What percentage sunlight reaches earth?

[Lukraak_Sisser]

This question is sheer pointless curiosity on my part really.

Wiki awnsers says about 2 billionth of all the sunlight generated actually hits earth, but doesnt explain how it got to this awnser.

In theory I suppose it can be calculated using cone calculations, but the problem is.
I am truly utterly and horribly bad at geometry based maths :)

Any mathematician here that can show me the calculations?

Cheers

[Careyp74]

interesting question. I know that there is a difference between how much sun light the earth gets in January and how much it gets in June, so the answer is not just one number.

[arthwollipot]

I'm really bad at maths, but you could work it out (an approximation, of course) by calculating the surface area of a sphere of radius 1AU (the orbit of the earth), and how much area the two-dimensional cross-section of the earth takes up, and comparing the two.

If the area of the cross-section of the earth is one billionth of the total surface area of the sphere, then wikianswers' calculation is accurate. I haven't done the sums.

[wollery]

My back of the envelope calculation gave 2.2 billionths of the total.

You take the cross sectional area of the Earth (pi*RE2) as a fraction of the surface area of a sphere of radius 1AU (4*pi*AU2).

The values are averages, so at some times the Earth will receive more light and at some times less light, but the difference is tiny.

[H3LL]

Just askin' ...

That Earth-sized patch on the Sun's surface must be generating light in all directions, not just at 90 degrees.

Surely only a small proportion is heading in the right direction.

How does that effect the calculations?

Does the non-90 degree light from elsewhere cancel it all out or add to it?

.

[nathan]

Just askin' ...

That Earth-sized patch on the Sun's surface must be generating light in all directions, not just at 90 degrees.


What Earth-sized patch?

[H3LL]

What Earth-sized patch?

You take the cross sectional area of the Earth (pi*RE2) as a fraction of the surface area of a sphere of radius 1AU (4*pi*AU2).

Did I misunderstand?

[RecoveringYuppy]

Yes, you said on the Sun's surface. The patch wollery is descrbing is the Earth's surface (in the Sun's sky so to speak). It's the patch of the Sun's "sky" (not surface) blocked by the Earth.

ETA:
surface area of a sphere of radius 1AU (4*pi*AU2).
That's not the surface of the Sun wollery is calculating, it's the surface area of a sphere as large as Earth's orbit.

[H3LL]

That's not the surface of the Sun wollery is calculating, it's the surface area of a sphere as large as Earth's orbit.

Gotcha.

Please ignore the small, embarrassed fellow hiding behind the curtain.

Carry on....

[RecoveringYuppy]

Wouldn't think it's anything to be embarrassed about. There is an alternative way to calculate this that would involve calculating a patch of the Sun's surface. If you drew a cone from the Sun's center out to the Earth you could determine what patch of the Sun surface is effectively producing Earth's share of sunlight and you'd get the same answer. That patch on the Sun's surface would be a much smaller circle than Earth is.


Surely only a small proportion is heading in the right direction.

How does that effect the calculations?

Does the non-90 degree light from elsewhere cancel it all out or add to it?

.
And the answer in that case would be "it cancels out".