Sound Intensity — Thunder (Worked Example)

Given

• P = 1.0 × 10⁷ W

• r = 2.0 km = 2000 m

Find

The intensity I.

Principle

Thunder obeys the same inverse-square spreading: I = P/(4πr²).

Formula

Solution

1. Convert: $r = 2000\ \mathrm{m}$

2. Substitute: $I = 10^7/(4\pi \times 2000^2)$

3. Denominator: $4\pi \times 4 \times 10^6 = 5.027 \times 10^7$

4. Divide: $I = 0.199\ \mathrm{W/m^2}$

Unit check

W ÷ m² = W/m². The inverse-square law keeps units consistent: power over area is intensity.

What this means physically

0.199 W/m² ≈ 113 dB — a startling crack even 2 km away. Real thunder is directional and absorbed by air, so this spherical estimate is an upper bound.