If sound traveled as a sphere, then it wouldn't be able to circle the Earth. If you imagine a sphere growing from a point on the Earth's surface, most of it will wind up in space, and none of it will reach the opposite side of the Earth. The 1/r^2 falloff is probably just an approximation that holds true for short distances or smaller intensities.
In this case, the soundwave was able to follow the curvature of the Earth, which implies that it wouldn't decay as a sphere but rather a plane, which would be a 1/r falloff. I wonder why sound of massive intensity will follow Earth's curvature?
Why don't you think that the sound waves would go in all directions, and be absorbed / redirected at the edges? What you see following the curvature of the Earth is the result of that, and once you factor in the second-order wave reflection, it's still on the order of 1/r^2 isn't it?
Nah, imagine a point on the earth. Now trace rays from that point in every possible direction. All of those rays will eventually lead to space or to the ground, and none will reach the other side of the Earth. Since molecules are physically further apart the higher you go in the atmosphere, it seems unlikely that the energy would be redirected at the edges, only dispersed. That must mean the wave is literally following wherever the atmosphere is thickest rather than simply being reflected.
When I hear "energy dispersed at the edges", I think "lost" (or effectively absorbed), not "reflected back the way it came". I can't think of any obvious mechanism that would lead to significant coherent reflection of sound waves off the upper atmosphere. Similarly, I think a lot of the sound intensity would be absorbed by material on the ground rather than reflected back up. I vote for predominantly 1/r^2 behavior.
[As a physics prof, I probably ought to be doing some sort of calculation to justify that, but I don't have the time. I'm sure it's been analyzed formally somewhere already. Just off the cuff, though, the speed of sound is lower at low temperature, so refraction of a sound wave in the upper atmosphere will tend to bend it outward, away from the Earth. I'm thus imagining that loud sounds may tend to locally push the atmosphere out from the planet a little bit, and I expect that coming back to equilibrium will be a lossy process that doesn't preserve the shape of the wave.]
Is diffraction the reason why soundwaves could travel around the whole Earth? My previous understanding of diffraction was that obstacles cause waves to propagate in different ways. But the thinning of the atmosphere isn't really an "obstacle." The molecules that soundwaves use to propagate are simply further apart from each other, meaning waves are more likely to disperse and lose energy than to keep traveling or bounce. That would imply the boom from the volcano should disperse into space and go silent rather than travel around the Earth. But since that doesn't happen, it seems like the waves follow wherever the atmosphere is thick.
I'm having trouble understanding how diffraction would cause that end result of "waves go where the atmosphere is." If waves could bounce off of the thin atmosphere near space, that would make total sense. But they can't bounce due to thin atmosphere, only disperse, so it seems like there's some other phenomenon in play.
Sound waves are points of high pressure and points of low pressure. At each point of high pressure the pressure tends to spread uniformly in all directions. Likewise at points of low pressure there is sucking from all directions. Opposite directions cancel, i.e. orthogonal to the wave direction. All in all the sum of all points of pressure creates a moving wavefront. Which naturally bends around obstacles. The earth is just a very big obstacle.
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?
In this case, the soundwave was able to follow the curvature of the Earth, which implies that it wouldn't decay as a sphere but rather a plane, which would be a 1/r falloff. I wonder why sound of massive intensity will follow Earth's curvature?