3}{2} V \]