A $50 M H z$ sky wave takes $4.04 m s$ to reach a receiver via re-transmission from a satellite $600 k m$ above the earth's surface. Assuming re-transmission time by satellite is negligible, the distance between source and receiver is about:
(Velocity of wave $= 3 \times 10^{8} {m s}^{- 1}$ and the radius of the earth $= 6400 k m$ ).

150 k m

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180 k m

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170 k m

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190 k m

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Solution

The correct option is C $170 k m$
$Speed = \frac{distance}{time} = \frac{2 x}{t}$
$x = \frac{3 \times 10^{8} \times 4.04 \times 10^{- 3}}{2}$
$x \approx 606 k m$

$d = \sqrt{x^{2} - h_{s}^{2}}$
$d = \sqrt{606^{2} - 600^{2}}$
$d \approx 85 k m$
Distance between source and receiver is $2 d = 170 k m$

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A 50MHz sky wave takes 4.04ms to reach a receiver via re-transmission from a satellite 600km above the earth's surface. Assuming re-transmission time by satellite is negligible, the distance between source and receiver is about:(Velocity of wave =3×108ms−1 and the radius of the earth =6400km).

The correct option is C 170kmSpeed=distancetime=2xtx=3×108×4.04×10−32x≈606 kmd=√x2−h2sd=√6062−6002d≈85 kmDistance between source and receiver is 2d=170 km

The correct option is C $170 k m$
$Speed = \frac{distance}{time} = \frac{2 x}{t}$
$x = \frac{3 \times 10^{8} \times 4.04 \times 10^{- 3}}{2}$
$x \approx 606 k m$

$d = \sqrt{x^{2} - h_{s}^{2}}$
$d = \sqrt{606^{2} - 600^{2}}$
$d \approx 85 k m$
Distance between source and receiver is $2 d = 170 k m$