Three spherical balls of masses 1 kg, 2kg and 3 kg are placed at the corners of an equilateral triangle of side 1 m. The magnitude of the gravitational force exerted by 2 kg and 3 kg masses on 1 kg mass.

Options

A. 19 G
B. `sqrt(17)G`
C. `sqrt(19)G`
D. `G/sqrt(17)`

Answer

To solve the problem of finding the magnitude of the gravitational force exerted by the 2 kg and 3 kg masses on the 1 kg mass placed at the corners of an equilateral triangle, we can follow these steps:

### Step 1: Identify the Forces
We need to calculate the gravitational forces exerted on the 1 kg mass (let's call it mass A) by the 2 kg mass (mass B) and the 3 kg mass (mass C).

### Step 2: Calculate the Force Exerted by Mass B on Mass A
Using Newton's law of gravitation, the gravitational force \( F_{AB} \) between mass A (1 kg) and mass B (2 kg) is given by:

\[ F_{AB} = \frac{G \cdot m_A \cdot m_B}{r^2} \]

Where:
- \( G \) is the universal gravitational constant (\( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))
- \( m_A = 1 \, \text{kg} \)
- \( m_B = 2 \, \text{kg} \)
- \( r = 1 \, \text{m} \)

Substituting the values:

\[ F_{AB} = \frac{6.674 \times 10^{-11} \cdot 1 \cdot 2}{1^2} = 1.3348 \times 10^{-10} \, \text{N} \]

### Step 3: Calculate the Force Exerted by Mass C on Mass A
Similarly, the gravitational force \( F_{AC} \) between mass A (1 kg) and mass C (3 kg) is given by:

\[ F_{AC} = \frac{G \cdot m_A \cdot m_C}{r^2} \]

Where:
- \( m_C = 3 \, \text{kg} \)

Substituting the values:

\[ F_{AC} = \frac{6.674 \times 10^{-11} \cdot 1 \cdot 3}{1^2} = 2.0022 \times 10^{-10} \, \text{N} \]

### Step 4: Determine the Resultant Force
Since the forces \( F_{AB} \) and \( F_{AC} \) are not in the same direction (they form an angle of 60 degrees), we need to calculate the resultant force using the cosine rule.

The formula for the resultant force \( F_{net} \) when two forces are at an angle \( \theta \) is:

\[ F_{net} = \sqrt{F_{AB}^2 + F_{AC}^2 + 2 \cdot F_{AB} \cdot F_{AC} \cdot \cos(60^\circ)} \]

Substituting the values:

\[ F_{net} = \sqrt{(1.3348 \times 10^{-10})^2 + (2.0022 \times 10^{-10})^2 + 2 \cdot (1.3348 \times 10^{-10}) \cdot (2.0022 \times 10^{-10}) \cdot \frac{1}{2}} \]

Calculating each term:

1. \( (1.3348 \times 10^{-10})^2 = 1.7825 \times 10^{-20} \)
2. \( (2.0022 \times 10^{-10})^2 = 4.0088 \times 10^{-20} \)
3. \( 2 \cdot (1.3348 \times 10^{-10}) \cdot (2.0022 \times 10^{-10}) \cdot \frac{1}{2} = 2.6740 \times 10^{-20} \)

Now, substituting these values back into the equation:

\[ F_{net} = \sqrt{1.7825 \times 10^{-20} + 4.0088 \times 10^{-20} + 2.6740 \times 10^{-20}} \]

\[ F_{net} = \sqrt{8.4653 \times 10^{-20}} \approx 2.91 \times 10^{-10} \, \text{N} \]

### Final Result
The magnitude of the gravitational force exerted by the 2 kg and 3 kg masses on the 1 kg mass is approximately \( 2.91 \times 10^{-10} \, \text{N} \).

---