## Calculus (3rd Edition)

$-2.3\times10^{-4}\,N/m$
We find the rate of change of the force as follows: $\frac{dF}{dr}=\frac{d}{dr}(\frac{2.99\times10^{16}}{r^{2}})=2.99\times10^{16}\times\frac{d}{dr}(\frac{1}{r^{2}})$ $=2.99\times10^{16}\times-\frac{2r}{r^{4}}=\frac{-5.98\times10^{16}}{r^{3}}$ $r$= radius of Earth$= 6378100\,m$ We plug in the radius: $\frac{dF}{dr}|_{r=6378100}=\frac{-5.98\times10^{16}}{(6378100)^{3}}$$=-2.3\times10^{-4}\,N/m$