## Calculus: Early Transcendentals 8th Edition

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