<p>OK, let start with some assumptions: spherical objects and no atmospheric effects.</p>
<p>Here's the relevant geometry with the object on the left representing a planet and the object on the right representing the sun:</p>
<p><a href="https://i.sstatic.net/YcYuq.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/YcYuq.png" alt="enter image description here" /></a></p>
<p>Where d is the distance between the two spheres, r<sub>1</sub> and r<sub>2</sub> are the radii of the spheres. <span class="math-container">$\theta_1$</span> is the angle showing the portion of the sphere illuminated.  We'll use it later to compute what fraction of the sphere is lit.</p>
<p>We can solve for <span class="math-container">$\theta_1$</span> easily by noticing that the solid line quadrilateral can be drawn as below.  We're flipping it so that the side labeled L is on the bottom.</p>
<p><a href="https://i.sstatic.net/bvRJD.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/bvRJD.png" alt="geometric modelisation" /></a></p>
<p>Noting that the upper triangle is a a right-triangle with sides known allows us to solve for the unknown angle <span class="math-container">$\theta_1$</span> as follows:</p>
<p><span class="math-container">$$d \sin\left(\frac{\pi}{2}-\theta_1\right) = r_2-r_1$$</span></p>
<p><span class="math-container">$$\sin\left(\frac{\pi}{2}-\theta_1\right) = \frac{r_2-r_1}{d}$$</span></p>
<p><span class="math-container">$$\frac{\pi}{2}-\theta_1 = \arcsin\left(\frac{r_2-r_1}{d}\right)$$</span></p>
<p><span class="math-container">$$\theta_1 = \arcsin\left(\frac{r_2-r_1}{d}\right) + \frac{\pi}{2}$$</span></p>
<p>Or, with a small angle approximation for when the distance is much larger than the difference in radii:</p>
<p><span class="math-container">$$\theta_1 \approx \frac{r_2-r_1}{d} + \frac{\pi}{2}$$</span></p>
<p>The <a href="https://en.wikipedia.org/wiki/Solid_angle" rel="nofollow noreferrer">solid angle subtended by a cone</a> with apex angle <span class="math-container">$2\theta$</span> is given by (see figure below for the definition of \theta relative to the apex angle):</p>
<p><span class="math-container">$$\Omega = 2\pi(1-\cos\theta)$$</span></p>
<p><a href="https://i.sstatic.net/esb60.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/esb60.png" alt="Diagram showing a section through the centre of a cone" /></a><BR>
Image from <a href="https://en.wikipedia.org/wiki/File:Steradian_cone_and_cap.svg" rel="nofollow noreferrer">Wikipedia</a>.</p>
<p>Substituting <span class="math-container">$\theta_1$</span> for <span class="math-container">$\theta$</span>:</p>
<p><span class="math-container">$$\Omega = 2\pi\left(1-\cos\left(\arcsin\left(\frac{r_2-r_1}{d}\right) + \frac{\pi}{2}\right)\right)$$</span></p>
<p><span class="math-container">$$\Omega = 2\pi\left(1+\sin\left(\arcsin\left(\frac{r_2-r_1}{d}\right)\right)\right)$$</span></p>
<p><span class="math-container">$$\Omega = 2\pi\left(1+\frac{r_2-r_1}{d}\right)$$</span></p>
<p>Given there are <span class="math-container">$4\pi$</span> <a href="https://en.wikipedia.org/wiki/Steradian#:%7E:text=A%20steradian%20can%20be%20defined,where" rel="nofollow noreferrer">steradians</a> in a sphere, the fraction, <span class="math-container">$f$</span>, of the planet illuminated is given by:</p>
<p><span class="math-container">$$f = \frac{\Omega}{4\pi}$$</span></p>
<p><span class="math-container">$$f = \frac{2\pi\left(1+\frac{r_2-r_1}{d}\right)}{4\pi}$$</span></p>
<p><span class="math-container">$$f = \frac{1}{2}\left(1+\frac{r_2-r_1}{d}\right)$$</span></p>
<p>As the distance between the objects gets larger, the term on the right becomes small and the fraction illuminated tends to 50%. As the term on the right is always positive, the fraction illuminated will always be more than half so long as the planet is smaller than the sun.</p>
<p>Using values for the mean radius and distance to the sun (from <a href="https://nssdc.gsfc.nasa.gov/planetary/factsheet/" rel="nofollow noreferrer">Planetary Fact Sheet - Metric</a>) and with the Sun's mean radius of 695,700 (from <a href="https://nssdc.gsfc.nasa.gov/planetary/factsheet/sunfact.html" rel="nofollow noreferrer">Sun Fact Sheet</a>):</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th>Planet</th>
<th style="text-align: right;">Mean Radius (km)</th>
<th style="text-align: right;">Mean Solar Distance (km)</th>
<th style="text-align: center;">Fraction Illuminated</th>
</tr>
</thead>
<tbody>
<tr>
<td>Mercury</td>
<td style="text-align: right;">2,439.5</td>
<td style="text-align: right;">5.79x10<sup>7</sup></td>
<td style="text-align: center;">50.60%</td>
</tr>
<tr>
<td>Venus</td>
<td style="text-align: right;">6,502.0</td>
<td style="text-align: right;">1.08x10<sup>8</sup></td>
<td style="text-align: center;">50.32%</td>
</tr>
<tr>
<td>Earth</td>
<td style="text-align: right;">6,378.0</td>
<td style="text-align: right;">1.50x10<sup>8</sup></td>
<td style="text-align: center;">50.23%</td>
</tr>
<tr>
<td>Mars</td>
<td style="text-align: right;">3,396.0</td>
<td style="text-align: right;">2.28x10<sup>8</sup></td>
<td style="text-align: center;">50.15%</td>
</tr>
<tr>
<td>Jupiter</td>
<td style="text-align: right;">71,492.0</td>
<td style="text-align: right;">7.79x10<sup>8</sup></td>
<td style="text-align: center;">50.04%</td>
</tr>
<tr>
<td>Saturn</td>
<td style="text-align: right;">60,268.0</td>
<td style="text-align: right;">1.43x10<sup>9</sup></td>
<td style="text-align: center;">50.02%</td>
</tr>
<tr>
<td>Uranus</td>
<td style="text-align: right;">25,559.0</td>
<td style="text-align: right;">2.87x10<sup>9</sup></td>
<td style="text-align: center;">50.01%</td>
</tr>
<tr>
<td>Neptune</td>
<td style="text-align: right;">24,764.0</td>
<td style="text-align: right;">4.52x10<sup>9</sup></td>
<td style="text-align: center;">50.01%</td>
</tr>
<tr>
<td>Pluto</td>
<td style="text-align: right;">1,188.0</td>
<td style="text-align: right;">5.90x10<sup>9</sup></td>
<td style="text-align: center;">50.01%</td>
</tr>
</tbody>
</table>
</div>
<p>If atmospheric effects are considered then there is the possibility of <a href="https://skyandtelescope.org/astronomy-resources/transparency-and-atmospheric-extinction/" rel="nofollow noreferrer">extinction</a> occurring reducing the effective fraction of the planet that is illuminated. This would need to have a definition of how much light counts as illuminated before an assessment could be made for this effect.</p>
<p>Another effect, which may be more significant, is <a href="https://en.wikipedia.org/wiki/Atmospheric_refraction" rel="nofollow noreferrer">refraction</a> by the atmosphere which causes a larger portion of the surface to be illuminated than if there is not atmosphere present. This same effect on Earth means that geometrically the sun is below the horizon while we can still see it rising or setting.</p>

CC BY-SA 4.0
2023-10-16T06:00:54.100Z
Limiting the "fraction illuminated" to only 4 digits hides the details near the bottom of the table. I get 50.022218 for Saturn, 50.011675 for Uranus, 50.007422 for Neptune, and 50.005886 for Pluto. (My figures probably have spurious precision because I've ignored eccentricity and asphericity, especially for Pluto; but the `50%` additive term should be ignored when educing the available precision from the precision of the solar system dimensions, so clearly more precision is available than in the table above.)
Martin Kealey
CC BY-SA 4.0
2023-10-16T07:33:53.127Z
The important point in all cases is: The stellar radius is always bigger than the planetary radius. Thus by simple geometric logic as illustrated in the sketches here, alway an area >= 50% must be illuminated. That is even true for any convex-shaped object orbiting the star.
planetmaker
CC BY-SA 4.0
2023-10-16T07:38:15.290Z
Well, technically one could have a gas giant orbiting a white dwarf, and being less than half illuminated by it.
Martin Kealey
CC BY-SA 4.0
2023-10-16T08:22:24.780Z
@planetmaker "... 50% must be illuminated. That is even true for any convex-shaped object orbiting the star"  

Practically: probably. Theoretically: no. Consider a pyramid (or a regular tetrahedon) orbitally locked with a flat side towards the star. There is more area pointing away than towards the star.
DrPhil
CC BY-SA 4.0
2023-10-16T09:53:56.153Z
@DrPhil that's not as far-fetched as you might suppose. Even our moon is slightly egg-shaped, as it's tidally locked to the earth. So it's quite plausible that a planet tidally locked to a star could likewise become egg-shaped with its "fat" end facing the star.
Martin Kealey
CC BY-SA 4.0
2023-10-16T13:18:52.247Z
So for the purpose of answering the question, the Sun will NOT illuminate a portion of the sky considerably larger than 50%, even when the distance between them is very close.
Michael Richardson
CC BY-SA 4.0
2023-10-16T15:11:54.263Z
Amazing answer, thank you. In the diagram we have taken only the upper portion as being lit. So, Could you please explain for the fraction why we divide omega by 4pi and not 2pi since we would also have to consider the bottom half for illumination? Or could you please provide a link/ video where this concept would be explained?
Adiyarkku
CC BY-SA 4.0
2023-10-16T15:23:26.263Z
I understand why a sphere has 4pi steradians. I just don’t understand why we took that and didn’t consider the bottom part of illumination. And how did the fraction directly come to to the top and bottom part being lit ( 50% + little bit more)
Adiyarkku
CC BY-SA 4.0
2023-10-18T03:12:53.307Z
@Adiyarkku the bottom is simply a mirror image of the top, with the same internal proportion, and therefore the same overall proportion. (If A is 51% of B then 2A is 51% of 2B.)
Martin Kealey
CC BY-SA 4.0
2023-10-18T08:46:49.630Z
@MartinKealey what use would the additional precision serve?
Will
CC BY-SA 4.0
2023-10-18T16:52:57.170Z
@planetmaker "alway an area >= 50% must be illuminated" No. This is probably not always the cause for the earth either. Because the moon can block sunlight. And when we talk about the surface, then the light may not reach the surface (but then we have to define what exactly lit means. When at least 10% of light energy reaches the surface? Or 0.001%? Do we only consider visible light?)
12431234123412341234123
CC BY-SA 4.0
2023-10-19T03:48:13.267Z
@Will I've probably given more digits than are necessary, but *some* additional digits would be helpful. The opening question indicates a that the questioner has failed to grasp that the assertion of 50% in their textbook is an *approximation*. The margin by which the true answer differs from 50% is therefore an important part of dispelling that confusion. In that context, rounding 0.0074% _and_ 0.0059% both to 0.01% is suboptimal, because it hides the fact that the numbers continue to get smaller as you move away from the sun.
Martin Kealey
CC BY-SA 4.0
2023-10-21T10:44:54.067Z
@Adiyarkku I've added a figure from wikipedia showing where theta is defined for the cone apex angle.  Hopefully this clarifies why I only considered the top half of illumination when computing the angle.
GrapefruitIsAwesome
CC BY-SA 4.0
2023-10-22T00:10:59.683Z
@Connor Garcia Thanks and what problem did you see in my Comment?
Robbie Goodwin
CC BY-SA 4.0
2023-10-15T17:39:30.863Z
2023-10-21T10:43:39.587Z
2023-10-21T10:43:39.587Z
GrapefruitIsAwesome
<p>Since the sun is much larger than the earth, it should ideally light up a considerable portion of the earth rather than only half of it.</p>
<p>However the sun lighting only half of the earth can be explained using the fact that the sun is at a considerable distance away from the earth. As the distance of a light source from the object increases the portion of the object lit up reduces.</p>
<p>Applying the same logic to other planets (especially the ones beyond the earth), I would like to know how much of the surface of these other planets is lit up by the sun?</p>
<p>Going by the distance thing, a portion less than half should be lit up, which would mean a fraction of the surface (mainly around the centre from north to south) would always be in darkness.</p>
<p>Would be nice if diagrams could be used to explain</p>

CC BY-SA 4.0
2023-10-15T06:19:44.780Z
I cannot follow your logic that distance decreases the lit surface to below 50%. The sun is bigger than any planet, so always 50% and a tiny bit more are illuminated. The tiny bit is very tiny as the sun is very distant
planetmaker
CC BY-SA 4.0
2023-10-15T06:55:26.997Z
If you see the earth’s lit up portion, it is always 50%, (not more not less). If it was close enough to the earth, it wouldn’t be just 50% lit up, it would be 50% + a tiny bit more. It is only 50% because of the distance? The tiny portion gets excluded because of the distance.
Adiyarkku
CC BY-SA 4.0
2023-10-15T06:58:22.913Z
The exclusion of the tiny portion can be attributed to the distance. Which means that planets that are more distant from the sun, would have even a reduction in the 50%? Had they been as close as the earth, yes they would be lit up 50%, but since they are farther away, they should be less than 50%. What is the truth and why? @planetmaker check earth light: https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcQkrLcT18bxkyKiTDSdo5_AMzZU9iJZvy5nFw&usqp=CAU
Adiyarkku
CC BY-SA 4.0
2023-10-15T06:59:35.563Z
We know that the earth is only half lit because of the timings of the day and the North Pole South Pole having 6 months day and night. If it was 50% + a tiny portion, then the north and south poles both would always have light rather than being 6 months light 6 months darkness.
Adiyarkku
CC BY-SA 4.0
2023-10-15T07:15:44.663Z
Well, there are two different effects at play here. Assuming a point source, closer distance is less illumination. For an extended source, closer distance is more illumination. 

For far away planets since the sun is effectively a point, they have effectively 50% illumination, whereas the earth is a fraction of a percent above 50% illuminated directly by the sun
Topcode
CC BY-SA 4.0
2023-10-15T13:16:03.463Z
Two different effects? Waaat? is there a YouTube video or diagram explaining this @Topcode
Adiyarkku
CC BY-SA 4.0
2023-10-16T09:29:43.793Z
@Adiyarkku why do you think that the Earth is _exactly_ 50% illuminated? (This sounds like a pedagogical simplification, like π=22/7, that's _close_ to the truth, but not _perfectly_ true.) Even discounting atmospheric effects, orbital eccentricity, and asphericity, it's clear that the sun illuminates measurably more than 50%: full night is on average 4 minutes shorter than daytime. As for the poles being illuminated for 6 months each, no that's not true either: each equinox there's a period of about 31 hours when _both_ poles are directly illuminated (plus a bit longer because of refraction).
Martin Kealey
CC BY-SA 4.0
2023-10-16T13:29:00.550Z
There's so many qualifications and technicalities here, like what does "lit" mean and what does "surface" mean?  In one (very technical) sense very little of Venus's surface is lit by the Sun.  In another sense, quite a bit more than 50% of the earth's surface is lit by the Sun.
RBarryYoung
CC BY-SA 4.0
2023-10-16T14:51:16.880Z
And there's different amounts of "lit" throughout the penumbral region, even ignoring atmospheric effects and indirect illumination (moonlight).
Toby Speight
CC BY-SA 4.0
2023-10-16T15:13:28.527Z
@MartinKealey from elementary levels, that’s what diagrams show. That half of the earth is illuminated and the other half is dark. That’s why.
Adiyarkku
CC BY-SA 4.0
2023-10-16T23:09:34.243Z
Where does the sun being larger than the Earth, it should ideally light up a considerable portion of the earth rather than only half of it come from?

What makes you think the Sun or any star could shine on the other side of the Earth or any planet?
Robbie Goodwin
CC BY-SA 4.0
2023-10-17T10:26:00.110Z
@RobbieGoodwin as demonstrated in the answer below, as long as the Sun is larger than the planet, more than 50% of the planet will be illuminated. If the Sun was very very close to the Earth, it could illuminate significantly more than 50%, (over 99% if Earth and Sun were touching, but still over 70% if the distance between the two was 1 million km). There is only a single point that is guaranteed never to be directly illuminated.
jcaron
CC BY-SA 4.0
2023-10-18T03:07:49.043Z
@Adiyarkku Ahh, from elementary school text books? So yes, they're approximations for children. If you put a diagram showing exactly half next to a diagram showing the correct proportion, they _look_ the same, and most people could not tell the difference. If you used a ruler to measure the diagram, the difference would be less than the thickness of a line.
Martin Kealey
CC BY-SA 4.0
2023-10-18T08:04:22.627Z
Consider a big circle and a small circle right next to each other, along with the 2 outer lines that are tangential to both circles (representing how much of the small circle is illuminated, which is > 50%). When the circles move apart, the lines become closer to parallel, but they cannot ever become parallel, because this can only happen if the 2 circles are the same size. Parallel means 50% illuminated, so it will tend to 50% illuminated, and always be more than 50% illuminated. (I would've posted an answer, but for that I probably need diagrams, which I didn't really feel like drawing.)
NotThatGuy
CC BY-SA 4.0
2023-10-21T04:35:01.450Z
Atmospheric composition causes deflection which has a march greater effect on insulation and also caused this lightning on the dark side
bandybabboon
<the-sun><planet><light>
How much of the surface of other planets is lit by the sun?
Adiyarkku
# How much of the surface of other planets is lit by the sun?

(Asked by: username_0 on 2023-10-15)

Since the sun is much larger than the earth, it should ideally light up a considerable portion of the earth rather than only half of it.

However the sun lighting only half of the earth can be explained using the fact that the sun is at a considerable distance away from the earth. As the distance of a light source from the object increases the portion of the object lit up reduces.

Applying the same logic to other planets (especially the ones beyond the earth), I would like to know how much of the surface of these other planets is lit up by the sun?

Going by the distance thing, a portion less than half should be lit up, which would mean a fraction of the surface (mainly around the centre from north to south) would always be in darkness.

Would be nice if diagrams could be used to explain

**Tags:** the-sun, planet, light

> **Comments:**
> **username_1**: I cannot follow your logic that distance decreases the lit surface to below 50%. The sun is bigger than any planet, so always 50% and a tiny bit more are illuminated. The tiny bit is very tiny as the sun is very distant  
> **username_0**: If you see the earth’s lit up portion, it is always 50%, (not more not less). If it was close enough to the earth, it wouldn’t be just 50% lit up, it would be 50% + a tiny bit more. It is only 50% because of the distance? The tiny portion gets excluded because of the distance.  
> **username_0**: The exclusion of the tiny portion can be attributed to the distance. Which means that planets that are more distant from the sun, would have even a reduction in the 50%? Had they been as close as the earth, yes they would be lit up 50%, but since they are farther away, they should be less than 50%. What is the truth and why? @username_1 check earth light: https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcQkrLcT18bxkyKiTDSdo5_AMzZU9iJZvy5nFw&usqp=CAU  
> **username_0**: We know that the earth is only half lit because of the timings of the day and the North Pole South Pole having 6 months day and night. If it was 50% + a tiny portion, then the north and south poles both would always have light rather than being 6 months light 6 months darkness.  
> **username_2**: Well, there are two different effects at play here. Assuming a point source, closer distance is less illumination. For an extended source, closer distance is more illumination.   For far away planets since the sun is effectively a point, they have effectively 50% illumination, whereas the earth is a fraction of a percent above 50% illuminated directly by the sun  
> **username_0**: Two different effects? Waaat? is there a YouTube video or diagram explaining this @username_2  
> **username_3**: @username_0 why do you think that the Earth is _exactly_ 50% illuminated? (This sounds like a pedagogical simplification, like π=22/7, that's _close_ to the truth, but not _perfectly_ true.) Even discounting atmospheric effects, orbital eccentricity, and asphericity, it's clear that the sun illuminates measurably more than 50%: full night is on average 4 minutes shorter than daytime. As for the poles being illuminated for 6 months each, no that's not true either: each equinox there's a period of about 31 hours when _both_ poles are directly illuminated (plus a bit longer because of refraction).  
> **username_4**: There's so many qualifications and technicalities here, like what does "lit" mean and what does "surface" mean?  In one (very technical) sense very little of Venus's surface is lit by the Sun.  In another sense, quite a bit more than 50% of the earth's surface is lit by the Sun.  
> **username_5**: And there's different amounts of "lit" throughout the penumbral region, even ignoring atmospheric effects and indirect illumination (moonlight).  
> **username_0**: @MartinKealey from elementary levels, that’s what diagrams show. That half of the earth is illuminated and the other half is dark. That’s why.  
> **username_6**: Where does the sun being larger than the Earth, it should ideally light up a considerable portion of the earth rather than only half of it come from?  What makes you think the Sun or any star could shine on the other side of the Earth or any planet?  
> **username_7**: @RobbieGoodwin as demonstrated in the answer below, as long as the Sun is larger than the planet, more than 50% of the planet will be illuminated. If the Sun was very very close to the Earth, it could illuminate significantly more than 50%, (over 99% if Earth and Sun were touching, but still over 70% if the distance between the two was 1 million km). There is only a single point that is guaranteed never to be directly illuminated.  
> **username_3**: @username_0 Ahh, from elementary school text books? So yes, they're approximations for children. If you put a diagram showing exactly half next to a diagram showing the correct proportion, they _look_ the same, and most people could not tell the difference. If you used a ruler to measure the diagram, the difference would be less than the thickness of a line.  
> **username_8**: Consider a big circle and a small circle right next to each other, along with the 2 outer lines that are tangential to both circles (representing how much of the small circle is illuminated, which is > 50%). When the circles move apart, the lines become closer to parallel, but they cannot ever become parallel, because this can only happen if the 2 circles are the same size. Parallel means 50% illuminated, so it will tend to 50% illuminated, and always be more than 50% illuminated. (I would've posted an answer, but for that I probably need diagrams, which I didn't really feel like drawing.)  
> **username_9**: Atmospheric composition causes deflection which has a march greater effect on insulation and also caused this lightning on the dark side  

---

## 1 answers

### Answer by username_10 on 2023-10-15. Score: 53 | Accepted answer

OK, let start with some assumptions: spherical objects and no atmospheric effects.

Here's the relevant geometry with the object on the left representing a planet and the object on the right representing the sun:

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

Where d is the distance between the two spheres, r1 and r2 are the radii of the spheres. $\theta\_1$ is the angle showing the portion of the sphere illuminated. We'll use it later to compute what fraction of the sphere is lit.

We can solve for $\theta\_1$ easily by noticing that the solid line quadrilateral can be drawn as below. We're flipping it so that the side labeled L is on the bottom.

[![geometric modelisation](https://i.sstatic.net/bvRJD.png)](https://i.sstatic.net/bvRJD.png)

Noting that the upper triangle is a a right-triangle with sides known allows us to solve for the unknown angle $\theta\_1$ as follows:

$$d \sin\left(\frac{\pi}{2}-\theta\_1\right) = r\_2-r\_1$$

$$\sin\left(\frac{\pi}{2}-\theta\_1\right) = \frac{r\_2-r\_1}{d}$$

$$\frac{\pi}{2}-\theta\_1 = \arcsin\left(\frac{r\_2-r\_1}{d}\right)$$

$$\theta\_1 = \arcsin\left(\frac{r\_2-r\_1}{d}\right) + \frac{\pi}{2}$$

Or, with a small angle approximation for when the distance is much larger than the difference in radii:

$$\theta\_1 \approx \frac{r\_2-r\_1}{d} + \frac{\pi}{2}$$

The [solid angle subtended by a cone](https://en.wikipedia.org/wiki/Solid_angle) with apex angle $2\theta$ is given by (see figure below for the definition of \theta relative to the apex angle):

$$\Omega = 2\pi(1-\cos\theta)$$

[![Diagram showing a section through the centre of a cone](https://i.sstatic.net/esb60.png)](https://i.sstatic.net/esb60.png)  
Image from [Wikipedia](https://en.wikipedia.org/wiki/File:Steradian_cone_and_cap.svg).

Substituting $\theta\_1$ for $\theta$:

$$\Omega = 2\pi\left(1-\cos\left(\arcsin\left(\frac{r\_2-r\_1}{d}\right) + \frac{\pi}{2}\right)\right)$$

$$\Omega = 2\pi\left(1+\sin\left(\arcsin\left(\frac{r\_2-r\_1}{d}\right)\right)\right)$$

$$\Omega = 2\pi\left(1+\frac{r\_2-r\_1}{d}\right)$$

Given there are $4\pi$ [steradians](https://en.wikipedia.org/wiki/Steradian#:%7E:text=A%20steradian%20can%20be%20defined,where) in a sphere, the fraction, $f$, of the planet illuminated is given by:

$$f = \frac{\Omega}{4\pi}$$

$$f = \frac{2\pi\left(1+\frac{r\_2-r\_1}{d}\right)}{4\pi}$$

$$f = \frac{1}{2}\left(1+\frac{r\_2-r\_1}{d}\right)$$

As the distance between the objects gets larger, the term on the right becomes small and the fraction illuminated tends to 50%. As the term on the right is always positive, the fraction illuminated will always be more than half so long as the planet is smaller than the sun.

Using values for the mean radius and distance to the sun (from [Planetary Fact Sheet - Metric](https://nssdc.gsfc.nasa.gov/planetary/factsheet/)) and with the Sun's mean radius of 695,700 (from [Sun Fact Sheet](https://nssdc.gsfc.nasa.gov/planetary/factsheet/sunfact.html)):

| Planet | Mean Radius (km) | Mean Solar Distance (km) | Fraction Illuminated |
| --- | --- | --- | --- |
| Mercury | 2,439.5 | 5.79x107 | 50.60% |
| Venus | 6,502.0 | 1.08x108 | 50.32% |
| Earth | 6,378.0 | 1.50x108 | 50.23% |
| Mars | 3,396.0 | 2.28x108 | 50.15% |
| Jupiter | 71,492.0 | 7.79x108 | 50.04% |
| Saturn | 60,268.0 | 1.43x109 | 50.02% |
| Uranus | 25,559.0 | 2.87x109 | 50.01% |
| Neptune | 24,764.0 | 4.52x109 | 50.01% |
| Pluto | 1,188.0 | 5.90x109 | 50.01% |

If atmospheric effects are considered then there is the possibility of [extinction](https://skyandtelescope.org/astronomy-resources/transparency-and-atmospheric-extinction/) occurring reducing the effective fraction of the planet that is illuminated. This would need to have a definition of how much light counts as illuminated before an assessment could be made for this effect.

Another effect, which may be more significant, is [refraction](https://en.wikipedia.org/wiki/Atmospheric_refraction) by the atmosphere which causes a larger portion of the surface to be illuminated than if there is not atmosphere present. This same effect on Earth means that geometrically the sun is below the horizon while we can still see it rising or setting.

> **Comments:**
> **username_3**: Limiting the "fraction illuminated" to only 4 digits hides the details near the bottom of the table. I get 50.022218 for Saturn, 50.011675 for Uranus, 50.007422 for Neptune, and 50.005886 for Pluto. (My figures probably have spurious precision because I've ignored eccentricity and asphericity, especially for Pluto; but the `50%` additive term should be ignored when educing the available precision from the precision of the solar system dimensions, so clearly more precision is available than in the table above.)  
> **username_1**: The important point in all cases is: The stellar radius is always bigger than the planetary radius. Thus by simple geometric logic as illustrated in the sketches here, alway an area >= 50% must be illuminated. That is even true for any convex-shaped object orbiting the star.  
> **username_3**: Well, technically one could have a gas giant orbiting a white dwarf, and being less than half illuminated by it.  
> **username_11**: @username_1 "... 50% must be illuminated. That is even true for any convex-shaped object orbiting the star"    Practically: probably. Theoretically: no. Consider a pyramid (or a regular tetrahedon) orbitally locked with a flat side towards the star. There is more area pointing away than towards the star.  
> **username_3**: @username_11 that's not as far-fetched as you might suppose. Even our moon is slightly egg-shaped, as it's tidally locked to the earth. So it's quite plausible that a planet tidally locked to a star could likewise become egg-shaped with its "fat" end facing the star.  
> **username_12**: So for the purpose of answering the question, the Sun will NOT illuminate a portion of the sky considerably larger than 50%, even when the distance between them is very close.  
> **username_0**: Amazing answer, thank you. In the diagram we have taken only the upper portion as being lit. So, Could you please explain for the fraction why we divide omega by 4pi and not 2pi since we would also have to consider the bottom half for illumination? Or could you please provide a link/ video where this concept would be explained?  
> **username_0**: I understand why a sphere has 4pi steradians. I just don’t understand why we took that and didn’t consider the bottom part of illumination. And how did the fraction directly come to to the top and bottom part being lit ( 50% + little bit more)  
> **username_3**: @username_0 the bottom is simply a mirror image of the top, with the same internal proportion, and therefore the same overall proportion. (If A is 51% of B then 2A is 51% of 2B.)  
> **username_13**: @MartinKealey what use would the additional precision serve?  
> **username_14**: @username_1 "alway an area >= 50% must be illuminated" No. This is probably not always the cause for the earth either. Because the moon can block sunlight. And when we talk about the surface, then the light may not reach the surface (but then we have to define what exactly lit means. When at least 10% of light energy reaches the surface? Or 0.001%? Do we only consider visible light?)  
> **username_3**: @username_13 I've probably given more digits than are necessary, but *some* additional digits would be helpful. The opening question indicates a that the questioner has failed to grasp that the assertion of 50% in their textbook is an *approximation*. The margin by which the true answer differs from 50% is therefore an important part of dispelling that confusion. In that context, rounding 0.0074% _and_ 0.0059% both to 0.01% is suboptimal, because it hides the fact that the numbers continue to get smaller as you move away from the sun.  
> **username_10**: @username_0 I've added a figure from wikipedia showing where theta is defined for the cone apex angle.  Hopefully this clarifies why I only considered the top half of illumination when computing the angle.  
> **username_6**: @Connor Garcia Thanks and what problem did you see in my Comment?  

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