Answer
$-2.3\times10^{-4}\,N/m$
Work Step by Step
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$
