3 ms·

I don't think this is correct. A sound wave propagates in three dimensions the same as an electrical field (spherically) and so should also obey the inverse squ

by abduhl 12y ago

I don't think this is correct. A sound wave propagates in three dimensions the same as an electrical field (spherically) and so should also obey the inverse square law.

femto 12y ago
In this case, the atmosphere is essentially a thin (compared to the radius of the Earth) two dimensional sheet wrapping the Earth, so it's probably valid to think of the sound as propagating on a surface. There wouldn't be much loss of energy to space. Would the atmospheric interface with the ground reflect or absorb? The divergence is probably even less that 1/r because the Earth is roughly sperical, and the sound will focus at the antipode of its source. There might even be a region, approaching the focus, where as the sound propagates its sound level is actually increasing, not decreasing! It would be interesting to know what history has to say about Medellin, in Colombia, in 1883. Being at coordinates 6N 75E, it is roughly at the antipode of Krakatoa and could well have had an awful lot of energy dumped in it general direction a few hours after the eruption.
- sillysaurus3 12y ago
  Would gravity affect soundwaves at all? If sound is traveling parallel to the Earth's surface, will it follow along the curvature of the Earth (implying gravity has an effect on it) or was it able to propagate around the Earth four times simply because the intensity was so great that it was able to still be heard despite the loss of energy in the "up-down" directions relative to the origin of the sound? If I had to completely guess, I would think that gravity might have something to do with its behavior in this case, because if sound decayed at 1/r^2, it seems highly unlikely that the sound could survive four trips around the Earth. Yet we also have to explain how the sound was able to preserve itself even though sound travels "sort of spherically" (if you're up in the sky and someone shoots at you, you can hear it, so sound must be going up-down as well as left-right and every other direction) but if it was simply spherical in this case then it would decay into the upper atmosphere. If you imagine a sphere that grows from any point on the surface of the Earth, most of it will wind up in space, but since it circled the Earth, it must not have traveled exactly as sphere, which implies it didn't fall off as 1/r^2. The most likely explanation seems to be that sound of huge intensity somehow follows the curvature of the Earth, which is very interesting. I wonder if gravity is influencing the soundwaves directly (which might actually be nonsense, since waves aren't matter, but something that travels through matter) or if gravity somehow sets up the propagation medium in a way that lets the sound curve around the earth without losing intensity as 1/r^2.
  - kruczek 12y ago
    "which might actually be nonsense, since waves aren't matter, but something that travels through matter" I don't think that's much of a nonsense. These are waves propagating through a medium - in this case mostly air. And particles of air are affected by gravity, therefore I would very much expect them to follow Earth's curvature.
    - lutusp 12y ago
      > And particles of air are affected by gravity, therefore I would very much expect them to follow Earth's curvature. Gravity isn't the reason sound waves curve around the earth. One reason is that the curved surface is the only avenue open to them. There are a number of other, less influential reasons like different temperatures at different altitudes, but gravity isn't on the list.
  - jschwartzi 12y ago
    You're partially correct about Gravity, but there's more to it. The radius of the Earth is roughly 6000 km, while the thickness of the atmosphere is about 120km. Thus, the ratio of the two heights is about 1/60. Because the ratio of atmosphere thickness to terrestrial thickness is so small, for the purposes of modeling wave propagation through the atmosphere we can approximate the atmosphere as a thin spherical shell around the Earth. Furthermore, the Earth's surface forms a hard boundary below the shell, further confining the sound to the atmosphere, while the force of gravity also prevents the sound wave from escaping into space via the atmosphere. As other posters have stated, this forms a 2-dimensional propagation layer which results in a linear falloff rather than an inverse square falloff.

by abduhl 12y ago

I don't think this is correct. A sound wave propagates in three dimensions the same as an electrical field (spherically) and so should also obey the inverse square law.

femto 12y ago

In this case, the atmosphere is essentially a thin (compared to the radius of the Earth) two dimensional sheet wrapping the Earth, so it's probably valid to think of the sound as propagating on a surface. There wouldn't be much loss of energy to space. Would the atmospheric interface with the ground reflect or absorb? The divergence is probably even less that 1/r because the Earth is roughly sperical, and the sound will focus at the antipode of its source. There might even be a region, approaching the focus, where as the sound propagates its sound level is actually increasing, not decreasing! It would be interesting to know what history has to say about Medellin, in Colombia, in 1883. Being at coordinates 6N 75E, it is roughly at the antipode of Krakatoa and could well have had an awful lot of energy dumped in it general direction a few hours after the eruption.

sillysaurus3 12y ago
Would gravity affect soundwaves at all? If sound is traveling parallel to the Earth's surface, will it follow along the curvature of the Earth (implying gravity has an effect on it) or was it able to propagate around the Earth four times simply because the intensity was so great that it was able to still be heard despite the loss of energy in the "up-down" directions relative to the origin of the sound? If I had to completely guess, I would think that gravity might have something to do with its behavior in this case, because if sound decayed at 1/r^2, it seems highly unlikely that the sound could survive four trips around the Earth. Yet we also have to explain how the sound was able to preserve itself even though sound travels "sort of spherically" (if you're up in the sky and someone shoots at you, you can hear it, so sound must be going up-down as well as left-right and every other direction) but if it was simply spherical in this case then it would decay into the upper atmosphere. If you imagine a sphere that grows from any point on the surface of the Earth, most of it will wind up in space, but since it circled the Earth, it must not have traveled exactly as sphere, which implies it didn't fall off as 1/r^2. The most likely explanation seems to be that sound of huge intensity somehow follows the curvature of the Earth, which is very interesting. I wonder if gravity is influencing the soundwaves directly (which might actually be nonsense, since waves aren't matter, but something that travels through matter) or if gravity somehow sets up the propagation medium in a way that lets the sound curve around the earth without losing intensity as 1/r^2.
- kruczek 12y ago
  "which might actually be nonsense, since waves aren't matter, but something that travels through matter" I don't think that's much of a nonsense. These are waves propagating through a medium - in this case mostly air. And particles of air are affected by gravity, therefore I would very much expect them to follow Earth's curvature.
  - lutusp 12y ago
    > And particles of air are affected by gravity, therefore I would very much expect them to follow Earth's curvature. Gravity isn't the reason sound waves curve around the earth. One reason is that the curved surface is the only avenue open to them. There are a number of other, less influential reasons like different temperatures at different altitudes, but gravity isn't on the list.
- jschwartzi 12y ago
  You're partially correct about Gravity, but there's more to it. The radius of the Earth is roughly 6000 km, while the thickness of the atmosphere is about 120km. Thus, the ratio of the two heights is about 1/60. Because the ratio of atmosphere thickness to terrestrial thickness is so small, for the purposes of modeling wave propagation through the atmosphere we can approximate the atmosphere as a thin spherical shell around the Earth. Furthermore, the Earth's surface forms a hard boundary below the shell, further confining the sound to the atmosphere, while the force of gravity also prevents the sound wave from escaping into space via the atmosphere. As other posters have stated, this forms a 2-dimensional propagation layer which results in a linear falloff rather than an inverse square falloff.