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I was thinking about why the poles are colder and I came out with three possible explanations.

1. The atmosphere blocks light beams coming to poles

atmosphere explanation

The Sun is seen only slightly over the horizon at the poles, so the light beams have longer way to the poles. If only this case were true, then the total difference between the poles and the equator would be the same as the energy that was absorbed by the atmosphere.

2. Sun is only slightly over the horizon at the poles.

location explanation

Poles are often in shadow of the parts of the Earth which are closer to the equator.

3. Poles are farther from the Sun

geometrical explanation

The longer is the way to Sun the lower is the chance to catch a light from the Sun.

Please, correct me wherever I'm wrong. I'd like to know

  • Which of these three is most correct?
  • How significant are the described effects of the other explanations?
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    $\begingroup$ Did you try typing "Why are the poles cold?" into Google? $\endgroup$ – ACuriousMind Dec 13 '15 at 15:42
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    $\begingroup$ hko.gov.hk/education/edu06nature/ele_srad_e.htm $\endgroup$ – user83548 Dec 13 '15 at 15:42
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    $\begingroup$ Note that light rays that reach us from the sun are almost perfectly parallel, so your second diagram in particular is rather misleading. (Obviously, none of them are to scale but the not-to-scale-ness of the second one results in a situation that's very different from reality.) $\endgroup$ – David Richerby Dec 13 '15 at 21:10
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    $\begingroup$ The relatively small differences in distances to the Sun is not substantial compared with other effects: the Earth's orbit takes it closest to the Sun at the beginning of January, but in most of the northern hemisphere temperatures are usually much higher in July when the Sun is further away. $\endgroup$ – Henry Dec 14 '15 at 0:21
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No, the main reason the poles are colder is because the surface is angled with respect to the sun rays. ray angle modifies the amount of energy

That is the same reason why in winter it is colder than in summer. You can think that in the poles the winters are harder and the summers are softer, while in the equator it is the other way around.

That is also the reason why some solar panel are motorized: to keep them perpendicular to the sun rays and get the most out of them.

About your hypothesis:

  1. If the atmosphere were to absorb the energy, there still would be heat at the poles, maybe in the air instead of in the ground, but still there.

  2. In polar winter there is constant darkness (24-hour night) up to 6 months in the geometrical pole. But in polar summer there is constant sun and it is still quite cold. So while the shadow actually makes a difference (polar winter is way colder than polar summer), that is not why the poles are colder than equator.

  3. The Earth is 150000000 km from the sun, and the radious of the Earth is about 6500 km, way to small to make a difference.

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    $\begingroup$ Why does the angle effect the heating difference? $\endgroup$ – Ulad Kasach Dec 13 '15 at 15:55
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    $\begingroup$ @VladK: Look at my graphic. In both cases the exposed surface and the light intensity are the same, only the angle changes, The first one receives the double amount of energy than the second one: 4 sun-rays instead of 2 sun-rays. Imagine the surface parallel to the rays and you will get not rays at all! $\endgroup$ – rodrigo Dec 13 '15 at 15:57
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    $\begingroup$ @VladK: If you want to go technical, the energy density is multiplied by the sine of the angle between the rays and the surface. In the first one it would be 90º, and in the second one 30º. $sin(90^o) = 1$, $sin(30^o) = 0.5$. $\endgroup$ – rodrigo Dec 13 '15 at 15:59
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    $\begingroup$ @VladK look at a sheet of paper straight on, then tilt the paper so that you're almost looking at the edge. It takes up a lot less visual space, doesn't it? Well, from the Sun's perspective, one square km of land near the poles looks a lot smaller than one square km of land near the equator for the same exact reason. And the Sun puts out equal sunlight in every direction, meaning the amount of sunlight received is proportional to angular size. A "tilted" bit of Earth gets that much less light. $\endgroup$ – hobbs Dec 14 '15 at 7:46
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    $\begingroup$ Also, you can do the experiment at home. Use a torch in a darkened room (to avoid the light bouncing off white walls illuminating everything diffusely), and illuminate the wall from the same distance, once perpendicularly, and once at an angle - you'll see that the wall gets darker, and also why: the beam is spread over a larger area. $\endgroup$ – orion Dec 14 '15 at 10:14
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It is mostly the angle of the light, but the atmosphere has some effect. When determining the amount of power per unit area hitting a surface (the Irradiance $E_e$), the intensity, albedo, and incident angle all play a role. $$E_e = \epsilon I \cos(\theta) $$ Here, I'm using $I$ for the intensity (time averaged Poynting vector), $\epsilon$ for the efficiency of absorption (blackbody is 1), and $\theta$ for the incident angle of the light (perpendicular is 0). https://en.wikipedia.org/wiki/Irradiance is a decent place to start reading online.

The third effect is fairly small. Since the intensity falls off as $1/R^2$, a 1% increase in distance from the sun causes a 2% decrease in intensity of sunlight. (This is a Taylor series approximation that works well under around 5%.) Check the radius of Earth as compared to the distance from the sun; it is a fraction of a percent.

The first effect is complicated. Certainly, sunlight dims as the sun gets lower, even for a detector facing the sun. But some of that lost energy warms the atmosphere. Does that count toward energy absorbed or get excluded? How much is it as compared to when the sun is overhead? Is it more pronounced on cloudy days?

Don't forget that polar regions spend a lot of time in shadow. The opposite is https://en.wikipedia.org/wiki/Midnight_sun.

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