Three equal masses of 1 kg each are placed at the vertices of an equilateral triangle of side 1 m, then the gravitational force on one of the masses due to other masses is (approx.)

Options

A. `{6.67(10^-11)}`N
B. `10 ^-3` N
C. `[11.5(10^-11)] N`
D. `[11.5(10^-3)]N`

Answer

To find the gravitational force on one of the masses due to the other two masses placed at the vertices of an equilateral triangle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Masses and Distance**:
- We have three equal masses \( m_1 = m_2 = m_3 = 1 \, \text{kg} \).
- The distance between any two masses (the side of the triangle) is \( d = 1 \, \text{m} \).

2. **Use Newton's Law of Gravitation**:
- The gravitational force \( F \) between two masses is given by:
\[ F = \frac{G m_1 m_2}{d^2} \]
- Here, \( G \) (the universal gravitational constant) is approximately \( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \).

3. **Calculate the Gravitational Force Between Two Masses**:
- Substituting the values into the formula:
\[ F = \frac{6.67 \times 10^{-11} \times 1 \times 1}{1^2} = 6.67 \times 10^{-11} \, \text{N} \]

4. **Determine the Resultant Force on One Mass**:
- We will calculate the resultant force on one mass (say \( m_1 \)) due to the other two masses (\( m_2 \) and \( m_3 \)).
- The forces \( F_{12} \) (from \( m_2 \) to \( m_1 \)) and \( F_{13} \) (from \( m_3 \) to \( m_1 \)) are equal and have the same magnitude \( F \).

5. **Calculate the Angle Between the Forces**:
- In an equilateral triangle, the angle between the lines of action of the forces \( F_{12} \) and \( F_{13} \) is \( 60^\circ \).

6. **Use the Formula for Resultant of Two Vectors**:
- The magnitude of the resultant force \( F_R \) can be calculated using the formula:
\[ F_R = \sqrt{F_{12}^2 + F_{13}^2 + 2 F_{12} F_{13} \cos \theta} \]
- Since \( F_{12} = F_{13} = F \):
\[ F_R = \sqrt{F^2 + F^2 + 2 F^2 \cos 60^\circ} \]
- Knowing that \( \cos 60^\circ = \frac{1}{2} \):
\[ F_R = \sqrt{F^2 + F^2 + 2 F^2 \cdot \frac{1}{2}} = \sqrt{F^2 + F^2 + F^2} = \sqrt{3F^2} = F\sqrt{3} \]

7. **Substituting the Value of \( F \)**:
- Now substituting \( F = 6.67 \times 10^{-11} \):
\[ F_R = 6.67 \times 10^{-11} \sqrt{3} \approx 6.67 \times 10^{-11} \times 1.732 \approx 1.1547 \times 10^{-10} \, \text{N} \]

### Final Answer:
The gravitational force on one of the masses due to the other two masses is approximately \( 1.15 \times 10^{-10} \, \text{N} \).

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