The gravitational potential at centre of earth is

Options

A. `-(2GM)/R`
B. `-(3GM)/(2R)`
C. `(3GM)/(2R)`
D. `-(GM)/(2R)`

Answer

To find the gravitational potential at the center of the Earth, we can follow these steps:

### Step-by-Step Solution:

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

2. **Consider a Thin Spherical Shell**:
Imagine a thin spherical shell of thickness \( dx \) at a distance \( x \) from the center of the Earth. The mass \( dm \) of this shell can be expressed in terms of its density \( \rho \) and volume.

3. **Mass of the Shell**:
The volume \( dV \) of the thin shell is given by:
\[ dV = 4 \pi x^2 dx \]
Therefore, the mass \( dm \) of the shell is:
\[ dm = \rho \cdot dV = \rho \cdot 4 \pi x^2 dx \]

4. **Gravitational Potential Contribution from the Shell**:
The potential \( dV \) due to this thin shell at the center of the Earth is:
\[ dV = -\frac{G \, dm}{x} = -\frac{G \, \rho \cdot 4 \pi x^2 dx}{x} \]
Simplifying this gives:
\[ dV = -G \rho \cdot 4 \pi x \, dx \]

5. **Integrate to Find Total Potential**:
To find the total gravitational potential \( V \) at the center of the Earth, integrate \( dV \) from \( x = 0 \) to \( x = R \) (where \( R \) is the radius of the Earth):
\[ V = \int_0^R -G \rho \cdot 4 \pi x \, dx \]
This integrates to:
\[ V = -G \rho \cdot 4 \pi \left[\frac{x^2}{2}\right]_0^R = -G \rho \cdot 4 \pi \cdot \frac{R^2}{2} \]

6. **Expressing in Terms of Mass**:
The mass \( M \) of the Earth can be expressed as:
\[ M = \rho \cdot \frac{4}{3} \pi R^3 \]
Substituting \( \rho \) in the potential equation gives:
\[ V = -\frac{3}{2} \frac{G M}{R} \]

7. **Final Result**:
At the center of the Earth, the gravitational potential is:
\[ V = -\frac{3}{2} \frac{G M}{R} \]

### Conclusion:
Thus, the gravitational potential at the center of the Earth is \( -\frac{3}{2} \frac{G M}{R} \).

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