Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 6 - Dynamics I: Motion Along a Line - Exercises and Problems - Page 155: 23


$a = 5.90\times 10^{-3}~m/s^2$

Work Step by Step

We can write the expression for the force of gravity as: $F_G = \frac{GM_sM_E}{r^2}$ We can use a force equation to find the acceleration of the earth toward the sun: $F_G = M_E~a$ $\frac{GM_sM_E}{r^2} = M_E~a$ $a = \frac{GM_s}{r^2}$ $a = \frac{(6.67\times 10^{-11}~m^3 / kg~s^2)(1.99\times 10^{30}~kg)}{(1.50\times 10^{11}~m)^2}$ $a = 5.90\times 10^{-3}~m/s^2$
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.