Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.7 - Rates of Change in the Natural and Social Sciences - 3.7 Exercises - Page 234: 20

Answer

(a) $\frac{dF}{dr} = \frac{-2GmM}{r^3}$ The minus sign indicates that the force decreases as $r$ increases. (b) $\frac{dF}{dr} = -16~N/km$

Work Step by Step

(a) $F = \frac{GmM}{r^2}$ $\frac{dF}{dr} = \frac{-2GmM}{r^3}$ The minus sign indicates that the force decreases as $r$ increases. (b) $\frac{dF}{dr} = \frac{-2GmM}{r^3}$ $\frac{-2GmM}{(20,000)^3} = -2$ $GmM = 20,000$ We can find $\frac{dF}{dr}$ when $r = 10,000~km$: $\frac{dF}{dr} = \frac{-2GmM}{r^3}$ $\frac{dF}{dr} = \frac{-2(20,000)^3}{(10,000)^3}$ $\frac{dF}{dr} = (-2)(2)^3$ $\frac{dF}{dr} = -16~N/km$
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