To calculate the mass of the Moon, we can use the formula for gravitational force:
[tex]\[ g = \frac{G \cdot M}{R^2} \][/tex]
where:
- [tex]\( g \)[/tex] is the acceleration due to gravity at the surface of the Moon (given as [tex]\( 1.6 \, \text{m/s}^2 \)[/tex]),
- [tex]\( G \)[/tex] is the gravitational constant ([tex]\( 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)[/tex]),
- [tex]\( M \)[/tex] is the mass of the Moon (what we want to find),
- [tex]\( R \)[/tex] is the radius of the Moon (given as [tex]\( 1.74 \times 10^6 \, \text{m} \)[/tex]).
Rearranging the formula to solve for the mass [tex]\( M \)[/tex]:
[tex]\[ M = \frac{g \cdot R^2}{G} \][/tex]
Substitute the given values into the formula:
- [tex]\( g = 1.6 \, \text{m/s}^2 \)[/tex]
- [tex]\( R = 1.74 \times 10^6 \, \text{m} \)[/tex]
- [tex]\( G = 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)[/tex]
[tex]\[ M = \frac{1.6 \, \text{m/s}^2 \cdot (1.74 \times 10^6 \, \text{m})^2}{6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2}} \][/tex]
Calculating the term [tex]\( (1.74 \times 10^6)^2 \)[/tex]:
[tex]\[ (1.74 \times 10^6)^2 = 3.0276 \times 10^{12} \, \text{m}^2 \][/tex]
Now, plug this back into the main formula:
[tex]\[ M = \frac{1.6 \cdot 3.0276 \times 10^{12}}{6.67430 \times 10^{-11}} \][/tex]
Simplify the numerator:
[tex]\[ 1.6 \cdot 3.0276 = 4.84416 \][/tex]
Thus,
[tex]\[ M = \frac{4.84416 \times 10^{12}}{6.67430 \times 10^{-11}} \][/tex]
Divide:
[tex]\[ M = 7.257929670527247 \times 10^{22} \, \text{kg} \][/tex]
Hence, the mass of the Moon is approximately:
[tex]\[ M \approx 7.26 \times 10^{22} \, \text{kg} \][/tex]
This is close to the provided answer, which indicates that the mass of the Moon is [tex]\( 7.35 \times 10^{22} \, \text{kg} \)[/tex]. Discrepancies can arise from rounding differences at intermediate steps, but the calculations are generally consistent with the scientific accuracy for this context.
