## College Physics (4th Edition)

We can use conservation of energy to find the initial speed $v_i$: $\frac{1}{2}mv_i^2-\frac{G~M_E~m}{R_E} = 0-\frac{G~M_E~m}{5R_E}$ $\frac{1}{2}mv_i^2 = \frac{4G~M_E~m}{5R_E}$ $v_i^2 = \frac{8G~M_E}{5R_E}$ $v_i = \sqrt{\frac{8G~M_E}{5R_E}}$ $v_i = \sqrt{\frac{(8)(6.67\times 10^{-11}~m^3/kg~s^2)~(5.97\times 10^{24}~kg)}{(5)(6.38\times 10^6~m)}}$ $v_i = 9.99\times 10^3~m/s = 9.99~km/s$ The projectile would need an initial speed of 9.99 km/s