Answer :
To solve this problem, we can use Newton's Law of Universal Gravitation, which states:
[tex]\[ F = G \frac{m_1 \cdot m_2}{r^2} \][/tex]
where:
- [tex]\( F \)[/tex] is the gravitational force between the two objects,
- [tex]\( G \)[/tex] is the gravitational constant ([tex]\( 6.67430 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \)[/tex]),
- [tex]\( m_1 \)[/tex] is the mass of the first object,
- [tex]\( m_2 \)[/tex] is the mass of the second object,
- [tex]\( r \)[/tex] is the distance between the centers of the two masses.
Given:
- [tex]\( F = 8.67 \times 10^{-2} \, \text{N} \)[/tex],
- [tex]\( G = 6.67430 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \)[/tex],
- [tex]\( m_1 = 5 \times 10^8 \, \text{kg} \)[/tex],
- [tex]\( r = 50000 \, \text{m} \)[/tex],
We need to find [tex]\( m_2 \)[/tex].
Rearrange the formula to solve for [tex]\( m_2 \)[/tex]:
[tex]\[ m_2 = \frac{F \cdot r^2}{G \cdot m_1} \][/tex]
Substitute the given values into the equation:
[tex]\[ m_2 = \frac{8.67 \times 10^{-2} \, \text{N} \cdot (50000 \, \text{m})^2}{6.67430 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \cdot 5 \times 10^8 \, \text{kg}} \][/tex]
Calculate the numerator and the denominator separately for clarity:
[tex]\[ \text{Numerator} = 8.67 \times 10^{-2} \, \text{N} \cdot 2500000000 \, \text{m}^2 \][/tex]
[tex]\[ \text{Numerator} = 8.67 \times 10^{-2} \cdot 2.5 \times 10^9 \][/tex]
[tex]\[ \text{Numerator} = 2.1675 \times 10^8 \][/tex]
[tex]\[ \text{Denominator} = 6.67430 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \cdot 5 \times 10^8 \, \text{kg} \][/tex]
[tex]\[ \text{Denominator} = 3.33715 \times 10^{-2} \, \text{kg} \][/tex]
Now, divide the numerator by the denominator:
[tex]\[ m_2 = \frac{2.1675 \times 10^8}{3.33715 \times 10^{-2}} \][/tex]
[tex]\[ m_2 \approx 6.495063152 \times 10^9 \, \text{kg} \][/tex]
So the mass of the other asteroid is approximately [tex]\( 6.495063152 \times 10^9 \, \text{kg} \)[/tex].
Therefore, the closest answer is:
B. [tex]\( 6.5 \times 10^9 \, \text{kg} \)[/tex]
[tex]\[ F = G \frac{m_1 \cdot m_2}{r^2} \][/tex]
where:
- [tex]\( F \)[/tex] is the gravitational force between the two objects,
- [tex]\( G \)[/tex] is the gravitational constant ([tex]\( 6.67430 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \)[/tex]),
- [tex]\( m_1 \)[/tex] is the mass of the first object,
- [tex]\( m_2 \)[/tex] is the mass of the second object,
- [tex]\( r \)[/tex] is the distance between the centers of the two masses.
Given:
- [tex]\( F = 8.67 \times 10^{-2} \, \text{N} \)[/tex],
- [tex]\( G = 6.67430 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \)[/tex],
- [tex]\( m_1 = 5 \times 10^8 \, \text{kg} \)[/tex],
- [tex]\( r = 50000 \, \text{m} \)[/tex],
We need to find [tex]\( m_2 \)[/tex].
Rearrange the formula to solve for [tex]\( m_2 \)[/tex]:
[tex]\[ m_2 = \frac{F \cdot r^2}{G \cdot m_1} \][/tex]
Substitute the given values into the equation:
[tex]\[ m_2 = \frac{8.67 \times 10^{-2} \, \text{N} \cdot (50000 \, \text{m})^2}{6.67430 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \cdot 5 \times 10^8 \, \text{kg}} \][/tex]
Calculate the numerator and the denominator separately for clarity:
[tex]\[ \text{Numerator} = 8.67 \times 10^{-2} \, \text{N} \cdot 2500000000 \, \text{m}^2 \][/tex]
[tex]\[ \text{Numerator} = 8.67 \times 10^{-2} \cdot 2.5 \times 10^9 \][/tex]
[tex]\[ \text{Numerator} = 2.1675 \times 10^8 \][/tex]
[tex]\[ \text{Denominator} = 6.67430 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \cdot 5 \times 10^8 \, \text{kg} \][/tex]
[tex]\[ \text{Denominator} = 3.33715 \times 10^{-2} \, \text{kg} \][/tex]
Now, divide the numerator by the denominator:
[tex]\[ m_2 = \frac{2.1675 \times 10^8}{3.33715 \times 10^{-2}} \][/tex]
[tex]\[ m_2 \approx 6.495063152 \times 10^9 \, \text{kg} \][/tex]
So the mass of the other asteroid is approximately [tex]\( 6.495063152 \times 10^9 \, \text{kg} \)[/tex].
Therefore, the closest answer is:
B. [tex]\( 6.5 \times 10^9 \, \text{kg} \)[/tex]
