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 center.

Explanation

To find 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 positions and distances
The four particles are located at the corners of a square of edge length \( a \). The center of the square is at a distance of \( \frac{a}{\sqrt{2}} \) from each corner (using the Pythagorean theorem).

### Step 2: Calculate the gravitational force from each corner mass
The gravitational force \( F \) between two masses is given by the formula:
\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]
where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the distance between them.

1. **From mass \( 4m \)** at corner A:
\[ F_{A} = \frac{G \cdot m \cdot 4m}{\left(\frac{a}{\sqrt{2}}\right)^2} = \frac{4Gm^2}{\frac{a^2}{2}} = \frac{8Gm^2}{a^2} \]

2. **From mass \( 3m \)** at corner B:
\[ F_{B} = \frac{G \cdot m \cdot 3m}{\left(\frac{a}{\sqrt{2}}\right)^2} = \frac{3Gm^2}{\frac{a^2}{2}} = \frac{6Gm^2}{a^2} \]

3. **From mass \( 2m \)** at corner C:
\[ F_{C} = \frac{G \cdot m \cdot 2m}{\left(\frac{a}{\sqrt{2}}\right)^2} = \frac{2Gm^2}{\frac{a^2}{2}} = \frac{4Gm^2}{a^2} \]

4. **From mass \( m \)** at corner D:
\[ F_{D} = \frac{G \cdot m \cdot m}{\left(\frac{a}{\sqrt{2}}\right)^2} = \frac{Gm^2}{\frac{a^2}{2}} = \frac{2Gm^2}{a^2} \]

### Step 3: Determine the direction of the forces
Each force acts along the line connecting the center of the square to the respective corner. The forces from opposite corners will have components that can be resolved into x and y directions.

### Step 4: Calculate the net force components
Since the square is symmetric, we can sum the x and y components separately.

1. **X-components**:
- \( F_{A} \) and \( F_{C} \) will have components in the positive x-direction.
- \( F_{B} \) and \( F_{D} \) will have components in the negative x-direction.

2. **Y-components**:
- \( F_{A} \) and \( F_{B} \) will have components in the positive y-direction.
- \( F_{C} \) and \( F_{D} \) will have components in the negative y-direction.

### Step 5: Calculate the resultant force
The total gravitational force acting on the mass \( m \) at the center is the vector sum of all the forces. Since the forces from opposite corners cancel each other out due to symmetry, we can calculate the resultant force as:
\[ F_{\text{net}} = F_{A} + F_{B} + F_{C} + F_{D} \]
Substituting the values:
\[ F_{\text{net}} = \left(\frac{8Gm^2}{a^2} + \frac{6Gm^2}{a^2} + \frac{4Gm^2}{a^2} + \frac{2Gm^2}{a^2}\right) = \frac{20Gm^2}{a^2} \]

### Final Result
The total gravitational force acting on the particle of mass \( m \) at the center of the square is:
\[ F_{\text{net}} = \frac{20Gm^2}{a^2} \]