Four particles having masses, m, 2m, 3m, and 4m are placed at the four corners of a square of edge a. Find the gravitational force acting on a particle of mass m placed at the centre.

Options

A. `(24m^2G)/a^2`
B. `(6m^2G)/a^2`
C. `(4sqrt2Gm^2)/a^2`
D. zero

Explanation

To find the gravitational force acting on a particle of mass \( m \) placed at the center of a square with particles of masses \( m, 2m, 3m, \) and \( 4m \) at its corners, we can follow these steps:

### Step 1: Identify the Configuration
We have a square with vertices labeled as \( A, B, C, D \) and a mass \( m \) at the center \( O \). The masses at the corners are:
- Mass at \( A \): \( m \)
- Mass at \( B \): \( 2m \)
- Mass at \( C \): \( 3m \)
- Mass at \( D \): \( 4m \)

### Step 2: Calculate the Distance from the Center to the Corners
The distance \( R \) from the center \( O \) to any corner of the square can be calculated using the Pythagorean theorem. Since the length of each side of the square is \( a \), the distance is given by:
\[ R = \frac{a}{\sqrt{2}} \]

### Step 3: Calculate the Gravitational Force from Each Mass
The gravitational force \( F \) exerted by a mass \( M \) on a mass \( m \) at distance \( R \) is given by:
\[ F = \frac{G \cdot M \cdot m}{R^2} \]
Substituting \( R = \frac{a}{\sqrt{2}} \):
\[ R^2 = \left(\frac{a}{\sqrt{2}}\right)^2 = \frac{a^2}{2} \]
Thus, the force from each mass becomes:
\[ F = \frac{G \cdot M \cdot m}{\frac{a^2}{2}} = \frac{2G \cdot M \cdot m}{a^2} \]

### Step 4: Calculate the Forces from Each Corner Mass
1. **Force from mass \( m \) at \( A \)**:
\[ F_1 = \frac{2G \cdot m \cdot m}{a^2} = \frac{2Gm^2}{a^2} \]

2. **Force from mass \( 2m \) at \( B \)**:
\[ F_2 = \frac{2G \cdot 2m \cdot m}{a^2} = \frac{4Gm^2}{a^2} \]

3. **Force from mass \( 3m \) at \( C \)**:
\[ F_3 = \frac{2G \cdot 3m \cdot m}{a^2} = \frac{6Gm^2}{a^2} \]

4. **Force from mass \( 4m \) at \( D \)**:
\[ F_4 = \frac{2G \cdot 4m \cdot m}{a^2} = \frac{8Gm^2}{a^2} \]

### Step 5: Resolve Forces into Components
The forces will have both x and y components. The forces \( F_1 \) and \( F_3 \) act diagonally towards the center, while \( F_2 \) and \( F_4 \) act similarly but with different magnitudes.

- **Forces along the x-axis**:
\[ F_{x1} = F_1 \cdot \frac{1}{\sqrt{2}}, \quad F_{x2} = F_2 \cdot \frac{1}{\sqrt{2}}, \quad F_{x3} = -F_3 \cdot \frac{1}{\sqrt{2}}, \quad F_{x4} = -F_4 \cdot \frac{1}{\sqrt{2}} \]

- **Forces along the y-axis**:
\[ F_{y1} = F_1 \cdot \frac{1}{\sqrt{2}}, \quad F_{y2} = -F_2 \cdot \frac{1}{\sqrt{2}}, \quad F_{y3} = -F_3 \cdot \frac{1}{\sqrt{2}}, \quad F_{y4} = F_4 \cdot \frac{1}{\sqrt{2}} \]

### Step 6: Calculate Net Forces in Each Direction
1. **Net force in the x-direction**:
\[ F_{net,x} = \left(\frac{2Gm^2}{a^2} + \frac{4Gm^2}{a^2}\right) \cdot \frac{1}{\sqrt{2}} - \left(\frac{6Gm^2}{a^2} + \frac{8Gm^2}{a^2}\right) \cdot \frac{1}{\sqrt{2}} \]
Simplifying gives:
\[ F_{net,x} = \frac{6Gm^2}{a^2\sqrt{2}} - \frac{14Gm^2}{a^2\sqrt{2}} = -\frac{8Gm^2}{a^2\sqrt{2}} \]

2. **Net force in the y-direction**:
\[ F_{net,y} = \left(\frac{2Gm^2}{a^2} - \frac{4Gm^2}{a^2}\right) \cdot \frac{1}{\sqrt{2}} - \left(\frac{6Gm^2}{a^2} - \frac{8Gm^2}{a^2}\right) \cdot \frac{1}{\sqrt{2}} \]
Simplifying gives:
\[ F_{net,y} = -\frac{2Gm^2}{a^2\sqrt{2}} - \left(-\frac{2Gm^2}{a^2}\right) \cdot \frac{1}{\sqrt{2}} = 0 \]

### Step 7: Calculate the Resultant Force
The resultant force can be calculated using the Pythagorean theorem:
\[ F_{net} = \sqrt{(F_{net,x})^2 + (F_{net,y})^2} = \sqrt{\left(-\frac{8Gm^2}{a^2\sqrt{2}}\right)^2 + 0^2} = \frac{8Gm^2}{a^2\sqrt{2}} \]

### Final Answer
Thus, the net gravitational force acting on the mass \( m \) at the center of the square is:
\[ F_{net} = \frac{8Gm^2}{a^2\sqrt{2}} \]