Question

A 50 MHz sky wave takes 4.04 ms to reach a receiver via re-transmission from a satellite 600 km above earth’s surface. Assuming re-transmission time by satellite negligible, find the distance between source and receiver. If communication between the two was to be done by Line of Sight (LOS) method, what should size and placement of receiving and transmitting antenna be?

Answer 1

`(2x)/"time"` = velocity

`2x = 3 xx 10^8  m/s xx 4.04 xx 10^-3s`

`x = (12.12 xx 10^5)/2 m`

= `6.06 xx 10^5 m`

= 606 km

`d^2 = x^2 - h_s^2 = (606)^2 - (600)^2`

= 7236; d = 85.06 km

Distance between source and receiver = 2d ≅170 km

`d_m = 2sqrt(2Rh_T), 2d = d_m, 4d^2 = 8 Rh_T`

`d^2/(2R) = h_T= 7236/(2 xx 6400)` ≈ 0.565 km = 565 m.