<p>While this might seem to be strictly a math problem, it's really loaded with Astronomy, let's see what we can learn!</p>
<p>If you're far enough away that you can see essentially a full hemisphere (which you can't if you're close), the apparent angular width is twice the half-width, and that's given by</p>
<p><span class="math-container">$$\theta_{HW} = \arcsin\frac{r}{R} \approx \frac{r}{R}$$</span></p>
<p>for small angles, where <span class="math-container">$r$</span> is the physical radius of the object and <span class="math-container">$R$</span> is the distance from the viewer to its center. This is also similar to or the same as <a href="https://en.wikipedia.org/wiki/Moon" rel="nofollow noreferrer">@peterh's method</a>.</p>
<p>There's (at least) two problems with trying to talk about the radius of the Sun;</p>
<ol>
<li>The Sun is oblate</li>
<li>The Sun is diffuse; it doesn't have a well defined edge</li>
</ol>
<p>Let's do #2 first. There is a formal working definition for the optical edge of the Sun, and <a href="https://astronomy.stackexchange.com/a/2106/7982">this excellent answer</a> to the question <em>How do you define the diameter of the Sun</em> states:</p>
<blockquote>
<p>Most literature will define the diameter of the Sun up to the photosphere, the layer of the solar atmosphere you would see if you were to observe the Sun in white light.</p>
<p>The base of the <a href="https://en.wikipedia.org/wiki/Photosphere" rel="nofollow noreferrer">Photosphere</a> is defined as the region where the <a href="https://en.wikipedia.org/wiki/Optical_depth" rel="nofollow noreferrer">optical depth</a> is around 2/3, or the region where the plasma becomes transparent to most optical light wavelengths.</p>
<p>Of course the true edge of the solar atmosphere could be considered as the heliopause, where the direct influence of the Sun's magnetic field and solar wind end and interstellar space begins.</p>
</blockquote>
<p>With that definition of the edge, let's look at the shape of the Sun.  <a href="https://en.wikipedia.org/wiki/Sun" rel="nofollow noreferrer">Wikipedia</a> gives both 695,700 and 696,392 km for the equatorial radius of the Sun, sourcing from <a href="https://arxiv.org/abs/1510.07674" rel="nofollow noreferrer">IAU 2015 Resolution B3...</a> and <a href="https://arxiv.org/abs/1203.4898" rel="nofollow noreferrer">Measuring the Solar Radius from Space during the 2003 and 2006 Mercury Transits</a> respectively.</p>
<p>Let's use 695,700 km because it's the one that I see most often and &quot;because IAU&quot;.</p>
<p>The Wikipedia article gives a <a href="https://astronomy.stackexchange.com/a/28069/7982">flattening</a> of 9E-06, which makes the polar radius only about 10 parts per million smaller, which is a <em>much smaller equator-to-pole difference</em> than I remembered it to be. I guess we can ignore it after all!</p>
<p>Seen from the Earth then, who's orbit takes it from 152.1 million to 147.1 million km from the Sun, the angular width of the Sun would vary from about 1887 to 1951 arcseconds (31.4 to 32.5 arcminutes).</p>
<p>At what distance would an object with a radius of 695,700 km have an apparent width of 7 arcseconds (which is 3.39E-05 radians)? Flipping the equation around gives 2×695700/3.39E-05 or 4.105E+10 kilometers about or 274 AU.</p>
<hr />
<h2>What is it like 274 AU from the Sun?</h2>
<p>In addition to appearing to be roughly 274 times smaller from here than from Earth, the Sun is 274×274 times dimmer. Using <span class="math-container">$2.5 \times \log_{10}(274×274)$</span> we get that the Sun would appear about 12.2 magnitudes dimmer than it's -26.8 magnitude brightness at 1 AU, or -14.6 magnitude, which is still 2 magnitudes brighter than a typical full moon.</p>
<p>Your orbital period around the Sun would be over 4,500 years, and you'd be moving at 1.8 km/sec in that orbit, as opposed to 30 km/sec for the Earth.</p>
<p>You'd be well past the <a href="https://en.wikipedia.org/wiki/Kuiper_belt" rel="nofollow noreferrer">Kuiper belt</a> but nowhere near the <a href="https://en.wikipedia.org/wiki/List_of_Solar_System_objects_most_distant_from_the_Sun_in_2018" rel="nofollow noreferrer">Oort cloud</a> (this is a simplification, but it will do for now), and you'd be way past the <a href="https://en.wikipedia.org/wiki/List_of_Solar_System_objects_most_distant_from_the_Sun_in_2018" rel="nofollow noreferrer">farthest known solar system body</a> in 2018! <a href="https://en.wikipedia.org/wiki/V774104" rel="nofollow noreferrer">V774104</a>.</p>
<p><a href="https://i.sstatic.net/ZmQGg.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ZmQGg.gif" alt="V774104" /></a></p>
<p><strong>above:</strong> from EarthSky.org's <a href="http://earthsky.org/space/new-most-distant-object-in-solar-system-v774104" rel="nofollow noreferrer">New most distant object in solar system</a> Image via S. Sheppard / C. Trujillo / D. Tholen / Subaru Telescope/ <a href="http://www.skyandtelescope.com/astronomy-news/v774104-most-distant-solar-system-object-11212015" rel="nofollow noreferrer">skyandtelescope.com</a>.</p>
<p>And at 274 AU, you'd be much farther from the Sun than any deep space probe as well. Voyagers 1 and 2 are &quot;only&quot; 118 and 142 AU from the Sun now, and New Horizons is only about 42 AU.</p>
<p>I don't know the name of the place you'd be, it looks pretty lonely though!</p>
<p><a href="https://i.sstatic.net/Rg9nC.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Rg9nC.jpg" alt="enter image description here" /></a></p>
<p><a href="https://commons.wikimedia.org/wiki/File:PIA17046_-_Voyager_1_Goes_Interstellar.jpg" rel="nofollow noreferrer">Source</a></p>

CC BY-SA 4.0
2018-10-19T13:02:55.350Z
I wish you could favorite answers, I love the picture there, that's an amazing representation of the Oort cloud. I never really pictured how "thick" it could be. Is that to scale? The Oort cloud is from 1,000AU to 100,000AU?
Magic Octopus Urn
CC BY-SA 4.0
2018-10-19T13:20:23.270Z
@MagicOctopusUrn this is a schematic/cartoon representation only, and the distance scale is logarithmic. Solar system structures this far away (especially the Oort cloud) are inferred and predicted, not seen or measured. It's educated speculation.
uhoh
CC BY-SA 4.0
2018-10-19T13:40:56.690Z
fair point, the source does confirm that the estimates are from 1,000AU to 100,000 AU though.
Magic Octopus Urn
CC BY-SA 4.0
2018-10-22T21:36:36.600Z
What are your thoughts on this proof, that falsifies Edmond Halley's proposition from 1720 that stars appear larger than they are because of an "optical illusion"? https://www.youtube.com/watch?v=ivhXqDnrV4o
pol0
CC BY-SA 4.0
2018-10-23T00:57:16.923Z
@pol0 consider posting that as a new question? It's not directly related to this answer and so comments here are not the right place to open up a new discussion. Thanks!
uhoh
CC BY-SA 4.0
2018-10-19T12:06:31.527Z
2018-10-19T12:29:10.503Z
2020-06-17T09:47:40.653Z
uhoh
<p>From us, the Sun is visible around a half grad, which is <span class="math-container">$\approx$</span> 30 arc min = 1800 arc seconds.</p>

<p>To get to 7 arc seconds, you need to go 1800/7 times farther away, so the result is around 250 AU.</p>

CC BY-SA 4.0
2018-10-19T06:17:46.847Z
Using Sun’s radius of 695,000 km, I get a slightly bigger figure. But OP will have to decide whether to use your handy estimate or do the calculations for an exact answer. It’s basic high school maths.
Chappo Hasn't Forgotten
CC BY-SA 4.0
2018-10-19T11:03:42.813Z
@Chappo Right - the goal of my answer was also to show, how can the OP calculate such things in head, in seconds. This is why also Jan Doggen's edit is sub-optimal, although I decided to let it as he did (probably also he didn't understand, the essence of the post also to make it clear, how to calculate it quickly).
peterh
CC BY-SA 4.0
2018-10-18T00:42:24.847Z
2018-10-18T08:55:33.310Z
2018-10-18T08:55:33.310Z
peterh
<p>If an observer travels away from our sun, at what distance would our sun occupy 7 arc seconds of space?</p>

CC BY-SA 4.0
2018-10-18T00:42:50.060Z
Why do you want 7 arc sec?
peterh
CC BY-SA 4.0
2018-10-18T22:35:02.810Z
Pro tip: it is really much better if you first try to do something for yourself, and then visit us with the problem you've found.
peterh
<the-sun><angular-diameter>
From what distance would our sun have an angular diameter of 7 arc seconds?
pol0
# From what distance would our sun have an angular diameter of 7 arc seconds?

(Asked by: username_0 on 2018-10-17)

If an observer travels away from our sun, at what distance would our sun occupy 7 arc seconds of space?

**Tags:** the-sun, angular-diameter

> **Comments:**
> **username_1**: Why do you want 7 arc sec?  
> **username_1**: Pro tip: it is really much better if you first try to do something for yourself, and then visit us with the problem you've found.  

---

## 2 answers

### Answer by username_2 on 2018-10-19. Score: 3

While this might seem to be strictly a math problem, it's really loaded with Astronomy, let's see what we can learn!

If you're far enough away that you can see essentially a full hemisphere (which you can't if you're close), the apparent angular width is twice the half-width, and that's given by

$$\theta\_{HW} = \arcsin\frac{r}{R} \approx \frac{r}{R}$$

for small angles, where $r$ is the physical radius of the object and $R$ is the distance from the viewer to its center. This is also similar to or the same as [@username_1's method](https://en.wikipedia.org/wiki/Moon).

There's (at least) two problems with trying to talk about the radius of the Sun;

1. The Sun is oblate
2. The Sun is diffuse; it doesn't have a well defined edge

Let's do #2 first. There is a formal working definition for the optical edge of the Sun, and [this excellent answer](https://astronomy.stackexchange.com/a/2106/7982) to the question *How do you define the diameter of the Sun* states:

> Most literature will define the diameter of the Sun up to the photosphere, the layer of the solar atmosphere you would see if you were to observe the Sun in white light.
>
> The base of the [Photosphere](https://en.wikipedia.org/wiki/Photosphere) is defined as the region where the [optical depth](https://en.wikipedia.org/wiki/Optical_depth) is around 2/3, or the region where the plasma becomes transparent to most optical light wavelengths.
>
> Of course the true edge of the solar atmosphere could be considered as the heliopause, where the direct influence of the Sun's magnetic field and solar wind end and interstellar space begins.

With that definition of the edge, let's look at the shape of the Sun. [Wikipedia](https://en.wikipedia.org/wiki/Sun) gives both 695,700 and 696,392 km for the equatorial radius of the Sun, sourcing from [IAU 2015 Resolution B3...](https://arxiv.org/abs/1510.07674) and [Measuring the Solar Radius from Space during the 2003 and 2006 Mercury Transits](https://arxiv.org/abs/1203.4898) respectively.

Let's use 695,700 km because it's the one that I see most often and "because IAU".

The Wikipedia article gives a [flattening](https://astronomy.stackexchange.com/a/28069/7982) of 9E-06, which makes the polar radius only about 10 parts per million smaller, which is a *much smaller equator-to-pole difference* than I remembered it to be. I guess we can ignore it after all!

Seen from the Earth then, who's orbit takes it from 152.1 million to 147.1 million km from the Sun, the angular width of the Sun would vary from about 1887 to 1951 arcseconds (31.4 to 32.5 arcminutes).

At what distance would an object with a radius of 695,700 km have an apparent width of 7 arcseconds (which is 3.39E-05 radians)? Flipping the equation around gives 2×695700/3.39E-05 or 4.105E+10 kilometers about or 274 AU.

---

## What is it like 274 AU from the Sun?

In addition to appearing to be roughly 274 times smaller from here than from Earth, the Sun is 274×274 times dimmer. Using $2.5 \times \log\_{10}(274×274)$ we get that the Sun would appear about 12.2 magnitudes dimmer than it's -26.8 magnitude brightness at 1 AU, or -14.6 magnitude, which is still 2 magnitudes brighter than a typical full moon.

Your orbital period around the Sun would be over 4,500 years, and you'd be moving at 1.8 km/sec in that orbit, as opposed to 30 km/sec for the Earth.

You'd be well past the [Kuiper belt](https://en.wikipedia.org/wiki/Kuiper_belt) but nowhere near the [Oort cloud](https://en.wikipedia.org/wiki/List_of_Solar_System_objects_most_distant_from_the_Sun_in_2018) (this is a simplification, but it will do for now), and you'd be way past the [farthest known solar system body](https://en.wikipedia.org/wiki/List_of_Solar_System_objects_most_distant_from_the_Sun_in_2018) in 2018! [V774104](https://en.wikipedia.org/wiki/V774104).

[![V774104](https://i.sstatic.net/ZmQGg.gif)](https://i.sstatic.net/ZmQGg.gif)

**above:** from EarthSky.org's [New most distant object in solar system](http://earthsky.org/space/new-most-distant-object-in-solar-system-v774104) Image via S. Sheppard / C. Trujillo / D. Tholen / Subaru Telescope/ [skyandtelescope.com](http://www.skyandtelescope.com/astronomy-news/v774104-most-distant-solar-system-object-11212015).

And at 274 AU, you'd be much farther from the Sun than any deep space probe as well. Voyagers 1 and 2 are "only" 118 and 142 AU from the Sun now, and New Horizons is only about 42 AU.

I don't know the name of the place you'd be, it looks pretty lonely though!

[![enter image description here](https://i.sstatic.net/Rg9nC.jpg)](https://i.sstatic.net/Rg9nC.jpg)

[Source](https://commons.wikimedia.org/wiki/File:PIA17046_-_Voyager_1_Goes_Interstellar.jpg)

> **Comments:**
> **username_3**: I wish you could favorite answers, I love the picture there, that's an amazing representation of the Oort cloud. I never really pictured how "thick" it could be. Is that to scale? The Oort cloud is from 1,000AU to 100,000AU?  
> **username_2**: @username_3 this is a schematic/cartoon representation only, and the distance scale is logarithmic. Solar system structures this far away (especially the Oort cloud) are inferred and predicted, not seen or measured. It's educated speculation.  
> **username_3**: fair point, the source does confirm that the estimates are from 1,000AU to 100,000 AU though.  
> **username_0**: What are your thoughts on this proof, that falsifies Edmond Halley's proposition from 1720 that stars appear larger than they are because of an "optical illusion"? https://www.youtube.com/watch?v=ivhXqDnrV4o  
> **username_2**: @username_0 consider posting that as a new question? It's not directly related to this answer and so comments here are not the right place to open up a new discussion. Thanks!  

---

### Answer by username_1 on 2018-10-18. Score: 3

From us, the Sun is visible around a half grad, which is $\approx$ 30 arc min = 1800 arc seconds.

To get to 7 arc seconds, you need to go 1800/7 times farther away, so the result is around 250 AU.

> **Comments:**
> **username_4**: Using Sun’s radius of 695,000 km, I get a slightly bigger figure. But OP will have to decide whether to use your handy estimate or do the calculations for an exact answer. It’s basic high school maths.  
> **username_1**: @username_4 Right - the goal of my answer was also to show, how can the OP calculate such things in head, in seconds. This is why also Jan Doggen's edit is sub-optimal, although I decided to let it as he did (probably also he didn't understand, the essence of the post also to make it clear, how to calculate it quickly).  

---

