Answer
a) $14,833 \ m/s$
b) $22,804 \ m/s$
Work Step by Step
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) $22,804 \ m/s$
