QuestionJune 15, 2025

If it takes a planet 2.8times 10^8 s to orbit a star with a mass of 6.2times 10^30 kg, what is the average distance between the planet and the star? 1.43times 10^9 m 9.36times 10^11 m 5.42times 10^13 m 9.06times 10^17 m

If it takes a planet 2.8times 10^8 s to orbit a star with a mass of 6.2times 10^30 kg, what is the average distance between the planet and the star? 1.43times 10^9 m 9.36times 10^11 m 5.42times 10^13 m 9.06times 10^17 m
If it takes a planet 2.8times 10^8 s to orbit a star with a mass of 6.2times 10^30 kg, what is the average distance between the
planet and the star?
1.43times 10^9 m
9.36times 10^11 m
5.42times 10^13 m
9.06times 10^17 m

Solution
4.4(203 votes)

Answer

9.36\times 10^{11} m Explanation 1. Use Kepler's Third Law Kepler's Third Law states T^2 = \frac{4\pi^2}{G M} a^3, where T is the orbital period, M is the mass of the star, and a is the semi-major axis (average distance). 2. Rearrange for Semi-Major Axis Solve for a: a^3 = \frac{G M T^2}{4\pi^2}. 3. Substitute Known Values Given T = 2.8\times 10^{8} s, M = 6.2\times 10^{30} kg, G = 6.674 \times 10^{-11} \text{ m}^3 \text{kg}^{-1} \text{s}^{-2}. Calculate a^3 = \frac{(6.674 \times 10^{-11})(6.2\times 10^{30})(2.8\times 10^{8})^2}{4\pi^2}. 4. Compute the Average Distance Calculate a = \sqrt[3]{a^3} to find the average distance.

Explanation

1. Use Kepler's Third Law<br /> Kepler's Third Law states $T^2 = \frac{4\pi^2}{G M} a^3$, where $T$ is the orbital period, $M$ is the mass of the star, and $a$ is the semi-major axis (average distance).<br />2. Rearrange for Semi-Major Axis<br /> Solve for $a$: $a^3 = \frac{G M T^2}{4\pi^2}$.<br />3. Substitute Known Values<br /> Given $T = 2.8\times 10^{8}$ s, $M = 6.2\times 10^{30}$ kg, $G = 6.674 \times 10^{-11} \text{ m}^3 \text{kg}^{-1} \text{s}^{-2}$.<br /> Calculate $a^3 = \frac{(6.674 \times 10^{-11})(6.2\times 10^{30})(2.8\times 10^{8})^2}{4\pi^2}$.<br />4. Compute the Average Distance<br /> Calculate $a = \sqrt[3]{a^3}$ to find the average distance.
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