The avarage intensity of solar radiation that falls normally on surface just out
ID: 1897242 • Letter: T
Question
The avarage intensity of solar radiation that falls normally on surface just outside the earth's atmosphere is 1.4 kW/m^2.a) What radiation pressure is exerted on this surface, assuming complete absorption?
b) How does this pressure compare with the earth's atmospheric pressure which is 10^5 N/m^2
Note) PLEASE GIVE ALL DETAILS OF SOLUTION.
Explanation / Answer
EM-radiation emitted = 0.15*250 = 37.5 J/s Fraction falling on a 1m² surface at radius 0.75m given by .. 1(m²)/sphere area(m²) = 1/4pr² = 1/4p 0.75² = 0.14 Amount of em-radiation energy falling per m² = 0.14*37.5 = 5.31 (J/s)m¯² According to Einstein E = mc².. E/c = mc = momentum equivalent (?p) for radiation Inputting E as (J/s)m¯² (instead of J) we get ?p/?t per m² = force/area = pressure (Pa) c = 3.0 E8 m/s E/c ? (J/s)m¯²/c .... N/m² > = 5.31/3.0 E8 Pressure = 1.77 E-8 N/m² (Pa) for the black surface (assuming complete absorption of the radiation) A mirror surface (reflecting 100%) produces a ?v = 2u, hence ?p = 2(black surface) mirror experiences a pressure 2*(1.77 E-8) = 3.54 E-8 N/m² (Pa)
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