A 50 MHz sky wave 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 be the size of transmitting antenna ?

Explanation

Here, total time taken, `t = 4.04 ms = 4.04 xx 10^(-3)`s
let x be the distance of satellite from the surface of earth.
Total tiem taken `(t) = ("total distance travelled"(2x))/("speed of em wave "(c)) :. x = (ct)/2 = ((3 xx 10^8)(4.04 xx 10^(-3)))/(2)`
` = 6.06 xx 10 ^5 m = 606 km`
Let T be the source of em wave (i.e., transmitter), R be receiver ans S
be satellite at locations shown in figure.
`d^2 = x^2 - h^2 = (606)^2 - (600)^2 = 7236 or d = 85.06 km`
Distance between source and receiver `= 2d - 2 xx 85.06 ~~ 170 km`
For the line of sight communication, let S be the source of
em, wave, R be receiver and `h_(T)` be the size of antenna used.
The maximum distance covered on ground from the
transmitter by the emitted em waves is
` d = sqrt (2Rh_t) or h_t = d^2/(2R) = (7236/( 2xx 6.4 xx 10^6)) = 0.565 km = 565m`.