# What unit of distance to use when calculating power loss?

80% of Earth's atmosphere is within 10 miles of Earth's surface. I know that power attenuates inversely as the square of the distance within the atmosphere so it occurs to me that a 50,000 watt signal at a frequency high enough to penetrate the atmosphere might be reduced 100 times if I calculate using 1/10^2.

But that distance is also 16Km; using Km the formula becomes 1/16^2 and the signal would be reduced to 1/256th it's original strength (195 watts) not 1/100th (500 watts).

What are the correct units to use to calculate loss of radio wave signal strength through Earth's atmosphere?

• Write down the full law by which you say that "power attenuates inversely as the square of the distance". You will see that it is not as simple as merely multiplying the original power with the inverse of the distance squared. (For one, the units are wrong, since W and W/m^2 are not both units of power) – ACuriousMind Dec 20 '14 at 0:57
• It doesn't matter what units you use so long as you use them properly. $50000$ Watts divided by $10$ miles squared is $500 W/mi^2$. The same amount divided by $16$ km squared is $195W/km^2$. But that is the same number. When we compute the power loss, we cancel out the units appropriately in the correct equation such that it's always the same no matter if you use imperial or metric. But everyone uses metric. (You should use metric too, it's far better than imperial) – Jim Dec 20 '14 at 19:45