Two particles each of mass 'm' are placed at `A` and `C` are such `AB=BC=L`. The gravitational force on the third particle placed at `D` at a distance `L` on the perpendicular bisector of the line `AC` is

Options

A. `(Gm^(2))/(L^(2))` along `BD`
B. `(Gm^(2))/(sqrt(2)L^(2))` along `DB`
C. `(Gm^(2))/(L^(2))` along `AC`
D. `(Gm^(2))/(sqrt(2)L^(2))` along `BD`

Explanation

To solve the problem, let's break it down step by step.

### Step 1: Understand the Setup
We have two particles, each of mass \( m \), located at points \( A \) and \( C \) such that the distance \( AB = BC = L \). A third particle is placed at point \( D \) on the perpendicular bisector of line \( AC \), at a distance \( L \) from the midpoint \( B \).

### Step 2: Determine the Positions
Let's place the points in a coordinate system:
- Let \( A \) be at \( (-\frac{L}{2}, 0) \)
- Let \( C \) be at \( (\frac{L}{2}, 0) \)
- The midpoint \( B \) will be at \( (0, 0) \)
- The point \( D \) will be at \( (0, L) \)

### Step 3: Calculate the Distance from \( D \) to \( A \) and \( C \)
Using the distance formula, we can find the distances \( AD \) and \( CD \):
- Distance \( AD = \sqrt{ \left( 0 - (-\frac{L}{2}) \right)^2 + (L - 0)^2 } = \sqrt{ \left( \frac{L}{2} \right)^2 + L^2 } = \sqrt{ \frac{L^2}{4} + L^2 } = \sqrt{ \frac{5L^2}{4} } = \frac{L\sqrt{5}}{2} \)
- Distance \( CD = \sqrt{ \left( 0 - \frac{L}{2} \right)^2 + (L - 0)^2 } = \sqrt{ \left( -\frac{L}{2} \right)^2 + L^2 } = \frac{L\sqrt{5}}{2} \)

### Step 4: Calculate the Gravitational Force on \( D \)
The gravitational force \( F \) exerted by each mass \( m \) at points \( A \) and \( C \) on mass \( m \) at point \( D \) is given by Newton's law of gravitation:
\[ F = \frac{G m^2}{r^2} \]
where \( r \) is the distance from \( D \) to \( A \) or \( C \). Thus, the force due to \( A \) and \( C \) is:
\[ F_{AD} = F_{CD} = \frac{G m^2}{\left( \frac{L\sqrt{5}}{2} \right)^2} = \frac{G m^2}{\frac{5L^2}{4}} = \frac{4G m^2}{5L^2} \]

### Step 5: Resolve the Forces into Components
Since the forces from \( A \) and \( C \) are symmetrical about the y-axis, we can resolve them into x and y components:
- The angle \( \theta \) between the line \( AD \) (or \( CD \)) and the vertical line is given by:
\[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{\frac{L}{2}}{L} = \frac{1}{2} \]
Thus, \( \theta = \tan^{-1}(0.5) \).

The components of the forces can be calculated as:
- The vertical component (y-direction) of each force:
\[ F_{y} = F \cos \theta = F \cdot \frac{L}{\sqrt{L^2 + \left(\frac{L}{2}\right)^2}} = F \cdot \frac{2}{\sqrt{5}} \]
- The horizontal component (x-direction) cancels out because they are equal and opposite.

### Step 6: Calculate the Net Force
The net force in the y-direction is:
\[ F_{net} = 2 F_y = 2 \cdot \frac{4G m^2}{5L^2} \cdot \frac{2}{\sqrt{5}} = \frac{16G m^2}{5L^2 \sqrt{5}} \]

### Final Result
The gravitational force on the third particle placed at point \( D \) is:
\[ F_{net} = \frac{16G m^2}{5L^2 \sqrt{5}} \]