Question

The radio horizon of a transmitting antenna of height 39.2 m is (Radius of the earth = 6400 km)

• A. 44.8 km
• B. 19.6 km
• C. 22.4 km
• D. 78.4 km

Hint:

The formula for the maximum line-of-sight distance between two antennas of heights \(h_T\) and \(h_R\) is \(d_{max} = \sqrt{2Rh_T} + \sqrt{2Rh_R}\). The radio horizon is just the case for one antenna (\(h_R=0\)). A useful approximation for calculations in km is \(d(\text{km}) \approx 3.57\sqrt{h(\text{m})}\). Here, \(3.57\sqrt{39.2} \approx 3.57 \times 6.26 \approx 22.35\) km.

Answer 1

The Correct Option is C

The radio horizon, or the maximum line-of-sight distance (\(d_T\)) from a transmitting antenna of height \(h_T\), is given by the formula:
\( d_T = \sqrt{2Rh_T} \), where R is the radius of the Earth.
It is important to ensure all units are consistent. Let's use meters.
Height of the antenna, \( h_T = 39.2 \) m.
Radius of the Earth, \( R = 6400 \text{ km} = 6400 \times 1000 \text{ m} = 6.4 \times 10^6 \) m.
Now, substitute these values into the formula.
\( d_T = \sqrt{2 \times (6.4 \times 10^6) \times 39.2} \).
\( d_T = \sqrt{12.8 \times 10^6 \times 39.2} = \sqrt{12.8 \times 39.2 \times 10^6} \).
Let's calculate the product \( 12.8 \times 39.2 \).
\( 12.8 \times 39.2 = 501.76 \).
\( d_T = \sqrt{501.76 \times 10^6} = \sqrt{501.76} \times 10^3 \) m.
We need to find the square root of 501.76. We can notice that \( 22^2 = 484 \) and \( 23^2 = 529 \). The value is close to 22. Let's try \(22.4^2\).
\( 22.4 \times 22.4 = 501.76 \).
So, \( \sqrt{501.76} = 22.4 \).
Therefore, \( d_T = 22.4 \times 10^3 \) m.
To convert this distance to kilometers, we divide by 1000.
\( d_T = 22.4 \) km.