<p>I tried to get a estimate for an upper-bound with currently deployed instruments. Assuming that, resolution-wise, the James Webb Space Telescope is the best telescope we have today, I looked into its FAQ at <a href="https://jwst.nasa.gov/content/about/faqs/faq.html#sharp" rel="nofollow noreferrer">https://jwst.nasa.gov/content/about/faqs/faq.html#sharp</a> to find its angular resolution:</p>
<blockquote>
<p><strong>What is Webb's angular resolution, and how will its images compare to Hubble's? Will they be as beautiful?</strong></p>
<p>Webb's angular resolution, or sharpness of vision, will be the same as
Hubble's, but in the near infrared. This means that Webb images will
appear just as sharp as Hubble's do.</p>
<p>Webb will have an angular resolution of somewhat better than 0.1
arc-seconds at a wavelength of 2 micrometers (one degree = 60
arc-minutes = 3600 arc-seconds). Seeing at a resolution of 0.1
arc-second means that Webb could see details the size of a US penny at
a distance of about 24 miles (40 km), or a regulation soccer ball at a
distance of 340 miles (550 km).</p>
</blockquote>
<p>Now imagine we can distribute stars at will around the celestial sphere. The best distribution I can imagine, to fit the maximum number of stars while still keeping angular resolution, is to put them evenly spaced at the celestial equator, at 0.1 arc-second intervals. And then do the same for every parallel circle above and bellow, separated by 0.1 arc-seconds of celestial &quot;latitude&quot; (as the parallel circles shrink by cos(latitude angle), its necessary to adjust the number of stars accordingly). The following <a href="https://pharo.org/" rel="nofollow noreferrer">Pharo 10</a> code calculates approximately how many stars we could fit this way:</p>
<pre><code>| resolution starsAtEquator starsInHemisphere |
Transcript clear.
resolution := 0.1. &quot;Resolution in arc-seconds&quot;
starsAtEquator := 360 * 60 * 60 / resolution.
starsInHemisphere := ((0 to: starsAtEquator // 4) collect: [ :each | 
                          (each * Float pi / (starsAtEquator // 2)) cos
                          * starsAtEquator round: 0 ]) sum.
Transcript
    show: 'Stars in whole celestial sphere: ';
    show: 2 * starsInHemisphere
</code></pre>
<p>The result is about 5.35e13, or about 50 trillion objects individually resolvable, at James Webb resolution.</p>
<p>Correction: In the first version of this answer, I said that If it took only one second to reposition the telescope to look at a different point, it would take more than one and a half million years to cover the entire sky, assuming it would have to reposition at every point. But as pointed by @planetmaker, the field of view of the James Webb telescope is 3' x 3', so the time to cover the entire sky would be much lower.</p>

CC BY-SA 4.0
2022-06-08T22:11:00.987Z
The FOV of the James Webb telescope is 3' x 3'.so the survey takes much less time, less than a year
planetmaker
CC BY-SA 4.0
2022-06-09T15:58:49.237Z
Would this be more of an upper bound, since stars aren't distributed this way?
Connor Garcia
CC BY-SA 4.0
2022-06-09T16:37:13.340Z
Yes, it's a upper bound, assuming the resolution from the best instrument currently deployed, and a sky packed full of stars in every direction according to this resolution.
ksousa
CC BY-SA 4.0
2022-06-08T22:02:40.737Z
2022-06-11T22:56:49.220Z
2022-06-11T22:56:49.220Z
ksousa
<p>Wikipedia estimates the number of stars in the <a href="https://en.wikipedia.org/wiki/Milky_Way" rel="noreferrer">Milky Way</a> at 100 to 400 billion.</p>
<p>One of our most ambitious astrometry projects, Gaia, has been able to resolve and map almost 2 billion stars in the Milky Way and the Small and Large Magellanic Clouds according to <a href="https://ras.ac.uk/news-and-press/news/gaia-most-accurate-data-ever-nearly-two-billion-stars" rel="noreferrer">this article</a>.</p>
<p>Still, that is only about 0.5% to 2% of the total number of stars in the Milky Way.  We can assume that given more time, with better telescopes, this list of stars will continue to grow.  But will some stars always be obscured by dust clouds or our dense galactic center?  For faraway galaxies, is the light too intermixed to resolve particular stars?</p>
<p><strong>Question: Is there an upper limit to the number of stars humans will be able to specifically resolve (without interstellar travel)?</strong></p>

CC BY-SA 4.0
2022-06-08T00:26:08.787Z
I don't think there's any star in observable Universe that couldn't me mapped, at least in principle. I don't think asking what *will* be possible is a good idea.
Mithoron
CC BY-SA 4.0
2022-06-08T01:56:35.380Z
@Mithoron Wouldn't you need a telescope wider than the solar system to map the individual stars in a tight binary system in a distant galaxy?
Connor Garcia
CC BY-SA 4.0
2022-06-08T12:39:15.393Z
Gravity waves, or wait for gravitational lensing ;)
Mithoron
CC BY-SA 4.0
2022-06-08T13:23:53.583Z
Oh, for waves it would be even worse, so just lensing, or whatever nigh-magical method might be invented in the future.
Mithoron
CC BY-SA 4.0
2022-06-12T14:40:12.033Z
I've never seen a star I couldn't resolve :-)
uhoh
<star><galaxy><gaia>
How many stars can we resolve?
Connor Garcia
# How many stars can we resolve?

(Asked by: username_0 on 2022-06-07)

Wikipedia estimates the number of stars in the [Milky Way](https://en.wikipedia.org/wiki/Milky_Way) at 100 to 400 billion.

One of our most ambitious astrometry projects, Gaia, has been able to resolve and map almost 2 billion stars in the Milky Way and the Small and Large Magellanic Clouds according to [this article](https://ras.ac.uk/news-and-press/news/gaia-most-accurate-data-ever-nearly-two-billion-stars).

Still, that is only about 0.5% to 2% of the total number of stars in the Milky Way. We can assume that given more time, with better telescopes, this list of stars will continue to grow. But will some stars always be obscured by dust clouds or our dense galactic center? For faraway galaxies, is the light too intermixed to resolve particular stars?

**Question: Is there an upper limit to the number of stars humans will be able to specifically resolve (without interstellar travel)?**

**Tags:** star, galaxy, gaia

> **Comments:**
> **username_1**: I don't think there's any star in observable Universe that couldn't me mapped, at least in principle. I don't think asking what *will* be possible is a good idea.  
> **username_0**: @username_1 Wouldn't you need a telescope wider than the solar system to map the individual stars in a tight binary system in a distant galaxy?  
> **username_1**: Gravity waves, or wait for gravitational lensing ;)  
> **username_1**: Oh, for waves it would be even worse, so just lensing, or whatever nigh-magical method might be invented in the future.  
> **username_2**: I've never seen a star I couldn't resolve :-)  

---

## 1 answers

### Answer by username_3 on 2022-06-08. Score: 5 | Accepted answer

I tried to get a estimate for an upper-bound with currently deployed instruments. Assuming that, resolution-wise, the James Webb Space Telescope is the best telescope we have today, I looked into its FAQ at <https://jwst.nasa.gov/content/about/faqs/faq.html#sharp> to find its angular resolution:

> **What is Webb's angular resolution, and how will its images compare to Hubble's? Will they be as beautiful?**
>
> Webb's angular resolution, or sharpness of vision, will be the same as
> Hubble's, but in the near infrared. This means that Webb images will
> appear just as sharp as Hubble's do.
>
> Webb will have an angular resolution of somewhat better than 0.1
> arc-seconds at a wavelength of 2 micrometers (one degree = 60
> arc-minutes = 3600 arc-seconds). Seeing at a resolution of 0.1
> arc-second means that Webb could see details the size of a US penny at
> a distance of about 24 miles (40 km), or a regulation soccer ball at a
> distance of 340 miles (550 km).

Now imagine we can distribute stars at will around the celestial sphere. The best distribution I can imagine, to fit the maximum number of stars while still keeping angular resolution, is to put them evenly spaced at the celestial equator, at 0.1 arc-second intervals. And then do the same for every parallel circle above and bellow, separated by 0.1 arc-seconds of celestial "latitude" (as the parallel circles shrink by cos(latitude angle), its necessary to adjust the number of stars accordingly). The following [Pharo 10](https://pharo.org/) code calculates approximately how many stars we could fit this way:

```
| resolution starsAtEquator starsInHemisphere |
Transcript clear.
resolution := 0.1. "Resolution in arc-seconds"
starsAtEquator := 360 * 60 * 60 / resolution.
starsInHemisphere := ((0 to: starsAtEquator // 4) collect: [ :each | 
                          (each * Float pi / (starsAtEquator // 2)) cos
                          * starsAtEquator round: 0 ]) sum.
Transcript
    show: 'Stars in whole celestial sphere: ';
    show: 2 * starsInHemisphere

```

The result is about 5.35e13, or about 50 trillion objects individually resolvable, at James Webb resolution.

Correction: In the first version of this answer, I said that If it took only one second to reposition the telescope to look at a different point, it would take more than one and a half million years to cover the entire sky, assuming it would have to reposition at every point. But as pointed by @username_4, the field of view of the James Webb telescope is 3' x 3', so the time to cover the entire sky would be much lower.

> **Comments:**
> **username_4**: The FOV of the James Webb telescope is 3' x 3'.so the survey takes much less time, less than a year  
> **username_0**: Would this be more of an upper bound, since stars aren't distributed this way?  
> **username_3**: Yes, it's a upper bound, assuming the resolution from the best instrument currently deployed, and a sky packed full of stars in every direction according to this resolution.  

---

