## Essential University Physics: Volume 1 (4th Edition) Clone

We need to make the asteroid so that the velocity of the jumper is below the escape speed of the asteroid. Thus, we first find how fast a jumper that goes 3 meters up on earth is jumping using conservation of energy: $\frac{1}{2}mv^2 = mgh \\ v=\sqrt{2gh}=\sqrt{2\times9.81\times3}=7.672 \ m/s$ We know the following equation for escape speed: $v_{esc}=\sqrt{\frac{2GM}{r}}$ $v_{esc}=\sqrt{\frac{2G(\frac{4}{3}\pi r^3)(2500)}{r}}$ $r = \sqrt{\frac{v_{esc}^2}{6666.67\pi G}}$ Plugging in the value for G (a constant) and v (found above) gives: $r=9491 m$ This means that: $\fbox{d=12,982 meters}$