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?

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    $\begingroup$ 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) $\endgroup$ – ACuriousMind Dec 20 '14 at 0:57
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    $\begingroup$ 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) $\endgroup$ – Jim Dec 20 '14 at 19:45

Propagation of electromagnetic waves on Earth is highly wavelength dependent, which is also true for the question whether the atmosphere has anything to do with it. What you are looking at is actually a formula for the geometric attenuation (aka free space or path loss), which has absolutely nothing to do with the atmosphere. As ACuriousMind points out, your units are wrong, and so is your mental model of wave propagation between antennas. Antennas have a near and a far field and both can be complicated. What you are looking for is the far-field pattern. By far-field we mean the radiation patter of the antenna that develops many wavelengths away from it. The effective far-field power density created by an antenna at a certain distance and angle is a function of the antenna's design. If you know this power density for a given antenna at a given distance, then you can estimate the further signal loss due to geometry for another distance.

The technological way to approach this properly is called "link budget" and I would suggest you start with that to understand the fundamental influences of different factors on the relationship between transmitter and receiver antenna power. http://en.wikipedia.org/wiki/Link_budget

  • $\begingroup$ Ahh link budgets. They are one of the few things I miss about being a space engineer. So simple, yet so significant $\endgroup$ – Jim Dec 20 '14 at 19:40

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