Conceptual Physics (12th Edition)

Published by Addison-Wesley
ISBN 10: 0321909100
ISBN 13: 978-0-32190-910-7

Chapter 9 - Plug and Chug - Page 178: 33


$$F= G \frac{m_{1} m_{2}}{d^{2}} = (6.67 \times 10^{-11} \frac {N \cdot m^{2}}{kg^{2}}) \frac{(1 kg)(6.0 \times 10^{24} kg)}{(6.4 \times 10^{6} m)^{2}} = 9.8 N $$ The result is not surprising - it is the weight of a 1-kg mass at the surface of the Earth.
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.