<p>WolframAlpha says (to the query "<a href="http://www.wolframalpha.com/input/?i=apparent%20magnitude%20of%20star%20with%20absolute%20magnitude%20%3D%208.23%20and%20distance%20%3D%201.1%20ly" rel="nofollow noreferrer">apparent magnitude of star with absolute magnitude = 8.23 and distance = 1.1 ly</a>") that the apparent magnitude of Gliese 170 would be about 0.87 on average. I took the value of the absolute magnitude (8.23) from Wikipedia. You can plug in different values for the distance in the formula and see what comes out. The formula used by WolframAlpha is </p>

<p>$$D_l = 10^{{m-M \over 5} +1} pc$$</p>

<p>where $D_l$ = luminosity distance, $m$ = apparent magnitude, $M$ = absolute magnitude, $pc$ = parsec.</p>

CC BY-SA 3.0
2017-01-06T16:12:56.200Z
2017-01-06T16:12:56.200Z
pablodf76
<p>I read <a href="http://www.iflscience.com/space/a-star-is-on-a-collision-course-with-the-edge-of-our-solar-system/" rel="nofollow noreferrer" title="this article">this article</a> on <a href="https://en.wikipedia.org/wiki/Gliese_710" rel="nofollow noreferrer">Gliese 710</a> that stated it would come within about 1.1 ly (+/- 0.577 ly) of earth in 1.35 million years. Wikipedia <a href="https://en.wikipedia.org/wiki/Gliese_710" rel="nofollow noreferrer">states</a> it would have an apparent magnitude of 0.5, and the <a href="http://www.aanda.org/articles/aa/abs/2016/11/aa29835-16/aa29835-16.html" rel="nofollow noreferrer">original paper abstract</a> states it would have a -2.7 apparent magnitude. </p>

<p>From reading the <a href="https://en.wikipedia.org/wiki/List_of_brightest_stars" rel="nofollow noreferrer" title="brightest stars">list of brightest stars</a>, apparent magnitude of 0.5 would make it about the 10th brightest star, and -2.7 would make it the brightest (besides the sun). </p>

<p>I'm curious about the discrepancy between these sources.</p>

<p>How bright would Gliese 710 appear if it were at the near end (0.523 ly) and far end (1.677 ly) of the expected closest approach to us?</p>

CC BY-SA 3.0
2017-01-06T14:30:11.940Z
Calculate it yourself:  you can treat it as a point source, so irradiance goes as the square of the distance; magnitudes are a log scale.
Carl Witthoft
<star><apparent-magnitude>
How bright could Gliese 710 appear in 1.35 million years?
joseph.hainline
# How bright could Gliese 710 appear in 1.35 million years?

(Asked by: username_0 on 2017-01-06)

I read [this article](http://www.iflscience.com/space/a-star-is-on-a-collision-course-with-the-edge-of-our-solar-system/ "this article") on [Gliese 710](https://en.wikipedia.org/wiki/Gliese_710) that stated it would come within about 1.1 ly (+/- 0.577 ly) of earth in 1.35 million years. Wikipedia [states](https://en.wikipedia.org/wiki/Gliese_710) it would have an apparent magnitude of 0.5, and the [original paper abstract](http://www.aanda.org/articles/aa/abs/2016/11/aa29835-16/aa29835-16.html) states it would have a -2.7 apparent magnitude.

From reading the [list of brightest stars](https://en.wikipedia.org/wiki/List_of_brightest_stars "brightest stars"), apparent magnitude of 0.5 would make it about the 10th brightest star, and -2.7 would make it the brightest (besides the sun).

I'm curious about the discrepancy between these sources.

How bright would Gliese 710 appear if it were at the near end (0.523 ly) and far end (1.677 ly) of the expected closest approach to us?

**Tags:** star, apparent-magnitude

> **Comments:**
> **username_1**: Calculate it yourself:  you can treat it as a point source, so irradiance goes as the square of the distance; magnitudes are a log scale.  

---

## 1 answers

### Answer by username_2 on 2017-01-06. Score: 2 | Accepted answer

WolframAlpha says (to the query "[apparent magnitude of star with absolute magnitude = 8.23 and distance = 1.1 ly](http://www.wolframalpha.com/input/?i=apparent%20magnitude%20of%20star%20with%20absolute%20magnitude%20%3D%208.23%20and%20distance%20%3D%201.1%20ly)") that the apparent magnitude of Gliese 170 would be about 0.87 on average. I took the value of the absolute magnitude (8.23) from Wikipedia. You can plug in different values for the distance in the formula and see what comes out. The formula used by WolframAlpha is

$$D\_l = 10^{{m-M \over 5} +1} pc$$

where $D\_l$ = luminosity distance, $m$ = apparent magnitude, $M$ = absolute magnitude, $pc$ = parsec.


---

