Two masses, `800 kg` and `600 kg` are at a distance `0.25 m` apart. The magnitude of total force experienced by a body of mass `1 kg` placed at a point distance `0.2 m` from the `800 kg` mass `0.15 m` from the `600 kg` mass :

Options

A. `3.4 xx 10^(-6) N`
B. `2.22 xx 10^(-6) N`
C. `3.22 xx 10^(-6) N`
D. `2.22 xx 10^(-8) N`

Explanation

To solve the problem, we need to calculate the gravitational force exerted on the 1 kg mass by both the 800 kg and 600 kg masses using Newton's law of gravitation. The formula for gravitational force is given by:

\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]

where:
- \( F \) is the gravitational force,
- \( G \) is the gravitational constant \( (6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2) \),
- \( m_1 \) and \( m_2 \) are the masses,
- \( r \) is the distance between the centers of the two masses.

### Step 1: Calculate the force between the 800 kg mass and the 1 kg mass.

Given:
- \( m_1 = 800 \, \text{kg} \)
- \( m_2 = 1 \, \text{kg} \)
- \( r = 0.2 \, \text{m} \)

Using the formula:

\[ F_{800} = \frac{G \cdot 800 \cdot 1}{(0.2)^2} \]

Calculating \( F_{800} \):

\[ F_{800} = \frac{6.674 \times 10^{-11} \cdot 800 \cdot 1}{0.04} = \frac{6.674 \times 10^{-11} \cdot 800}{0.04} \]

\[ F_{800} = \frac{5.3392 \times 10^{-8}}{0.04} = 1.3348 \times 10^{-6} \, \text{N} \]

### Step 2: Calculate the force between the 600 kg mass and the 1 kg mass.

Given:
- \( m_1 = 600 \, \text{kg} \)
- \( m_2 = 1 \, \text{kg} \)
- \( r = 0.15 \, \text{m} \)

Using the formula:

\[ F_{600} = \frac{G \cdot 600 \cdot 1}{(0.15)^2} \]

Calculating \( F_{600} \):

\[ F_{600} = \frac{6.674 \times 10^{-11} \cdot 600 \cdot 1}{0.0225} = \frac{6.674 \times 10^{-11} \cdot 600}{0.0225} \]

\[ F_{600} = \frac{4.0044 \times 10^{-8}}{0.0225} = 1.7787 \times 10^{-6} \, \text{N} \]

### Step 3: Calculate the total force experienced by the 1 kg mass.

The total gravitational force \( F_{total} \) is the vector sum of \( F_{800} \) and \( F_{600} \). Since both forces act in the same direction (towards their respective masses), we can simply add them:

\[ F_{total} = F_{800} + F_{600} \]

\[ F_{total} = 1.3348 \times 10^{-6} + 1.7787 \times 10^{-6} = 3.1135 \times 10^{-6} \, \text{N} \]

### Final Answer:
The magnitude of the total force experienced by the 1 kg mass is approximately \( 3.1135 \times 10^{-6} \, \text{N} \).

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