Four particles having masses, m, wm, 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.

Explanation

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

### Step-by-Step Solution:

1. **Identify the Setup**:
We have a square with side length \( a \). The masses at the corners are \( m \), \( 2m \), \( 3m \), and \( 4m \). The mass \( m \) is placed at the center of the square.

2. **Calculate the Distance from the Center to the Corners**:
The distance \( R \) from the center of the square to any corner is given by:
\[ R = \frac{a}{\sqrt{2}} \]

3. **Calculate the Gravitational Force from Each Mass**:
The gravitational force \( F \) between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by:
\[ F = \frac{G m_1 m_2}{r^2} \]
For each corner mass, we can calculate the force acting on the mass \( m \) at the center.

- **Force due to mass \( m \) at corner**:
\[ F_1 = \frac{G m \cdot m}{R^2} = \frac{G m^2}{\left(\frac{a}{\sqrt{2}}\right)^2} = \frac{2G m^2}{a^2} \]

- **Force due to mass \( 2m \) at corner**:
\[ F_2 = \frac{G (2m) \cdot m}{R^2} = \frac{G \cdot 2m^2}{\left(\frac{a}{\sqrt{2}}\right)^2} = \frac{4G m^2}{a^2} \]

- **Force due to mass \( 3m \) at corner**:
\[ F_3 = \frac{G (3m) \cdot m}{R^2} = \frac{G \cdot 3m^2}{\left(\frac{a}{\sqrt{2}}\right)^2} = \frac{6G m^2}{a^2} \]

- **Force due to mass \( 4m \) at corner**:
\[ F_4 = \frac{G (4m) \cdot m}{R^2} = \frac{G \cdot 4m^2}{\left(\frac{a}{\sqrt{2}}\right)^2} = \frac{8G m^2}{a^2} \]

4. **Determine the Resultant Forces**:
The forces \( F_1 \) and \( F_3 \) act in opposite directions along one diagonal, and \( F_2 \) and \( F_4 \) act along the other diagonal. Since they are symmetrical, we can calculate the net force in one direction and then combine them.

- **Net force along one diagonal**:
\[ F_{\text{net, diagonal 1}} = F_4 - F_2 = \frac{8G m^2}{a^2} - \frac{4G m^2}{a^2} = \frac{4G m^2}{a^2} \]

- **Net force along the other diagonal**:
\[ F_{\text{net, diagonal 2}} = F_3 - F_1 = \frac{6G m^2}{a^2} - \frac{2G m^2}{a^2} = \frac{4G m^2}{a^2} \]

5. **Calculate the Resultant Force**:
Since both net forces are equal and act at right angles to each other, we can use the Pythagorean theorem to find the resultant force \( F_{\text{net}} \):
\[ F_{\text{net}} = \sqrt{(F_{\text{net, diagonal 1}})^2 + (F_{\text{net, diagonal 2}})^2} = \sqrt{\left(\frac{4G m^2}{a^2}\right)^2 + \left(\frac{4G m^2}{a^2}\right)^2} \]
\[ = \sqrt{2 \left(\frac{4G m^2}{a^2}\right)^2} = \frac{4G m^2}{a^2} \sqrt{2} \]

6. **Final Result**:
Thus, the gravitational force acting on the particle of mass \( m \) at the center is:
\[ F_{\text{net}} = \frac{4\sqrt{2} G m^2}{a^2} \]