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 13 - Newton's Theory of Gravity - Exercises and Problems: 11

Answer

The free-fall acceleration on the surface of Titan is $1.35~m/s^2$

Work Step by Step

We can find the free-fall acceleration $g_T$ at the surface of Titan. $g_T = \frac{G~M}{R^2}$ $g_T = \frac{(6.67\times 10^{-11}~m^3/kg~s^2)(1.35\times 10^{23}~kg)}{(2.58\times 10^6~m)^2}$ $g_T = 1.35~m/s^2$ The free-fall acceleration on the surface of Titan is $1.35~m/s^2$.
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