Certainly! Let's solve this problem step by step.
### Given Information:
1. Gravitational Force (F) between two asteroids: [tex]\( F = 6.2 \times 10^8 \, \text{N} \)[/tex]
2. Distance (d) between the asteroids: [tex]\( d = 2100 \, \text{km} = 2100 \times 10^3 \, \text{m} \)[/tex]
3. Gravitational Constant (G): [tex]\( G = 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)[/tex]
4. Mass Relationship: Mass of asteroid [tex]\( Y \)[/tex] [tex]\( (M_y) \)[/tex] is 3 times the mass of asteroid [tex]\( Z \)[/tex] [tex]\( (M_z) \)[/tex], i.e., [tex]\( M_y = 3 \times M_z \)[/tex]
### Gravitational Force Formula:
The formula for the gravitational force between two masses is given by:
[tex]\[ F = \frac{G \cdot M_y \cdot M_z}{d^2} \][/tex]
### Substituting [tex]\( M_y = 3 \times M_z \)[/tex] into the formula:
[tex]\[ F = \frac{G \cdot (3 \times M_z) \cdot M_z}{d^2} \][/tex]
[tex]\[ F = \frac{3 \cdot G \cdot M_z^2}{d^2} \][/tex]
### Solving for [tex]\( M_z \)[/tex]:
1. Multiply both sides by [tex]\( d^2 \)[/tex]:
[tex]\[ F \cdot d^2 = 3 \cdot G \cdot M_z^2 \][/tex]
2. Isolate [tex]\( M_z^2 \)[/tex]:
[tex]\[ M_z^2 = \frac{F \cdot d^2}{3 \cdot G} \][/tex]
3. Plugging in the known values:
[tex]\[ M_z^2 = \frac{6.2 \times 10^8 \times (2100 \times 10^3)^2}{3 \times 6.67430 \times 10^{-11}} \][/tex]
### Calculating [tex]\( M_z \)[/tex]:
[tex]\[ M_z^2 \approx 1.3655364607524388 \times 10^{31} \][/tex]
So, the mass of asteroid [tex]\( Z \)[/tex]:
[tex]\[ M_z \approx \sqrt{1.3655364607524388 \times 10^{31}} \][/tex]
[tex]\[ M_z \approx 3695316577442911.0 \, \text{kg} \][/tex]
### Mass of Asteroid [tex]\( Y \)[/tex]:
Since [tex]\( M_y = 3 \times M_z \)[/tex]:
[tex]\[ M_y = 3 \times 3695316577442911.0 \, \text{kg} \][/tex]
[tex]\[ M_y \approx 1.1085949732328732 \times 10^{16} \, \text{kg} \][/tex]
Thus, the mass of asteroid [tex]\( Y \)[/tex] is [tex]\( 1.1 \times 10^{16} \, \text{kg} \)[/tex].
### Answer:
[tex]\[ \boxed{1.1 \times 10^{16} \, \text{kg}} \][/tex]
