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a pretty good approximation) like blackbodies (hypothetical bodies that absorb and emit radiation perfectly). To apply this analysis, we assume that the Sun radiates the same amount of energy as an identically sized spherical black body having a temperature of 5800 K [Recall the Kelvin temperature or absolute temperature is T (Kelvin)= T (°C) + 273.15]. Using a solar radius (Ro) of 6.96x108 m, calculate the surface area of the Sun, using A = 4πRo2 (recall π = 3.14).

2. Using the Stefan-Boltzmann law for the luminosity (L= [J/s]), calculate L, the energy radiated per unit time [J/s] by the Sun treated as a black body at temperature T. The formula is L = sAT4 (s, the Stefan-Boltzmann constant = 5.67 X 10-8 J/ m2 K4 s).

II. The Solar Energy Absorbed by the Earth

3. Although the Sun radiates the amount of energy per second calculated above, only a small portion of this total energy crosses each square meter of area at the distance of the Earth from the Sun. Assume the entire Sun's luminosity crosses a sphere with a radius as big as the Earth's orbit around the Sun (or d, the distance from the Earth to the Sun). Now calculate how much energy will hit each square meter of surface of this imaginary sphere (the top of the earths atmosphere) each second (we'll call this the solar flux, F).

Take the distance of the Earth from the Sun (d) = 1 AU = 1.5 x 1011 m. Calculate F (units J/m2 s or W/m2) using the following equation: F = L/(4πd2)

4. Of the area of the large imaginary sphere, only the cross-sectional area of the Earth will intercept the light from the Sun. That is, of the Sun's luminosity, the Earth will intercept only the amount given by multiplying the cross-sectional area of Earth by the flux you just found. Thus the energy hitting the Earth, E = πFRE2, where RE = the radius of the Earth, 6.38 x 106 m). Calculate E. So this value is the energy per unit time (units are Joules/second or Watts where 1 W= 1 J/s) that impinges upon the top of the Earth's atmosphere each second.

5. The Earth does not absorb 100 % of the incoming radiation. We must account for the light directly reflected back into space. The albedo (a), or fraction of the incoming radiation reflected back into space, is measured (by satellites) to be about 0.35. This value is the average value for Earth. It accounts for the different surface reflectivity of Earth. For example, light striking fresh snow reflects about 90% of the incoming visible radiation. Typical non-vegetated land surfaces reflect about 35%, forests about 18%, and clouds about 75%. The oceans exhibit a reflectivity of about 10%. Overall, the average fraction of 0.35 (or 35%) of visible incident solar radiation is reflected back into space and does NOT contribute to heating. Thus, the fraction of energy that actually enters the Earth's atmosphere (i.e., is absorbed) is a = (1- a) = 0.65. Correct your estimate of E in question 4 to get the final value of the solar energy absorbed (EA) by the Earth each second (EA = a E). 5b. By the way, how would the albedo change if the Earth experienced a new ice age? 5c.Would such a change be a positive or negative feedback with respect to global warming? 5d. What if the fraction of polar ice did not change but the continents sank and were covered by oceans so that there was 90% ocean and only 10% exposed land. Would that be a positive or negative feedback?