If in a system the force of attraction between two point masses of `1kg` each situated `1km` apart is taken as a unit force and is called notwen (newton written in reverse order) If `G = 6.67 xx 10^(-11) N-m^(2)kg^(-2)` in `\SI` units, the relation of newton and nowton is

Options

A. 1 notwen`=6.67xx10^(-11)` newton
B. 2 notwen`=6.67xx10^(-17)` newton
C. 3 notwen`=6.67xx10^(-17)` newton
D. 4 notwen`=6.67xx10^(-12)` newton

Explanation

To solve the problem, we need to find the relationship between the unit of force called "notwen" and the standard unit of force, the newton (N).

### Step-by-Step Solution:

1. **Understanding the Gravitational Force Formula**:
The gravitational force \( F \) between two point masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by the formula:
\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]
where \( G \) is the gravitational constant.

2. **Assigning Values**:
According to the problem, we have:
- \( m_1 = 1 \, \text{kg} \)
- \( m_2 = 1 \, \text{kg} \)
- \( r = 1 \, \text{km} = 1000 \, \text{m} \)

3. **Substituting Values into the Formula**:
Substitute the values into the gravitational force formula:
\[ F = \frac{G \cdot 1 \cdot 1}{(1000)^2} \]
Simplifying this gives:
\[ F = \frac{G}{10^6} \]

4. **Setting the Force Equal to 1 Notwen**:
According to the problem, this force is defined as 1 notwen. Therefore, we can write:
\[ 1 \, \text{notwen} = \frac{G}{10^6} \]

5. **Substituting the Value of G**:
We know that \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \). Substituting this value into the equation gives:
\[ 1 \, \text{notwen} = \frac{6.67 \times 10^{-11}}{10^6} \]

6. **Calculating the Value of 1 Notwen**:
Simplifying the right side:
\[ 1 \, \text{notwen} = 6.67 \times 10^{-11} \times 10^{-6} = 6.67 \times 10^{-17} \, \text{N} \]

7. **Final Relation**:
Therefore, the relationship between notwen and newton is:
\[ 1 \, \text{notwen} = 6.67 \times 10^{-17} \, \text{N} \]