- Use Kepler's Third Law, \(T^2 = \frac{4\pi^2 d^3}{GM}\), to relate orbital period to distance.
- Perform the numerical calculation using the provided physical constants to find the period in seconds, resulting in approximately \(9.323 \times 10^8\) seconds.
- Convert the result into Earth years using the factor \(1 \text{ year} \approx 3.156 \times 10^7\) seconds.
- Conclude that Saturn's orbital period is approximately \(\boxed{29.6}\) Earth years.
Explanation
- Analysis of the Problem
The goal is to calculate the orbital period \(T\) of Saturn in Earth years, given its orbital distance \(d = 1.43 \times 10^{12} \, \text{m}\), the mass of the Sun \(M = 1.99 \times 10^{30} \, \text{kg}\), and the gravitational constant \(G = 6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2\). The relationship between orbital period and distance is given by Kepler's Third Law: \(T^2 = \left(\frac{4\pi^2}{GM}\right) d^3\). Once we determine \(T\) in seconds, we will convert it into Earth years using the conversion factor that one year is approximately 365.25 days.
- Calculating the Orbital Period in Seconds
First, we substitute the given values into the formula for \(T^2\):
\[T^2 = \frac{4 \times \pi^2 \times (1.43 \times 10^{12})^3}{6.674 \times 10^{-11} \times 1.99 \times 10^{30}}\]
Calculating the cube of the distance:
\((1.43 \times 10^{12})^3 \approx 2.924 \times 10^{36} \, \text{m}^3\)
Multiplying by \(4\pi^2\):
\(4 \times \pi^2 \times 2.924 \times 10^{36} \approx 1.155 \times 10^{38} \, \text{m}^3\)
Next, the denominator \(GM\):
\(6.674 \times 10^{-11} \times 1.99 \times 10^{30} \approx 1.328 \times 10^{20} \, \text{N} \cdot \text{m}^2\)
Now, find \(T^2\):
\(T^2 \approx \frac{1.155 \times 10^{38}}{1.328 \times 10^{20}} \approx 8.697 \times 10^{17} \, \text{s}^2\)
Taking the square root gives:
\(T \approx \sqrt{8.697 \times 10^{17}} \approx 9.323 \times 10^8 \, \text{s}\)
- Conversion to Earth Years and Conclusion
To convert the orbital period \(T\) from seconds to Earth years, we use the fact that there are \(365.25\) days in a year, and each day has \(24 \times 3600 = 86400\) seconds.
One Earth year in seconds is:
\(1 \, \text{year} \approx 365.25 \times 86400 \approx 3.15576 \times 10^7 \, \text{s}\)
Now, convert \(T\):
\(T_{\text{years}} = \frac{9.323 \times 10^8 \, \text{s}}{3.15576 \times 10^7 \, \text{s/year}} \approx 29.54 \, \text{years}\)
Rounding our result to the nearest option, we find that the value is approximately \(29.6\) Earth years.
Examples
Kepler's laws are foundational in modern astrophysics and space exploration. By calculating orbital periods, scientists can determine the mass of central stars around which planets orbit or predict the travel time for missions to outer planets. For example, understanding that Saturn takes approximately 29.6 Earth years to complete one orbit around the Sun is crucial for planning long-term space probes like the Cassini-Huygens mission, which required complex trajectories and multiple gravitational assists over several years to reach its destination. Applying these physical formulas is essential for precise navigation in the solar system.