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A TV transmitting antenna is $128 \mathrm{~m}$, tall. If the receiving antenna is at the ground level, the maximum distance between, them for satisfactory communication in line of sight mode is (radius of the earth $6.4 \times 10^6 \mathrm{~m}$ )
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Verified Answer
The correct answer is:
$\frac{128}{\sqrt{10}} \mathrm{~km}$
Height of transmitting antenna $d \approx \sqrt{2 R_e h_T}$
Given
$\begin{aligned}
R_e & =6.4 \times 10^6 \mathrm{~m} \\
h_T & =128 \mathrm{~m} \\
d & =?
\end{aligned}$
The maximum distance
$\begin{aligned}
& d=\sqrt{2 \times 6.4 \times 10^6 \times 128} \\
& d=\sqrt{128 \times 10^5 \times 128} \\
& d=\frac{128 \times 10^2 \sqrt{10}}{\sqrt{10}} \times \sqrt{10} \\
& d=\frac{128 \times 10^3}{\sqrt{10}} \mathrm{~m}=\frac{128}{\sqrt{10}} \mathrm{~km}
\end{aligned}$
Given
$\begin{aligned}
R_e & =6.4 \times 10^6 \mathrm{~m} \\
h_T & =128 \mathrm{~m} \\
d & =?
\end{aligned}$
The maximum distance
$\begin{aligned}
& d=\sqrt{2 \times 6.4 \times 10^6 \times 128} \\
& d=\sqrt{128 \times 10^5 \times 128} \\
& d=\frac{128 \times 10^2 \sqrt{10}}{\sqrt{10}} \times \sqrt{10} \\
& d=\frac{128 \times 10^3}{\sqrt{10}} \mathrm{~m}=\frac{128}{\sqrt{10}} \mathrm{~km}
\end{aligned}$
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