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Applications and Investigations in Earth Science (9th Edition)
9th Edition
ISBN:9780134746241
Author:Edward J. Tarbuck, Frederick K. Lutgens, Dennis G. Tasa
Publisher:Edward J. Tarbuck, Frederick K. Lutgens, Dennis G. Tasa
Chapter1: The Study Of Minerals
Section: Chapter Questions
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I just received this answer from Bartleby but I don't know how the person got to the final answer. Could you please explain further, any conversions that were made. 

Step 1
Kepler's third law states that the planets' orbital periods' square is directly proportional to the
cube of the semi-major axes of their orbits. So it means as the radius of the orbit of the planet
gets increased, the planet's period to orbit the sun also increases.
Step 2
We know that the Earth-Sun distance is 1 AU.
It is given that the distance of the planet from the sun is 18AU.
According to the third law of Kepler,
T2=R3
T2= (18)3
T2= 5832
T=76 years.
The orbital period of the planet would be 76 Earth-years.
Transcribed Image Text:Step 1 Kepler's third law states that the planets' orbital periods' square is directly proportional to the cube of the semi-major axes of their orbits. So it means as the radius of the orbit of the planet gets increased, the planet's period to orbit the sun also increases. Step 2 We know that the Earth-Sun distance is 1 AU. It is given that the distance of the planet from the sun is 18AU. According to the third law of Kepler, T2=R3 T2= (18)3 T2= 5832 T=76 years. The orbital period of the planet would be 76 Earth-years.
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