Doing the math

An initial goal of verifying calculations that were mentioned in Al Gore's recent speech:

"Scientists have confirmed that enough solar energy falls on the surface of the earth every 40 minutes to meet 100 percent of the entire world's energy needs for a full year. Tapping just a small portion of this solar energy could provide all of the electricity America uses.

And enough wind power blows through the Midwest corridor every day to also meet 100 percent of US electricity demand."

Calculating the disk area of the Earth, as seen by incoming solar energy: Area circle is pi r squared so frontal area of the Earth in sunshine is pi times 6.38e8 meters = 8.159e20 m squared.

"Through one square meter of space facing the Sun pass 1390 W of sunlight
"http://www.nas.nasa.gov/About/Education/SpaceSettlement/75SummerStudy/Chapt.2.html

"this is nearly twice the maximum of 747 W striking a square meter normal to the Sun at the Earth's surface"

Thus solar energy striking the disk of the earth at any given moment is 1390 w/msq times earth disk area of 8.159e20 m squared = 11341e20 or 1.13e24 watts. In 40 minutes that would be 2/3 that in watt-hours, or 7.56e23 Wh. If a terawatt is 1e12 W, that is 7.56e11 terawatts of solar energy hitting the disk of the earth every 40 minutes.

That is not all energy usable to power our civilization's needs, of course; it is just for comparison purposes, to bracket in the potentials.

So how much energy does civilization currently use in a full year? That part of this effort needs to be done later, the evening's wine has relaxed and dinner needs to be prepared.

I will conclude, for the moment, with another quote from the abovementioned reference, "If the Sun's energy is converted with 10 percent efficiency to electrical power which is sold at a rate of $.012/kW-hr, a square kilometer of space would return more than $14,000,000 each year."