3 ms·

It's not that high. You have to use the radar equation with the moon being a very lossy but huge target cross section. The path loss at 432 MHz and average moon

• BenjiWiebe 3y ago
  It's still not going to happen accidentally.
• extraduder_ire 3y ago
  Would it be closer to the other value if you factored in the atmosphere, and whatever else is in the way? As an aside, would radio via the moon ever be practical? Without pumping out so much transmit power that you pick up the signal anyway, without pointing a receiver at the moon. I think it would be neat, if only for the sake of novelty.
  • drmpeg 3y ago
    No, there's very little loss from the atmosphere at 432 MHz. The 390 dB figure is a misunderstanding of how the path loss is calculated. The poster "wl" took the one way path loss (195 dB) and doubled it. But the moon is a huge reflector and provides "gain". Here's a C program with the correct equation. #include <stdio.h> #include <stdlib.h> #include <math.h> int main(int argc, char **argv) { double d, f, lambda, loss; if (argc != 3) { fprintf(stderr, "usage: moon <km> <frequency(MHz)>\n"); exit(-1); } d = atof(argv[1]) * 1000.0; f = atof(argv[2]); lambda = 299792458.0 / (f * 1000000.0); loss = (0.065 * (1.738e6 * 1.738e6) * (lambda * lambda)) / (631.65468167 * (d * d * d * d)); printf("EME path loss = %f dB\n", 10 * log10(loss)); return 0; } 1.738e6 is the radius of the moon in meters and 0.065 is the reflection efficiency. Satellites make moonbounce impracticable.