The gravitational potential at the centre of a square of side a when four point masses m each are kept at its vertices will be

Options

A. `4sqrt(2)(Gm)/(a)`
B. `-4sqrt(2)(Gm)/(a)`
C. `2sqrt(2)(Gm)/(a)`
D. `-2sqrt(2)(Gm)/(a)`

Answer

To find the gravitational potential at the center of a square with side length \( a \) where four point masses \( m \) are placed at each vertex, we can follow these steps:

### Step-by-step Solution:

1. **Identify the Geometry**:
- We have a square with vertices labeled as \( A, B, C, D \). Each vertex has a mass \( m \). The center of the square is denoted as point \( O \).

2. **Calculate the Distance from the Center to a Vertex**:
- The distance from the center \( O \) to any vertex (say \( A \)) can be calculated using the Pythagorean theorem. Since the square has a side length \( a \), the distance \( AO \) is given by:
\[ AO = \frac{a}{\sqrt{2}} \]

3. **Gravitational Potential Due to One Mass**:
- The gravitational potential \( V \) due to a point mass \( m \) at a distance \( r \) is given by the formula:
\[ V = -\frac{Gm}{r} \]
- For one mass \( m \) located at a distance \( AO = \frac{a}{\sqrt{2}} \), the potential \( V_A \) at the center \( O \) is:
\[ V_A = -\frac{Gm}{\frac{a}{\sqrt{2}}} = -\frac{Gm \sqrt{2}}{a} \]

4. **Calculate the Total Gravitational Potential**:
- Since there are four identical masses \( m \) at each vertex, the total gravitational potential \( V_{\text{net}} \) at the center \( O \) is the sum of the potentials due to all four masses:
\[ V_{\text{net}} = V_A + V_B + V_C + V_D = 4 \left(-\frac{Gm \sqrt{2}}{a}\right) \]
- Thus, we can simplify this to:
\[ V_{\text{net}} = -\frac{4Gm \sqrt{2}}{a} \]

5. **Final Result**:
- Therefore, the gravitational potential at the center of the square is:
\[ V_{\text{net}} = -\frac{4Gm \sqrt{2}}{a} \]

### Conclusion:
The gravitational potential at the center of a square of side \( a \) with four point masses \( m \) at its vertices is:
\[ -\frac{4Gm \sqrt{2}}{a} \]