Ok...I've struggled with this. Let's say I'm dropping a mass from height h from the earth's surface. Since I'm considering this as a non-inertial reference frame, I understand that there is an easternly deflection. I've done the calculation and end up with de=(1/3)(w)cos(lamda)((8h^2)/g)^(1/2) where omega is the rate of rotation of the earth and lamda is the degree latitude. Problem. "Repeat the calculation to second order in omega and find the southernly deflection." The easternly deflection I'm good with. I don't understand how there is any southern deflection. The calculation and explination is eluding me at this point.