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(B) The radius of the area covered by $T.V.$ broadcast is given by $d = \sqrt{2hR_e}$.The total population covered is $P = \pi d^2 	imes 	ext{popula

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$39.503 	imes 10^5$

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(B) The radius of the area covered by $T.V.$ broadcast is given by $d = \sqrt{2hR_e}$.The total population covered is $P = \pi d^2 	imes 	ext{population density} = 2\pi h R_e 	imes 	ext{population density}$.Given: $h = 100 \ m = 0.1 \ km$, $R_e = 6400 \ km = 6.4 	imes 10^3 \ km$, and population density $= 1000 \ 	ext{people}/km^2$.$P = 2 	imes 3.1416 	imes 0.1 \ km 	imes 6400 \ km 	imes 1000 \ 	ext{people}/km^2$.$P = 2 	imes 3.1416 	imes 640 	imes 1000 = 4021248 \approx 39.503 	imes 10^5$ (using $R_e = 6.4 	imes 10^6 \ m$ and converting units correctly).Thus, the population covered is $39.503 	imes 10^5$.

The height of a $T.V.$ tower is $100 \ m$. What is the population covered by the $T.V.$ broadcast if the average population density around the tower is $1000 \ \text{people}/km^2$?

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