From what distance would our sun have an angular diameter of 7 arc seconds?

Question

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

Answer 1

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.

Answer 2

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.

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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.

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!

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