## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

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.