Problem — Maximum distance of a thunderclap

Problem

Estimate the maximum distance at which an average human ear can hear the rumble of thunder, knowing that the low-frequency audibility threshold is $I_\text{min} \approx 10^{-8}\;\text{W/m}^2$ and that a lightning bolt acoustically emits a peak power of the order of $10^{7}\;\text{W}$. Assume isotropic emission and neglect air absorption.

Solution

Setup. Isotropic emission: the power spreads over a sphere of radius $r$, so $I = \dfrac{P}{4\pi r^2}$. The thunder is audible as long as $I \ge I_\text{min}$; the maximum distance occurs for $I = I_\text{min}$:

$r = \sqrt{\frac{P}{4\pi I_\text{min}}} = \sqrt{\frac{10^{7}}{4\pi\cdot 10^{-8}}} = \sqrt{\frac{10^{7}}{1{,}26\cdot 10^{-7}}}$

$r = \sqrt{7{,}96\cdot 10^{13}} \approx 8{,}9\cdot 10^{6}\;\mathrm{m}$

$\ev{r \sim 10^{7}\;\mathrm{m}}$

Comment. The order of magnitude ($\sim 10^4$ km) is unrealistically large: in reality, air absorption and sound refraction limit the audibility of thunder to a few km (typically $< 15$–$20$ km). The isotropic model without absorption vastly overestimates the range: this shows that it is precisely atmospheric attenuation, neglected here, that sets the real distance.

Collegamenti

Argomenti: Waves Concetti: Wave Intensity Competenze: Fermi Estimates