To determine the gravitational force between two masses, we use Newton's law of universal gravitation, which is given by the formula:
[tex]\[ F = G \frac{m_1 \cdot m_2}{r^2} \][/tex]
where:
- [tex]\( F \)[/tex] is the gravitational force between the two masses,
- [tex]\( G \)[/tex] is the gravitational constant, [tex]\( 6.67 \times 10^{-11} \: \text{N} \cdot (\text{m}^2 / \text{kg}^2) \)[/tex],
- [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] are the masses of the two objects, each 8 kg in this case,
- [tex]\( r \)[/tex] is the distance between the centers of the two masses, which is 2 meters in this case.
Given:
- [tex]\( m_1 = 8 \: \text{kg} \)[/tex]
- [tex]\( m_2 = 8 \: \text{kg} \)[/tex]
- [tex]\( r = 2 \: \text{m} \)[/tex]
Substitute these values into the formula:
[tex]\[ F = 6.67 \times 10^{-11} \frac{8 \: \text{kg} \cdot 8 \: \text{kg}}{(2 \: \text{m})^2} \][/tex]
First, calculate the numerator:
[tex]\[ 8 \: \text{kg} \cdot 8 \: \text{kg} = 64 \: \text{kg}^2 \][/tex]
Next, calculate the denominator:
[tex]\[ (2 \: \text{m})^2 = 4 \: \text{m}^2 \][/tex]
Now, substitute these into the equation:
[tex]\[ F = 6.67 \times 10^{-11} \frac{64 \: \text{kg}^2}{4 \: \text{m}^2} \][/tex]
Simplify the fraction:
[tex]\[ \frac{64 \: \text{kg}^2}{4 \: \text{m}^2} = 16 \: \text{kg}^2 / \text{m}^2 \][/tex]
So the force now is:
[tex]\[ F = 6.67 \times 10^{-11} \times 16 \: \text{N} \][/tex]
Multiplying these values:
[tex]\[ F = 6.67 \times 16 \times 10^{-11} \][/tex]
[tex]\[ F = 106.72 \times 10^{-11} \][/tex]
[tex]\[ F = 1.0672 \times 10^{-9} \: \text{N} \][/tex]
Hence, the gravitational force between the two bowling balls is:
[tex]\[ \boxed{1.07 \times 10^{-9} \: \text{N}} \][/tex]
So, the correct answer is:
A. [tex]\( 1.07 \times 10^{-9} \: \text{N} \)[/tex]
