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

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    $\begingroup$ Why do you want 7 arc sec? $\endgroup$ – user259412 Oct 18 '18 at 0:42
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    $\begingroup$ 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. $\endgroup$ – user259412 Oct 18 '18 at 22:35
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    $\begingroup$ @RobJeffries Please don't do that. There were already answers that now no longer match the question. $\endgroup$ – Jan Doggen Oct 25 '18 at 8:52
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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.

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    $\begingroup$ 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. $\endgroup$ – Chappo Oct 19 '18 at 6:17
  • $\begingroup$ @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). $\endgroup$ – user259412 Oct 19 '18 at 11:03
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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 @peterh's method.

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 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 is defined as the region where the 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 gives both 695,700 and 696,392 km for the equatorial radius of the Sun, sourcing from IAU 2015 Resolution B3... and Measuring the Solar Radius from Space during the 2003 and 2006 Mercury Transits 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 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 but nowhere near the Oort cloud (this is a simplification, but it will do for now), and you'd be way past the farthest known solar system body in 2018! V774104.

V774104

above: from EarthSky.org's New most distant object in solar system Image via S. Sheppard / C. Trujillo / D. Tholen / Subaru Telescope/ skyandtelescope.com.

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

Source

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    $\begingroup$ 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? $\endgroup$ – Magic Octopus Urn Oct 19 '18 at 13:02
  • $\begingroup$ @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. $\endgroup$ – uhoh Oct 19 '18 at 13:20
  • $\begingroup$ fair point, the source does confirm that the estimates are from 1,000AU to 100,000 AU though. $\endgroup$ – Magic Octopus Urn Oct 19 '18 at 13:40
  • $\begingroup$ 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"? youtube.com/watch?v=ivhXqDnrV4o $\endgroup$ – pol0 Oct 22 '18 at 21:36
  • $\begingroup$ @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! $\endgroup$ – uhoh Oct 23 '18 at 0:57

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