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.

Explanation

To solve the problem of finding the gravitational force acting on a particle of mass \( m \) placed at the center of a square with four 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 corners labeled as follows:
- Mass \( m \) at corner A
- Mass \( 2m \) at corner B
- Mass \( 3m \) at corner C
- Mass \( 4m \) at corner D

The side length of the square is \( a \).

### Step 2: Determine the Distance from the Center
The distance from the center of the square to any corner (say point A) can be calculated using the Pythagorean theorem. The distance \( OA \) is given by:
\[ OA = \frac{a}{\sqrt{2}} \]
This is true for all corners since they are equidistant from the center.

### Step 3: Calculate the Gravitational Forces
The gravitational force \( F \) between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by:
\[ F = \frac{G m_1 m_2}{r^2} \]
We will calculate the gravitational force acting on the mass \( m \) at the center due to each of the four corner masses.

1. **Force due to mass \( m \) at A**:
\[ F_{OA} = \frac{G m \cdot m}{\left(\frac{a}{\sqrt{2}}\right)^2} = \frac{G m^2}{\frac{a^2}{2}} = \frac{2G m^2}{a^2} \]

2. **Force due to mass \( 2m \) at B**:
\[ F_{OB} = \frac{G (2m) \cdot m}{\left(\frac{a}{\sqrt{2}}\right)^2} = \frac{G (2m) m}{\frac{a^2}{2}} = \frac{4G m^2}{a^2} \]

3. **Force due to mass \( 3m \) at C**:
\[ F_{OC} = \frac{G (3m) \cdot m}{\left(\frac{a}{\sqrt{2}}\right)^2} = \frac{G (3m) m}{\frac{a^2}{2}} = \frac{6G m^2}{a^2} \]

4. **Force due to mass \( 4m \) at D**:
\[ F_{OD} = \frac{G (4m) \cdot m}{\left(\frac{a}{\sqrt{2}}\right)^2} = \frac{G (4m) m}{\frac{a^2}{2}} = \frac{8G m^2}{a^2} \]

### Step 4: Determine the Resultant Forces
Next, we need to find the resultant forces along the diagonals of the square.

- Forces \( F_{OA} \) and \( F_{OC} \) act along diagonal AC.
- Forces \( F_{OB} \) and \( F_{OD} \) act along diagonal BD.

1. **Resultant force along AC**:
\[ F_{AC} = F_{OC} - F_{OA} = \frac{6G m^2}{a^2} - \frac{2G m^2}{a^2} = \frac{4G m^2}{a^2} \]

2. **Resultant force along BD**:
\[ F_{BD} = F_{OD} - F_{OB} = \frac{8G m^2}{a^2} - \frac{4G m^2}{a^2} = \frac{4G m^2}{a^2} \]

### Step 5: Calculate the Magnitude of the Net Force
Since \( F_{AC} \) and \( F_{BD} \) are equal and perpendicular to each other, we can use the Pythagorean theorem to find the resultant force \( F \):
\[ F = \sqrt{F_{AC}^2 + F_{BD}^2} = \sqrt{\left(\frac{4G m^2}{a^2}\right)^2 + \left(\frac{4G m^2}{a^2}\right)^2} \]
\[ = \sqrt{2 \left(\frac{4G m^2}{a^2}\right)^2} = \frac{4G m^2}{a^2} \sqrt{2} = \frac{4\sqrt{2} G m^2}{a^2} \]

### Step 6: Determine the Direction
The direction of the net force will be at a 45-degree angle downward from the center towards the line connecting the corners.

### Final Answer
The gravitational force acting on the mass \( m \) at the center of the square is:
\[ F = \frac{4\sqrt{2} G m^2}{a^2} \]
and it acts at a 45-degree angle towards the corner masses.