## College Physics (4th Edition)

We can find the orbital speed: $v_o = \sqrt{\frac{G~M_E}{R}}$ $v_o = \sqrt{\frac{(6.67\times 10^{-11}~m^3/kg~s^2)~(5.97\times 10^{24}~kg)}{(4)(6.38\times 10^6~m)}}$ $v_o = 3.95\times 10^3~m/s = 3.95~km/s$ We can find the escape speed: $v_e = \sqrt{\frac{2G~M_E}{R}}$ $v_e = \sqrt{2}\times \sqrt{\frac{G~M_E}{R}}$ $v_e = \sqrt{2}\times v_o$ $v_e = \sqrt{2}\times 3.95~km/s$ $v_e = 5.59~km/s$ We can find the difference between the two speeds: $5.59~km/s - 3.95~km/s = 1.64~km/s$ To be able to escape Earth, the speed should be increased by 1.64 km/s.