Answer: A) \( 6.36 \times 10^{24} \, \text{kg} \)
Solution:
The mass of the Earth can be calculated using Kepler's third law:
\[
T^2 = \frac{4 \pi^2 r^3}{G M}
\]
Rearranging for \( M \) (mass of the Earth):
\[
M = \frac{4 \pi^2 r^3}{G T^2}
\]
Step 1: Convert given values to SI units
- Radius of orbit: \( r = 3.9 \times 10^5 \, \text{km} = 3.9 \times 10^8 \, \text{m} \)
- Orbital period: \( T = 27.3 \, \text{days} = 27.3 \times 24 \times 3600 \, \text{seconds} = 2.36 \times 10^6 \, \text{s} \)
Step 2: Substitute the values into the formula
\[
M = \frac{4 \pi^2 (3.9 \times 10^8)^3}{6.67 \times 10^{-11} (2.36 \times 10^6)^2}
\]
Calculate \( r^3 \):
\[
r^3 = (3.9 \times 10^8)^3 = 5.93 \times 10^{25} \, \text{m}^3
\]
Calculate \( T^2 \):
\[
T^2 = (2.36 \times 10^6)^2 = 5.57 \times 10^{12} \, \text{s}^2
\]
Substitute these into the formula:
\[
M = \frac{2.35 \times 10^{26}}{3.71 \times 10^{2}} \approx 6.36 \times 10^{24} \, \text{kg}
\]
Final Answer:
- The mass of the Earth is approximately \( 6.36 \times 10^{24} \, \text{kg} \).
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