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Question: A 50 MHz sky wave takes 4.04 ms to reach a receiver via re-transmission from a satellite 600 km abov...

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.

A

606 km

B

170 km

C

340 km

D

280 km

Answer

170 km

Explanation

Solution

: here

Total time taken

t=4.04ms=4.04×10−3st = 4.04ms = 4.04 \times 10^{- 3}s

Let x be the distance of satellite from the surface of earth

Total time taken (t)

=Totaldistan⁡cetravelled(2x)Speedofelectromagneticwave(c)= \frac{Totaldis\tan cetravelled(2x)}{Speedofelectromagneticwave(c)}

∴x=ct2=(3×108)(4.04×10−3)2=6.06×105m=606km\therefore x = \frac{ct}{2} = \frac{(3 \times 10^{8})(4.04 \times 10^{- 3})}{2} = 6.06 \times 10^{5}m = 606kmLet T be the source of electromagnetic waves (i.e. transmitter), R be receiver and S be satellite at locations as shown in figure

}{= 7236 }{\therefore d = 85.06km}$$ Distance between source and receiver <img src="https://cdn.pureessence.tech/canvas_521.png?top_left_x=203&top_left_y=600&width=148&height=170" style="width:0.96528in;height:1.125in" /> $$= 2d = 2 \times 85.06 = 170km$$