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 7 - Newton's Third Law - Exercises and Problems - Page 177: 10


$a = 4.92\times 10^{-21}~m/s^2$

Work Step by Step

The gravitational force between the earth and the meteorite will cause both objects to accelerate toward each other. We can find the acceleration of the earth just before impact. Let $M_E$ be the mass of the earth. Then; $F_G = M_E~a$ $\frac{G~M_E~M_M}{R^2} = M_E~a$ $a = \frac{G~M_M}{R^2}$ $a = \frac{(6.67\times 10^{-11}~m^3/kg~s^2)(3000~kg)}{(6.38\times 10^6~m)^2}$ $a = 4.92\times 10^{-21}~m/s^2$
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.