For a spherical celestial object of radius R, the acceleration due to gravity g at a distance x from the center of the object is g = (g0)(R^2)/(x^2), where g0 is the acceleration due to gravity at the object's surface and x > R. For the moon, take g0 = 1.63 m/s2 and R = 3200 km. If a rock is released from rest at a height of 3R above the lunar surface, with what speed does the rock impact the moon? Hint: Its acceleration is a function of position and increases as the object falls. So do not use constant acceleration free-fall equations, but go back to basics.
.5(v^2-v0^2) = (g0)(R^2)(1/x-1/x0)
The Attempt at a Solution
i said that v0=0 and solved for v, x=3R x0=R, i got .899km/s. apparently that is incorrect.