UNIVERSE LL W/SAPLINGPLUS MULTI SEMESTER
UNIVERSE LL W/SAPLINGPLUS MULTI SEMESTER
11th Edition
ISBN: 9781319278670
Author: Freedman
Publisher: MAC HIGHER
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Chapter 4, Problem 51Q
To determine

The sidereal period of an Earth-like planet revolving around a star with four times the mass of the Sun, given that the semi-major axis of the planet is 1 au.

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If G = 6.674 ⨉ 10 −11m3/kg/s 2and M Earth= 5.972 ⨉ 10 24kg and the sidereal period of the Earth is 27.32 days, then, from Kepler’s third law in #4, what is the expected orbital distance of the Moon?
What would be the period of revolution of a hypothetical planet whose circular orbit around the sun has a radius of 1.75 AU? (Hint: 1 AU = 1 Astronomical Unit = 1.5*1011) a) 2.3 yrs                                              b) 1.45 yrs                                       c) 2.9 yrs
Neptune orbits the Sun with an orbital radius of 4.495 x 10^12 m. If the earth to sun distance 1A.U. = 1.5 x 10^11 m, a) Determine how many A.U.'s is Neptune's orbital radius (Round to the nearest tenth). b) Given the Sun's mass is 1.99 x10^30 kg, use Newton's modified version of Kepler's formula T^2 = (4pi^2/Gm(star)) x d^3 to find the period in seconds using scientific notation. (Round to the nearest thousandth). C) Convert the period in part b) to years (Round to the nearest tenth)

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UNIVERSE LL W/SAPLINGPLUS MULTI SEMESTER

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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