Answer
$v = (1.6\times 10^6)~m/s$
Work Step by Step
We can find the gravitational acceleration at the surface:
$a_g = \frac{GM}{R^2}$
$a_g = \frac{(6.67\times 10^{-11}~N~m^2/kg^2)(2.0\times 10^{30}~kg)}{(10^4~m)^2}$
$a_g = 1.3\times 10^{12}~m/s^2$
We can find the speed of an object after falling $1.0~m$:
$v^2 = v_0^2+2ay$
$v^2 = 0+2ay$
$v = \sqrt{2ay}$
$v = \sqrt{(2)(1.3\times 10^{12}~m/s^2)(1.0~m)}$
$v = (1.6\times 10^6)~m/s$
