What percentage of the energy radiated by he sun reaches the earths surface?

Answer 1

Assume the Sun radiates electromagnetic radiation and solar wind particles equally in all directions.

Then the portion of its radiation that strikes the Earth (or Earth's magnetic shield) is: the cross-sectional area of the Earth divided by the surface area of the sphere centered on the Sun, at 1AU.

So, that's:

"pi" x (Earth radius)2/ 4"pi" x (Earth-Sun distance)2 = 4.62 x 10-10

That's 0.0000000462 percent of it.

That's a pretty good answer, without going into minor complications.

(Or 93.3 dB less.)

Answer 2

About 0.000000045 percent of all of the radiation from the Sun reaches Earth.

When it gets to Earth, there is about 1.4 kilowatts per square meter just before the sunlight hits the Earth's atmosphere. Just over a third of that actually falls on the surface of the Earth because of scattering.

Answer 3

The energy is radiated equally in all directions into a sphere with a radius of 150 million kilometres, which has a surface area. On that sphere sits the Earth with a radius of 6378 kilometres, which has a circular cross-section area which intercepts part of the total energy. The ratio of the two areas answers the question.

Answer 4

That fraction is nominally

(1/2 the Earth's surface area) / (the area of a sphere whose radius is the Earth-Sun distance)

Use the following approximations:

Earth radius = 4,000 miles
Earth-Sun distance = 93 million miles

then the fraction is

2 (pi) (4,000)^2 / 4 (pi) (93 million) ^2 =

(1/2) (4 / 93000)^2 = 9.25 x 10-10

The percentage is 0.0000000925 % of the sun's total output that reaches
the sunward-facing side of the Earth.

How much actually reaches the surface is another matter. (It would be less.)

Answer 5

Well, I don't know percentages, or even where to find the information. But it must be a fairly low amount. The great majority of the Sun's energy is either reflected directly, or it heats up things that later radiate the heat back into space.