Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 8 - Dynamics II: Motion in a Plane - Exercises and Problems: 11

Answer

The free-fall acceleration on the surface of the moon is $1.58~m/s^2$

Work Step by Step

Let $g_m$ be the free-fall acceleration on the surface of the moon; $g_m = \frac{G~M_m}{r^2}$ To find an expression for $g_m$, we can use the equation of the period $T$ for an object's orbit. $T^2 = \frac{4\pi^2~r^3}{G~M_m}$ $\frac{G~M_m}{r^2} = \frac{4\pi^2~r}{T^2}$ $g_m = \frac{4\pi^2~r}{T^2}$ $g_m = \frac{(4\pi^2)(1.74\times 10^6~m)}{[(110~min)(60~s/min)]^2}$ $g_m = 1.58~m/s^2$ The free-fall acceleration on the surface of the moon is $1.58~m/s^2$.
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