Question regarding the earths energy.

Maraxus · Post by **Maraxus** » 2005-12-31 04:43pm

I have a question for any of the physicists that roam these boards.

How much energy in joules hits the earth at the equator on a day to day basis?

any help would be greatly appreciated.

Wyrm · Post by **Wyrm** » 2005-12-31 05:00pm

D'ya mean solar energy?

Sriad · Post by **Sriad** » 2005-12-31 05:22pm

Secondary clarification: How big is the "equator" you're talking about? The literal division between Northern and Souther hemispheres is a line, hence 0 area, hence 0 energy. OTOH, the region between the Tropics of Cancer and Capricorn has... uh, significant area.

Maraxus · Post by **Maraxus** » 2005-12-31 05:26pm

A square meter at 0 degrees latitude.

Kuroneko · Post by **Kuroneko** » 2005-12-31 06:00pm

The Sun's luminosity is P = 3.827e26W, with a distance of AU = 149597870691m, the incident radiation for a surface normal is I = P/(4πAU²) = 1.361e3W/m². As per Lambert's law, the intensity is I(θ) = [I/2][cos θ], where θ is the angle of incidence. Since the ecliptic is at 23.5°relative to the equator, the intensity of at the equator at high noon is S = 1.248e3W/m², without adjusting for atmospheric albedo or absorption. One can apply Lambert's law again to adjust for the varying angle of incidence throughout the day, but there's an easier way. Since the equatorial radius is R = 6378135m, a band of thickness δ<<R and length 2R will have incident energy 2δSR, but since this energy is spread throughout the whole band, a δ-long piece will have only δ/(2πR) of it, so that the average power on the is δ²S/π = 397.2W if δ = 1m. Over a twenty-four hour period, this makes 3.43e7J. Again, this is before adjusting for atmospheric reflection and absorption, but since I'm unsure of those parameters, I'll simply guesstimate 2.9-3.4eJ.

Note on Edit: I'm not sure what my addled mind was thinking yesterday, but that factor of 1/2 on the Lambert's cosine law should not be there (I think I was compensating for night, nevermind that this way it would have been done twice). Hence, the figures should be doubled to 3.44e7J per twenty-four hours. How much the atmosphere affects this depends on the ('average') weather conditions, obviously, but I still have no data regarding that.