In a line of sight radio communication,a dist | vedclass

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(A) The range $d$ for line of sight communication is given by the formula: $d = \sqrt{2Rh_T} + \sqrt{2Rh_R}$.Given: $d = 50 	imes 10^3\, m$,$R = 6.4

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$32$

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(A) The range $d$ for line of sight communication is given by the formula: $d = \sqrt{2Rh_T} + \sqrt{2Rh_R}$.Given: $d = 50 	imes 10^3\, m$,$R = 6.4 	imes 10^6\, m$,$h_R = 70\, m$.Substituting the values:$50 	imes 10^3 = \sqrt{2 	imes 6.4 	imes 10^6 	imes h_T} + \sqrt{2 	imes 6.4 	imes 10^6 	imes 70}$.$50000 = \sqrt{12.8 	imes 10^6 	imes h_T} + \sqrt{896 	imes 10^6}$.$50000 = \sqrt{12.8 	imes 10^6} 	imes \sqrt{h_T} + 29933$.$50000 - 29933 = 3577.7 	imes \sqrt{h_T}$.$20067 = 3577.7 	imes \sqrt{h_T}$.$\sqrt{h_T} \approx 5.61$.$h_T \approx 31.47\, m \approx 32\, m$.

In a line of sight radio communication,a distance of about $50\, km$ is kept between the transmitting and receiving antennas. If the height of the receiving antenna is $70\, m$,then the minimum height of the transmitting antenna should be........$m$ (Radius of the Earth $= 6.4 \times 10^6\, m$)

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