<p>James K's answer is right on, and probably what you're after. I just wanted to mention an effect which comes into play if you look really, really far away:</p>

<p>Because light moves at a finite speed, we see galaxies (and other things) as they were in the past. And because the Universe expands, everything was closer together in the past, so we see distant galaxies as they were when they were closer and hence spanned a larger angle on the sky.</p>

<p>Thus, we have two competing effects: On the one hand, galaxies become smaller and smaller with distance, as expected and as decribed in James K's answer. On the other hand, we look further and further into the past, and hence see galaxies at a time when they looked larger and larger. The result — which is obtained by integrating the <a href="https://en.wikipedia.org/wiki/Friedmann_equations" rel="nofollow noreferrer">Friedmann equation</a> — is that galaxies become smaller and smaller out to a certain distance (roughly 15 billion lightyears), after which the second effect starts to dominate and they start to grow in size.</p>

<p>Or, in other words, an arcseconds spans a larger and larger physical size (as described by James K) out to ~15 Gly (where it spans roughly 28,000 lightyears), after which it spans a smaller and smaller size. The relation looks like this:</p>

<p><a href="https://i.sstatic.net/NMPox.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/NMPox.png" alt="scale"></a></p>

CC BY-SA 3.0
2017-04-13T08:44:06.070Z
I understand that we see the galaxies as they were in the past, so we see them closer than their actual distance to us. But i dont unserstand the last paragraph :/
Dan Wears Prada
CC BY-SA 3.0
2017-04-13T20:02:28.970Z
@DanWearsPrada: I edited the last two paragraphs. Do they make sense now?
pela
CC BY-SA 3.0
2017-04-13T08:35:51.903Z
2017-04-13T20:01:40.733Z
2017-04-13T20:01:40.733Z
pela
<p>You can do your own calculations easily, no need to ask anyone. Just change the distance units in the examples below:</p>

<p><a href="http://www.wolframalpha.com/input/?i=tan(1%20arcsec)%20*%2010%20km" rel="nofollow noreferrer">tan(1 arcsec) * 10 km</a></p>

<p><a href="http://www.wolframalpha.com/input/?i=tan(1%20arcsec)%20*%204%20light%20years" rel="nofollow noreferrer">tan(1 arcsec) * 4 light years</a></p>

CC BY-SA 3.0
2017-04-14T09:10:25.190Z
To be 100% exact, shouldn’t it be 2*tan(0.5 arcsec)*distance? But for all practical purposes, approximating it by the arc as in James K’s answer is probably sufficient.
chirlu
CC BY-SA 3.0
2017-04-13T17:51:42.277Z
2017-04-13T17:51:42.277Z
Florin Andrei
<p><strong>How big is one arcsecond at various distances?</strong></p>

<p>An arcsecond is a small angle, 1/3600 of a degree or about 5 millionths of a radian ($4.85\times10^{-6}$). To estimate the size of something that appears 1 arcsecond across you can use the small angle approximation to trigonometry: Multiply the distance to the object by $4.85\times10^{-6}$</p>

<p>Examples:</p>

<ul>
<li>One arcsecond on the moon is 1.87km</li>
<li>One arcsecond on the sun is 727km</li>
<li>One arcsecond on Mars (when it is closest to Earth) is 237km</li>
<li>One arcsecond at distance of one parsec is one astronomical unit (AU), by definition</li>
<li>One arcsecond on Alpha Centauri is 200 million km</li>
<li>One arcsecond on the Andromeda galaxy is 100 trillion km</li>
</ul>

<p>The atmosphere limits how much detail you can see. Typically the smallest detail you can see is about 3 or 4 arcseconds, though professional telescopes do better by being built on the top of mountains and using various tricks, like adaptive optics.</p>

CC BY-SA 3.0
2017-04-13T08:06:43.650Z
Ys this is exactly what i was looking for thanks! Now i feel we are really short-sighted...

I will edit my question to your bold
Dan Wears Prada
CC BY-SA 3.0
2017-04-12T15:11:33.010Z
2017-04-13T12:38:16.223Z
2017-04-13T12:38:16.223Z
James K
<p>I've been reading this article
<a href="https://en.wikipedia.org/wiki/Celestial_coordinate_system" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Celestial_coordinate_system</a>
and a question came to my mind.</p>

<p>The minimum units are seconds ( ″ ). I picture an angle of 1 second as being like a "cone" going out from Earth. It's really small at the beginning but as you go further its width increases.</p>

<p>How wide does this cone get as you get further into the universe? How big is an arcsecond on different astronomical bodies?</p>

CC BY-SA 3.0
2017-04-11T15:47:36.723Z
The coordinate system is as precise as you want to make it. It's like any other coordinate system: add more decimal places and you have a more precise location.
Asher
CC BY-SA 3.0
2017-04-11T22:26:37.670Z
What @Asher says is not exactly true. The precision of your coordinate system is limited by how precisely you can define its zero-point. However the positions of things in this coordinate system can have finite precision, which is determined by your telescope/measurment method.
AtmosphericPrisonEscape
CC BY-SA 3.0
2017-04-12T07:19:21.067Z
so it depends on the telescope you are using? how can that be the rule?
Dan Wears Prada
CC BY-SA 3.0
2017-04-13T08:28:55.507Z
Edited to reflect the comment below my answer.
James K
<coordinate><size>
How big is one arcsecond at various distances?
Dan Wears Prada
# How big is one arcsecond at various distances?

(Asked by: username_0 on 2017-04-11)

I've been reading this article
<https://en.wikipedia.org/wiki/Celestial_coordinate_system>
and a question came to my mind.

The minimum units are seconds ( ″ ). I picture an angle of 1 second as being like a "cone" going out from Earth. It's really small at the beginning but as you go further its width increases.

How wide does this cone get as you get further into the universe? How big is an arcsecond on different astronomical bodies?

**Tags:** coordinate, size
> **Comments:**
> **username_1**: The coordinate system is as precise as you want to make it. It's like any other coordinate system: add more decimal places and you have a more precise location.  
> **username_2**: What @Asher says is not exactly true. The precision of your coordinate system is limited by how precisely you can define its zero-point. However the positions of things in this coordinate system can have finite precision, which is determined by your telescope/measurment method.  
> **username_0**: so it depends on the telescope you are using? how can that be the rule?  
> **username_3**: Edited to reflect the comment below my answer.  

---

## 3 answers

### Answer by username_3 on 2017-04-12. Score: 10 | Accepted answer

**How big is one arcsecond at various distances?**

An arcsecond is a small angle, 1/3600 of a degree or about 5 millionths of a radian ($4.85\times10^{-6}$). To estimate the size of something that appears 1 arcsecond across you can use the small angle approximation to trigonometry: Multiply the distance to the object by $4.85\times10^{-6}$

Examples:

* One arcsecond on the moon is 1.87km
* One arcsecond on the sun is 727km
* One arcsecond on Mars (when it is closest to Earth) is 237km
* One arcsecond at distance of one parsec is one astronomical unit (AU), by definition
* One arcsecond on Alpha Centauri is 200 million km
* One arcsecond on the Andromeda galaxy is 100 trillion km

The atmosphere limits how much detail you can see. Typically the smallest detail you can see is about 3 or 4 arcseconds, though professional telescopes do better by being built on the top of mountains and using various tricks, like adaptive optics.

> **Comments:**
> **username_0**: Ys this is exactly what i was looking for thanks! Now i feel we are really short-sighted...  I will edit my question to your bold  

---

### Answer by username_4 on 2017-04-13. Score: 4

James K's answer is right on, and probably what you're after. I just wanted to mention an effect which comes into play if you look really, really far away:

Because light moves at a finite speed, we see galaxies (and other things) as they were in the past. And because the Universe expands, everything was closer together in the past, so we see distant galaxies as they were when they were closer and hence spanned a larger angle on the sky.

Thus, we have two competing effects: On the one hand, galaxies become smaller and smaller with distance, as expected and as decribed in James K's answer. On the other hand, we look further and further into the past, and hence see galaxies at a time when they looked larger and larger. The result — which is obtained by integrating the [Friedmann equation](https://en.wikipedia.org/wiki/Friedmann_equations) — is that galaxies become smaller and smaller out to a certain distance (roughly 15 billion lightyears), after which the second effect starts to dominate and they start to grow in size.

Or, in other words, an arcseconds spans a larger and larger physical size (as described by James K) out to ~15 Gly (where it spans roughly 28,000 lightyears), after which it spans a smaller and smaller size. The relation looks like this:

[![scale](https://i.sstatic.net/NMPox.png)](https://i.sstatic.net/NMPox.png)

> **Comments:**
> **username_0**: I understand that we see the galaxies as they were in the past, so we see them closer than their actual distance to us. But i dont unserstand the last paragraph :/  
> **username_4**: @DanWearsPrada: I edited the last two paragraphs. Do they make sense now?  

---

### Answer by username_5 on 2017-04-13. Score: 0

You can do your own calculations easily, no need to ask anyone. Just change the distance units in the examples below:

[tan(1 arcsec) \* 10 km](http://www.wolframalpha.com/input/?i=tan(1%20arcsec)%20*%2010%20km)

[tan(1 arcsec) \* 4 light years](http://www.wolframalpha.com/input/?i=tan(1%20arcsec)%20*%204%20light%20years)

> **Comments:**
> **username_6**: To be 100% exact, shouldn’t it be 2*tan(0.5 arcsec)*distance? But for all practical purposes, approximating it by the arc as in James K’s answer is probably sufficient.  

---

