Certainly! Let's solve this step-by-step using the gravitational force formula.
The gravitational force between two masses [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] separated by a distance [tex]\( r \)[/tex] is given by Newton's law of gravitation:
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/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] 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 the values:
- [tex]\( F = 1.05 \times 10^{-4} \, \text{N} \)[/tex]
- [tex]\( r = 100,000 \, \text{m} \)[/tex]
- [tex]\( m_1 = 3.5 \times 10^6 \, \text{kg} \)[/tex]
We need to find [tex]\( m_2 \)[/tex].
First, rearrange the formula to solve for [tex]\( m_2 \)[/tex]:
[tex]\[ m_2 = \frac{F \cdot r^2}{G \cdot m_1} \][/tex]
Now substitute the given values into the equation:
[tex]\[ m_2 = \frac{1.05 \times 10^{-4} \cdot (100,000)^2}{6.67430 \times 10^{-11} \cdot 3.5 \times 10^6} \][/tex]
Perform the calculations step-by-step:
1. Calculate [tex]\( r^2 \)[/tex]:
[tex]\[ (100,000)^2 = 10^{10} \][/tex]
2. Multiply [tex]\( F \)[/tex] by [tex]\( r^2 \)[/tex]:
[tex]\[ 1.05 \times 10^{-4} \times 10^{10} = 1.05 \times 10^6 \][/tex]
3. Calculate [tex]\( G \cdot m_1 \)[/tex]:
[tex]\[ 6.67430 \times 10^{-11} \times 3.5 \times 10^6 = 2.335005 \times 10^{-4} \][/tex]
4. Divide the results:
[tex]\[ m_2 = \frac{1.05 \times 10^6}{2.335005 \times 10^{-4}} \][/tex]
5. Simplify the division:
[tex]\[ m_2 \approx 4494853392.8651705 \][/tex]
Therefore, the mass [tex]\( m_2 \)[/tex] of the second asteroid is approximately [tex]\( 4.49 \times 10^9 \, \text{kg} \)[/tex].
Looking at the options provided:
A. [tex]\( 4.1 \times 10^9 \, \text{kg} \)[/tex]
B. [tex]\( 4.5 \times 10^9 \, \text{kg} \)[/tex]
C. [tex]\( 4.1 \times 10^8 \, \text{kg} \)[/tex]
D. [tex]\( 4.5 \times 10^8 \, \text{kg} \)[/tex]
The closest value is B. [tex]\( 4.5 \times 10^9 \, \text{kg} \)[/tex].
Thus, the mass of the other asteroid is [tex]\( 4.5 \times 10^9 \, \text{kg} \)[/tex].
