Three masses, each equal to M, are placed at the three corners of a square of side a. the force of attraction on unit mass at the fourth corner will be

Options

A. `(GM)/(3a^(2))`
B. `(GM)/(a^(2)) sqrt(3)`
C. `(3GM)/(a^(2))`
D. `(GM)/(a^(2))[1/2+sqrt(2)]`

Explanation

To solve the problem of finding the force of attraction on a unit mass placed at the fourth corner of a square with three masses \( M \) at the other corners, we can follow these steps:

### Step-by-Step Solution

1. **Identify the Setup**:
We have three masses \( M \) located at corners A, B, and C of a square with side length \( a \). The fourth corner D has a unit mass \( m = 1 \).

2. **Calculate Forces from Each Mass**:
- **Force from Mass at A (F_AD)**:
The distance between mass A and the unit mass at D is \( a \). Using the formula for gravitational force:
\[ F_{AD} = \frac{M \cdot 1 \cdot G}{a^2} = \frac{MG}{a^2} \]
- **Force from Mass at B (F_BD)**:
The distance between mass B and the unit mass at D is also \( a \):
\[ F_{BD} = \frac{M \cdot 1 \cdot G}{a^2} = \frac{MG}{a^2} \]
- **Force from Mass at C (F_CD)**:
The distance between mass C and the unit mass at D is the diagonal of the square, which is \( \sqrt{2}a \):
\[ F_{CD} = \frac{M \cdot 1 \cdot G}{(\sqrt{2}a)^2} = \frac{MG}{2a^2} \]

3. **Determine the Direction of Forces**:
- The forces \( F_{AD} \) and \( F_{BD} \) act along the sides of the square (horizontal and vertical).
- The force \( F_{CD} \) acts along the diagonal of the square.

4. **Resolve Forces into Components**:
- The forces \( F_{AD} \) and \( F_{BD} \) are perpendicular to each other. Their resultant can be found using the Pythagorean theorem:
\[ F_{resultant} = \sqrt{F_{AD}^2 + F_{BD}^2} = \sqrt{\left(\frac{MG}{a^2}\right)^2 + \left(\frac{MG}{a^2}\right)^2} = \sqrt{2} \cdot \frac{MG}{a^2} = \frac{\sqrt{2}MG}{a^2} \]

5. **Combine with the Force from Mass C**:
- The resultant force from A and B acts at a \( 45^\circ \) angle to the axes, and we need to add \( F_{CD} \) which acts at \( 45^\circ \) as well:
- Therefore, the total force \( F_{total} \) is:
\[ F_{total} = F_{resultant} + F_{CD} = \frac{\sqrt{2}MG}{a^2} + \frac{MG}{2a^2} \]

6. **Factor Out Common Terms**:
- Factor out \( \frac{MG}{a^2} \):
\[ F_{total} = \frac{MG}{a^2} \left( \sqrt{2} + \frac{1}{2} \right) \]

7. **Final Expression**:
- Thus, the force of attraction on the unit mass at the fourth corner is:
\[ F_{total} = \frac{MG}{a^2} \left( \sqrt{2} + \frac{1}{2} \right) \]