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A 50MHz sky wave takes 4.04 ms to reach ...

A 50MHz sky wave takes 4.04 ms to reach a receiver via retransmission from a satellite 600km above earht's surface. Assuming re-transmission time by satellite negligible, find the distance between source and reciever.

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The correct Answer is:
A
Let receiver is at point `A` and source is at `B`.
velocity of waves `= 3xx 10^8 m//s`
Time to reach a receiver `= 4.04 ms =4.04xx 10^-3s`
Let the height of satellite is.
`h_s = 600km`
Radius of earth = 6400 km
Size of transmitting antenna = `h_T`
We know that
`("Distance travelled by wave")/("Time") =` Velocity of waves
`(2x)/(4.04 xx10^-3)=3 xx 10^8`
or `x=(3xx 10^8xx4.04xx10^-3)/2`
`= 6.06 xx 10^5=606km`
Using pythgoras 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`
`= 2xx 85.06`
`= 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`.
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