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

Let receiver is at point A and source is at B.

Velocity of waves = `3xx10^(8)m//s`
Time to reach a receiver = 4.04 ms = `4.04 xx 10^(-3)s`
Let the height of satellite is `h_(s)=600` km
Radius of earth = 6400 km
Size of transmitting antenna = `h_(T)`
We know that `("Distance travelled by wave")/("Time")="Velocity of waves"`
`(2x)/(4.04xx10^(-3))=3xx10^(8)`
or `x = (3xx10^(8)xx4.04 xx 10^(-3))/(2)`
`=6.06 xx 10^(5) = 606 km` Using Phythagoras theorem,
`d^(2)=x^(2)-h_(s)^(2)=(606)^(2)-(600)^(2)=7236`
or d = 85.06 km
So, the distance between source and receiver = 2d
`=2xx85.06 = 170 km `
The maximum distance covered on ground from the transmitter by emitted EM waves
`d=sqrt(2Rh_(T))`
or `(d^(2))/(2R)=h_(T)`
or size of antenna `h_(T)=(7236)/(2xx6400)`
`=0.565 km = 565 m`