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Background

Obviously, being next to explosions is bad. The farther away one is from the explosion, the larger the hemisphere of the shock wave is, and so the more the energy from the explosion is dissipated. Typically, this rate of loss of energy density is modeled as $1/r^2$, where $r$ is the distance from the detonation.

However, this model misses an interesting effect that occurs (at least hypothetically) for sufficiently large explosions. Supposing that the explosion in question occurred at the north pole, once the shock wave moves past the equator, the size of the circle formed by the shock wave moving along the ground decreases, and eventually converges on the south pole (i.e. the antipodal point), resulting in constructive interference that could seemingly pose more of a danger to someone who's there than someone who's much closer to the explosion.

Question

Can this phenomenon occur in practice in cases of extremely large explosions, such as the Tsar bomb? If so, then how close to the north pole would one have to be to feel the same effects that someone on the south pole feels?

Research

I know this question is somewhat silly, but I don't think it's completely absurd, either. The blast wave from the Tsar bomb circled the Earth three times, and the atmosphere can focus blast waves in such a way as to make them more deadly to somebody who's far away from the explosion. The difference in this case is that the focusing mechanism is the shape of the atmosphere, not local/regional differences in atmospheric density. Finally, I'm aware of this question, but it has a different focus than mine, and the question and answers don't address the antipodal focusing principle that I'm asking about.

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  • $\begingroup$ Why would it make much difference if it were detonated at one of the poles? The earth is spherically symmetric so a detonation anywhere would have this same effect you propose. $\endgroup$
    – joseph h
    Commented Jul 31, 2021 at 10:13
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    $\begingroup$ It doesn't, but "north pole" and "south pole" is the only pair of (approximately) antipodal points on the Earth's surface that people know off the top of their head. $\endgroup$
    – Matthew Milone
    Commented Jul 31, 2021 at 14:21
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    $\begingroup$ I'm new here, so I could be mistaken, but I don't think that the nuclear physics tag is appropriate. This question doesn't really depend on the mechanics or effects of nuclear bombs in particular--it would be completely unchanged if the Tsar bomb was literally 55 megatons of TNT instead of a nuclear bomb. Nevertheless, thank you for adding the atmospheric science tag. $\endgroup$
    – Matthew Milone
    Commented Jul 31, 2021 at 14:30
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    $\begingroup$ @josephh - the Earth is actually not spherically symmetric. It is ellipsoidal, fatter around the equator than the great circle through the poles. $\endgroup$
    – Jon Custer
    Commented Jul 31, 2021 at 16:34
  • $\begingroup$ I know it’s not a perfect sphere. My point is that anywhere on the surface, the effect will be roughly the same. $\endgroup$
    – joseph h
    Commented Jul 31, 2021 at 21:14

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Can this phenomenon occur in practice in cases of extremely large explosions, such as the Tsar bomb?

Basically yes.

Supposing that the explosion in question occurred at the north pole, once the shock wave moved past the equator, the size of the circle formed by the shock wave moving along the ground decreases, and eventually converges on the south pole (i.e. the antipodal point), resulting in constructive interference that could seemingly pose more of a danger to someone who's there than someone who's much closer to the explosion.

If so, then how close to the north pole would one have to be to feel the same effects that someone on the south pole feels?

You would not feel the same effect.

While the impulse will travel and converge it will lose energy as it travels as it has to affect the medium it travels through.

It is also not travelling on a 2-D surface (the surface of Earth), but will also radiate some energy into the interior of the Earth. That will also disipate energy reaching the opposite pole.

So the energy reaching the opposite pole will be significantly reduced.

Also consider that if your hypothesis was correct then every nuclear detonation ever made would have refocused at the opposite side of the globe, which clearly they did not.

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  • $\begingroup$ I know that an observer wouldn't feel the same effect at the south pole as they would at the north pole, due to energy dissipation. My question relies on the assumption that there are certain circles on the Earth's surface (such as the equator) that would experience a milder pressure change than the south pole. If that's correct, then by the intermediate value theorem, there must be another circle north of the first one at which the pressure change equals the pressure change experienced at the south pole. $\endgroup$
    – Matthew Milone
    Commented Aug 1, 2021 at 18:18
  • $\begingroup$ The intermediate value theorem (IVT) is about values between two known positions. You are talking about a value outside that range of positions. The IVT does not apply. $\endgroup$
    – StephenG - Help Ukraine
    Commented Aug 1, 2021 at 18:26
  • $\begingroup$ No, I'm talking about a pressure value inside the interval of positions on Earth's surface. The fact that I find the value significant because it matches a pressure value at a position outside that position interval is irrelevant to whether the IVT is applicable. Let $x$ be an observer's latitude in degrees (with southern latitudes being negative), and let $f(x)$ be the pressure the observer experiences from the shock wave. Let $a=90, b=0, s=-90$. It's obvious that $f(a)>f(b), f(a)>f(s)$. If $f(b)<f(s)$, then $f(a)>f(s)>f(b)$, and so $∃c: f(c)=f(s), b<c<a$. $\endgroup$
    – Matthew Milone
    Commented Aug 1, 2021 at 23:59
  • $\begingroup$ You argument assumes (without reason) that $f(b)<f(s)$ and ignores the 3-D and temporal nature of the problem and all the physics. Good luck with that. $\endgroup$
    – StephenG - Help Ukraine
    Commented Aug 3, 2021 at 6:53
  • $\begingroup$ I didn't assume that--at least not initially. I asked whether the pressure at the south pole could be greater than the pressure at the equator ("Can this phenomenon occur in practice[..]?"), and your answer was "Basically yes.", so I took that point for granted in my comments. Regardless, I'm not ignoring the damping effect; I'm hypothesizing that the focusing effect can overcome the damping effect. This does happen for sufficiently low damping coefficients. My question hinges on what the value of the damping coefficient actually is. $\endgroup$
    – Matthew Milone
    Commented Aug 8, 2021 at 14:39

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