Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 8 - Section 8.2 - Universal Gravitation - Example - Page 135: 8.1

Answer

The answer is below.

Work Step by Step

We use the given equation to find: a) $ a = \frac{GM_E}{R_E^2} = \frac{6.67\times 10^{-11}\times5.67\times10^{24}}{(6.37 \times 10^6)^2} = 9.81 \ m/s^2$ b) $ a = \frac{GM_E}{r^2} = \frac{6.67\times 10^{-11}\times5.67\times10^{24}}{(6.37 \times 10^6+380 \times 10^3)^2} = 8.74\ m/s^2$
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