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?

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Answered by

alok kumar singh | Contributor-Level 10

4 months ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

height of satellite hs= 600km
As we know velocity = distance/time

2x/4.0410^-3 = 3⁸

So x=606 km after solving
According to Pythagoras theorem d²=x²-h²= 606²-600²= 7236

So d= 85.06km so total distance will be double from receiver to transmitter = 170.1km
And d = √2Rh
h=7236/2×6400 = 565m

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