Answer :
To find the gravitational force between two astronauts, we use Newton's Law of Universal Gravitation, which can be stated mathematically as:
[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 (both [tex]\( 100 \, \text{kg} \)[/tex] in this case),
- [tex]\( r \)[/tex] is the distance between the centers of the two masses ( [tex]\( 2 \, \text{m} \)[/tex] in this case).
Substitute the given values into the formula:
[tex]\[ F = (6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2) \frac{100 \, \text{kg} \cdot 100 \, \text{kg}}{(2 \, \text{m})^2} \][/tex]
Calculate the denominator:
[tex]\[ (2 \, \text{m})^2 = 4 \, \text{m}^2 \][/tex]
Then the formula becomes:
[tex]\[ F = (6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2) \frac{10000 \, \text{kg}^2}{4 \, \text{m}^2} \][/tex]
Simplify the fraction:
[tex]\[ \frac{10000 \, \text{kg}^2}{4 \, \text{m}^2} = 2500 \, \text{kg}^2 / \text{m}^2 \][/tex]
Then multiply by the gravitational constant [tex]\( G \)[/tex]:
[tex]\[ F = 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \cdot 2500 \, \text{kg}^2 / \text{m}^2 \][/tex]
[tex]\[ F = 6.67 \times 2500 \times 10^{-11} \, \text{N} \][/tex]
[tex]\[ F = 16675 \times 10^{-11} \, \text{N} \][/tex]
[tex]\[ F = 1.6675 \times 10^{-7} \, \text{N} \][/tex]
Therefore, the force of gravity between the two astronauts is:
[tex]\[ \boxed{1.67 \times 10^{-7} \, \text{N}} \][/tex]
Thus, the correct answer is:
C. [tex]\(1.67 \times 10^{-7} \, \text{N} \)[/tex]
[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 (both [tex]\( 100 \, \text{kg} \)[/tex] in this case),
- [tex]\( r \)[/tex] is the distance between the centers of the two masses ( [tex]\( 2 \, \text{m} \)[/tex] in this case).
Substitute the given values into the formula:
[tex]\[ F = (6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2) \frac{100 \, \text{kg} \cdot 100 \, \text{kg}}{(2 \, \text{m})^2} \][/tex]
Calculate the denominator:
[tex]\[ (2 \, \text{m})^2 = 4 \, \text{m}^2 \][/tex]
Then the formula becomes:
[tex]\[ F = (6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2) \frac{10000 \, \text{kg}^2}{4 \, \text{m}^2} \][/tex]
Simplify the fraction:
[tex]\[ \frac{10000 \, \text{kg}^2}{4 \, \text{m}^2} = 2500 \, \text{kg}^2 / \text{m}^2 \][/tex]
Then multiply by the gravitational constant [tex]\( G \)[/tex]:
[tex]\[ F = 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \cdot 2500 \, \text{kg}^2 / \text{m}^2 \][/tex]
[tex]\[ F = 6.67 \times 2500 \times 10^{-11} \, \text{N} \][/tex]
[tex]\[ F = 16675 \times 10^{-11} \, \text{N} \][/tex]
[tex]\[ F = 1.6675 \times 10^{-7} \, \text{N} \][/tex]
Therefore, the force of gravity between the two astronauts is:
[tex]\[ \boxed{1.67 \times 10^{-7} \, \text{N}} \][/tex]
Thus, the correct answer is:
C. [tex]\(1.67 \times 10^{-7} \, \text{N} \)[/tex]
