Three points A , B and C each of mass m is placed in a line with AB=BC=d. Find the gravitational force on a fourth particle P of the same mass placed at a distance d from particle B on the perpendicular bisector of the line AC.

Options

A. `((sqrt(2) + 1)/(sqrt(2))) (Gm^(2))/(d^(2))`
B. `((sqrt(3) + 1)/(sqrt(2))) (Gm^(2))/(d^(2))`
C. `((sqrt(2) + 1)/(sqrt(3))) (Gm^(2))/(d^(2))`
D. `((sqrt(2))/(sqrt(5))) (Gm^(2))/(d^(2))`

Explanation

To solve the problem, we need to find the gravitational force acting on a particle P due to three other particles A, B, and C, each of mass m, placed in a line with distances AB = BC = d. The particle P is located at a distance d from B on the perpendicular bisector of the line AC.

### Step-by-Step Solution:

1. **Diagram Setup**:
- Place points A, B, and C on a horizontal line. Let A be at (-d, 0), B at (0, 0), and C at (d, 0).
- Point P is located at (0, d) on the perpendicular bisector of line AC.

2. **Calculate Distances**:
- The distance from P to A (denoted as PA) can be calculated using the distance formula:
\[ PA = \sqrt{(0 - (-d))^2 + (d - 0)^2} = \sqrt{d^2 + d^2} = d\sqrt{2} \]
- Similarly, the distance from P to C (denoted as PC) is also:
\[ PC = \sqrt{(0 - d)^2 + (d - 0)^2} = \sqrt{d^2 + d^2} = d\sqrt{2} \]
- The distance from P to B (denoted as PB) is simply:
\[ PB = d \]

3. **Calculate Gravitational Forces**:
- The gravitational force between two masses is given by:
\[ F = \frac{G m_1 m_2}{r^2} \]
- The force exerted on P by A (FA):
\[ F_A = \frac{G m^2}{(d\sqrt{2})^2} = \frac{G m^2}{2d^2} \]
- The force exerted on P by B (FB):
\[ F_B = \frac{G m^2}{d^2} \]
- The force exerted on P by C (FC):
\[ F_C = \frac{G m^2}{(d\sqrt{2})^2} = \frac{G m^2}{2d^2} \]

4. **Resolve Forces into Components**:
- The forces FA and FC will have components along the vertical and horizontal directions. The angles with respect to the vertical are 45 degrees (since P is directly above the midpoint of AC).
- The vertical components of FA and FC are:
\[ F_{A_y} = F_A \cdot \cos(45^\circ) = \frac{G m^2}{2d^2} \cdot \frac{1}{\sqrt{2}} = \frac{G m^2}{2\sqrt{2} d^2} \]
\[ F_{C_y} = F_C \cdot \cos(45^\circ) = \frac{G m^2}{2d^2} \cdot \frac{1}{\sqrt{2}} = \frac{G m^2}{2\sqrt{2} d^2} \]
- The total vertical force from A and C:
\[ F_{A_y} + F_{C_y} = \frac{G m^2}{2\sqrt{2} d^2} + \frac{G m^2}{2\sqrt{2} d^2} = \frac{G m^2}{\sqrt{2} d^2} \]

5. **Total Gravitational Force**:
- The total gravitational force acting on P is the sum of the vertical components from A and C plus the force from B:
\[ F_{\text{net}} = F_B + F_{A_y} + F_{C_y} = \frac{G m^2}{d^2} + \frac{G m^2}{\sqrt{2} d^2} \]
- Factor out \( \frac{G m^2}{d^2} \):
\[ F_{\text{net}} = \frac{G m^2}{d^2} \left( 1 + \frac{1}{\sqrt{2}} \right) \]

### Final Answer:
The gravitational force on particle P is:
\[ F_{\text{net}} = \frac{G m^2}{d^2} \left( 1 + \frac{1}{\sqrt{2}} \right) \]