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When confronting people with this question, I often get back something along the lines of "the higher the latitude, the smaller/wider the angle at which the sun hits the surface. As a result the same energy is spread across a larger surface, causing the insolation (W/m^2) to be lower at higher latitudes."

See image below. (This is basically the explanation for why temperatures tend to be higher at the equator.)

But isn't the angle of the sun "corrected for" by simply placing the panels at the right angle?

If that is indeed the case, then what is the real reason why solar panels capture less energy at higher latitudes?

enter image description here

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    $\begingroup$ This question already has been asked: earthscience.stackexchange.com/questions/12396/… Essentially the thing you are missing in your model is attenuation of the sunlight... since the light that reaches the surface at high latitudes had to travel through more atmosphere, thereby reducing it's intensity via scattering and absorption. $\endgroup$
    – f.thorpe ♦
    May 12 at 3:30
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    $\begingroup$ @f.thorpe None of the answers in that question (which was also closed as a duplicate) mention the concept of air mass. $\endgroup$
    – David Hammen
    May 12 at 8:07
  • $\begingroup$ @f.thorpe Thanks, but that's different question. The main answer to that question was as follows: "The very short, non-technical version (tl;dr): Each unit (think "beam of sunlight") is spread over a larger area." As I point out with the drawing that is exactly the type of answer that I usually get, and that I think is incorrect for explaining the lower energy output for solar panels at higher latitude. $\endgroup$
    – shamiv
    May 13 at 11:21
  • $\begingroup$ @Jean-MariePrival Thanks but that's a different question. In fact, it is the answer to that question that I usually (I would say wrongly) get when I ask my question. But thanks anyway! $\endgroup$
    – shamiv
    May 13 at 11:30

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If that is indeed the case, then what is the real reason why solar panels capture less energy at higher latitudes?

In a nutshell, air mass.

You aren't going to get a nice suntan (or a not so nice sunburn) if you sit unprotected (no sunscreen, no hat, minimal clothing) in the Sun for ten minutes just before sunset. You might well get a nice suntan (or a not so nice sunburn) if you sit unprotected in the Sun for ten minutes at solar noon at the equator.

The reason is that the radiation from the Sun has to pass through ever more atmosphere as the Sun's zenith angle increases. The zenith angle is 90° at sunrise and sunset, 0° when the Sun is directly overhead. The minimum zenith angle occurs at solar noon. At high latitudes, this is not all that low. The Sun is not directly overhead at high latitudes even at solar noon.

Even without clouds, the Earth's atmosphere is not perfectly transparent. The atmosphere reflects/bends some of the incoming sunlight away from the Earth, and also absorbs some of that incoming sunlight. As I mentioned at the start of my answer, this concept is encapsulated by the somewhat ad hoc principal of air mass. The amount of solar radiation that does reach the ground decreases with increasing solar zenith angle. Even at the northern hemisphere summer solstice, sunlight in Iceland is attenuated compared to sunlight in the tropics.

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  • $\begingroup$ The "not getting a sunburn at higher latitudes"-example only works to explain the effect without the difference in radiation per area (geometrically), if you lie on the ground on the equator and stand perpendicular to the sun before sunset. But I agree with the air mass explanation. $\endgroup$
    – hschoell
    May 12 at 10:41
  • $\begingroup$ Thanks, that makes sense! $\endgroup$
    – shamiv
    May 13 at 11:32

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