## Essential University Physics: Volume 1 (3rd Edition)

Using conservation of energy, we see: $\frac{-GM_Em}{R_E}+\frac{1}{2}mv_0^2=K_0+U_0$ From this, we use simplification using known equations to find: $h = \frac{R_Eh'}{(R_E-h')} \\ h = \frac{.99hR_E}{(R_E-.99h)}\\ h = \frac{R_E}{99}=\frac{6.4\times10^6}{99}\approx \fbox{64,000 meters}$