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Something like 9,100 stars are visible with the naked eye across the entire planet. Each appears in a particular constellation and shines with a specific magnitude, unless it’s a variable star like Algol, in which case its brightness varies. From each one of these stars we might imagine looking back towards the Sun, which shines in each of their skies as surely as they do in ours.

Depending on that star’s distance from us the Sun may appear bright or faint. From the neighboring star system Alpha Centauri, 4.4 light-years distant, it shines at magnitude +0.5, nearly as bright as Procyon in Canis Minor. But from far-flung Deneb at about 2,600 light-years you’d need dark skies and an 8-inch telescope to spot the 14.3-magnitude speck.

Winter can be confining. Maybe that’s what inspired my desire to figuratively leave the solar system and see it from other points of view. This got me wondering how bright the Sun would be and in what constellation it would appear when viewed from a planet — imaginary or real — orbiting one or more of the familiar bright stars?

## Magnitude mastery

First, we’ll need to calculate the Sun’s apparent magnitude from a specific distance. For that you’ll use the formula for determining a star’s absolute magnitude but solve for *apparent magnitude* instead. Here’s one iteration of this gem of mathematical magic:

**m-M = 5 log(d/10)**

We’ll have to plug in *M* = absolute magnitude of the Sun and *d* = distance to a star of our choosing in parsecs. Once we know these values, we can do the math to find *m*, the Sun’s apparent magnitude at the distance of that selected star.

The Sun’s absolute magnitude, *M*, or how bright it would appear at a standard distance of 10 parsecs (32.6 light-years), is the easy one to plug in since we already know it. Absolute magnitudes provide a standard reference for directly comparing stars’ true luminosities. The Sun has an absolute magnitude of 4.8, a considerable step down from its apparent brightness of 26.7!

## Please pass the parsecs

It’s easy to look up a star’s distance (*d*), but depending on your source, you might find it quoted in light-years — and so you’ll have to convert to parsecs. A parsec is equal to 3.26 light-years, which is the distance at which the radius of Earth’s orbit subtends an angle of one arc-second. A star located one parsec away would have a parallax of one arcsecond. Proxima Centauri, the closest star to the Sun, has the greatest parallax, equal to 0.77 arcsecond.

Many online converters will help you make parsecs out of light-years. I found this one straightforward, mostly ad-free, and with a host of fascinating options, including a conversion from light-years to *cubits*! (If you’ve never heard of a cubit, it’s an ancient unit of measure equaling the length of the forearm from the elbow to the tip of the middle finger . . . it’s derived from the Latin for “elbow,” in fact.)

Good one-stop shops for stellar distances include the Abrams Planetarium list of the 100 brightest stars or any recent volume of the Royal Astronomical Society of Canada’s Observer’s Handbook.

Now you have everything you need to solve for the Sun’s apparent brightness from your star of choice. Copy and paste the equation into an equation solver (OK, I’m lazy), put in the values for *M* and *d* (remember, it’s parsecs only), and out comes *m*. So much fun. I use MathPapa, but there are many others.

## Sunshine at Arcturus

Here are some examples of the Sun’s apparent magnitude as seen from some familiar stars:

- Alpha Centauri = +0.5
- Sirius = 1.9
- Procyon = 2.5
- Altair = 3.4
- Vega = 4.2
- Pollux = 4.9
- Arcturus = 5.1
- Capella = 5.4
- Aldebaran = 6.3
- Regulus = 6.7
- Spica = 9.3
- Antares = 11.0

Once we know the Sun’s apparent brightness, we’ll next want to recreate its appearance from an alien star system’s sky. For that we’ll need two things: its celestial coordinates as seen from the star *and* the positions of familiar stars and constellations from the point of view of an observer at that location.

Let’s choose Sirius, located at R.A. 6^{h} 45^{m}, Dec. –16° 43′. To find the point in the sky directly opposite Sirius — where you’d see the Sun from that star — just add 12 hours of R.A. and reverse the declination sign. That puts the Sun (now 2nd magnitude) at R.A. 18^{h} 45^{m}, Dec. +16° 43′ in southeastern Hercules. If adding 12 hours to the R.A. shoots it past 24 hours, subtract 24 from that number for the correct right ascension.

## A crushed teapot?

Although stars appear pasted on a flat sky, space is obviously three-dimensional — 4D really, but three will do for our purposes. When you jump from one star to another, its neighbors also shift around. This can seriously alter a constellation’s outline compared to the view from Earth. The farther you travel, the more bent out of shape the constellations become. To prevent warp-overload I chose only relatively close stars. Once you get beyond 50 light-years or so, depending on the region of the sky, star patterns and their magnitudes quickly become unfamiliar. In fact, I was amazed at how malleable the patterns were given the relatively modest distances I “traveled.”

For my mode of transportation I chose the free space simulator Celestia, which is available for Windows, Linux, macOS, iOS and Android. I had to play around with it a bit and add a few of my own enhancements, but it does a great job plotting the Sun’s location from when you’re far at home. Be aware that its range of “dot” sizes to depict stars is limited, so I tried my best to represent the Sun and other key stars as you’d see them visually. Please take these representations with a grain of interstellar dust.

The jaunts were not only fun and instructive, but if there was any Sun-centered, lingering Copernicanism left in my veins when I began this journey, I arrived home more unmoored than ever. Not that that’s a bad thing. Picturing the Sun’s place in the firmaments of its fellow suns opened up my mind to what it means to be a star among many.

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