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 will use Newton's law of universal gravitation, which states that 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:
- \( F \) is the gravitational force between the masses,
- \( G \) is the gravitational constant, approximately \( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \),
- \( m_1 \) and \( m_2 \) are the masses (in kg),
- \( r \) is the distance between the centers of the two masses (in meters).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Masses \( m_1 = 1 \, \text{kg} \) and \( m_2 = 1 \, \text{kg} \)
- Gravitational force \( F = 10^{-9} \, \text{kg-wt} \) (Note: 1 kg-wt = 9.8 N, so we convert it to Newtons)
- \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)

2. **Convert the Force to Newtons:**
\[ F = 10^{-9} \, \text{kg-wt} = 10^{-9} \times 9.8 \, \text{N} = 9.8 \times 10^{-9} \, \text{N} \]

3. **Rearranging the Formula:**
To find the distance \( r \), we rearrange the formula:
\[ r^2 = \frac{G \cdot m_1 \cdot m_2}{F} \]
Therefore,
\[ r = \sqrt{\frac{G \cdot m_1 \cdot m_2}{F}} \]

4. **Substituting the Values:**
Substitute the known values into the equation:
\[ r = \sqrt{\frac{6.67 \times 10^{-11} \cdot 1 \cdot 1}{9.8 \times 10^{-9}}} \]

5. **Calculating the Value:**
\[ r = \sqrt{\frac{6.67 \times 10^{-11}}{9.8 \times 10^{-9}}} \]
\[ r = \sqrt{6.81 \times 10^{-3}} \approx 0.0826 \, \text{m} \]

6. **Convert to Centimeters:**
\[ r \approx 0.0826 \, \text{m} \times 100 \approx 8.26 \, \text{cm} \]

### Final Answer:
The distance between the two point masses is approximately \( 8.26 \, \text{cm} \).