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

Question

If surface area of the world is A, and the solar energy per unit of surface are is B, could humans ever harness more energy than AB?
What I mean is, though solar energy seems an infinite, really, we could only get as much as hits earth. Right?

Answer 1

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.

Answer 2

not really

first of all, there are some minor energy sources not related to solar radiation (primarily nuclear energy)

and, due to some efficiency limitations, the maximum energy that harnassable at the earth's surface is less that AB

and finally, there is a lot more solar radiation that does not strike the earth's surface, and technology already exists to harness that, although not necessarily cost effectively

Answer 3

Yes and no. We could never harness more than hits Earth by using collectors on the Earth, but we could put solar collectors in space and collect energy that would not normally hit Earth. The energy could be beamed to Earth by laser or other means, or it could be used in space to manufacture things needed on Earth and the things themselves shipped to Earth.

Answer 4

Makes sense to me although you might have to consider the medium that you are actually collecting the solar energy with.

Answer 5

you acquire the quantity genuine, yet for the exterior section you forgot to count quantity a number of of the surfaces. To calculate floor section, you will possibly desire to upload up the component of ALL surfaces of the forged. for this reason, that consists of the two semicircular bases (genuine and backside), the curved outdoors (25.a million in^2), and the quickly part opposite the curve. genuine and backside A = (a million/2 Pi r^2) = 2Pi, circumstances 2 = 4Pi = 12.6 in^2 Curved outdoors A = a million/2 (2Pi r) h = 8Pi = 25.a million in^2 quickly part = 4in * 4in = 16in^2 finished = fifty 3.7 in^2 wish this helped!