Question

The T.V transmission tower in Delhi has a height of $240\text{m}$. The distance up to which the broadcast can be received (taking the radius of earth to be $6.4\times {10}^{6}m)$ is (in km) :

Answer 1

$x\approx R_e\theta$
$\cos \theta =\sqrt{1-{\theta }^2}=\frac{R_e}{R_e+h}={(1+\frac{h}{R_e})}^{-1}$
$1-{\theta }^2={(1+\frac{h}{R_e})}^{-2}$
$\Rightarrow 1-{\theta }^2\approx 1-\frac{2h}{R_e}$
$\therefore \theta =\sqrt{\frac{2h}{R_e}}$
$\therefore x=R_e\sqrt{\frac{2h}{R_e}}\approx \sqrt{2h{R}_e}$
$x=\sqrt{2\times \frac{240}{1000}\times 6400\text{km}}$
$x=8\times 4\sqrt{3}\text{km}\approx 32\times 1.7\approx 55\text{km}$