Prove that for propagation of space waves, the transmission range of a TV transmitter antenna, situated at height $h$ above the Earth's surface is $d = \sqrt{2 R h}$ , where $R$ is the radius of the Earth.

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Solution

Let $R =$ radius of the earth
$h =$ height of the TV tower
and d will be the maximum distance upto which signal can be received
also $d = B C = A C$ (aprox.)
Now, in right angle triangle $Δ A C O$
$(O A)^{2} = (A C)^{2} + (O C)^{2}$
$(R + h)^{2} = d^{2} + R^{2}$
$d = \sqrt{h^{2} + 2 h R}$
and as we know $h << R$
then $d = \sqrt{2 h R}$ .

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Prove that for propagation of space waves, the transmission range of a TV transmitter antenna, situated at height h above the Earth's surface is d=√2Rh, where R is the radius of the Earth.

Let R= radius of the earthh= height of the TV tower and d will be the maximum distance upto which signal can be received also d=BC=AC (aprox.)Now, in right angle triangle ΔACO(OA)2=(AC)2+(OC)2(R+h)2=d2+R2d=√h2+2hRand as we know h<<Rthen d=√2hR.

Prove that for propagation of space waves, the transmission range of a TV transmitter antenna, situated at height $h$ above the Earth's surface is $d = \sqrt{2 R h}$ , where $R$ is the radius of the Earth.