PV Gap

Peggy Said:

What percentage of the world’s energy needs could be met using solar energy?

We Answered:

100%

It is possible to fit 1,858,560 solar modules in a square mile. An area of solar panels 102 miles to a side would be sufficient to generate 4,000,000,000,000 kWh of electricity or enough to power the entire US. --Source The Solar Living Source Book by John Schaeffer

I can't find the link that states that the amount of sunlight that hits the Earth in like 1 hour is enough to power the planet for a year...

Sorry

Brett Said:

Can you use solar energy anywhere in the world?

We Answered:

only if the sun is there

Warren Said:

what is the potential world energy - solar energy?

We Answered:

by fifty if you mean suns they would scorch the earth. best power is wind.

Kurt Said:

What will it be like if our daily world ran on solar energy?

We Answered:

The world would be very dark. In order to provide enough energy to meet global demands, the panels would have to cover the entire world's surface (and then more). There won't be any sunlight left below the panels.

(Which, incidentally, would require the use of artificial lighting and increase the demand for energy even more)

Clinton Said:

If surface area of world is A, solar energy per unit of surface area is B, is AB max eng harnessable on earth?

We Answered:

Well, yeah, except for a few things, some of which have been mentioned. First, the number is only that high if efficiency is 100%. Second, we could theoretically collect energy at places other than on our surface, such as on the planet Mercury. Finally, we can only collect energy hitting the sunny side of Earth, so really 1/2 of AB.

Unfortunately, even 1/2 AB is too much. Because of the curvature of the Earth, we are not getting the full energy in a ray of light because there is the rays don't hit the Earth directly (except for a single point). The actual amount of sunlight that hits Earth is proportional not to AB, but to about A times the surface integral over the solar-visible surface of the cosine of the angular distance from the point of direct sunlight. I haven't really worked out the math, that's just an approximate intuitive estimate of the problem.