I am sorry but still confused. We didn't get H=Vo^2R( 2gR-Vo^2) did we? R is the earth's radius. If the gravitational attraction at the earth's surface is mg, then the attraction at some height r above the surface will be mgR^2/(R+r)^2. The attraction can be integrated from r = 0 to r = H to get the total work involved, which must equal the kinetic energy mVo^2/2. However, when we integral from r=0 to r=H of mgR^2/(R+r)^2 and set it equal to mVo^2/2, H= does not give the value Vo^2R/(2gR-Vo^2) ??