21st Century Astronomy
21st Century Astronomy
6th Edition
ISBN: 9780393428063
Author: Kay
Publisher: NORTON
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Chapter 7, Problem 35QP
To determine

The ratio of angular momenta of jupiter and earth.

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Neptune orbits the Sun with an orbital radius of 4.495 x 10^12 m. If the earth to sun distance 1 A.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 x 10^30 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).
Calculate the angular momentum (in kg · m2/s) of Mercury in its orbit around the Sun. (The mass of Mercury is 3.300 ✕ 1023 kg, the orbital radius is 5.790 ✕ 107 km and the orbital period is 0.241 y.) kg · m2/s (b) Compare this angular momentum with the angular momentum of Mercury on its axis. (The radius of Mercury is 2.440 ✕ 103km and the rotation period is 1408 h.) Lorbital Lrotation =
Calculate the angular momentum (in kg · m2/s) of Mercury in its orbit around the Sun. (The mass of Mercury is 3.300 ✕ 1023 kg, the orbital radius is 5.790 ✕ 107 km and the orbital period is 0.241 y.) 9.213e32 is wrong
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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