3 ms·
Well, 100% of the emitted radiation hits the ground in terms of total power, but the radar is looking for the _closest_ reflection, which is some relatively sma
by lutorm 5y ago
Well, 100% of the emitted radiation hits the ground in terms of total power, but the radar is looking for the _closest_ reflection, which is some relatively small solid angle straight down. It doesn't change your conclusion though, that solid angle is independent of the distance to the ground.
- nwallin 5y ago> Well, 100% of the emitted radiation hits the ground in terms of total power, but the radar is looking for the _closest_ reflection, which is some relatively small solid angle straight down. This fixed angle paints a larger and larger circle as altitude increases. If you double your altitude, you quadruple the size of the surface that's reflecting radar back at you. A bunch of stuff cancels out and you end up with 1/r^2 at the receiver, which is what we started with.
- steerablesafe 5y agoAn alternative explanation is that you treat the ground as a flat mirror, and the radar detects the distance of your mirror image. The distance between you and your mirror image is (2r), so the signal strength you receive from your mirror image is ~1/(2r)^2, still follows 1/r^2 .