What percentage of the sun energy that reaches the earth is absorb by earths surface?

Answer 1

• About 50 percent of the Sun's incoming energy is absorbed by the Earth's surface. In addition, about 9 percent is reflected.

• About 22 percent of the Sun's incoming energy is reflected by clouds and the atmosphere.

• About 19 percent of the Sun's incoming energy is absorbed by clouds and the atmosphere.

Answer 2

The Earth receives about .00000005% of the Sun's energy (but only absorbs about 70% of what it receives), which ends up being about 174 petawatts.

Answer 3

The Earth receives 174 petawatts (PW) of incoming solar radiation (insolation) at the upper atmosphere. Approximately 30% is reflected back to space while the rest is absorbed by clouds, oceans and land masses. The spectrum of solar light at the Earth's surface is mostly spread across the visible and near-infrared ranges with a small part in the near-ultraviolet.

Retrieved from http://en.wikipedia.org/wiki/Sun_energy#Energy_from_the_Sun

Answer 4

The Earth receives 0.000000045% of the Sun's Energy


1) You must first calculate the surface area of a sphere with a circumference the size of the orbit of the Earth. Ignore pi because it will cancel out anyway in division.


4 x 93,000,000^2 = 34,596,000,000,000,000 sq. miles


2) Then calculate the area of a circle with the Earth's mean radius. The Earth is not a perfect sphere so you must use the mean radius.

NOTE: It would be a mistake to use half of the Earth's total surface area because the Sun's light does not shine on all parts of a sphere evenly. You must use just the circular area of the Earth's mean radius. Again ignore pi because it will cancel out in the division equation anyway.


3958.7^2 = 15,671,305.6 sq. miles


3) Now divide the first product by the second product and you will get the ratio of the Sun's light that Earth receives.


Ans1/Ans2 = 1/2,207,601,643.6 = 4.52980275 x 10^-10 = 0.00000000045

This may seem small but put this into perspective: If the Sun were the size of a junior Basketball, the Earth would be a pinhead 76 feet away.

Answer 5

That would be something in the neighborhood of 0.0000000454 percent, reaching the Earth from the Sun. Of that, roughly half reaches the surface.

Using:

Radius of the orbit of Earth = Ro = 93 million miles

Area of the sphere centered on the Sun with Earth's orbital radius = 4 pi Ro2

Radius of the Earth = Re = 3,963 miles

Cross sectional area of Earth facing the Sun = pi Re2

Therefore, fraction intercepted by Earth = pi Re2/4 pi Ro2 = 0.25 x (Re/Ro)2

= 0.25 x (3,963/93,000,000)2 = 4.54 x 10-10 approximately.

That's 4.54 x 10-8 percent.

That's 0.0000000454 percent, reaching the Earth. The amount reaching the surface would be less, depending on cloud cover, etc.

On average, just over half of the incoming energy (from the top of the atmosphere) reaches the surface.

Answer 6

Roughly 0.000000045 percent of the Sun's energy hits the Earth's atmosphere.

Some of that never makes it to the surface, and of the part that does, much of

it just bounces back into space.

Answer 7

Use:

Earth radius . . . 4,000 miles

Orbital radius . . . 93,000,000 miles

Assume that the sun's output is isotropically distributed.

Then the fraction of the sun's total output that's intercepted by the Earth is

(area of 4,000-mile-radius circle)/(area of 93-million-mile-radius sphere)

(pi) x (4,000)2/(4 pi) x (93 million)2 = 4.625 x 10-10 = 0.00000004625 percent = 93.3 dB down

Answer 8

That can be calculated quite easily. Earth has a radius of 6367 km (I took the average of the "polar radius" and "equatorial radius" on Wikipedia; though it really doesn't matter much which one you take). That makes for a cross-sectional area of pi x radius squared = 12.74 million (square kilometers).A sphere, centered on the Sun, with a radius of 149.6 million kilometers (the mean distance from Earth to Sun) would be 4 x pi x radius squared = 281200 (trillion square kilometers); the entire radiation of the Sun spreads over this area; if you divide this by the previous number, you get about 22100 million. That is, one part out of 22100 million, of the Sun's total power output, reaches Earth.