# Is this the correct way to apply an inverse square law to radio wave propagation?

This question is vaguely a follow up to this one and this one.

If I have a completely flat surface with a radio transmitter with 500 W ($\approx$ 57 dBm) of power, broadcasting on a frequency such that groundwave propagation isn't terribly relevant, how can I use an inverse square law to estimate the power of the transmission at certain distances? My initial thought is this:

1. Assume that 1 m away from the transmitter, the signal is basically undiminished.

2. Apply inverse square law to get the power at, say, 10 km.

\begin{align} \frac{I_1}{I_2} &= \frac{d_2^2}{d_1^2} \\ \\ \frac{500\ W}{I_2} &= \frac{10 000^2 \ m^2}{1^2 \ m^2} \\ \\ I_2 &= 5\times 10^{-6} \ W \\ \\ &\approx -23 \ \text{dBm} \end{align}

Am I barking up the wrong tree, or is this close to the correct approach?

(I know this sounds like a homework question, but I promise it's not; just a question from someone with an embarrassingly small amount of knowledge of physics!)

• when you say "flat surface" do you mean the aperture of a horn, for example? – hyportnex Sep 13 '18 at 20:32
• I guess I mean a flat, two-dimensional plane with no changes in elevation. This question derives from some map data I have, so I guess for now I'm ignoring any curvature (I realize that may be a silly idea). – Michael A Sep 13 '18 at 20:34

The inverse square law is applicable to a lossless spherical propagating wave (non directional antenna, no obstructions etc..). This is an insufficient initial physics approximation to a much more complicated communication engineering problem.

However, under this assumption, your antenna is supplying $500W$ of power in all directions. At a distance of $10km$, that power is uniformly distributed over a sphere of radius $10km$. In other words, the power per $m^2$ will be

$\frac{500}{4 \pi r^2}$ in units of $\frac{watt}{m^2}$.

There is no need to use a nearby intermediate point at $1m$ for this approximation.

• And then it's trivial to convert watts per square meter to dbm per square meter. Thank you! – Michael A Sep 14 '18 at 15:09

This is more an Electrical Engineering question than a physics question. Your basic idea is correct, though there are some differences from what actually happenes. For an antenna working with wave length around 1m (say, TV antenna), The EM field at 1 meter is considered "near field", where electrical field and magnetic field are not proportional. In "far field", however, they are proportional. You also need to consider the directionality of the antenna. A dipole antenna has directionality like a donut. Let's assume a directionless antenna (which seems not exist). In this case, you have power density of $500W/10000^2m^2$ at 10km. How much power (in W) you actually receive at 10km, depends on directionality and efficiency of your receiver antenna.

• Thank you for this; I was vaguely aware that the receiver matters, but I just wanted to check that my math was somewhat in the right direction first. – Michael A Sep 13 '18 at 20:53