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?


I wrote https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-solve-astro-13115.m to solve this with the full results at: https://github.com/barrycarter/bcapps/blob/master/ASTRO/brightest-stars-from-other-stars.txt.bz2

As viewed from RigelKentaurusB, the Sun has magnitude 0.48, and is the 10th brightest star in the sky. This is the only star system with known exoplanets where the Sun even makes the top 10.

From EpsilonEridani, the sun shines at magnitude 2.37, making it the 79th brightest star in their sky. Things go rapidly downhill from there:

79 EpsilonEridani Sun 2.36646
207 HIP113020 Sun 3.18865
208 HIP85523 Sun 3.11216
247 HIP106440 Sun 3.29606
396 HIP15510 Sun 3.74083
423 HIP74995 Sun 3.81444
685 Fomalhaut Sun 4.25746
878 HIP64924 Sun 4.48202
949 HIP109388 Sun 4.54456
972 HIP99825 Sun 4.55679

This is based on Mathematica's list of 552 exoplanets orbiting 464 stars, and a total list of 88,637 stars.

If the naked eye visibility limit is magnitude 5.5, the Sun would only be visible from 28 of these stars:

10 RigelKentaurusB Sun 0.476137
79 EpsilonEridani Sun 2.36646
208 HIP85523 Sun 3.11216
207 HIP113020 Sun 3.18865
247 HIP106440 Sun 3.29606
396 HIP15510 Sun 3.74083
423 HIP74995 Sun 3.81444
685 Fomalhaut Sun 4.25746
878 HIP64924 Sun 4.48202
949 HIP109388 Sun 4.54456
972 HIP99825 Sun 4.55679
1086 HIP57443 Sun 4.65677
1126 HIP21932 Sun 4.69871
1381 HIP57087 Sun 4.87838
1426 HIP83043 Sun 4.89689
1409 Pollux Sun 4.90049
1617 HIP10138 Sun 5.01833
1649 HIP57050 Sun 5.04143
1676 HIP3093 Sun 5.05658
1716 HIP48331 Sun 5.06528
1831 Gl317 Sun 5.13093
2104 HIP22627 Sun 5.24572
2264 Rho1Cancri Sun 5.31849
2263 HIP40693 Sun 5.32723
2397 HIP27887 Sun 5.36943
2355 HIP80337 Sun 5.37669
2382 GJ1214 Sun 5.3879
2689 UpsilonAndromedae Sun 5.47503

If the limit is magnitude 6.5, 33 more star systems can see the Sun:

2838 Alrai Sun 5.52682
2926 HIP53721 Sun 5.571
3399 HIP79431 Sun 5.69393
3637 MuArae Sun 5.74863
3679 HIP113357 Sun 5.76061
3783 TauBootis Sun 5.7949
3941 HIP98767 Sun 5.83457
4071 HIP85647 Sun 5.86725
4318 HIP71395 Sun 5.92513
4580 HIP6379 Sun 5.95739
4783 IotaHorologii Sun 6.01138
4835 HIP7978 Sun 6.02527
4867 RhoCoronaeBorealis Sun 6.03887
5016 HIP1292 Sun 6.0583
4949 NN3634 Sun 6.08757
4993 HIP55848 Sun 6.10352
5252 HIP49699 Sun 6.11214
5300 HIP83389 Sun 6.11214
5031 HIP65721 Sun 6.11803
5343 HIP79248 Sun 6.12236
5385 PiMensae Sun 6.12986
5453 EpsilonReticulii Sun 6.13303
6158 HIP98505 Sun 6.251
6351 HIP113421 Sun 6.30305
6436 Nu2CanisMajoris Sun 6.31593
6343 HIP99711 Sun 6.32197
6737 Hamal Sun 6.35637
6516 HIP64457 Sun 6.37975
6867 HIP58451 Sun 6.4205
7614 HIP25110 Sun 6.43775
7525 HIP109378 Sun 6.46941
7538 HIP54906 Sun 6.47034
7957 HIP96901 Sun 6.48193

OLD ANSWER FOR REFERENCE (note that Sirius does not have known exoplanets, so the Sirius solution I propose below wouldn't work):

As others have noted, since Sirius is 25 times more luminous than our Sun (though still not luminous as Canopus or Rigel), so it would be the brightest star for most nearby star systems, and all the exoplanets we've found are fairly closeby.

The only possible exception would be an exoplanet orbiting Sirius itself, since Sirius would be considered their sun, and not a star in the night sky.

Unfortunately, Sirius' smaller companion, Sirius B, would be much brighter than our Sun there. I'm also pretty sure Sirius B would be in the night sky (not just in the daytime sky), so it would count.

I'd like to poke around a bit more before declaring this answer complete. I think I can show that any point where the Sun is brighter than Sirius must be less than 4.3 light years (and probably much less) away from our solar system, so, unless we discover an star system MUCH closer than Proxima Centauri, there are no exoplanets where our Sun is the brightest star.

  • 1
    $\begingroup$ I just saw this, very nice! 3.5 years later I wonder if the results are any different... $\endgroup$ – uhoh Jul 27 '19 at 10:07
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    $\begingroup$ It would be interesting to do this for all nearby stars, even those w/o exoplanets (or currently undiscovered exoplanets). I've put it on my project list at github.com/barrycarter/bcapps/tree/master/README-projects.txt which means I may someday get around to it, but probably not. $\endgroup$ – user21 Jul 30 '19 at 17:59

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

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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    $\begingroup$ I like this. I was going to use the triangle inequality to show that if you are distance d1 from the Sun, and the Sun is distance d2 from Sirius, you are necessarily less than distance d1+d2 from Sirius, but your approach is better. $\endgroup$ – user21 Jan 8 '16 at 5:30

You can use Celestia to get a view of the sun in the night sky from other solar systems, just like this guy did, although the post is a little old.

  • $\begingroup$ This might be better as a comment than an answer (really cool link, though!). $\endgroup$ – HDE 226868 Jan 10 '16 at 15:52

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 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.

  • $\begingroup$ I object to this answer. Sirius is only about 1.7 times as large as the Sun, weighs only twice as much, and is still bright outside our atmosphere. Sirius is bright because of a combination of its inherent luminosity (ie, brightness) AND its distance. I sort of understand your atmosphere argument (Sirius' peak light wavelength is different from the Sun's), but I think you need to flesh it out a bit. $\endgroup$ – user21 Jan 7 '16 at 17:30

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