Answer
$v = 2.4\times 10^4~m/s$
Work Step by Step
We can use conservation of energy to find the meteorite's speed at the surface:
$K_f+U_f = K_i+U_i$
$K_f+U_f = 0$
$K_f = -U_f$
$\frac{1}{2}mv^2 = -(-\frac{GMm}{r})$
$v^2 = \frac{2GM}{r}$
$v = \sqrt{\frac{2GM}{r}}$
$v = \sqrt{\frac{(2)(6.67\times 10^{-11}~N~m^2/kg^2)(7.0\times 10^{24}~kg)}{1.6\times 10^6~m}}$
$v = 2.4\times 10^4~m/s$
