<p>Both solar altitude above the horizon and length of day influence total received solar energy. On December 21st both decrease with increasing latitude (in the Northern hemisphere), both dropping to zero at the Arctic Circle. On June 21st solar altitude above the horizon decreases with increasing latitude north of the Tropic of Cancer, but length of day increases with increasing latitude. Thus on June 21st maximum total daily solar energy is received at latitude 30 to 35 degrees North assuming average cloudiness, and at latitude 35 to 40 degrees North with clear skies. That's one reason why Death Valley gets so hot in summer --- it is at or near the latitude of maximum total daily solar energy receipt considering both sun altitude above the horizon and length of day.</p>

CC BY-SA 3.0
2017-10-03T19:36:35.020Z
2017-10-03T19:36:35.020Z
Jack Denur
<p>As noted in the comments, this answer applies to things like sun-bathing and solar panels, but it does not apply so much to a specific point-receptor like an eyeball. If all objects in question are pointing directly at the sun, then the angle of incidence is equal for all of them and this answer does not apply.</p>

<p>For an optic facing its target, the amount of atmosphere that the light passes through is a very large influencer. At higher latitudes, the sun is not directly overhead, and so the light is not coming straight down through the path of least atmosphere. Instead, it comes in at an angle, passing through more of the atmosphere before it gets to you.</p>

<p>For sun-bathers, solar panels, and the ground in general, the sunlight absorbed and reflected does depend very much on what is described in this answer. For that reason, more expensive solar panels are mounted on devices which alter their angle to face the sun for increased light exposure. And a sun-bather could likewise increase their exposure by mounting their platform at an angle. This is the direction the rest of the answer will take.</p>

<p>The answer is similar to the answer to some other questions, such as "Why does the solar power intensity change with the season?" and "Why does the solar intensity change with the height of the sun in the sky (ie: with the time of day)?"</p>

<h2>The very short, non-technical version (tl;dr)</h2>

<p>Each unit (think "beam of sunlight") is spread over a larger area.</p>

<p>That might not seem intuitive at first, but that is the answer in a nutshell. To see why, continue to the long version.</p>

<h2>The mechanics of the situation</h2>

<p>The sun is huge, but at great distances it can be thought of as a <strong>point-source</strong> of light. The sun emits light in every direction, so its energy, the electromagnetic radiation, can be thought of as beams of sunlight coming out from it in every direction. However, the farther away you get, the fewer of those sunlight beams cross your path, and if you get far away you can see, from the below illustration, that the lines which cross your path are becoming ever more parallel and <strong>directional</strong>.</p>

<p><a href="https://i.sstatic.net/OWGhu.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/OWGhu.png" alt="Sun as point source of light"></a></p>

<p>However, the sun is <em>very, very, very</em> far away - many millions of miles (and kilometers) away. At this distance, the electromagnetic radiation is all arriving at roughly parallel trajectories. That is, the light reaching us can be viewed as coming from a <strong>directional light source</strong>; it's all going in the same direction. You will see my crude illustration of that coming up after the next picture, but first...</p>

<p>When this happens, the angle at which the light arrives relative to the surface of the earth that it is arriving at, known as the <strong>angle of incidence</strong>, is what makes all the difference here. In the image below, we have light arriving from the left, perhaps from a flashlight(s), and two objects we want to see. The surface of the object below will be illuminated more, but the surface of the object above will be illuminated less.</p>

<p><a href="https://i.sstatic.net/YdreY.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/YdreY.png" alt="Surface area illumination of triangles at different angles of incidence"></a></p>

<p>Notice the object above gets 2 units of light to illuminate its entire surface, whereas the object below gets 5 units of light for its surface.</p>

<p>Looked at another way, the object above has to spread 2 units of light across its entire surface, where the object below gets that same amount of light concentrated on only a smaller part of it.</p>

<h3>Now back to Earth</h3>

<p>We see the same thing happen. Notice in the below picture that I have labeled the diameter of the area covered by my arbitrary unit of light. The distance is literally measured in image pixels. Near the equator, 1 unit of light illuminates 27 pixels, but closer to the pole the same amount of light needs to be spread out over 61 pixels of Earth's surface, so it's illuminated less.</p>

<p><a href="https://i.sstatic.net/7ndE1.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/7ndE1.png" alt="enter image description here"></a></p>

<p>And that is the answer to the question, and the related questions mentioned in my introduction.</p>

<h2>Some specific calculations</h2>

<p>The specific ratio of illumination difference for the two areas in the above image would be (<em>approximately</em>) simply the division of the two numbers.</p>

<p>The area closer to the pole is getting <code>27/61</code> as much light per longitudinal unit cross-section (a north/south strip of land) as the area closer to the equator, or 0.4426 times as much light.</p>

<p>Or, in the other direction, the area at the equator is getting <code>61/27</code> as much light, or 2.26 times as much light. So the equator area is getting more than twice as much light.</p>

<p>We can easily see from this why the sun might appear very much brighter or dimmer depending on the angle, and therefore depending on the latitude. This also explains the temperature difference: the equator is receiving a lot more energy, so it gets hotter.</p>

<h3>Answer applied to seasons</h3>

<p>I said earlier that this also answers the question about the illumination change for the seasons. The only extra bit of information that we need to understand that is this: Earth's pole around which it revolves is not perfectly perpendicular to the direction that the light is arriving from. Because of this, the specific angle changes as Earth orbits around the sun. The orbit around the sun takes a year (That's what a year <em>is</em>, by definition), so this change of angle over time has a yearly cycle during which the angle of incidence (the fancy term from above again, which just means "the difference of angle of the light to the object it is hitting") changes. This change is large enough to further change the sunlight that we receive, in the exact same way as discussed above.</p>

<h3>Further notes about temperature</h3>

<p>Just like before, this also affects temperature. The temperature, however, cannot be calculated by a simple ratio of area coverage. The temperature is affected by multiple other complicated systems. One of the major temperature influencers is the <strong>weather</strong>. Another of which is the difference between energy absorbed and energy lost (a physics thing: Earth is not a "<strong>black body</strong>" radiator).</p>

<p>Weather affect on temperature is very complicated and would require its own post much longer than this one, but to sum it up super succinctly: stuff like the wind and rain and such change the temperature and also move it around.</p>

<p>The physics black-body thing likewise would require its own long post, but again, super succinctly: Earth is also putting out its own energy into space at the same time it is receiving energy from the sun, and the affect this has on temperature is a fine balancing act. Once that balancing act tips to one side, the temperature swings way over in that direction: during the spring when temps are rising the energy absorption is enough higher than the energy radiation that the temperature rises, and the opposite is true for autumn when temps are falling.</p>

<p>The above descriptions for both weather and block-body radiation are way oversimplified.</p>

<h3>That's it</h3>

<p>And now the answer applies fully to seasonal changes as well, and it applies not only to illumination but also to temperature. I will leave it as an exercise for the reader to think about the daily affect (Why is it so much easier to look at the sun when it is low on the horizon just before it sets?)</p>

<p>And, as stated before, there are other influencers for these things; this just happens to be the biggest one.</p>

CC BY-SA 3.0
2017-09-29T03:53:57.887Z
From a technical/physical point of view the answer contains several inaccuracies: 1) better talk of irradiance than intensity, Though not completely wrong intensity is often used source based (=outgoing) and measured perpendicular to the direction of propagation, while irradiance is the radiant flux (power) **received** by a surface per unit area 2) the brightness of the sun doesn't really depend on the latitude but the relative thickness and the nature of the atmosphere (attenuation). Under clear sky/atmosphere the sun at the north pole has almost the same brightness as at the equator.
klanomath
CC BY-SA 3.0
2017-09-29T04:28:47.857Z
2b) Without atmosphere the sun would have almost (± 0.007%)  the same "brightness" seen from the north pole or from the equator in the summer half-year. The only difference are the +6375 km distance from the sun at the north pole at equinox (even less in the north summer).
klanomath
CC BY-SA 3.0
2017-09-29T06:38:07.060Z
I agree with @klanomath. OP correctly answers the question: how does the elevation angle of sun affect the irradiance per surface area of earth. What is (I think) fairly as important to notice is that if this was the only mechanism affecting then a person sunbathing by standing in the pole would get as much 'sun' as a person lying down on the equator. The key here is that the 'amount of atmosphere' attenuating the sunlight is larger in the poles.
Communisty
CC BY-SA 3.0
2017-09-29T19:40:40.643Z
@klanomath You do bring up a good point. The effect I outline deals with exposure over a surface area. For any given singular point in space on the surface of Earth does (if all other variables are equal) theoretically is in the path of approximately the same amount of sunlight; the +/- a few thousand miles difference (equator is a few thousand miles closer to the sun than the pole) makes little difference, and a point cares not about angle since the point is not 2D. However, the sun bather or solar panel (2D illuminated space) cares very much, though if you angled your bed or solar panel(1/2)
Aaron
CC BY-SA 3.0
2017-09-29T19:50:48.903Z
(2/2) differently so that it points toward the sun and cares not for the angle of Earth's surface, then you have just changed the angle. For an optic or a properly angled panel, amount of atmosphere is a large factor, yes. So it all depends on what exactly we are talking about. Perhaps the original question is not specific enough, though it was based on another question at a different SE and was crafted to mimic that other question closely. I'll probably update the answer with this thought process. Thank you.
Aaron
CC BY-SA 3.0
2017-09-28T20:28:45.300Z
2017-10-02T15:17:15.300Z
2017-10-02T15:17:15.300Z
Aaron
<p>People at the equator get to bask in more sunlight than Santa Clause and other inhabitants of the arctic regions. Not quite as pronounced, but they get more than me too.</p>

<p>Why is the sunlight more intense closer to the equator and less intense farther away from it?</p>

<p>When I posted this question, I was not thinking about the possible ambiguities, such as "Are you talking about the exposure across a surface area with some non-perpendicular angle to the sun," or "Are you talking about the light gathered by an optic facing the sun?" There is a difference. Since "basking in sunlight" was the example use case, let us assume exposure across a surface area which is lying on the ground.</p>

CC BY-SA 3.0
2017-09-28T20:31:19.880Z
I had posted this originally as an answer to a question on a different Stack Exchange site (The Great Outdoors), then I realized that I misread the other question and that my answer did not really answer it. To save the answer, someone suggested I post a Q&A here on earthscience.stackexchange. I hope it fares better here.
Aaron
CC BY-SA 3.0
2017-10-04T13:58:04.930Z
@Spencer Thank you. The questions are not identical, but yes, they are close enough to be marked duplicate, and the answers are the same. The answer on the other question also has better images and discusses the atmospheric effect in more detail. Thank you.
Aaron
CC BY-SA 3.0
2017-10-04T16:22:58.177Z
My mistake -- I was the one who advised reposting this from outdoors.se. I did search for duplicates first, but somehow missed that one.
Pont
<temperature><sun><light><equator><insolation>
Why does the intensity of sunlight depend on your latitude?
Aaron
# Why does the intensity of sunlight depend on your latitude?

(Asked by: username_0 on 2017-09-28)

People at the equator get to bask in more sunlight than Santa Clause and other inhabitants of the arctic regions. Not quite as pronounced, but they get more than me too.

Why is the sunlight more intense closer to the equator and less intense farther away from it?

When I posted this question, I was not thinking about the possible ambiguities, such as "Are you talking about the exposure across a surface area with some non-perpendicular angle to the sun," or "Are you talking about the light gathered by an optic facing the sun?" There is a difference. Since "basking in sunlight" was the example use case, let us assume exposure across a surface area which is lying on the ground.

**Tags:** temperature, sun, light, equator, insolation

> **Comments:**
> **username_0**: I had posted this originally as an answer to a question on a different Stack Exchange site (The Great Outdoors), then I realized that I misread the other question and that my answer did not really answer it. To save the answer, someone suggested I post a Q&A here on earthscience.stackexchange. I hope it fares better here.  
> **username_0**: @Spencer Thank you. The questions are not identical, but yes, they are close enough to be marked duplicate, and the answers are the same. The answer on the other question also has better images and discusses the atmospheric effect in more detail. Thank you.  
> **username_1**: My mistake -- I was the one who advised reposting this from outdoors.se. I did search for duplicates first, but somehow missed that one.  

---

## 2 answers

### Answer by username_0 on 2017-09-28. Score: 14 | Accepted answer

As noted in the comments, this answer applies to things like sun-bathing and solar panels, but it does not apply so much to a specific point-receptor like an eyeball. If all objects in question are pointing directly at the sun, then the angle of incidence is equal for all of them and this answer does not apply.

For an optic facing its target, the amount of atmosphere that the light passes through is a very large influencer. At higher latitudes, the sun is not directly overhead, and so the light is not coming straight down through the path of least atmosphere. Instead, it comes in at an angle, passing through more of the atmosphere before it gets to you.

For sun-bathers, solar panels, and the ground in general, the sunlight absorbed and reflected does depend very much on what is described in this answer. For that reason, more expensive solar panels are mounted on devices which alter their angle to face the sun for increased light exposure. And a sun-bather could likewise increase their exposure by mounting their platform at an angle. This is the direction the rest of the answer will take.

The answer is similar to the answer to some other questions, such as "Why does the solar power intensity change with the season?" and "Why does the solar intensity change with the height of the sun in the sky (ie: with the time of day)?"

## The very short, non-technical version (tl;dr)

Each unit (think "beam of sunlight") is spread over a larger area.

That might not seem intuitive at first, but that is the answer in a nutshell. To see why, continue to the long version.

## The mechanics of the situation

The sun is huge, but at great distances it can be thought of as a **point-source** of light. The sun emits light in every direction, so its energy, the electromagnetic radiation, can be thought of as beams of sunlight coming out from it in every direction. However, the farther away you get, the fewer of those sunlight beams cross your path, and if you get far away you can see, from the below illustration, that the lines which cross your path are becoming ever more parallel and **directional**.

[![Sun as point source of light](https://i.sstatic.net/OWGhu.png)](https://i.sstatic.net/OWGhu.png)

However, the sun is *very, very, very* far away - many millions of miles (and kilometers) away. At this distance, the electromagnetic radiation is all arriving at roughly parallel trajectories. That is, the light reaching us can be viewed as coming from a **directional light source**; it's all going in the same direction. You will see my crude illustration of that coming up after the next picture, but first...

When this happens, the angle at which the light arrives relative to the surface of the earth that it is arriving at, known as the **angle of incidence**, is what makes all the difference here. In the image below, we have light arriving from the left, perhaps from a flashlight(s), and two objects we want to see. The surface of the object below will be illuminated more, but the surface of the object above will be illuminated less.

[![Surface area illumination of triangles at different angles of incidence](https://i.sstatic.net/YdreY.png)](https://i.sstatic.net/YdreY.png)

Notice the object above gets 2 units of light to illuminate its entire surface, whereas the object below gets 5 units of light for its surface.

Looked at another way, the object above has to spread 2 units of light across its entire surface, where the object below gets that same amount of light concentrated on only a smaller part of it.

### Now back to Earth

We see the same thing happen. Notice in the below picture that I have labeled the diameter of the area covered by my arbitrary unit of light. The distance is literally measured in image pixels. Near the equator, 1 unit of light illuminates 27 pixels, but closer to the pole the same amount of light needs to be spread out over 61 pixels of Earth's surface, so it's illuminated less.

[![enter image description here](https://i.sstatic.net/7ndE1.png)](https://i.sstatic.net/7ndE1.png)

And that is the answer to the question, and the related questions mentioned in my introduction.

## Some specific calculations

The specific ratio of illumination difference for the two areas in the above image would be (*approximately*) simply the division of the two numbers.

The area closer to the pole is getting `27/61` as much light per longitudinal unit cross-section (a north/south strip of land) as the area closer to the equator, or 0.4426 times as much light.

Or, in the other direction, the area at the equator is getting `61/27` as much light, or 2.26 times as much light. So the equator area is getting more than twice as much light.

We can easily see from this why the sun might appear very much brighter or dimmer depending on the angle, and therefore depending on the latitude. This also explains the temperature difference: the equator is receiving a lot more energy, so it gets hotter.

### Answer applied to seasons

I said earlier that this also answers the question about the illumination change for the seasons. The only extra bit of information that we need to understand that is this: Earth's pole around which it revolves is not perfectly perpendicular to the direction that the light is arriving from. Because of this, the specific angle changes as Earth orbits around the sun. The orbit around the sun takes a year (That's what a year *is*, by definition), so this change of angle over time has a yearly cycle during which the angle of incidence (the fancy term from above again, which just means "the difference of angle of the light to the object it is hitting") changes. This change is large enough to further change the sunlight that we receive, in the exact same way as discussed above.

### Further notes about temperature

Just like before, this also affects temperature. The temperature, however, cannot be calculated by a simple ratio of area coverage. The temperature is affected by multiple other complicated systems. One of the major temperature influencers is the **weather**. Another of which is the difference between energy absorbed and energy lost (a physics thing: Earth is not a "**black body**" radiator).

Weather affect on temperature is very complicated and would require its own post much longer than this one, but to sum it up super succinctly: stuff like the wind and rain and such change the temperature and also move it around.

The physics black-body thing likewise would require its own long post, but again, super succinctly: Earth is also putting out its own energy into space at the same time it is receiving energy from the sun, and the affect this has on temperature is a fine balancing act. Once that balancing act tips to one side, the temperature swings way over in that direction: during the spring when temps are rising the energy absorption is enough higher than the energy radiation that the temperature rises, and the opposite is true for autumn when temps are falling.

The above descriptions for both weather and block-body radiation are way oversimplified.

### That's it

And now the answer applies fully to seasonal changes as well, and it applies not only to illumination but also to temperature. I will leave it as an exercise for the reader to think about the daily affect (Why is it so much easier to look at the sun when it is low on the horizon just before it sets?)

And, as stated before, there are other influencers for these things; this just happens to be the biggest one.

> **Comments:**
> **username_3**: From a technical/physical point of view the answer contains several inaccuracies: 1) better talk of irradiance than intensity, Though not completely wrong intensity is often used source based (=outgoing) and measured perpendicular to the direction of propagation, while irradiance is the radiant flux (power) **received** by a surface per unit area 2) the brightness of the sun doesn't really depend on the latitude but the relative thickness and the nature of the atmosphere (attenuation). Under clear sky/atmosphere the sun at the north pole has almost the same brightness as at the equator.  
> **username_3**: 2b) Without atmosphere the sun would have almost (± 0.007%)  the same "brightness" seen from the north pole or from the equator in the summer half-year. The only difference are the +6375 km distance from the sun at the north pole at equinox (even less in the north summer).  
> **username_4**: I agree with @username_3. OP correctly answers the question: how does the elevation angle of sun affect the irradiance per surface area of earth. What is (I think) fairly as important to notice is that if this was the only mechanism affecting then a person sunbathing by standing in the pole would get as much 'sun' as a person lying down on the equator. The key here is that the 'amount of atmosphere' attenuating the sunlight is larger in the poles.  
> **username_0**: @username_3 You do bring up a good point. The effect I outline deals with exposure over a surface area. For any given singular point in space on the surface of Earth does (if all other variables are equal) theoretically is in the path of approximately the same amount of sunlight; the +/- a few thousand miles difference (equator is a few thousand miles closer to the sun than the pole) makes little difference, and a point cares not about angle since the point is not 2D. However, the sun bather or solar panel (2D illuminated space) cares very much, though if you angled your bed or solar panel(1/2)  
> **username_0**: (2/2) differently so that it points toward the sun and cares not for the angle of Earth's surface, then you have just changed the angle. For an optic or a properly angled panel, amount of atmosphere is a large factor, yes. So it all depends on what exactly we are talking about. Perhaps the original question is not specific enough, though it was based on another question at a different SE and was crafted to mimic that other question closely. I'll probably update the answer with this thought process. Thank you.  

---

### Answer by username_2 on 2017-10-03. Score: 0

Both solar altitude above the horizon and length of day influence total received solar energy. On December 21st both decrease with increasing latitude (in the Northern hemisphere), both dropping to zero at the Arctic Circle. On June 21st solar altitude above the horizon decreases with increasing latitude north of the Tropic of Cancer, but length of day increases with increasing latitude. Thus on June 21st maximum total daily solar energy is received at latitude 30 to 35 degrees North assuming average cloudiness, and at latitude 35 to 40 degrees North with clear skies. That's one reason why Death Valley gets so hot in summer --- it is at or near the latitude of maximum total daily solar energy receipt considering both sun altitude above the horizon and length of day.


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