<p>Fortunately, Neptune is in an almost circular orbit that is almost in the same plane as that of the Earth. The <a href="https://nssdc.gsfc.nasa.gov/planetary/factsheet/neptunefact.html" rel="noreferrer">NASA planetary fact sheet</a> has Neptune at an apparent magnitude of 7.7 when it is <span class="math-container">$4.35\times 10^9$</span> km from Earth and the Sun, Earth and Neptune are in line.</p>
<p>If you were to move closer to Neptune along that line, it would grow brighter (but still appearing as a point source to the naked eye) as the inverse square of the distance to Neptune. Thus
<span class="math-container">$$7.7 - m = 5\log_{10}\left(\frac{4.35\times 10^9}{d}\right)\ ,$$</span>
where <span class="math-container">$d$</span> is the distance you are looking for (in km) and <span class="math-container">$m$</span> is the apparent magnitude you think you can see with the naked eye - typically, <span class="math-container">$m=6$</span> for good vision in a dark sky.</p>
<p>This equation can be rearranged as
<span class="math-container">$$ d = 4.35\times 10^9\times 10^{(m-7.7)/5}\ {\rm km}. $$</span></p>
<p>e.g., for <span class="math-container">$m= 6$</span>, <span class="math-container">$d = 1.99\times 10^9$</span> km.</p>
<p>Possibly you can do a bit better than <span class="math-container">$m=6$</span> out of the Earth's atmosphere. For <span class="math-container">$m=6.5$</span>, <span class="math-container">$d = 2.50\times 10^9$</span> km.</p>

CC BY-SA 4.0
2023-04-14T23:20:56.080Z
That means Uranus comes close enough if it gets into proper alignment, so you might see Neptune from hovering above Uranus or a Uranian moon.
Oscar Lanzi
CC BY-SA 4.0
2023-04-12T07:28:11.980Z
2023-04-12T07:28:11.980Z
ProfRob
<p>I am in space. What is the maximum distance that I could be from Neptune and see it with the naked eye. You can choose the optimal position. Also, assume that I have 20/20 vision.</p>

CC BY-SA 4.0
2023-04-12T01:57:29.060Z
What is the point of the question? Are to doing any calculation or is it a worldbuilding question?
Nilay Ghosh
CC BY-SA 4.0
2023-04-14T10:33:30.127Z
Curiosity @NilayGhosh
Starship
<neptune>
How close to Neptune would I have to be to see it with the naked eye?
Starship
# How close to Neptune would I have to be to see it with the naked eye?

(Asked by: username_0 on 2023-04-11)

I am in space. What is the maximum distance that I could be from Neptune and see it with the naked eye. You can choose the optimal position. Also, assume that I have 20/20 vision.

**Tags:** neptune
> **Comments:**
> **username_1**: What is the point of the question? Are to doing any calculation or is it a worldbuilding question?  
> **username_0**: Curiosity @NilayGhosh  

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## 1 answers

### Answer by username_2 on 2023-04-12. Score: 5 | Accepted answer

Fortunately, Neptune is in an almost circular orbit that is almost in the same plane as that of the Earth. The [NASA planetary fact sheet](https://nssdc.gsfc.nasa.gov/planetary/factsheet/neptunefact.html) has Neptune at an apparent magnitude of 7.7 when it is $4.35\times 10^9$ km from Earth and the Sun, Earth and Neptune are in line.

If you were to move closer to Neptune along that line, it would grow brighter (but still appearing as a point source to the naked eye) as the inverse square of the distance to Neptune. Thus
$$7.7 - m = 5\log\_{10}\left(\frac{4.35\times 10^9}{d}\right)\ ,$$
where $d$ is the distance you are looking for (in km) and $m$ is the apparent magnitude you think you can see with the naked eye - typically, $m=6$ for good vision in a dark sky.

This equation can be rearranged as
$$ d = 4.35\times 10^9\times 10^{(m-7.7)/5}\ {\rm km}. $$

e.g., for $m= 6$, $d = 1.99\times 10^9$ km.

Possibly you can do a bit better than $m=6$ out of the Earth's atmosphere. For $m=6.5$, $d = 2.50\times 10^9$ km.

> **Comments:**
> **username_3**: That means Uranus comes close enough if it gets into proper alignment, so you might see Neptune from hovering above Uranus or a Uranian moon.  

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