#### Answer

Since the total energy in the system is positive, the person will be able to escape from the asteroid.

#### Work Step by Step

We first find the initial speed of the jump on the earth.
$v^2 = v_0^2+2gy$
$v_0^2= 0 - 2gy$
$v_0 = \sqrt{-(2)(-9.80~m/s^2)(0.50~m)}$
$v_0 = 3.13~m/s$
We then find an expression for the initial kinetic energy when jumping on the earth. Let $M_p$ be the person's mass.
$K_0 = \frac{1}{2}M_p~v_0^2$
Note that on the asteroid, the initial kinetic energy will be the same. We can find the total energy in the system. Let $M_a$ be the asteroid's mass.
$E = K_0+U_0$
$E = \frac{1}{2}M_p~v_0^2-\frac{G~M_a~M_p}{R}$
$E = \frac{1}{2}M_p~(3.13~m/s)^2-\frac{(6.67\times 10^{-11}~m^3/kg~s^2)(1.0\times 10^{14}~kg)~M_p}{2000~m}$
$E = (4.90~m^2/s^2)~M_p-(3.34~m^2/s^2)~M_p$
$E = (1.56~M_p)~J$
Since the total energy in the system is positive, the person will be able to escape from the asteroid.