Conceptual Physics (12th Edition)

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

Chapter 9 - Plug and Chug: 34

Answer

$$F= G \frac{m_{1} m_{2}}{d^{2}} = (6.67 \times 10^{-11} \frac {N \cdot m^{2}}{kg^{2}}) \frac{(1 kg)(6.0 \times 10^{24} kg)}{(1.28 \times 10^{7} m)^{2}} = 2.4 N $$ The result is not surprising - by the inverse-square law, it is one-quarter the weight of a 1-kg mass at the surface.
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