Three identical point mass each of mass 1kg lie in the x-y plane at point (0,0), (0,0.2m) and (0.2m, 0). The net gravitational force on the mass at the origin is

Options

A. `1.67xx10^(-11)(hati+hatj)N`
B. `3.34xx10^(-10)(hati+hatj)N`
C. `1.67xx10^(-9)(hati+hatj)N`
D. `3.34xx10^(-10)(hati-hatj)N`

Explanation

To solve the problem of finding the net gravitational force on the mass at the origin due to the other two masses, we can follow these steps:

### Step 1: Identify the positions of the masses
- Mass 1 (m1 = 1 kg) is located at the origin (0, 0).
- Mass 2 (m2 = 1 kg) is located at (0, 0.2 m).
- Mass 3 (m3 = 1 kg) is located at (0.2 m, 0).

### Step 2: Calculate the gravitational forces acting on the mass at the origin
The gravitational force between two point masses is given by the formula:
\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]
where:
- \( G \) is the gravitational constant \( \approx 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \),
- \( m_1 \) and \( m_2 \) are the masses,
- \( r \) is the distance between the masses.

### Step 3: Calculate the force due to mass 2
- The distance \( r_{12} \) between mass 1 and mass 2 is 0.2 m.
- Using the formula:
\[ F_{12} = \frac{G \cdot m_1 \cdot m_2}{r_{12}^2} = \frac{6.67 \times 10^{-11} \cdot 1 \cdot 1}{(0.2)^2} \]
\[ F_{12} = \frac{6.67 \times 10^{-11}}{0.04} = 1.6675 \times 10^{-9} \, \text{N} \]
- This force acts downward (negative y-direction).

### Step 4: Calculate the force due to mass 3
- The distance \( r_{13} \) between mass 1 and mass 3 is also 0.2 m.
- Using the formula:
\[ F_{13} = \frac{G \cdot m_1 \cdot m_3}{r_{13}^2} = \frac{6.67 \times 10^{-11} \cdot 1 \cdot 1}{(0.2)^2} \]
\[ F_{13} = \frac{6.67 \times 10^{-11}}{0.04} = 1.6675 \times 10^{-9} \, \text{N} \]
- This force acts to the left (negative x-direction).

### Step 5: Determine the net gravitational force
- The net gravitational force \( \vec{F}_{net} \) on the mass at the origin is the vector sum of \( \vec{F}_{12} \) and \( \vec{F}_{13} \):
\[ \vec{F}_{net} = -F_{12} \hat{j} - F_{13} \hat{i} \]
\[ \vec{F}_{net} = -1.6675 \times 10^{-9} \hat{j} - 1.6675 \times 10^{-9} \hat{i} \]

### Step 6: Express the net force in component form
- The net force can be expressed as:
\[ \vec{F}_{net} = -1.6675 \times 10^{-9} \hat{i} - 1.6675 \times 10^{-9} \hat{j} \]

### Step 7: Final result
- The magnitude of the net gravitational force can be calculated as:
\[ |\vec{F}_{net}| = \sqrt{(-1.6675 \times 10^{-9})^2 + (-1.6675 \times 10^{-9})^2} \]
\[ |\vec{F}_{net}| = \sqrt{2 \cdot (1.6675 \times 10^{-9})^2} = 1.6675 \times 10^{-9} \sqrt{2} \]
- The final expression for the net gravitational force is:
\[ \vec{F}_{net} = -1.6675 \times 10^{-9} (\hat{i} + \hat{j}) \]