Two point masses each equal to 1 kg attract one another with a force of `10^(-9)` kg-wt. the distance between the two point masses is approximately `(G = 6.6 xx 10^(-11) "MKS units")`

Options

A. 8 cm
B. 0.8 cm
C. 80 cm
D. 0.08 cm

Explanation

To solve the problem, we need to find the distance between two point masses that attract each other with a given gravitational force. We will use the formula for gravitational force:

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

Where:
- \( F \) is the gravitational force between the two masses,
- \( G \) is the gravitational constant,
- \( m_1 \) and \( m_2 \) are the masses,
- \( r \) is the distance between the two masses.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of each point mass, \( m_1 = m_2 = 1 \, \text{kg} \)
- Gravitational force, \( F = 10^{-9} \, \text{kg-wt} \) (Note: 1 kg-wt = 9.8 N, so we convert this to Newtons)
- Gravitational constant, \( G = 6.67 \times 10^{-11} \, \text{m}^3/\text{kg} \cdot \text{s}^2 \)

First, convert the force from kg-wt to Newtons:
\[ F = 10^{-9} \, \text{kg-wt} \times 9.8 \, \text{N/kg-wt} = 9.8 \times 10^{-9} \, \text{N} \]

2. **Rearranging the Gravitational Force Formula:**
We need to solve for \( r \), so we rearrange the formula:
\[ r^2 = \frac{G \cdot m_1 \cdot m_2}{F} \]
\[ r = \sqrt{\frac{G \cdot m_1 \cdot m_2}{F}} \]

3. **Substituting the Values:**
Substitute \( G \), \( m_1 \), \( m_2 \), and \( F \) into the equation:
\[ r = \sqrt{\frac{6.67 \times 10^{-11} \, \text{m}^3/\text{kg} \cdot \text{s}^2 \cdot 1 \, \text{kg} \cdot 1 \, \text{kg}}{9.8 \times 10^{-9} \, \text{N}}} \]

4. **Calculating the Value:**
Calculate the numerator:
\[ 6.67 \times 10^{-11} \, \text{m}^3/\text{kg} \cdot \text{s}^2 \]
The denominator is:
\[ 9.8 \times 10^{-9} \, \text{N} \]
Now, compute:
\[ r = \sqrt{\frac{6.67 \times 10^{-11}}{9.8 \times 10^{-9}}} \]

5. **Final Calculation:**
Performing the division:
\[ \frac{6.67 \times 10^{-11}}{9.8 \times 10^{-9}} \approx 6.81 \times 10^{-3} \, \text{m}^2 \]
Now take the square root:
\[ r \approx \sqrt{6.81 \times 10^{-3}} \approx 0.0826 \, \text{m} \]

6. **Convert to Centimeters:**
To convert meters to centimeters:
\[ r \approx 0.0826 \, \text{m} \times 100 \approx 8.26 \, \text{cm} \]
Rounding this gives approximately \( 8 \, \text{cm} \).

### Conclusion:
The distance between the two point masses is approximately \( 8 \, \text{cm} \).