Question

Problem 4: Radiation from the Sun The intensity of the radiation from the Sun is 1360 W/m2 and frequency is f60 MHz. The distance between the Earth and the Sun is 1.5 x 1011m a) Assuming it radiates uniformly in all directions what is the total power output of the Sun? b) If the frequency increases by 1 MHz what would be the relative change in the power output?

Answer 1

Intensity is the ratio of power transmitted per unit area. therefore I = P/A I = intensity P = power A = area therefore Total power radiated from the sun in all directions P = IA {therefore I = P/A P = IA} intensity of power I = 1360 w/m^2 Area of a sphere A = 4 pi r^2 therefore P = (1360 w/m^2)(4pi r^2) = (1360 w/m^2)(4 times (1.5 times 10^11 m)^2) {therefore r = 1.5 times 10^11 m} P = (1.360 w/m^2)(4 times 2.25 times 10^22 m^2) p = equivalent 3.85 times 10^26 w \r\n output power is proportional to the frequency(f) therefore delta P/P = delta f/f Relative charge in output is delta P/P delta P/P = 1.0 MHZ/60 MHZ delta P/P = 0.017 {f = 60 MHZ P = 60 MHZ} charge in output power is (AP) delta P = (0.017)(3.85) times 10^26 w delta p/3.85 times 10^26 w = 0.017 delta P = (0.017)(3.85 times