## Essential University Physics: Volume 1 (4th Edition)

a) $14,833 \ m/s$ b) $22,804 \ m/s$
We use the law of conservation of energy to obtain: $\frac{1}{2}mv^2-\frac{1}{2}mv_0^2 =GMm(\frac{1}{r_1}-\frac{1}{r_2})$ $\frac{1}{2}v^2-\frac{1}{2}v_0^2 =GM(\frac{1}{r_1}-\frac{1}{r_2})$ $v^2-v_0^2 =2GM(\frac{1}{r_1}-\frac{1}{r_2})$ $v^2 =2GM(\frac{1}{r_1}-\frac{1}{r_2})+v_0^2$ $v =\sqrt{2GM(\frac{1}{r_1}-\frac{1}{r_2})+v_0^2}$ Using this, we find that the two speeds of impact are: a) $14,833 \ m/s$ b) $22,804 \ m/s$