## College Physics (4th Edition)

We can assume that kinetic energy and the total gravitational potential energy when the object is far away are both equal to zero. We can use conservation of energy to find the escape speed for an object near the Earth's surface: $KE+U_g +U_s = 0$ $\frac{1}{2}mv_e^2-\frac{G~M_E~m}{R_E}-\frac{G~M_s~m}{R_s} = 0$ $\frac{1}{2}v_e^2 = \frac{G~M_E}{R_E} + \frac{G~M_s}{R_s}$ $v_e^2 = 2G~(\frac{M_E}{R_E} + \frac{M_s}{R_s})$ $v_e = \sqrt{2G~(\frac{M_E}{R_E} + \frac{M_s}{R_s})}$ $v_e = \sqrt{(2)(6.67\times 10^{-11}~m^3/kg~s^2)~(\frac{5.97\times 10^{24}~kg}{6.38\times 10^6~m} + \frac{1.989\times 10^{30}~kg}{6.96\times 10^8~m})}$ $v_e = 6.18\times 10^4~m/s = 61.8~km/s$ To escape from the Earth and the sun, the minimum speed is 61.8 km/s.