- Kepler's third law relates orbital periods and distances as \(\frac{T_M^2}{T_E^2} = \frac{r_M^3}{r_E^3}\).
- By rearranging for \(r_E\), we get the formula \(r_E = r_M \left(\frac{T_E}{T_M}\right)^{2/3}\).
- Substituting the given values \(T_M=687\), \(T_E=365.25\), and \(r_M=2.279 \times 10^{11}\) yields \(r_E \approx 1.49566 \times 10^{11}\) m.
- The correct choice corresponds to \(\boxed{1.49 \times 10^{11} \text{ m}}\).
Explanation
- Analyzing Kepler's Third Law and Given Data
Kepler's third law of planetary motion establishes a fundamental relationship between the orbital period \(T\) of a planet and its average distance \(r\) from the Sun. It states that the square of the orbital period is directly proportional to the cube of the semi-major axis, represented by the ratio:
\[\frac{T_M^2}{T_E^2} = \frac{r_M^3}{r_E^3}\]
where \(T_M\) and \(r_M\) are the orbital period and distance of Mars, and \(T_E\) and \(r_E\) are those of Earth. We are given the following values:
- Mars's orbital period: \(T_M = 687\) days.
- Mars's distance from the Sun: \(r_M = 2.279 \times 10^{11}\) m.
- Earth's orbital period: \(T_E = 365.25\) days.
Our goal is to solve for Earth's orbital distance \(r_E\).
- Solving for Earth's Orbital Distance
To find Earth's distance \(r_E\), we rearrange the ratio formula:
\[r_E^3 = r_M^3 \cdot \frac{T_E^2}{T_M^2}\]
Taking the cube root of both sides, we isolate \(r_E\):
\[r_E = r_M \cdot \left(\frac{T_E}{T_M}\right)^{2/3}\]
Substituting the provided values into the equation:
\[r_E = 2.279 \times 10^{11} \cdot \left(\frac{365.25}{687}\right)^{2/3}\]
Performing the division:
\[\frac{365.25}{687} \approx 0.53166\]
Raising this to the power of \(2/3\):
\[(0.53166)^{2/3} \approx 0.65628\]
Multiplying this by \(r_M\):
\[r_E \approx 2.279 \times 10^{11} \cdot 0.65628 \approx 1.49566 \times 10^{11} \text{ m}\]
Rounding this result to three significant figures, we obtain \(1.49 \times 10^{11}\) m.
- Conclusion
By comparing our calculated value \(r_E \approx 1.49 \times 10^{11}\) m with the given options, we identify that the correct distance is \(1.49 \times 10^{11}\) m. This confirms the accuracy of applying Kepler's third law in this planetary context. The final answer is \(\boxed{1.49 \times 10^{11} \text{ m}}\).
Examples
Kepler's third law is essential in modern astrophysics for determining the distances of exoplanets from their host stars. By measuring the orbital period of a distant planet as it passes in front of its star (transit method), astronomers use this law to calculate how far the planet is from its sun. This helps scientists understand the temperature and potential habitability of planets outside our solar system, essentially allowing them to map out the geography of distant star systems from light-years away.