Answer
0.223 m/$s^{2}$
Work Step by Step
Using known equations, we find:
g(h) = $\frac{F(h)}{m}$ = $\frac{GM_{E}}{(R_{E}+h)^{2}}$
= $\frac{(6.67\times10^{-11}N\frac{m^{2}}{kg^{2}})(5.972\times10^{24}kg)}{(6371×10^{3}m+3.59×10^{7}m)^{2}}$ = 0.223 m/$s^{2}$
