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 th

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.