<p>As Mark Adler mentioned, the best way is to compare the brightness to other nearby stars. I'm going to assume that you have instantaneous travel time, and also take into account that you are actually getting closer to stars depending on the direction you go. I'm using <a href="http://en.wikipedia.org/wiki/List_of_nearest_stars" rel="noreferrer">this table</a> from Wikipedia. I'm going to go no further on the list than Sirius, and assume in each instance we are heading strait towards the Star. The formula for calculating the apperant magnitude given absolute magnitude, which is provided, is:</p>

<p>$$m = M - 5 (1-\log_{10}{d})$$</p>

<p>Setting up for our situation, the problem becomes:</p>

<p>$$4.85 - 5(1-\log_{10}({d_{☉})})= M - 5 (1-\log_{10}({d-d_{☉}}))$$</p>

<p>Or:</p>

<p>$$\frac{M-4.85}{5}=\log_{10}{\frac{d_{☉}}{d-d_{☉}}}$$</p>

<p>Continuing to solve for $d_{☉}$</p>

<p>$$d_{☉}=\frac{d\cdot10^{\frac{M-4.85}{5}}}{10^{\frac{M-4.85}{5}}+1}$$</p>

<p>Plugging that into a spreadsheet gives the following distances where the two stars are equally bright (Only including the strongest contenders)</p>

<ul>
<li>α Centauri A- 1.94 Light Years</li>
<li>α Centauri B- 2.61 Light Years</li>
<li>Sirius A- 1.46 light years</li>
</ul>

<p>Bottom line, heading 1.46 light years towards Sirius, you'll see both Sirius and the Sun as equally bright. This is approximately the edge of the <a href="http://en.wikipedia.org/wiki/Oort_cloud" rel="noreferrer">Oort Cloud</a>, and still within the gravitational influence of the Sun, but is well on our way to another star system.</p>

CC BY-SA 3.0
2016-02-09T18:12:52.530Z
How does this equation change, if we move tangentially to the star, rather than moving directly towards it?
Chris Koknat
CC BY-SA 3.0
2016-02-09T18:54:57.700Z
Different question entirely, but the distance is the thing of note. Moving in a direction that isn't directly towards the object would just change the distance formula.
PearsonArtPhoto
CC BY-SA 3.0
2016-02-09T20:05:31.627Z
I'm guessing that this accounts for the difference between Mark's 1.8 light years and your 1.46 light years.  They're both correct, but answer somewhat different questions.
Chris Koknat
CC BY-SA 3.0
2013-12-04T20:48:16.427Z
2016-08-28T18:54:06.067Z
2016-08-28T18:54:06.067Z
PearsonArtPhoto
<p>One way to answer would be to consider the brightest star in our sky (other than the Sun), which is Sirius.  Then determine how far you would have to be from our Sun for it to be as bright as Sirius is from here.</p>

<p>That turns out to be 1.8 light years.  That's not even halfway to the nearest star, so if you're in any other star system, then our Sun is just another star.  If you're anywhere in our solar system, even way out in the Oort cloud, then our Sun is way brighter than anything else.</p>

CC BY-SA 3.0
2013-12-04T16:38:39.687Z
2013-12-04T17:48:56.210Z
2013-12-04T17:48:56.210Z
Mark Adler
<p>When I stand outside looking at the night sky, to my untrained eye, everything except the moon looks like a star.  I know intellectually that some are planets circling our sun, and some are entire galaxies far away, but they all look basically the same.</p>

<p>How far out do you have be be from our sun for it to appear the same as any other 'star' in the sky? </p>

<p><em>Edit to clarify</em></p>

<p>As we migrate out through the solar system, when we look up to the sky the sun will get dimmer the farther away we get. On earth there is no doubt which star is our sun. </p>

<p>As we occupy bodies in the solar system farther from the sun, where will we be when the sun appears to be the same <a href="https://en.wikipedia.org/wiki/List_of_brightest_stars">brightness</a> as any other star in the sky? </p>

<the-sun><observational-astronomy><apparent-magnitude>
At what distance from Earth would our Sun be the same apparent magnitude as the next brightest star in the sky?
James Jenkins
# At what distance from Earth would our Sun be the same apparent magnitude as the next brightest star in the sky?

(Asked by: username_0 on 2013-12-04)

When I stand outside looking at the night sky, to my untrained eye, everything except the moon looks like a star. I know intellectually that some are planets circling our sun, and some are entire galaxies far away, but they all look basically the same.

How far out do you have be be from our sun for it to appear the same as any other 'star' in the sky?

*Edit to clarify*

As we migrate out through the solar system, when we look up to the sky the sun will get dimmer the farther away we get. On earth there is no doubt which star is our sun.

As we occupy bodies in the solar system farther from the sun, where will we be when the sun appears to be the same [brightness](https://en.wikipedia.org/wiki/List_of_brightest_stars) as any other star in the sky?

**Tags:** the-sun, observational-astronomy, apparent-magnitude

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## 2 answers

### Answer by username_3 on 2013-12-04. Score: 27 | Accepted answer

One way to answer would be to consider the brightest star in our sky (other than the Sun), which is Sirius. Then determine how far you would have to be from our Sun for it to be as bright as Sirius is from here.

That turns out to be 1.8 light years. That's not even halfway to the nearest star, so if you're in any other star system, then our Sun is just another star. If you're anywhere in our solar system, even way out in the Oort cloud, then our Sun is way brighter than anything else.


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### Answer by username_1 on 2013-12-04. Score: 19

As Mark Adler mentioned, the best way is to compare the brightness to other nearby stars. I'm going to assume that you have instantaneous travel time, and also take into account that you are actually getting closer to stars depending on the direction you go. I'm using [this table](http://en.wikipedia.org/wiki/List_of_nearest_stars) from Wikipedia. I'm going to go no further on the list than Sirius, and assume in each instance we are heading strait towards the Star. The formula for calculating the apperant magnitude given absolute magnitude, which is provided, is:

$$m = M - 5 (1-\log\_{10}{d})$$

Setting up for our situation, the problem becomes:

$$4.85 - 5(1-\log\_{10}({d\_{☉})})= M - 5 (1-\log\_{10}({d-d\_{☉}}))$$

Or:

$$\frac{M-4.85}{5}=\log\_{10}{\frac{d\_{☉}}{d-d\_{☉}}}$$

Continuing to solve for $d\_{☉}$

$$d\_{☉}=\frac{d\cdot10^{\frac{M-4.85}{5}}}{10^{\frac{M-4.85}{5}}+1}$$

Plugging that into a spreadsheet gives the following distances where the two stars are equally bright (Only including the strongest contenders)

* α Centauri A- 1.94 Light Years
* α Centauri B- 2.61 Light Years
* Sirius A- 1.46 light years

Bottom line, heading 1.46 light years towards Sirius, you'll see both Sirius and the Sun as equally bright. This is approximately the edge of the [Oort Cloud](http://en.wikipedia.org/wiki/Oort_cloud), and still within the gravitational influence of the Sun, but is well on our way to another star system.

> **Comments:**
> **username_2**: How does this equation change, if we move tangentially to the star, rather than moving directly towards it?  
> **username_1**: Different question entirely, but the distance is the thing of note. Moving in a direction that isn't directly towards the object would just change the distance formula.  
> **username_2**: I'm guessing that this accounts for the difference between Mark's 1.8 light years and your 1.46 light years.  They're both correct, but answer somewhat different questions.  

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