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 find the net gravitational force on the mass at the origin (0,0) due to the other two masses located at (0,0.2m) and (0.2m,0), we can follow these steps:

### Step 1: Identify the masses and their positions
We have three identical point masses, each of mass \( m = 1 \, \text{kg} \):
- Mass A at (0, 0) (the origin)
- Mass B at (0, 0.2m)
- Mass C at (0.2m, 0)

### Step 2: Calculate the gravitational force exerted on mass A by mass B
The gravitational force \( F_{AB} \) 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 \( (6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2) \)
- \( m_1 = m_2 = 1 \, \text{kg} \)
- \( r \) is the distance between the two masses. The distance between A and B is \( 0.2 \, \text{m} \).

Calculating \( F_{AB} \):
\[ F_{AB} = \frac{G \cdot 1 \cdot 1}{(0.2)^2} = \frac{6.67 \times 10^{-11}}{0.04} = 1.6675 \times 10^{-9} \, \text{N} \]
The direction of this force is downward (in the negative y-direction), so we can express it as:
\[ F_{AB} = -1.6675 \times 10^{-9} \, \hat{j} \]

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

Calculating \( F_{AC} \):
\[ F_{AC} = \frac{G \cdot 1 \cdot 1}{(0.2)^2} = \frac{6.67 \times 10^{-11}}{0.04} = 1.6675 \times 10^{-9} \, \text{N} \]
The direction of this force is to the right (in the positive x-direction), so we can express it as:
\[ F_{AC} = 1.6675 \times 10^{-9} \, \hat{i} \]

### Step 4: Calculate the net gravitational force on mass A
The net gravitational force \( F_{net} \) on mass A is the vector sum of \( F_{AB} \) and \( F_{AC} \):
\[ F_{net} = F_{AB} + F_{AC} = (-1.6675 \times 10^{-9} \, \hat{j}) + (1.6675 \times 10^{-9} \, \hat{i}) \]
Thus,
\[ F_{net} = 1.6675 \times 10^{-9} \, \hat{i} - 1.6675 \times 10^{-9} \, \hat{j} \]

### Step 5: Express the final answer
The net gravitational force on the mass at the origin can be expressed as:
\[ F_{net} = 1.6675 \times 10^{-9} \hat{i} - 1.6675 \times 10^{-9} \hat{j} \, \text{N} \]