At the surface of the Earth, the Sun delivers an estimated 1.43 kW/m 2 of energy
ID: 2011570 • Letter: A
Question
At the surface of the Earth, the Sun delivers an estimated 1.43 kW/m2 of energy. Suppose sunlight hits a 10.7 m by 44.9 m roof at an angle of 90.0°.
a) Estimate the total power incident on the roof.
b) Find the radiation pressure on the roof. (Assume the roof is black and totally absorbs the sunlight!)
Explanation / Answer
(a) At an angle of 90 degrees, the radiation is incident normal to the roof. area of the roof: 10.7x44.9=480.43m^2 to find the total power incident on the roof (in Watts) --> 1.43x10^3 W/m^2 * 480.43 m^3 (we multiplied the constant given by the Sun and the area of the roof) =687kW (b) Equation: P = S/c S = Sun constant given (1.43 KW/m^2) P = radiation pressure c = 3 x 10^8 m/s (speed of light constant) P = [1.43x10^3]/[3x10^8]=4.767x10^-6 N/m^2 (pressure is force over area) This answer makes sense, as radiation pressures are small for sunlight on earth (around 5x10^-6 N/m^2)
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