Question

A transmitting antenna at the top of a tower 32 m and height of receiving antenna is 200 m . What is the maximum distance between them for satisfactory communication in LOS mode ? Radius of earth 6.4xx10^6 m ?

Answer 1

Answer:

Given:

Transmitting antenna height = 32 m

Receiving antenna height = 200 m

Radius of Earth = 6.4 × 10^6 m

To find:

Maximum distance between the antennas in LOS mode.

Concept:

LOS means Line Of Sight. In this the waves from transmitting antenna goes along the line of sight towards the receiving antenna.

Calculation:

$\large{ \sf{d_{max} = \sqrt{2Rh_{t}} + \sqrt{2Rh_{r}}}}$

$\sf{ = > d_{max} = \sqrt{(2 \times 6.4 \times {10}^{6} \times 32)} + \sqrt{(2 \times 6.4 \times {10}^{6} \times 200)}}$

$\sf{ = > d_{max} = (20.23 \times {10}^{3}) + (50.59 \times {10}^{3} ) }$

$\sf{ = > d_{max} = 70.82 \times {10}^{3} m}$

$\sf{ = > d_{max} = 70.82 \: km}$

So final approx answer :

$\boxed{ \sf{ \huge{ \green{ d_{max} \approx 70 \: km}}}}$

Answer 2

Transmitting antenna height = 32 m

Receiving antenna height = 200 m

Radius of Earth = 6.4 × 10^6 m

To find:

Maximum distance between the antennas in LOS mode.

Concept:

LOS means Line Of Sight. In this the waves from transmitting antenna goes along the line of sight towards the receiving antenna.

Calculation:

$\large{ \sf{d_{max} = \sqrt{2Rh_{t}} + \sqrt{2Rh_{r}}}}d$

max

=

2Rh

t

+

2Rh

r

$\sf{ = > d_{max} = \sqrt{(2 \times 6.4 \times {10}^{6} \times 32)} + \sqrt{(2 \times 6.4 \times {10}^{6} \times 200)}}=>d$

max

=

(2×6.4×10

6

×32)

+

(2×6.4×10

6

×200)

$\sf{ = > d_{max} = (20.23 \times {10}^{3}) + (50.59 \times {10}^{3} ) }=>d$

max

=(20.23×10

3

)+(50.59×10

3

)

$\sf{ = > d_{max} = 70.82 \times {10}^{3} m}=>d$

max

=70.82×10

3

m

$\sf{ = > d_{max} = 70.82 \: km}=>d$

max

=70.82km