College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 8 - Problems - Page 311: 8

Answer

The ratio of the Earth's rotational inertia about its own axis to the Earth's rotational inertia about the sun is $7.25\times 10^{-10}$

Work Step by Step

We can find the rotational inertia $I_a$ of the Earth about its own axis: $I_a = \frac{2}{5}MR_E^2$ $I_a = \frac{2}{5}(5.97\times 10^{24}~kg)(6.38\times 10^6~m)^2$ $I_a = 9.72\times 10^{37}~kg~m^2$ We can find the rotational inertia $I_s$ of the Earth about the sun: $I_s = MR^2$ $I_s = (5.97\times 10^{24}~kg)(1.50\times 10^{11}~m)^2$ $I_s = 1.34\times 10^{47}~kg~m^2$ We can find the ratio: $\frac{I_a}{I_s} = \frac{9.72\times 10^{37}~kg~m^2}{1.34\times 10^{47}~kg~m^2} = 7.25\times 10^{-10}$ The ratio of the Earth's rotational inertia about its own axis to the Earth's rotational inertia about the sun is $7.25\times 10^{-10}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.