Two spherical balls each of mass 1 kg are placed 1 cm apart. The gravitational force of attraction between them - newtons

Options

A. `6.67 xx10^(-9)`
B. `6.67 xx10^(-7)`
C. `6.67 xx10^(-8)`
D. `6.67 xx10^(-5)`

Answer

To find the gravitational force of attraction between two spherical balls each of mass 1 kg and placed 1 cm apart, we can use Newton's law of universal gravitation, which is given by the formula:

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

Where:
- \( F \) is the gravitational force,
- \( G \) is the universal gravitational constant (\( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)),
- \( m_1 \) and \( m_2 \) are the masses of the two objects (in this case, both are 1 kg),
- \( r \) is the distance between the centers of the two masses (1 cm, which we need to convert to meters).

### Step-by-Step Solution:

1. **Identify the values**:
- Mass of the first ball, \( m_1 = 1 \, \text{kg} \)
- Mass of the second ball, \( m_2 = 1 \, \text{kg} \)
- Distance between the balls, \( r = 1 \, \text{cm} = 0.01 \, \text{m} \)
- Gravitational constant, \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)

2. **Plug the values into the formula**:
\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]
Substituting the values:
\[ F = \frac{6.67 \times 10^{-11} \cdot 1 \cdot 1}{(0.01)^2} \]

3. **Calculate \( r^2 \)**:
\[ r^2 = (0.01)^2 = 0.0001 \, \text{m}^2 \]

4. **Substitute \( r^2 \) back into the equation**:
\[ F = \frac{6.67 \times 10^{-11}}{0.0001} \]

5. **Perform the division**:
\[ F = 6.67 \times 10^{-11} \div 0.0001 = 6.67 \times 10^{-11} \div 10^{-4} = 6.67 \times 10^{-7} \, \text{N} \]

6. **Final answer**:
The gravitational force of attraction between the two balls is:
\[ F = 6.67 \times 10^{-7} \, \text{N} \]