## College Physics (4th Edition)

Note that the gravitational field around at a planet's surface is: $\sqrt{\frac{GM_p}{R_p^2}}$ We can find the escape speed from the planet: $v = \sqrt{\frac{2GM_p}{R_p}}$ $v = \sqrt{\frac{(GM_p)~(2R_p)}{R_p^2}}$ $v = \sqrt{(30.0~m/s^2)(2)(6.00\times 10^7~m)}$ $v = 6.00\times 10^4~m/s = 60.0~km/s$ The escape speed from the planet is 60.0 km/s.