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 find the gravitational potential at the center of the Earth, we start with the gravitational potential at the surface of the Earth, which is given as \( V = -\frac{GM}{R} \), where:

- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( R \) is the radius of the Earth.

### Step-by-Step Solution:

1. **Gravitational Potential at the Surface**:
The gravitational potential \( V \) at the surface of the Earth is given by:
\[ V = -\frac{GM}{R} \]

2. **Gravitational Potential Inside the Earth**:
Inside the Earth, the gravitational potential \( V \) varies with distance \( r \) from the center. The formula for gravitational potential inside a uniform sphere (like Earth) is:
\[ V = -\frac{GM}{2R} \left(3 - \frac{r^2}{R^2}\right) \]
where \( r \) is the distance from the center of the Earth.

3. **Substituting \( r = 0 \)**:
To find the gravitational potential at the center of the Earth, we substitute \( r = 0 \) into the equation:
\[ V = -\frac{GM}{2R} \left(3 - \frac{0^2}{R^2}\right) = -\frac{GM}{2R} \cdot 3 = -\frac{3GM}{2R} \]

4. **Relating to Surface Potential**:
We know that the potential at the surface \( V \) is \( -\frac{GM}{R} \). We can express the potential at the center in terms of \( V \):
\[ V_{\text{center}} = -\frac{3GM}{2R} = 3 \left(-\frac{GM}{2R}\right) = \frac{3}{2} V \]

5. **Conclusion**:
Thus, the gravitational potential at the center of the Earth is:
\[ V_{\text{center}} = \frac{3}{2} V \]

### Final Answer:
The value of the gravitational potential at the center of the Earth is \( \frac{3}{2} V \).

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