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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.

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.

Applications and Investigations in Earth Science (9th Edition)
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
Problem 1LR
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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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