Answer
$v = 1700~m/s$
Work Step by Step
We can write a general expression for the gravitational acceleration at the surface:
$g = \frac{GM}{R^2}$
We can write a general expression for the escape speed:
$v = \sqrt{\frac{2GM}{R}}$
We can find the escape speed:
$v = \sqrt{\frac{2GM}{R}}$
$v = \sqrt{2~g~R}$
$v = \sqrt{(2)~(3.0~m/s^2)(5.0\times 10^5~m)}$
$v = 1700~m/s$
