If four different masses `m_1, m_2,m_3` and `m_4` are placed at the four corners of a square of side `a`, the resultant gravitational force on a mass `m` kept at the centre is

Answer

To find the resultant gravitational force on a mass \( m \) placed at the center of a square with four different masses \( m_1, m_2, m_3, \) and \( m_4 \) located at the corners, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Setup**:
- We have a square with side length \( a \).
- The masses \( m_1, m_2, m_3, \) and \( m_4 \) are placed at the corners of the square.
- 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 can be calculated using the Pythagorean theorem.
- Since the center divides the square into two equal halves, the distance from the center to a corner is given by:
\[ r = \frac{a}{\sqrt{2}} \]

3. **Calculate the Gravitational Force from Each Mass**:
- The gravitational force \( F \) exerted by a mass \( m_i \) on mass \( m \) is given by Newton's law of gravitation:
\[ F_i = \frac{G m_i m}{r^2} \]
- Substituting \( r = \frac{a}{\sqrt{2}} \):
\[ F_i = \frac{G m_i m}{\left(\frac{a}{\sqrt{2}}\right)^2} = \frac{G m_i m \cdot 2}{a^2} \]

4. **Determine the Direction of Each Force**:
- The forces due to \( m_1 \) and \( m_3 \) will act along one diagonal of the square, while the forces due to \( m_2 \) and \( m_4 \) will act along the other diagonal.
- The force \( F_1 \) from \( m_1 \) will be directed towards the center, and similarly for \( m_3 \). The same applies for \( m_2 \) and \( m_4 \).

5. **Calculate the Resultant Forces Along Each Diagonal**:
- The resultant force along one diagonal (let's say from \( m_1 \) and \( m_3 \)):
\[ F_{13} = F_1 - F_3 = \frac{G m m_1}{\left(\frac{a}{\sqrt{2}}\right)^2} - \frac{G m m_3}{\left(\frac{a}{\sqrt{2}}\right)^2} = \frac{G m}{\frac{a^2}{2}} (m_1 - m_3) \]
- The resultant force along the other diagonal (from \( m_2 \) and \( m_4 \)):
\[ F_{24} = F_2 - F_4 = \frac{G m m_2}{\left(\frac{a}{\sqrt{2}}\right)^2} - \frac{G m m_4}{\left(\frac{a}{\sqrt{2}}\right)^2} = \frac{G m}{\frac{a^2}{2}} (m_2 - m_4) \]

6. **Combine the Resultant Forces**:
- Since \( F_{13} \) and \( F_{24} \) are perpendicular to each other, we can find the magnitude of the resultant gravitational force \( F_R \) using the Pythagorean theorem:
\[ F_R = \sqrt{F_{13}^2 + F_{24}^2} \]
- Substituting the expressions for \( F_{13} \) and \( F_{24} \):
\[ F_R = \sqrt{\left(\frac{G m}{\frac{a^2}{2}} (m_1 - m_3)\right)^2 + \left(\frac{G m}{\frac{a^2}{2}} (m_2 - m_4)\right)^2} \]
- Simplifying this gives:
\[ F_R = \frac{G m}{\frac{a^2}{2}} \sqrt{(m_1 - m_3)^2 + (m_2 - m_4)^2} \]

### Final Result:
The resultant gravitational force on mass \( m \) at the center of the square is:
\[ F_R = \frac{2G m}{a^2} \sqrt{(m_1 - m_3)^2 + (m_2 - m_4)^2} \]