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^(-9)(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
We have three identical point masses, each with a mass of 1 kg, located at:
- Mass A at (0, 0) (the origin)
- Mass B at (0, 0.2 m)
- Mass C at (0.2 m, 0)

### Step 2: Calculate the gravitational force exerted on the mass at the origin by Mass B
The gravitational force \( F \) between two point masses is given by the formula:
\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]
Where:
- \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \) (gravitational constant)
- \( m_1 = 1 \, \text{kg} \) (mass at the origin)
- \( m_2 = 1 \, \text{kg} \) (mass B)
- \( r \) is the distance between the two masses, which is \( 0.2 \, \text{m} \).

Calculating the force due to Mass B:
\[ F_B = \frac{6.67 \times 10^{-11} \cdot 1 \cdot 1}{(0.2)^2} = \frac{6.67 \times 10^{-11}}{0.04} = 1.6675 \times 10^{-9} \, \text{N} \]

This force acts in the negative y-direction (downward).

### Step 3: Calculate the gravitational force exerted on the mass at the origin by Mass C
Now, we calculate the gravitational force exerted by Mass C:
- The distance \( r \) between the mass at the origin and Mass C is also \( 0.2 \, \text{m} \).

Calculating the force due to Mass C:
\[ F_C = \frac{6.67 \times 10^{-11} \cdot 1 \cdot 1}{(0.2)^2} = \frac{6.67 \times 10^{-11}}{0.04} = 1.6675 \times 10^{-9} \, \text{N} \]

This force acts in the negative x-direction (to the left).

### Step 4: Determine the net gravitational force on the mass at the origin
The net gravitational force \( \vec{F}_{net} \) on the mass at the origin is the vector sum of the forces due to Mass B and Mass C:
\[ \vec{F}_{net} = \vec{F}_B + \vec{F}_C \]
Expressing this in vector form:
\[ \vec{F}_{net} = -1.6675 \times 10^{-9} \hat{j} - 1.6675 \times 10^{-9} \hat{i} \]

### Step 5: Final Result
Thus, the net gravitational force on the mass at the origin is:
\[ \vec{F}_{net} = -1.6675 \times 10^{-9} \hat{i} - 1.6675 \times 10^{-9} \hat{j} \, \text{N} \]