<p>The sun is about 100 times the size of the Earth (in diameter), and the distance from the sun to Earth is about 100 times the diameter of the sun. <strong><em>Below</em></strong> is an image showing the sun, Earth, and the distance between them to scale. It looks at first like nothing more than a black bar. This image in the original is <strong><em>4000</em></strong> pixels (the limit of my Pixlr Editor) by 60 pixels, and Earth is less than a pixel. To see it full size, <a href="https://i.sstatic.net/YTpMm.png" rel="nofollow noreferrer">click here</a> or on the image, then when it comes up in your web browser click on it again to magnify it to full size. Use your scroll bar to see the sun at one end and Earth at the other. You will need to look <strong><em>really</em></strong> closely to see Earth.</p>

<p><a href="https://i.sstatic.net/YTpMm.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/YTpMm.png" alt="the sun, Earth, and the distance between them to scale - public domain image"></a></p>

CC BY-SA 3.0
2015-11-23T04:43:54.017Z
How did you draw 'less than a pixel'?
Scott
CC BY-SA 3.0
2015-11-23T04:57:13.867Z
@Scott I just made the one pixel darker
Eubie Drew
CC BY-SA 3.0
2015-11-25T22:36:25.537Z
@RickyDemer lighter than the background, but darker than it would be with Earth filling the pixel.
Eubie Drew
CC BY-SA 3.0
2015-11-22T20:19:45.623Z
2015-11-22T21:23:26.250Z
2015-11-22T21:23:26.250Z
Eubie Drew
<blockquote>
  <p>But if you scale up the Sun to it's real size compared to the Earth,</p>
</blockquote>

<p>Is pretty big. But then to be realistic in that manner you would also have to scale up the distance from the earth to the sun.</p>

<p>But you don't have to even do that, as there is a much simpler way of seeing how large the sun is from the perspective of a position on the earth.</p>

<ol>
<li>Be on or very near to the earth (you are probably already achieving this).</li>
<li>Look at the sky.</li>
</ol>

<p>If it's daytime and not cloudy you should see the sun up there somewhere (please don't keep looking…). It's not very big, about 32′ (varies slightly) which is just slightly bigger than 0.5°.</p>

<p>That's about the same as the moon (or else transits wouldn't be the interesting-looking eclipses that we get). Or about the size of the princess' thumb at arm's length in the story of the princess who wanted the moon for her birthday.</p>

<p>It's not enough to bathe most of the earth in the sun's rays. It's just about enough for almost half to be lit, almost half to be dark, and to have a period where a given point of the earth is illuminated by some, but not all, of the sun's visible disc. You can see this by the same look-at-the-sky experiment, by looking toward the horizon during sunrise or sunset.</p>

CC BY-SA 3.0
2015-11-23T11:11:00.900Z
2015-11-23T16:36:02.973Z
2015-11-23T16:36:02.973Z
Jon Hanna
<p>Given the known average distance to the Sun, and the radii of Sun and Earth, the basic trigonometry is simple. 
If the Earth and Sun were exactly the same size, and there was no atmospheric refraction, then exactly half the planet, or 180 degrees, would be illuminated. But since the Sun is so much bigger, again assuming no atmospheric refraction, then exactly 180.522 degrees of the Earth would be illuminated. Calculating the additional illumination caused by atmospheric refraction of sunlight into the zone of darkness is fraught with difficulty because of variable atmospheric haze, temperature-controlled differences in air density, and cloud distribution. I'm guessing you could add a degree or two more if you were to include refracted twilight.</p>

CC BY-SA 3.0
2015-11-22T19:11:03.257Z
2015-11-22T19:11:03.257Z
Gordon Stanger
<p>Because no matter how big the Sun is compared to the Earth, only half the Earth can be facing the Sun at any given time.</p>

<p>Try drawing a diagram where the Sun is much bigger with respect to the Earth. Make it ten times or a hundred times bigger or go all out and draw it to scale. Suppose it to be so big that it doesn't even look like a circle,  but just a straight line of infinite length. Still, only half the Earth is facing it at any given time. Unless sunlight can penetrate the Earth, or somehow leaks around the edges, half the Earth will still be dark.</p>

<p><em>Update</em></p>

<p>Ahhh, I see where you're coming from now. Okay, true, if the Sun was infinitely large, or at least extremely large and extremely close, and if light left any given point on the surface of the Sun travelling in all directions, then yes, light rays could reach parts of the Earth not facing toward the Sun. They couldn't actually reach the point on the exact opposite side, but I guess close enough. I confess, I was thinking of light rays travelling from the Sun to the Earth in a direction parallel to a straight line from the center of the Earth to the center of the Sun. </p>

<p>I suppose that all light leaving the surface of the Sun is not travelling directly outward from the center, so to that extent, your model is plausible. Frankly I don't know enough about the physics to say how close that resembles reality. I'd think more light is travelling directly out from the center than at tangents, because the Sun is not a dark core with a thin glowing surface, but rather produces light from within. I have no idea what the relative numbers are though. But I guess the question is not the amount of light, but any light. So let's accept your model.</p>

<p>So yes, if the Sun was big enough and close enough, light from, say, the north pole of the Sun could hit a point on the "far side" of the Earth at a tangent. As in my diagram.</p>

<p>In real life, though, it's not THAT big or THAT close. Let's run the numbers.</p>

<p><a href="https://i.sstatic.net/p7iPL.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/p7iPL.gif" alt="enter image description here"></a></p>

<p>Let $s$ be the distance from the Earth to the Sun, $r$ the radius of the Sun, and $A$ the angle along the curve of the Earth from the north pole. Then we have the triangle shown below, where $h$ is the height of the triangle. The radius of the Earth is very small compared to all the other numbers here, and so basically gets lost in the rounding. As $s$ is very large compared to $r$, $h$ is very close to $r$, so just to keep the geometry simple, let's assume $h=r$. (As $r \ge h$, calculating the real value for $h$ hurts your case.)</p>

<p>So geometry tells us that $r=s * \sin A$.</p>

<p>The Sun is about 93 million miles away. So for light rays in this scenario to reach 45 degrees past the Earth's north pole, we'd have:</p>

<p>$$
r = 93,000,000 * \sin(45) \\
r = 93,000,000 * \frac{\sqrt 2}{2} \\
r \approx 65,000,000 \\
$$</p>

<p>That is, the radius of the sun would have to be more than 65 million miles. In real life it's nowhere near that, it's more like 432,000.</p>

<p>So how far past the pole can the light really bleed?</p>

<p>$$
h \ge s * \sin A \\
\sin A \le h / s \\
\sin A \le \frac {432,000}{93,000,000} \\
\sin A \le .004645 \\
A \le 0.266 °\\
$$</p>

<p>That is, sunlight can "bleed" past the pole by about 1/4 of 1 degree.</p>

CC BY-SA 3.0
2015-11-23T19:36:40.720Z
I think the OP was speculating that the Sun should be big enough that the light would leak around the edges, so to speak.
Matt Hall
CC BY-SA 3.0
2015-11-23T22:04:20.100Z
@kwinkunks Draw the diagram. It doesn't work like that. No matter how big the Sun is relative to the Earth, light rays can only directly impinge on half the surface. I suppose you could say that some light might reflect off objects in the atmosphere or near space to give light on the other side of the planet (like the Moon does to some extent), but it's not like making the Sun big enough means that the entire Earth would then be in direct sunlight. That's just not how the geometry works.
Jay
CC BY-SA 3.0
2015-11-23T22:43:18.387Z
I did [draw the diagram](http://imgur.com/JtQuF3d), and it does work like that.
Matt Hall
CC BY-SA 3.0
2015-11-23T22:46:52.897Z
@kwinkunks [You beat me to it!](http://i.imgur.com/BZ4rf1Q.png).
Pont
CC BY-SA 3.0
2015-11-26T00:10:33.270Z
@Jay I was a bit confused about this initially but I think it is because I was thinking of the sun as a point source which emits light radially and so you would be correct. However, the sun is not a point source and in fact emits light in all directions randomly. So in fact it is possible to have a light beam illuminating the other side of the earth as shown in the diagrams
Isopycnal Oscillation
CC BY-SA 3.0
2015-11-28T07:19:58.310Z
@kwinkunks et al: Ah, I see where you're coming from. See my update.
Jay
CC BY-SA 3.0
2015-11-23T14:55:44.473Z
2015-11-29T21:31:36.357Z
2015-11-29T21:31:36.357Z
Jay
<p>The sun is <a href="https://en.wikipedia.org/wiki/Astronomical_unit" rel="nofollow noreferrer">really far away</a>. Thus its rays are <a href="https://physics.stackexchange.com/questions/155075/what-makes-suns-light-travel-as-parallel-beams-towards-earth">essentially parallel</a> at the earth's orbit. So, while the diagram you posted is clearly a bit off in terms of the relative size and distance between the sun and the earth, the parallel rays are about right.</p>

<p><a href="https://i.sstatic.net/L4pXa.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/L4pXa.png" alt="Sun and Earth - public domain image"></a></p>

CC BY-SA 3.0
2015-11-22T15:59:41.220Z
So you mean that in the 3D perspective, the diagram is correct? Also imagine you are looking at the Earth from the Sun. I think the result might change, as uncountable amount (as far as I know) are shooting from the Sun (an unscientific way to say that) and every spot on Earth is possible. For example from the top of the sun to the top of the Earth? Probably the lit area will be larger than the hemisphere.
diamondburned
CC BY-SA 3.0
2015-11-22T17:03:38.953Z
The 2D solution to your misunderstanding is still all about the extreme distance between the sun and Earth. The diagram in your Q is essentially a cartoon.
Eubie Drew
CC BY-SA 3.0
2015-11-22T19:53:40.257Z
@dattutbrus this one looks about to scale if interested.   http://www.takayaiwamoto.com/Earth_Moon_Sun/solar_eclipse_model_1.gif    I like the diagram Kwinkunks chose better though.   The earth is about 150 solar diamaters away from the sun.   Imagine a basketball about 120 feet away from you.   That's about the equivalent.
userLTK
CC BY-SA 3.0
2015-11-23T13:56:46.313Z
@dattutbrus the lit area *will* be larger than the hemisphere, but just a tiny bit larger (fraction of a percent), so an approximation of a hemisphere is close enough for everyday life. However, saying "every spot on Earth is possible" is not true, even if "the sun" was extremely large (e.g. approaching infinity) and extremely close (but still being *outside*), there will still be an exact opposite area that is not lit.
Peteris
CC BY-SA 3.0
2015-11-22T13:25:26.667Z
2015-11-22T14:00:12.443Z
2017-04-13T12:40:15.470Z
Matt Hall
<p>The question is clear enough in my opinion but, I've found out that if you redraw this picture here: </p>

<p><a href="https://upload.wikimedia.org/wikipedia/commons/thumb/1/12/Seasons.svg/1920px-Seasons.svg.png" rel="nofollow noreferrer"><img src="https://upload.wikimedia.org/wikipedia/commons/thumb/1/12/Seasons.svg/1920px-Seasons.svg.png" alt="&quot;Seasons&quot;, as from Wikipedia"></a></p>

<p>You'll notice that the Sun is as big as the Earth. But if you scale up the Sun to it's real size compared to the Earth, you would realize that the Earth would covered in daylight ALL THE TIME. So what's going on here?</p>

CC BY-SA 3.0
2015-11-22T13:19:06.903Z
You should try drawing it, or finding such an image. Remember to scale the distance as well as the size.
Matt Hall
CC BY-SA 3.0
2015-11-22T22:23:14.423Z
You already got a couple excellent answers. Another way to visualize this: if your reasoning applied, the sun, as seen from earth, would also have to fill half the sky. Instead, you can pretty much cover it with your thumb (don't actually try that, but you can use the moon, which appears as roughly the same size as the sun).
Kevin Keane
CC BY-SA 3.0
2015-11-23T19:03:43.420Z
I can mentally interpret that image as if the Sun is actually far off in the distance even though the image shows it as being off to the right. (The shadowing on Earth forces left/right orientation.) If the Sun was far off, the rays of light would still be drawn almost exactly as they are. They wouldn't actually be parallel, but the angles between them would be very small. The Earth would be drawn as lit on only one side, as it actually is in "real life".
user2338816
<earth-observation><solar-terrestrial-physics><seasons><diurnal-cycle><scaling>
Why wouldn't the earth be "always daytime" if the "daily cycle of sunlight" was drawn full-scale?
diamondburned
# Why wouldn't the earth be "always daytime" if the "daily cycle of sunlight" was drawn full-scale?

(Asked by: username_0 on 2015-11-22)

The question is clear enough in my opinion but, I've found out that if you redraw this picture here:

[!["Seasons", as from Wikipedia](https://upload.wikimedia.org/wikipedia/commons/thumb/1/12/Seasons.svg/1920px-Seasons.svg.png)](https://upload.wikimedia.org/wikipedia/commons/thumb/1/12/Seasons.svg/1920px-Seasons.svg.png)

You'll notice that the Sun is as big as the Earth. But if you scale up the Sun to it's real size compared to the Earth, you would realize that the Earth would covered in daylight ALL THE TIME. So what's going on here?

**Tags:** earth-observation, solar-terrestrial-physics, seasons, diurnal-cycle, scaling

> **Comments:**
> **username_1**: You should try drawing it, or finding such an image. Remember to scale the distance as well as the size.  
> **username_2**: You already got a couple excellent answers. Another way to visualize this: if your reasoning applied, the sun, as seen from earth, would also have to fill half the sky. Instead, you can pretty much cover it with your thumb (don't actually try that, but you can use the moon, which appears as roughly the same size as the sun).  
> **username_3**: I can mentally interpret that image as if the Sun is actually far off in the distance even though the image shows it as being off to the right. (The shadowing on Earth forces left/right orientation.) If the Sun was far off, the rays of light would still be drawn almost exactly as they are. They wouldn't actually be parallel, but the angles between them would be very small. The Earth would be drawn as lit on only one side, as it actually is in "real life".  

---

## 5 answers

### Answer by username_1 on 2015-11-22. Score: 16 | Accepted answer

The sun is [really far away](https://en.wikipedia.org/wiki/Astronomical_unit). Thus its rays are [essentially parallel](https://physics.stackexchange.com/questions/155075/what-makes-suns-light-travel-as-parallel-beams-towards-earth) at the earth's orbit. So, while the diagram you posted is clearly a bit off in terms of the relative size and distance between the sun and the earth, the parallel rays are about right.

[![Sun and Earth - public domain image](https://i.sstatic.net/L4pXa.png)](https://i.sstatic.net/L4pXa.png)

> **Comments:**
> **username_0**: So you mean that in the 3D perspective, the diagram is correct? Also imagine you are looking at the Earth from the Sun. I think the result might change, as uncountable amount (as far as I know) are shooting from the Sun (an unscientific way to say that) and every spot on Earth is possible. For example from the top of the sun to the top of the Earth? Probably the lit area will be larger than the hemisphere.  
> **username_4**: The 2D solution to your misunderstanding is still all about the extreme distance between the sun and Earth. The diagram in your Q is essentially a cartoon.  
> **username_11**: @dattutbrus this one looks about to scale if interested.   http://www.takayaiwamoto.com/Earth_Moon_Sun/solar_eclipse_model_1.gif    I like the diagram Kwinkunks chose better though.   The earth is about 150 solar diamaters away from the sun.   Imagine a basketball about 120 feet away from you.   That's about the equivalent.  
> **username_12**: @dattutbrus the lit area *will* be larger than the hemisphere, but just a tiny bit larger (fraction of a percent), so an approximation of a hemisphere is close enough for everyday life. However, saying "every spot on Earth is possible" is not true, even if "the sun" was extremely large (e.g. approaching infinity) and extremely close (but still being *outside*), there will still be an exact opposite area that is not lit.  

---

### Answer by username_4 on 2015-11-22. Score: 11

The sun is about 100 times the size of the Earth (in diameter), and the distance from the sun to Earth is about 100 times the diameter of the sun. ***Below*** is an image showing the sun, Earth, and the distance between them to scale. It looks at first like nothing more than a black bar. This image in the original is ***4000*** pixels (the limit of my Pixlr Editor) by 60 pixels, and Earth is less than a pixel. To see it full size, [click here](https://i.sstatic.net/YTpMm.png) or on the image, then when it comes up in your web browser click on it again to magnify it to full size. Use your scroll bar to see the sun at one end and Earth at the other. You will need to look ***really*** closely to see Earth.

[![the sun, Earth, and the distance between them to scale - public domain image](https://i.sstatic.net/YTpMm.png)](https://i.sstatic.net/YTpMm.png)

> **Comments:**
> **username_5**: How did you draw 'less than a pixel'?  
> **username_4**: @username_5 I just made the one pixel darker  
> **username_4**: @RickyDemer lighter than the background, but darker than it would be with Earth filling the pixel.  

---

### Answer by username_6 on 2015-11-23. Score: 7

> But if you scale up the Sun to it's real size compared to the Earth,

Is pretty big. But then to be realistic in that manner you would also have to scale up the distance from the earth to the sun.

But you don't have to even do that, as there is a much simpler way of seeing how large the sun is from the perspective of a position on the earth.

1. Be on or very near to the earth (you are probably already achieving this).
2. Look at the sky.

If it's daytime and not cloudy you should see the sun up there somewhere (please don't keep looking…). It's not very big, about 32′ (varies slightly) which is just slightly bigger than 0.5°.

That's about the same as the moon (or else transits wouldn't be the interesting-looking eclipses that we get). Or about the size of the princess' thumb at arm's length in the story of the princess who wanted the moon for her birthday.

It's not enough to bathe most of the earth in the sun's rays. It's just about enough for almost half to be lit, almost half to be dark, and to have a period where a given point of the earth is illuminated by some, but not all, of the sun's visible disc. You can see this by the same look-at-the-sky experiment, by looking toward the horizon during sunrise or sunset.


---

### Answer by username_7 on 2015-11-22. Score: 6

Given the known average distance to the Sun, and the radii of Sun and Earth, the basic trigonometry is simple.
If the Earth and Sun were exactly the same size, and there was no atmospheric refraction, then exactly half the planet, or 180 degrees, would be illuminated. But since the Sun is so much bigger, again assuming no atmospheric refraction, then exactly 180.522 degrees of the Earth would be illuminated. Calculating the additional illumination caused by atmospheric refraction of sunlight into the zone of darkness is fraught with difficulty because of variable atmospheric haze, temperature-controlled differences in air density, and cloud distribution. I'm guessing you could add a degree or two more if you were to include refracted twilight.


---

### Answer by username_8 on 2015-11-23. Score: 3

Because no matter how big the Sun is compared to the Earth, only half the Earth can be facing the Sun at any given time.

Try drawing a diagram where the Sun is much bigger with respect to the Earth. Make it ten times or a hundred times bigger or go all out and draw it to scale. Suppose it to be so big that it doesn't even look like a circle, but just a straight line of infinite length. Still, only half the Earth is facing it at any given time. Unless sunlight can penetrate the Earth, or somehow leaks around the edges, half the Earth will still be dark.

*Update*

Ahhh, I see where you're coming from now. Okay, true, if the Sun was infinitely large, or at least extremely large and extremely close, and if light left any given point on the surface of the Sun travelling in all directions, then yes, light rays could reach parts of the Earth not facing toward the Sun. They couldn't actually reach the point on the exact opposite side, but I guess close enough. I confess, I was thinking of light rays travelling from the Sun to the Earth in a direction parallel to a straight line from the center of the Earth to the center of the Sun.

I suppose that all light leaving the surface of the Sun is not travelling directly outward from the center, so to that extent, your model is plausible. Frankly I don't know enough about the physics to say how close that resembles reality. I'd think more light is travelling directly out from the center than at tangents, because the Sun is not a dark core with a thin glowing surface, but rather produces light from within. I have no idea what the relative numbers are though. But I guess the question is not the amount of light, but any light. So let's accept your model.

So yes, if the Sun was big enough and close enough, light from, say, the north pole of the Sun could hit a point on the "far side" of the Earth at a tangent. As in my diagram.

In real life, though, it's not THAT big or THAT close. Let's run the numbers.

[![enter image description here](https://i.sstatic.net/p7iPL.gif)](https://i.sstatic.net/p7iPL.gif)

Let $s$ be the distance from the Earth to the Sun, $r$ the radius of the Sun, and $A$ the angle along the curve of the Earth from the north pole. Then we have the triangle shown below, where $h$ is the height of the triangle. The radius of the Earth is very small compared to all the other numbers here, and so basically gets lost in the rounding. As $s$ is very large compared to $r$, $h$ is very close to $r$, so just to keep the geometry simple, let's assume $h=r$. (As $r \ge h$, calculating the real value for $h$ hurts your case.)

So geometry tells us that $r=s \* \sin A$.

The Sun is about 93 million miles away. So for light rays in this scenario to reach 45 degrees past the Earth's north pole, we'd have:

$$
r = 93,000,000 \* \sin(45) \\
r = 93,000,000 \* \frac{\sqrt 2}{2} \\
r \approx 65,000,000 \\
$$

That is, the radius of the sun would have to be more than 65 million miles. In real life it's nowhere near that, it's more like 432,000.

So how far past the pole can the light really bleed?

$$
h \ge s \* \sin A \\
\sin A \le h / s \\
\sin A \le \frac {432,000}{93,000,000} \\
\sin A \le .004645 \\
A \le 0.266 °\\
$$

That is, sunlight can "bleed" past the pole by about 1/4 of 1 degree.

> **Comments:**
> **username_1**: I think the OP was speculating that the Sun should be big enough that the light would leak around the edges, so to speak.  
> **username_8**: @kwinkunks Draw the diagram. It doesn't work like that. No matter how big the Sun is relative to the Earth, light rays can only directly impinge on half the surface. I suppose you could say that some light might reflect off objects in the atmosphere or near space to give light on the other side of the planet (like the Moon does to some extent), but it's not like making the Sun big enough means that the entire Earth would then be in direct sunlight. That's just not how the geometry works.  
> **username_1**: I did [draw the diagram](http://imgur.com/JtQuF3d), and it does work like that.  
> **username_9**: @kwinkunks [You beat me to it!](http://i.imgur.com/BZ4rf1Q.png).  
> **username_10**: @username_8 I was a bit confused about this initially but I think it is because I was thinking of the sun as a point source which emits light radially and so you would be correct. However, the sun is not a point source and in fact emits light in all directions randomly. So in fact it is possible to have a light beam illuminating the other side of the earth as shown in the diagrams  
> **username_8**: @kwinkunks et al: Ah, I see where you're coming from. See my update.  

---

