Question

The maximum distance between the transmitting and receiving antennas for satisfactory communication in line of sight mode is 57.6 km. If the height of the receiving antenna is 80 m, the height of the transmitting antenna is (Radius of Earth = \( 6.4 \times 10^6 \) m):

• A. \( 28.8 \) m
• B. \( 51.2 \) m
• C. \( 25.6 \) m
• D. \( 14.4 \) m

Hint:

For line-of-sight communication, use the formula \( d = \sqrt{2Rh_t} + \sqrt{2Rh_r} \). Make sure to convert distances into the same units before calculations.

Answer 1

The Correct Option is B

Step 1: Recall the line-of-sight communication formula The maximum line-of-sight distance \( d \) is given by: \[ d = \sqrt{2 R h_1} + \sqrt{2 R h_2} \] Where: - \( d = 57.6 \times 10^3 \) m (distance in meters) - \( R = 6.4 \times 10^6 \) m (radius of Earth) - \( h_1 = 80 \) m (height of the receiving antenna) - \( h_2 = ? \) (height of the transmitting antenna)
Step 2: Substitute known values into the equation \[ 57.6 \times 10^3 = \sqrt{2 \times 6.4 \times 10^6 \times 80} + \sqrt{2 \times 6.4 \times 10^6 \times h_2} \] \[ 57.6 \times 10^3 = \sqrt{1.024 \times 10^9} + \sqrt{1.28 \times 10^7 \times h_2} \] \[ 57.6 \times 10^3 = 3.2 \times 10^4 + \sqrt{1.28 \times 10^7 \times h_2} \]
Step 3: Isolating the unknown term \[ 57.6 \times 10^3 - 3.2 \times 10^4 = \sqrt{1.28 \times 10^7 \times h_2} \] \[ 25.6 \times 10^3 = \sqrt{1.28 \times 10^7 \times h_2} \]
Step 4: Solving for \( h_2 \) Squaring both sides: \[ (25.6 \times 10^3)^2 = 1.28 \times 10^7 \times h_2 \] \[ 6.5536 \times 10^8 = 1.28 \times 10^7 \times h_2 \] \[ h_2 = \frac{6.5536 \times 10^8}{1.28 \times 10^7} = 51.2 \]
Step 5: Final Answer The height of the transmitting antenna is \( 51.2 \) m.