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

a) $4.59\ km/s$ b) $14.2\ km/s$
We use the law of conservation of energy to obtain: $\frac{1}{2}mv^2-\frac{1}{2}mv_0^2 =GMm(\frac{1}{r_1}-\frac{1}{r_2})$ $\frac{1}{2}v^2-\frac{1}{2}v_0^2 =GM(\frac{1}{r_1}-\frac{1}{r_2})$ $v^2-v_0^2 =2GM(\frac{1}{r_1}-\frac{1}{r_2})$ $v^2 =2GM(\frac{1}{r_1}-\frac{1}{r_2})+v_0^2$ $v =\sqrt{2GM(\frac{1}{r_1}-\frac{1}{r_2})+v_0^2}$ We know that the moon's orbit is $3.85\times10^8$ meters away from the center of the earth. Thus, we find: a) $v =\sqrt{2(6.67\times10^{-11})(5.97\times10^{24})(\frac{1}{6.37\times10^6}-\frac{1}{3.85\times10^8})+(12,000)^2}=4.59\ km/s$ b) $v =\sqrt{2(6.67\times10^{-11})(5.97\times10^{24})(\frac{1}{6.37\times10^6}-\frac{1}{3.85\times10^8})+(18,000)^2}=14.2\ km/s$