21st Century Astronomy
21st Century Astronomy
6th Edition
ISBN: 9780393428063
Author: Kay
Publisher: NORTON
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Chapter 3, Problem 36QP
To determine

Prove that the planet with period of 3 years and distance from 2 AU is not possible.

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A)At what altitude would a geostationary sattelite need to be above the surface of Mars? Assume the mass of Mars is 6.39 x 1023 kg, the length of a martian solar day is 24 hours 39minutes 35seconds, the length of the sidereal day is 24hours 37minutes 22seconds, and the equatorial radius is 3396 km. The answer can be calculated using Newton's verison of Kepler's third law.
(a) Jupiter's third-largest natural satellite, Io, follows an orbit with a semimajor axis of 422,000 km (4.22 ✕ 105 km) and a period of 1.77 Earth days (PIo = 1.77 d). To use Kepler's Third Law, we first must convert Io's orbital semimajor axis to astronomical units. One AU equals 150 million km (1 AU = 1.50 ✕ 108 km). Convert Io's a value to AU and record the result. aIo =  AU (b) One Earth year is about 365 days. Convert Io's orbital period to Earth years and record the result. PIo =  yr (c) Use the Kepler's Third Law Calculator to calculate Jupiter's mass in solar units. Record the result. MJup(Io) =  MSun (d) Based on this result, Jupiter's mass is about       that of the Sun. Jupiter has a similar fraction of the Sun's volume. The two objects therefore have rather similar density! In fact, Jupiter has a fairly similar composition as well: most of its mass is in the form of hydrogen and helium.
Congratulations! You just derived a version of Kepler's Third Law for Mars! Using the mass of Mars in kilograms and converting the 4.5 hours to seconds, calculate the distance from the center of the planet. GM kg 4π² ]s)² 3 = And then determine the distance (in km) from the surface. r = rm + rs rs km = km
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Kepler's Three Laws Explained; Author: PhysicsHigh;https://www.youtube.com/watch?v=kyR6EO_RMKE;License: Standard YouTube License, CC-BY