A TV transmission tower antenna is at a height of $20m$ .How much service area can it cover if the receiving antenna is i) at ground level ii) at a height of $25m$. Calculate the percentage increase in area covered in case if the receiving antenna is at ground level to that at a height of $25m$.
Answer
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Hint: First calculate the range of the antenna by using the height of the antenna and then calculate the area covered by the antenna for the different range and after that calculate the percentage change in the area due to the change in the height of the antenna.
Complete step by step answer:
We know the radius of Earth is $64 \times {10^5}m$.
Now the height of the TV transmission tower antenna is ${H_1} = 20m$ and now the height is increased to ${H_2} = 25m$ .
Now we calculate the range of the antenna having height ${H_1} = 20m$ .
We know the formula of the range of an antenna having height $H$ is $d = \sqrt {2RH} $
$R$ is the radius of the earth.
Hence at, ${H_1} = 20m$ we can calculate the range of the antenna is-
$ \Rightarrow {d_1} = \sqrt {2 \times 64 \times {{10}^5} \times 20} = 16 \times {10^3}m = 16km$.
Now calculate the range of the antenna having height ${H_2} = 25m$.
Hence at, ${H_2} = 25m$ the range of the antenna is,
$ \Rightarrow {d_2} = \sqrt {2 \times 64 \times {{10}^5} \times 25} = 17.9 \times {10^3}m = 17.9km$
Now we have to calculate the area covered by the antenna at different heights.
i) So for ${d_1}$ range the area which the antenna covered is ${A_1} = \pi {d_1}^2 = \pi \times {\left( {16} \right)^2} = 804.6k{m^2}$
II) And for ${d_2}$ range the area which the antenna covered is ${A_2} = \pi {d_2}^2 = \pi \times {\left( {17.9} \right)^2} = 3611.8k{m^2}$
Now we have to calculate the percentage change in area due to the change in height,
Percentage change in area$ = \dfrac{{{A_2} - {A_1}}}{{{A_1}}} \times 100 = \dfrac{{3611.8 - 804.6}}{{804.6}} \times 100 = 348.9$
Hence the percentage change in the area covered by the TV transmission antenna due to the change in height at ground level to that of height $25m$ is $348.9\% $.
Note: The line-of-sight communication means that there is no physical obstruction between the transmitter and the receiver. In such communication it is not necessary for the transmitting and receiving antennas to be at the same height.
Complete step by step answer:
We know the radius of Earth is $64 \times {10^5}m$.
Now the height of the TV transmission tower antenna is ${H_1} = 20m$ and now the height is increased to ${H_2} = 25m$ .
Now we calculate the range of the antenna having height ${H_1} = 20m$ .
We know the formula of the range of an antenna having height $H$ is $d = \sqrt {2RH} $
$R$ is the radius of the earth.
Hence at, ${H_1} = 20m$ we can calculate the range of the antenna is-
$ \Rightarrow {d_1} = \sqrt {2 \times 64 \times {{10}^5} \times 20} = 16 \times {10^3}m = 16km$.
Now calculate the range of the antenna having height ${H_2} = 25m$.
Hence at, ${H_2} = 25m$ the range of the antenna is,
$ \Rightarrow {d_2} = \sqrt {2 \times 64 \times {{10}^5} \times 25} = 17.9 \times {10^3}m = 17.9km$
Now we have to calculate the area covered by the antenna at different heights.
i) So for ${d_1}$ range the area which the antenna covered is ${A_1} = \pi {d_1}^2 = \pi \times {\left( {16} \right)^2} = 804.6k{m^2}$
II) And for ${d_2}$ range the area which the antenna covered is ${A_2} = \pi {d_2}^2 = \pi \times {\left( {17.9} \right)^2} = 3611.8k{m^2}$
Now we have to calculate the percentage change in area due to the change in height,
Percentage change in area$ = \dfrac{{{A_2} - {A_1}}}{{{A_1}}} \times 100 = \dfrac{{3611.8 - 804.6}}{{804.6}} \times 100 = 348.9$
Hence the percentage change in the area covered by the TV transmission antenna due to the change in height at ground level to that of height $25m$ is $348.9\% $.
Note: The line-of-sight communication means that there is no physical obstruction between the transmitter and the receiver. In such communication it is not necessary for the transmitting and receiving antennas to be at the same height.
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