<p>Leveraging Pierre Paquette's <a href="https://astronomy.stackexchange.com/questions/47295/could-neptune-be-viewed-with-the-naked-eye-from-uranus">excellent answer</a> and reference to <a href="https://arxiv.org/ftp/arxiv/papers/1808/1808.01973.pdf" rel="nofollow noreferrer">Hilton and Mallama</a>, the magnitude of Saturn can be estimated by:</p>
<p><span class="math-container">$$ V = 5 \log_{10} (rd) - 8.95 - 3.7\times10^{-4} \alpha + 6.16\times10^{-4} \alpha^2 $$</span></p>
<p>Here, <span class="math-container">$r\approx9.5$</span> AU is the distance from Saturn to the Sun, <span class="math-container">$d$</span> is the distance from Saturn to the observer, and <span class="math-container">$\alpha$</span> is the angle of the Sun/Saturn/Observer triangle.</p>
<p>If the observer is located on a planet with orbit <em>inside</em> of Saturn's orbit, it should be clear that the maximum apparent brightness of Saturn should occur when the observer is closest to Saturn, since both <span class="math-container">$d$</span> and <span class="math-container">$\alpha$</span> are at a minimum.</p>
<p>However, if an observer is on a planet with orbit <em>outside</em> of Saturn's orbit, it certainly isn't the case that Saturn is brightest when closest, since it will be backlit by the Sun from the perspective of the observer.</p>
<p>Using the <a href="https://en.wikipedia.org/wiki/Law_of_cosines#Applications" rel="nofollow noreferrer">Law of Cosines</a>, we can compute the distance between the observer and Saturn as a function of <span class="math-container">$\alpha$</span>:</p>
<p><span class="math-container">$$d(\alpha) = r\cos{\alpha}\pm \sqrt{c^2-r^2\sin^2{\alpha}}$$</span></p>
<p>Here, <span class="math-container">$c \approx 19.2$</span> AU is the distance from Uranus to the Sun. This allows us to formulate the expression of magnitude only as a function of <span class="math-container">$\alpha$</span>.</p>
<p><span class="math-container">$$ V(\alpha) = 5 \log_{10} (r^2\cos{\alpha}\pm r\sqrt{c^2-r^2\sin^2{\alpha}}) - 8.95 - 3.7\times10^{-4} \alpha + 6.16\times10^{-4} \alpha^2 $$</span></p>
<p>Plugging in values from 0 to 180 degrees for alpha, we get the following magnitude phase curve:</p>
<p><a href="https://i.sstatic.net/ZDIn2.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ZDIn2.png" alt="enter image description here" /></a></p>
<p>Here the x-axis is <span class="math-container">$\alpha$</span> in degrees, and the y-axis is apparent magnitude. It shows that the maximum apparent magnitude of Saturn as seen from Uranus is at opposition (perhaps unintuitively since it is also the maximum distance between the two planets' orbital paths).  <strong>The value at opposition is about 3.228.</strong>  As a comparison, Saturn is always much more bright from Earth, <a href="https://en.wikipedia.org/wiki/Apparent_magnitude" rel="nofollow noreferrer">varying between -0.55 and 1.17</a>.</p>
<p><strong>Notes:</strong></p>
<ol>
<li><p>For convenience, we assume in this answer that the orbits are circular and coplanar.  Since both planets have low inclination and eccentricity, we wouldn't expect a very different phase curve from a more sophisticated model.</p>
</li>
<li><p>The above answer is for the apparent magnitude of the sphere of Saturn only.  The rings of Saturn can contribute significantly to the brightness of the overall system.  Hilton and Mallama provide a more sophisticated equation involving <span class="math-container">$\beta = \sqrt{\beta_1\beta_2}$</span>, where <span class="math-container">$\beta_1$</span> and <span class="math-container">$\beta_2$</span> are the inclinations of Saturn's rings with respect to the Sun and the observer respectively. One could get a higher fidelity answer using this equation, perhaps assuming that the maximum ring illumination can occur simultaneously in opposition with <span class="math-container">$\beta_1=\beta_2 \approx 26.7^\circ$</span></p>
</li>
</ol>
<p><span class="math-container">$$ V = 5 \log_{10} (rd) - 8.914 - 1.825 \sin{\beta}+ 0.026 α - 0.378 \sin{\beta} e^{-2.25\alpha} $$</span></p>

CC BY-SA 4.0
2021-11-15T23:16:26.033Z
Thx. So are you saying that the answer to my question is +3.88 in apparent magnitude?
Constantthin
CC BY-SA 4.0
2021-11-16T00:01:34.277Z
@Constantthin no, +3.228
Connor Garcia
CC BY-SA 4.0
2021-11-16T03:00:22.037Z
I don’t understand your chart. Does it say that the smallest apparent magnitude is measured at zero angle?
Constantthin
CC BY-SA 4.0
2021-11-16T03:10:24.110Z
yes, the smallest mag is at zero angle
Connor Garcia
CC BY-SA 4.0
2021-11-16T09:15:13.333Z
I don’t understand how that can be, Venus’ lowest apparent magnitude is not at zero angle, but at 39 degrees from the sun. Shouldn’t Saturn be in a similar situation in relation to Uranus?
Constantthin
CC BY-SA 4.0
2021-11-16T12:55:42.057Z
https://www.st-andrews.ac.uk/~bds2/ltsn/ljm/JAVA/VENUS/Venusmot.htm
Constantthin
CC BY-SA 4.0
2021-11-16T12:59:05.920Z
And isn’t it contradicting your earlier statement: “However, if an observer is on a planet with orbit outside of Saturn's orbit, it certainly isn't the case that Saturn is brightest when closest, since it will be backlit by the Sun from the perspective of the observer.” Or am I missing something?
Constantthin
CC BY-SA 4.0
2021-11-16T14:57:09.843Z
@Constantthin  Observing Venus from Earth is not very analogous to Saturn from Uranus.  Venus/Earth closest distance is 0.28 AU, or 1/6 the furthest distance of 1.72 AU. For Saturn/Uranus, the closest distance is about 10 AU or only 1/3 the furthest distance of 30 AU.  Also, Venus  has some phase curve inflections due to scattering of sunlight by H2SO4 in the Venus atmosphere.
Connor Garcia
CC BY-SA 4.0
2021-11-16T15:02:14.433Z
@Constantthin Also, recall $\alpha$ is the angle in the Sun/planet/observer triangle.  So a zero angle for an observer interior to Saturn's orbit occurs when the observer and Saturn are closest.  But a zero angle for an observer outside of Saturn's orbit occurs when Saturn and the observer are furthest.
Connor Garcia
CC BY-SA 4.0
2021-11-16T22:52:34.300Z
By being furthest away from eachother would that mean to have the sun in the middle?
Constantthin
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2021-11-16T22:56:32.203Z
Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/131510/discussion-between-constantthin-and-connor-garcia).
Constantthin
CC BY-SA 4.0
2021-11-15T22:52:17.287Z
2021-11-15T22:52:17.287Z
Connor Garcia
<p>Considering that the apparent magnitude scale is an inverted scale; what is the lowest apparent magnitude of Saturn that can be seen from Uranus. We are of course talking about when Saturn is viewed from one of the moons of Uranus.</p>
<p>Saturn is in a similar relationship with Uranus as Venus is to Earth. To know this could be helpful when attempting to answer this question.</p>

CC BY-SA 4.0
2021-11-11T12:46:03.937Z
The answer to this is far more complex because of the phase angle complications.
ProfRob
CC BY-SA 4.0
2021-11-11T13:34:53.117Z
This is very similar to [your previous question,](https://astronomy.stackexchange.com/questions/47295/31410) if not a duplicate.
WarpPrime
CC BY-SA 4.0
2021-11-12T04:49:02.027Z
I provided all the elements to calculate that as well in the first reference of my first answer to your question about Uranus and Neptune. Please read it before asking more related questions; a lot is in there.
Pierre Paquette
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2021-11-12T12:16:35.327Z
additionally, the resource uloh linked you to in his answer, that also is able to answers you this question.
planetmaker
CC BY-SA 4.0
2021-11-12T16:21:45.160Z
Still a duplicate of your previous question.
WarpPrime
CC BY-SA 4.0
2021-11-13T05:09:02.250Z
It is not a duplicate because that other question did not address this specific issue.
Constantthin
CC BY-SA 4.0
2021-11-13T14:24:24.203Z
How could a similar answer make the question a duplicate? A similar question surely can, but hardly a similar answer. It wouldn’t be right to close a question just because an answer touches on similar stuff, would it?
Constantthin
CC BY-SA 4.0
2021-11-14T19:18:33.813Z
It is the same question as your previous one. They both come down to “how bright is planet A from planet B?” The identity of the planet changing from one question to another is not significant in this context—not enough to justify two questions. Also, as I said, the first reference in my first reply to your first question gives you the tools to find the answer yourself. READ IT. And USE THE TOOLS.
Pierre Paquette
CC BY-SA 4.0
2021-11-15T00:26:59.383Z
You may be right to a certain degree, but overall I think you are wrong, because the answer to this  new and different question would be a lot more complicated, like ProfRob also pointed out. Only a few cleaver people like you can compute a thing like this.
Constantthin
CC BY-SA 4.0
2021-11-15T00:56:19.903Z
I just took a look at that answer and I am bewildered about what “phase angle” has to do with “eccentricities and orbital inclinations”.
Constantthin
CC BY-SA 4.0
2021-11-15T02:44:05.160Z
If your old answer answers this new question feel free to repost it here,
Constantthin
<distances><saturn><apparent-magnitude><uranus>
What is Saturn’s apparent magnitude at its brightest when viewed from Uranus?
Constantthin
# What is Saturn’s apparent magnitude at its brightest when viewed from Uranus?

(Asked by: username_0 on 2021-11-11)

Considering that the apparent magnitude scale is an inverted scale; what is the lowest apparent magnitude of Saturn that can be seen from Uranus. We are of course talking about when Saturn is viewed from one of the moons of Uranus.

Saturn is in a similar relationship with Uranus as Venus is to Earth. To know this could be helpful when attempting to answer this question.

**Tags:** distances, saturn, apparent-magnitude, uranus
> **Comments:**
> **username_1**: The answer to this is far more complex because of the phase angle complications.  
> **username_2**: This is very similar to [your previous question,](https://astronomy.stackexchange.com/questions/47295/31410) if not a duplicate.  
> **username_3**: I provided all the elements to calculate that as well in the first reference of my first answer to your question about Uranus and Neptune. Please read it before asking more related questions; a lot is in there.  
> **username_4**: additionally, the resource uloh linked you to in his answer, that also is able to answers you this question.  
> **username_2**: Still a duplicate of your previous question.  
> **username_0**: It is not a duplicate because that other question did not address this specific issue.  
> **username_0**: How could a similar answer make the question a duplicate? A similar question surely can, but hardly a similar answer. It wouldn’t be right to close a question just because an answer touches on similar stuff, would it?  
> **username_3**: It is the same question as your previous one. They both come down to “how bright is planet A from planet B?” The identity of the planet changing from one question to another is not significant in this context—not enough to justify two questions. Also, as I said, the first reference in my first reply to your first question gives you the tools to find the answer yourself. READ IT. And USE THE TOOLS.  
> **username_0**: You may be right to a certain degree, but overall I think you are wrong, because the answer to this  new and different question would be a lot more complicated, like ProfRob also pointed out. Only a few cleaver people like you can compute a thing like this.  
> **username_0**: I just took a look at that answer and I am bewildered about what “phase angle” has to do with “eccentricities and orbital inclinations”.  
> **username_0**: If your old answer answers this new question feel free to repost it here,  

---

## 1 answers

### Answer by username_5 on 2021-11-15. Score: 1

Leveraging Pierre Paquette's [excellent answer](https://astronomy.stackexchange.com/questions/47295/could-neptune-be-viewed-with-the-naked-eye-from-uranus) and reference to [Hilton and Mallama](https://arxiv.org/ftp/arxiv/papers/1808/1808.01973.pdf), the magnitude of Saturn can be estimated by:

$$ V = 5 \log\_{10} (rd) - 8.95 - 3.7\times10^{-4} \alpha + 6.16\times10^{-4} \alpha^2 $$

Here, $r\approx9.5$ AU is the distance from Saturn to the Sun, $d$ is the distance from Saturn to the observer, and $\alpha$ is the angle of the Sun/Saturn/Observer triangle.

If the observer is located on a planet with orbit *inside* of Saturn's orbit, it should be clear that the maximum apparent brightness of Saturn should occur when the observer is closest to Saturn, since both $d$ and $\alpha$ are at a minimum.

However, if an observer is on a planet with orbit *outside* of Saturn's orbit, it certainly isn't the case that Saturn is brightest when closest, since it will be backlit by the Sun from the perspective of the observer.

Using the [Law of Cosines](https://en.wikipedia.org/wiki/Law_of_cosines#Applications), we can compute the distance between the observer and Saturn as a function of $\alpha$:

$$d(\alpha) = r\cos{\alpha}\pm \sqrt{c^2-r^2\sin^2{\alpha}}$$

Here, $c \approx 19.2$ AU is the distance from Uranus to the Sun. This allows us to formulate the expression of magnitude only as a function of $\alpha$.

$$ V(\alpha) = 5 \log\_{10} (r^2\cos{\alpha}\pm r\sqrt{c^2-r^2\sin^2{\alpha}}) - 8.95 - 3.7\times10^{-4} \alpha + 6.16\times10^{-4} \alpha^2 $$

Plugging in values from 0 to 180 degrees for alpha, we get the following magnitude phase curve:

[![enter image description here](https://i.sstatic.net/ZDIn2.png)](https://i.sstatic.net/ZDIn2.png)

Here the x-axis is $\alpha$ in degrees, and the y-axis is apparent magnitude. It shows that the maximum apparent magnitude of Saturn as seen from Uranus is at opposition (perhaps unintuitively since it is also the maximum distance between the two planets' orbital paths). **The value at opposition is about 3.228.** As a comparison, Saturn is always much more bright from Earth, [varying between -0.55 and 1.17](https://en.wikipedia.org/wiki/Apparent_magnitude).

**Notes:**

1. For convenience, we assume in this answer that the orbits are circular and coplanar. Since both planets have low inclination and eccentricity, we wouldn't expect a very different phase curve from a more sophisticated model.
2. The above answer is for the apparent magnitude of the sphere of Saturn only. The rings of Saturn can contribute significantly to the brightness of the overall system. Hilton and Mallama provide a more sophisticated equation involving $\beta = \sqrt{\beta\_1\beta\_2}$, where $\beta\_1$ and $\beta\_2$ are the inclinations of Saturn's rings with respect to the Sun and the observer respectively. One could get a higher fidelity answer using this equation, perhaps assuming that the maximum ring illumination can occur simultaneously in opposition with $\beta\_1=\beta\_2 \approx 26.7^\circ$

$$ V = 5 \log\_{10} (rd) - 8.914 - 1.825 \sin{\beta}+ 0.026 α - 0.378 \sin{\beta} e^{-2.25\alpha} $$

> **Comments:**
> **username_0**: Thx. So are you saying that the answer to my question is +3.88 in apparent magnitude?  
> **username_5**: @Constantthin no, +3.228  
> **username_0**: I don’t understand your chart. Does it say that the smallest apparent magnitude is measured at zero angle?  
> **username_5**: yes, the smallest mag is at zero angle  
> **username_0**: I don’t understand how that can be, Venus’ lowest apparent magnitude is not at zero angle, but at 39 degrees from the sun. Shouldn’t Saturn be in a similar situation in relation to Uranus?  
> **username_0**: https://www.st-andrews.ac.uk/~bds2/ltsn/ljm/JAVA/VENUS/Venusmot.htm  
> **username_0**: And isn’t it contradicting your earlier statement: “However, if an observer is on a planet with orbit outside of Saturn's orbit, it certainly isn't the case that Saturn is brightest when closest, since it will be backlit by the Sun from the perspective of the observer.” Or am I missing something?  
> **username_5**: @Constantthin  Observing Venus from Earth is not very analogous to Saturn from Uranus.  Venus/Earth closest distance is 0.28 AU, or 1/6 the furthest distance of 1.72 AU. For Saturn/Uranus, the closest distance is about 10 AU or only 1/3 the furthest distance of 30 AU.  Also, Venus  has some phase curve inflections due to scattering of sunlight by H2SO4 in the Venus atmosphere.  
> **username_5**: @Constantthin Also, recall $\alpha$ is the angle in the Sun/planet/observer triangle.  So a zero angle for an observer interior to Saturn's orbit occurs when the observer and Saturn are closest.  But a zero angle for an observer outside of Saturn's orbit occurs when Saturn and the observer are furthest.  
> **username_0**: By being furthest away from eachother would that mean to have the sun in the middle?  
> **username_0**: Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/131510/discussion-between-constantthin-and-connor-garcia).  

---

