The /r^2 factor is due to the fact that the surface of a sphere grows proportional to r^2. That will hold for the wavefront half way around the earth, but from there it will be reverse. The energy will converge to full strength at the point opposite from the origin. That is if there were no loss. A significant portion of the energy is converted to heat and otherwise dispersed from being a compact wavefront.
The energy is dispersed in all directions through air molecules. At the top it dissipates and at the bottom it is absorbed by the earth. The only extra boost you'd get along vectors parallel to the surface would be reflections of the energy, which other commenters have said is small ... so this term doesn't change enough to get the function out of O(r^-2) does it?
You do get a boost from the wave contracting and converging towards the opposite point from halfway around the earth. I think it is not at all obvious what the order of damping will be.
