Question

One goal of the Russian space program is to illuminate dark northern cities with sunlight reflected to the Earth from a 200 m diameter mirrored surface in orbit. Several smaller prototypes have already been constructed and put into orbit. (a) Assume that sunlight with intensity 1340 W/m2 falls on the mirror nearly perpendicularly and that the atmosphere of the Earth allows 78.3% of the energy of sunlight to pass though it in clear weather. What is the power received by a city when the space mirror is reflecting light to it? W (b) The plan is for the reflected sunlight to cover a circle of diameter 8.00 km. What is the intensity of light (the average magnitude of the Poynting vector) received by the city? W/m2 (c) This intensity is what percentage of the vertical component of sunlight at St. Petersburg in winter, when the sun reaches an angle of 7.80° above the horizon at noon? Need Help? Read It Viewing Saved Work Revert to Last Response Submit Answer

Answer 1

A)

Area of mirror =$\pi$d²/4 =$\pi$*200²/4 =31415 m²

>Net power received at mirror position =1340*31415 =42096100 W (or)42.1 MW

78.3 % received => 0.783*42096100 =32961246.3 W or 32.96124 MW received

B)

Total area to be covered =$\pi$d²/4 =$\pi$*(8000²)/4 =5.03*10⁷ m²

=>   32961246.3/(5.03*10^7) =0.655293167 W/m²

C)

0.783*1340 = 1049.22  W/m² by the time radiation reaches Earth

Vertical component =1049.22*sin7.8 =142.023 W/m²

0.655293167*100/142.023 =0.4601939   %

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