<p>Another way of thinking about the question is, "how close do you need to be for a dim star (our Sun) to outshine a brighter one?"</p>

<p>Consider our Sun, and another star a distance $d$ away that is $m$ times brighter than the Sun (in terms of luminosity or absolute magnitude).  Let's look a location in space that is a distance $x$ from the Sun in the direction of the second star, and a distance $y$ in the perpendicular direction.</p>

<p><img src="https://i.sstatic.net/knbdP.png" alt="enter image description here"></p>

<p>The amount of light seen at the location is proportional to the inverse square of the distance.  The relative amount of light from each star is therefore just:</p>

<p>$$
\frac{1}{x^2+y^2} \qquad\qquad \frac{m}{(d-x)^2+y^2}.
$$</p>

<p>We want to know when the first is greater than the second:</p>

<p>$$
\frac{1}{x^2+y^2} &gt; \frac{m}{(d-x)^2+y^2} \\
(d-x)^2+y^2 &gt; m x^2 + m y^2 \\
d^2 - 2dx &gt; (m-1)x^2 + (m-1)y^2 \\
\left(1+\frac{1}{m-1}\right)d^2 &gt; \left(\frac{1}{m-1}\right)d^2 + 2dx + (m-1)\left(x^2 + y^2\right) \\
\left(\frac{m-1+1}{m-1}\right)d^2 &gt; (m-1)\left(\frac{d^2}{(m-1)^2} + 2x\frac{d}{m-1} + x^2 + y^2\right) \\
m\left(\frac{d}{m-1}\right)^2 &gt; \left(\frac{d}{m-1} + x\right)^2 + y^2 \\
$$</p>

<p>This inequality describes a sphere of radius $r=d \sqrt{m}/(m-1)$, centered around the point $x=-d/(m-1),\ y=0$.</p>

<hr>

<p>Let's take the specific case of Sirius, for which $m=25.4$ and $d=8.60~\text{ly}$.</p>

<p>Applying the above equations, we get $r=1.77~\text{ly}$ and $x=-0.35~\text{ly}$.  Therefore, even at half the distance of the next-closest star Sirius is brighter than the Sun.  This means that the Sun is the brightest star in the sky for exactly eight planets.</p>

CC BY-SA 3.0
2016-01-08T04:41:55.413Z
2016-01-08T06:06:31.893Z
2016-01-08T06:06:31.893Z
2012rcampion
<p>You can use Celestia to get a view of the sun in the night sky from other solar systems, <a href="http://astroblogger.blogspot.ie/2010/03/tour-of-exoplanets-in-celestia.html" rel="nofollow">just like this guy did</a>, although the post is a little old.</p>

CC BY-SA 3.0
2016-01-10T15:52:28.113Z
This might be better as a comment than an answer (really cool link, though!).
HDE 226868
CC BY-SA 3.0
2016-01-07T16:14:20.397Z
2016-01-07T16:14:20.397Z
StephenG - Help Ukraine
<p>Dependents. Unless blocked or outshone by a neighboring star Sirius A would continue to be the brightest star visible, Sirius A is bright due to its <a href="http://hubblesite.org/newscenter/archive/releases/2005/36/full/" rel="nofollow">size</a> and our atmosphere, not because of distance, if the planet had a different atmosphere that broke up wavelengths of light differently than earth other stars may show brighter.</p>

CC BY-SA 3.0
2016-01-07T13:05:01.043Z
2016-01-07T13:05:01.043Z
Zac Grierson
<p>The brightest star on the night sky is Sirius, here in the Earth. But are there any exoplanets, where the brightest star on the night sky is our sun?</p>

CC BY-SA 3.0
2016-01-07T12:39:22.297Z
My guess would be 'none'. From the distance of the closest exoplanet, Sol would be just another of the brighter stars - no more.
Andrew Thompson
<exoplanet>
From which exoplanets is our sun the brightest star on the night sky?
Iter Ator
# From which exoplanets is our sun the brightest star on the night sky?

(Asked by: username_0 on 2016-01-07)

The brightest star on the night sky is Sirius, here in the Earth. But are there any exoplanets, where the brightest star on the night sky is our sun?

**Tags:** exoplanet

> **Comments:**
> **username_1**: My guess would be 'none'. From the distance of the closest exoplanet, Sol would be just another of the brighter stars - no more.  

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## 3 answers

### Answer by username_2 on 2016-01-08. Score: 7

Another way of thinking about the question is, "how close do you need to be for a dim star (our Sun) to outshine a brighter one?"

Consider our Sun, and another star a distance $d$ away that is $m$ times brighter than the Sun (in terms of luminosity or absolute magnitude). Let's look a location in space that is a distance $x$ from the Sun in the direction of the second star, and a distance $y$ in the perpendicular direction.

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

The amount of light seen at the location is proportional to the inverse square of the distance. The relative amount of light from each star is therefore just:

$$
\frac{1}{x^2+y^2} \qquad\qquad \frac{m}{(d-x)^2+y^2}.
$$

We want to know when the first is greater than the second:

$$
\frac{1}{x^2+y^2} > \frac{m}{(d-x)^2+y^2} \\
(d-x)^2+y^2 > m x^2 + m y^2 \\
d^2 - 2dx > (m-1)x^2 + (m-1)y^2 \\
\left(1+\frac{1}{m-1}\right)d^2 > \left(\frac{1}{m-1}\right)d^2 + 2dx + (m-1)\left(x^2 + y^2\right) \\
\left(\frac{m-1+1}{m-1}\right)d^2 > (m-1)\left(\frac{d^2}{(m-1)^2} + 2x\frac{d}{m-1} + x^2 + y^2\right) \\
m\left(\frac{d}{m-1}\right)^2 > \left(\frac{d}{m-1} + x\right)^2 + y^2 \\
$$

This inequality describes a sphere of radius $r=d \sqrt{m}/(m-1)$, centered around the point $x=-d/(m-1),\ y=0$.

---

Let's take the specific case of Sirius, for which $m=25.4$ and $d=8.60~\text{ly}$.

Applying the above equations, we get $r=1.77~\text{ly}$ and $x=-0.35~\text{ly}$. Therefore, even at half the distance of the next-closest star Sirius is brighter than the Sun. This means that the Sun is the brightest star in the sky for exactly eight planets.


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### Answer by username_3 on 2016-01-07. Score: 1

You can use Celestia to get a view of the sun in the night sky from other solar systems, [just like this guy did](http://astroblogger.blogspot.ie/2010/03/tour-of-exoplanets-in-celestia.html), although the post is a little old.

> **Comments:**
> **username_4**: This might be better as a comment than an answer (really cool link, though!).  

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### Answer by username_5 on 2016-01-07. Score: 0

Dependents. Unless blocked or outshone by a neighboring star Sirius A would continue to be the brightest star visible, Sirius A is bright due to its [size](http://hubblesite.org/newscenter/archive/releases/2005/36/full/) and our atmosphere, not because of distance, if the planet had a different atmosphere that broke up wavelengths of light differently than earth other stars may show brighter.


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