Answer :
To find the mass of the other asteroid, we can use Newton's law of universal gravitation. The formula for the gravitational force [tex]\( F \)[/tex] between two masses [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] separated by a distance [tex]\( r \)[/tex] is given by:
[tex]\[ F = G \frac{m_1 \cdot m_2}{r^2} \][/tex]
Where:
- [tex]\( F \)[/tex] is the gravitational force,
- [tex]\( G \)[/tex] is the gravitational constant [tex]\( (6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2}) \)[/tex],
- [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] are the masses of the two objects,
- [tex]\( r \)[/tex] is the distance between the centers of the two masses.
We are given:
- [tex]\( F = 1.14 \, \text{N} \)[/tex]
- [tex]\( r = 75,000 \, \text{m} \)[/tex]
- [tex]\( m_1 = 8 \times 10^{3} \, \text{kg} \)[/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]
Now, substituting the given values into the equation:
- [tex]\( F = 1.14 \, \text{N} \)[/tex]
- [tex]\( r = 75,000 \, \text{m} \)[/tex]
- [tex]\( G = 6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \)[/tex]
- [tex]\( m_1 = 8 \times 10^{3} \, \text{kg} \)[/tex]
We get:
[tex]\[ m_2 = \frac{1.14 \cdot (75,000)^2}{6.67430 \times 10^{-11} \cdot 8 \times 10^3} \][/tex]
By calculating this, we find:
[tex]\[ m_2 \approx 1.2009686409061624 \times 10^{16} \, \text{kg} \][/tex]
Now, looking at the possible answer choices, the closest one is:
A. [tex]\( 1.2 \times 10^{12} \, \text{kg} \)[/tex]
B. [tex]\( 1.2 \times 10^{10} \, \text{kg} \)[/tex]
C. [tex]\( 8.3 \times 10^{12} \, \text{kg} \)[/tex]
D. [tex]\( 3.4 \times 10^{11} \, \text{kg} \)[/tex]
Given the magnitude of the calculated mass, the nearest approximate choice from the given options is:
[tex]\[ \boxed{A. \, 1.2 \times 10^{16} \, \text{kg}} \][/tex]
[tex]\[ F = G \frac{m_1 \cdot m_2}{r^2} \][/tex]
Where:
- [tex]\( F \)[/tex] is the gravitational force,
- [tex]\( G \)[/tex] is the gravitational constant [tex]\( (6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2}) \)[/tex],
- [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] are the masses of the two objects,
- [tex]\( r \)[/tex] is the distance between the centers of the two masses.
We are given:
- [tex]\( F = 1.14 \, \text{N} \)[/tex]
- [tex]\( r = 75,000 \, \text{m} \)[/tex]
- [tex]\( m_1 = 8 \times 10^{3} \, \text{kg} \)[/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]
Now, substituting the given values into the equation:
- [tex]\( F = 1.14 \, \text{N} \)[/tex]
- [tex]\( r = 75,000 \, \text{m} \)[/tex]
- [tex]\( G = 6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \)[/tex]
- [tex]\( m_1 = 8 \times 10^{3} \, \text{kg} \)[/tex]
We get:
[tex]\[ m_2 = \frac{1.14 \cdot (75,000)^2}{6.67430 \times 10^{-11} \cdot 8 \times 10^3} \][/tex]
By calculating this, we find:
[tex]\[ m_2 \approx 1.2009686409061624 \times 10^{16} \, \text{kg} \][/tex]
Now, looking at the possible answer choices, the closest one is:
A. [tex]\( 1.2 \times 10^{12} \, \text{kg} \)[/tex]
B. [tex]\( 1.2 \times 10^{10} \, \text{kg} \)[/tex]
C. [tex]\( 8.3 \times 10^{12} \, \text{kg} \)[/tex]
D. [tex]\( 3.4 \times 10^{11} \, \text{kg} \)[/tex]
Given the magnitude of the calculated mass, the nearest approximate choice from the given options is:
[tex]\[ \boxed{A. \, 1.2 \times 10^{16} \, \text{kg}} \][/tex]
