Answer
$v = 1400~m/s$
Work Step by Step
We can write a general expression for the gravitational acceleration at the surface:
$g = \frac{GM}{R^2}$
Then: $~~GM = gR^2$
We can use conservation of energy to find the speed at the surface of the asteroid:
$K_2+U_2 = K_1+U_1$
$K_2 = 0+U_1-U_2$
$\frac{1}{2}mv^2 = -\frac{GMm}{3R} -(-\frac{GMm}{R})$
$v^2 = \frac{2GM}{R}-\frac{2GM}{3R}$
$v^2 = \frac{2gR^2}{R}-\frac{2gR^2}{3R}$
$v^2 = 2gR-\frac{2gR}{3}$
$v^2 = \frac{4gR}{3}$
$v = \sqrt{\frac{4gR}{3}}$
$v = \sqrt{\frac{(4)(3.0~m/s^2)(5.0\times 10^5~m)}{3}}$
$v = 1400~m/s$
