The sun’s diameter is 1.39 x 10^9 m, the earth’s diameter is 1.27 x 10^7 m, and

Question

The sun’s diameter is 1.39 x 10^9 m, the earth’s diameter is 1.27 x 10^7 m, and the radius of the earth’s orbit is 1.50 x 10^11 m. Assume that the sun is a black body emitter at 8500 K.

a) What is the total flux (W/m^2) incident on a surface parallel to the earth’s surface at 23 degrees north latitude, at noon on the summer solstice (June 21)?

b) What is the total flux (W/m^2) incident on a surface parallel to the earth’s surface at the north pole on the summer solstice?

c) Integrate the incident flux over the surface of the earth to obtain the total power (W) incident on the earth from the sun?

Explanation / Answer

given, sun's diameter, D = 1.39*10^9 m

earth's diameter, d = 1.27*10^7 m

radius of earth's orbit around sun, r = 1.5*10^11 m

surface temperature of sun T = 8500 K

sun is a black body emitter

a. Power emitted by sun, P = sigma*A*T^4 [ where sigma is stefans constant =5.67*10^-8 ]

Area of sun, A = pi*D^2 = 6.0698*10^18

P = 5.67*10^-8 * 6.0698*10^18 * 8500^4 = 1.7965*10^27 W

flux incident on surface parallel to earth's surface at 23 deg N on summer solistice = Psin(23 + phi)/a

where phi is thje angle of earth's axis with vertical wrt to solar plane

phi = 23.4 deg

where a = 4*pi*r^2

I = P*sin(23+23.4)/4*pi*(1.5*10^11)^2 = 4601.373 W/m^2

b. at north pole,

flux incident on surface parallel to earth's surface at 0 N on summer solistice = Psin(phi)/a

where phi is thje angle of earth's axis with vertical wrt to solar plane

phi = 23.4 deg

where a = 4*pi*r^2

I = P*sin(23.4)/4*pi*(1.5*10^11)^2 = 2523.46 W/m^2

c. total power incident on earth = P*pi*d^2/4*pi*r^2*4 = P*d^2/16r^2 = 0.80488*10^18 W/

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