What should be the height of transmitting antenna and the population covered if the television telecast is to cover a radius of 150 km? The average population density around the tower is 2000 km$$^{-2}$$ and the value of $$R_e = 6.5 \times 10^6$$ m.

The range of a TV transmission tower of height $$h$$ above the Earth's surface is given by: $$d = \sqrt{2R_e h}$$

We need $$d = 150 \text{ km} = 150 \times 10^3 \text{ m}$$ with $$R_e = 6.5 \times 10^6 \text{ m}$$.

Solving for $$h$$: $$h = \frac{d^2}{2R_e} = \frac{(150 \times 10^3)^2}{2 \times 6.5 \times 10^6} = \frac{2.25 \times 10^{10}}{1.3 \times 10^7} = \frac{2.25}{1.3} \times 10^3 = 1730.8 \text{ m} \approx 1731 \text{ m}$$

The area covered is a circle of radius $$d = 150 \text{ km}$$: $$A = \pi d^2 = \pi \times (150)^2 = \pi \times 22500 \approx 70686 \text{ km}^2$$

With population density $$\rho = 2000 \text{ km}^{-2}$$, the population covered is: $$P = \rho \times A = 2000 \times 70686 \approx 1.413 \times 10^8 = 1413 \times 10^5$$

Therefore, the height of the transmitting antenna is $$1731 \text{ m}$$ and the population covered is $$1413 \times 10^5$$.