Power of solar radiation is very high is 3.8 * 10 to 20 degrees MW.At the Earth gets a tiny part of the solar energy.It is about half a billionth of.those.0.5 * 10 degree( -9).Consequently, E = 3.8 * 10 20 0.5 * * degree to 10 degree( -9) = 2 * 10 MW at 11 degrees, which is a percentage of 0.5 * 10 degree( -7)%.
Continue the themeSince the calculations have been made, then I add only your opinion.
I think our planet gets much sunlight - much as it is necessary for a harmonious life.All "created" with the mind.A little more or less already violate our ordinary life and entail quite different conditions.
Our location in relation to the Sun, as well as possible.
You presented the so-called anthropic principle: if the conditions were different( not necessarily those, for example, changing some global constant), then there was no one to discuss them! - More than a year ago continue the themeProbably very little.You can certainly count area of a sphere of radius 150 million kilometers( the distance from the Sun to the Earth), and then the area of a circle with a radius of 6000 km( radius of the Earth).And it is possible a different way: to take from the directory, how much energy reaches the Earth from the Sun( about 1350 watts per 1 sq. M), multiplied by the Earth's surface( about 500 million square kilometers, converted into square meters - a million times more.), Will7.10 ^ 17 watts.And the Sun( also from the directory) emits a second 3,6.10 ^ 26 watts.It turns out about two billionths of a ten-or two percent.It remains to compare this result with geometric calculation.But I leave this task to future researchers.For question 1.
I have to add to the answer.Said earth's surface area must be reduced at least by half, because the sun does not illuminate simultaneously the entire Earth.A "at least" - because in different places lit terrestrial hemisphere is a different energy obtained from the sun.This is in addition to that of the sunlight to reach the Earth's surface is not, but it does not take into account.
Now the second method is much simpler.No need to calculate the surface areas of "heavenly spheres": because this area is proportional to the square of the linear dimension.We obtain the radius of Earth's orbit of 150 million km, the Earth's radius of 6000 km, the first more than 25 thousand times.A different area is already at 625 million times.That will be the return value and that same share, namely about 1,5.10 ^ -9.That is pretty close to the first evaluation.
One about the second option did not think, thought enough to estimate the range of the area ratio( which is a circle instead of a ball) radius of the Earth - 6370 km radius of the sphere surface area of 150 million km. - More than a year ago with a radius easier, I decided without a calculator: 150 divided by 6 without remainder. - More than a year ago right, only it is necessary to take into account that the area of a sphere of radius R = 150 000 000 km, in which the energy of solar radiation is distributed is calculated according to the formula 4 * Pi * R ^ 2, and the area of a circle with a radius r =6000 km, roughly speaking, the projection of the Earth's area in the sphere expressed above is calculated by the formula pi * r ^ 2.So the earth gets no part 1/625000000 and 4 times less, ie,1/2500000000 or 4 * 10 ^( - 10) or part of 4 * 10 ^( - 8)%. - More than a year ago is so much energy would accrue to the world if some of it is not reflected back into space atmosphere, but still so much energy wasted is lost - more than a year ago continue the theme