8
$\begingroup$

I'm reading the book Cosmos by Carl Sagan, and in it he states:

Striking the Earth's atmosphere, a modest cometary fragment would produce a great radiant fireball and a mighty blast wave, which would burn trees, level forests and be heard around the world.

The 'heard around the world' part may have simply been written expression, but I was wondering how loud a sound would have to be to be heard around the world.

My naive attempt:

I seem to remember being told once that due to the effects of diffraction, lower frequency sound waves travel the furthest distance (and that is why thunder sounds so deep to our ears). This site states that the weakest sound wave audible at the relatively low frequency of 3kHz is 0 dB.

enter image description here

To find the minimum intensity sound source that could be heard across the globe, I assume that the sound source is on the other side of the Earth from me. I'm not too sure how well sound travels through the earth, so I assumed that the sound wave has to travel around the earth, a path length of $\pi R$. From the definition of intensity of a sound wave:

$$I(r) = \frac{P}{4\pi r^2}$$

I find the power of the sound wave to be:

$$P = I(r)4\pi r^2 = (1*10^{-12})(4)(\pi)(6371000 * pi)^2 = 5034 \text{W}$$

That doesn't seem like a very powerful sound wave to me. According to this wiki, a turbojet engine gives off a sound wave that's $100,000\text{W}$, and I most certainly don't hear them all the time from my room. Evidently I've done something majorly wrong, would someone mind giving a better estimate? I'm particularly interested in how large an event would have to be to be heard across the world (for example how large a meteor). (roughly).

Improve this question
$\endgroup$
3
  • 5
    $\begingroup$ You are, for one thing, neglecting the atmospheric attenuation, which is much larger than the purely geometric attenuation. See e.g. en.wikibooks.org/wiki/Engineering_Acoustics/…. Realistically I would assume, at least, 0.01dB/km even at the lower end of the spectrum. For 20,000km path length that's at least 200dB of additional attenuation, which might get you and Sagan into the right ball park. $\endgroup$
    – CuriousOne
    Commented Nov 1, 2015 at 8:09
  • $\begingroup$ Just one rifle, in Massachusetts, in the late 18th century;-) $\endgroup$
    – Carl Witthoft
    Commented Nov 1, 2015 at 13:39
  • 1
    $\begingroup$ @CuriousOne - The Krakatoa eruption in 1883 produced 180 dB sound waves ~100 miles from the explosion (e.g., https://en.wikipedia.org/wiki/1883_eruption_of_Krakatoa) and was heard over ~3000 miles away. The pressure wave was measured to have encompassed the world at least 5 times (don't know whether people heard it around the world though). Regardless, it is arguably one of the loudest things ever heard by human ears that was also recorded. $\endgroup$
    – honeste_vivere
    Commented Sep 30, 2016 at 15:18

1 Answer 1

Reset to default
1
$\begingroup$

Let's do some rough estimation. Considering:

  • CuriousOne's very good and fitting comment
  • with regards to geometrical estimation given in question
  • and the last but not least the human physiology ($10^{-12} \ W$ can hear only a full healthy man, with perception much less than a whisper and only on a frequency range improbable due to the diffraction)

we would need ca. $300 \ \mathrm{dB}$ at the source area to produce fine audible sound all around the globe.

Now, since the medium is air only, we can use $I \propto p^2$ for conversion from intensity to sound pressure. Therefore:

$$ 20 \log \frac{p}{p_0} = 300 \ \mathrm{dB} $$

With usual reference value $p_0 = 2\cdot10^{-5} \ \mathrm{Pa}$ we easily get:

$$ p \approx 10 \ \mathrm{GPa} $$

that's a lot even for a static pressure in solid bodies and here it should be time dependent changing pressure in a gas. It would immediately result in state changes of the air.

"Disussion":

  • We haven't used any reflections here (from the solid body of the earth and from "open end of the atmosphere" as well). It should work for us in the meaning of reconcentration of spreaded energy. The final value should therefore be lower. On the other hand, we would need to take at least big mountain ridges into account.
  • The given estimation is of course invalid, because it is based on linear acoustics. It could provide only the very rough guesstimate at its best. The nonlinear effects will be very pronounced.

"Conclusion": "Audible all around the world" is moreless a figure of speech. In my opinion, the trick would not be in a focused sound source area of insane strength but in a large atmospheric breakdown (i.e. large and spreading source area).

Improve this answer
$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.