Answer

Verified

54.3k+ views

**Hint:**The maximum line of sight is the maximum distance between the transmitting and the receiving antenna beyond which they would no longer be able to send or receive signals from one another. The curvature of the earth is what causes the distance to be finite i.e. a flat earth would have no maximum line of sight.

**Formula used:**In this solution we will be using the following formula;

\[hy{p^2} = op{p^2} + ad{j^2}\] where \[hyp\] is the hypotenuse of a right angled triangle, \[opp\] is the opposite side and \[adj\] is the adjacent side

**Complete Step-by-Step Solution:**

We shall draw two antennae on the surface of a round earth as shown in the figure

The maximum line of sight by diagram is given by

\[S = {x_t} + {x_r}\]

From Pythagoras theorem which states that

\[hy{p^2} = op{p^2} + ad{j^2}\] where \[hyp\] is the hypotenuse of a right angled triangle, \[opp\] is the opposite side and \[adj\] is the adjacent side

Hence,

\[{\left( {R + {h_t}} \right)^2} = {R^2} + x_t^2\]

\[ \Rightarrow x_t^2 = {\left( {R + {h_t}} \right)^2} - {R^2}\]

By opening the bracket and subtracting in the right hand side, we have

\[x_t^2 = {R^2} + {h_t}^2 + 2{h_t}R - {R^2}\]

\[ \Rightarrow x_t^2 = {h_t}^2 + 2{h_t}R\]

Now, if the height of the transmission antennae is taken to be small with respect to the radius of the earth, then the square of the height can be dropped as in

For \[{h_t} < < R\]. Then \[{h_t}^2 + 2{h_t}R = 2{h_t}R\].

Then the transmission antennae height can be given as

\[x_t^2 = 2{h_t}R\],

By finding the square root of both sides, we have

\[{x_t} = \sqrt {2{h_t}R} \]

Similarly, for the receiving antennae, we have that

\[{x_r} = \sqrt {2{h_R}R} \].

Then, the maximum line of sight is given by

\[S = {x_t} + {x_r} = \sqrt {2{h_t}R} + \sqrt {2{h_R}R} \]

**Note:**For clarity, the dropping of the square of the height can be justified through the following analysis; The equation \[{h_t}^2 + 2{h_t}R\] can be written as \[{R^2}\left( {\dfrac{{{h^2}}}{{{R^2}}} + \dfrac{{2h}}{R}} \right)\], now since, the height is small relative to the radius of the earth, then \[\dfrac{h}{R} < < 0\]. Now, from maths, we observe that when a number is less than zero, then the square of the number is even far less than zero. Hence, the first term in the bracket \[{\left( {\dfrac{h}{R}} \right)^2} \to 0\] and hence can be neglected. If we multiply \[{R^2}\], we have

\[{R^2}\left( {0 + \dfrac{{2{h_R}}}{R}} \right)\]

\[ \Rightarrow 2hR\]

Recently Updated Pages

Why do we see inverted images in a spoon class 10 physics JEE_Main

A farsighted man who has lost his spectacles reads class 10 physics JEE_Main

State whether true or false The outermost layer of class 10 physics JEE_Main

Which is not the correct advantage of parallel combination class 10 physics JEE_Main

State two factors upon which the heat absorbed by a class 10 physics JEE_Main

What will be the halflife of a first order reaction class 12 chemistry JEE_Main

Other Pages

Which of the following distance time graph is representing class 11 physics JEE_Main

The resultant of vec A and vec B is perpendicular to class 11 physics JEE_Main

when an object Is placed at a distance of 60 cm from class 12 physics JEE_Main

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

The angle between the hands of a clock when the ti-class-11-maths-JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main