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

We can find the acceleration due to gravity $g_m$ on the surface of Mars as: $g_m = \frac{G~M_m}{R_m^2}$ $g_m = \frac{(6.67\times 10^{-11}~m^3/kg~s^2)(6.42\times 10^{23}~kg)}{(3.39\times 10^6~m)^2}$ $g_m = 3.73~m/s^2$ The astronaut can give the ball a certain amount of kinetic energy. This initial kinetic energy is the same on the earth as it is on Mars. Then the potential energy at the highest point will be the same on earth and Mars. We can find the height $h_m$ that the ball reaches on Mars. $m~g_m~h_m = mgh$ $h_m = \frac{gh}{g_m}$ $h_m = \frac{(9.80~m/s^2)(15~m)}{3.73~m/s^2}$ $h_m = 39.4~m$ On Mars, the astronaut can throw the ball to a height of 39.4 meters.