Brightness of the Local Group galaxies

My tickets to Andromeda Way Station III were confirmed last night; I leave in 2 days. After several days of superduperhyperluminal travel I will be reaching Andromeda’s galactic halo.

Now, how bright will Andromeda appear to my unaided eye? With 1 trillion stars, will Andromeda, seen face-on from within its own halo, outshine the Sun as seen on earth?

What about other local group galaxies, such as Triangulum, LMC, SMC etc? I might visit them before returning.

I need to know to pack my eye gear.

Wouldn’t it be about the same as the Milky Way viewed from earth (i.e. not even close to as bright as the sun)? The two galaxies are roughly comparable in most respects.

Don’t think so - as we enter the galactic plane, most of the stars in the galaxy become hidden away, obscured by nebulas, dust clouds and such. But with no obstructions from the galactic halo (which contains very little or no dust), viewing the galaxy face-on, with 1 trillion stars, a massive spiral like Andromeda should look stunningly impressive, nothing like our night sky.

Question is, how bright?

I think TimeWinder is right on this one. While it’s true that much is obscured by dust etc when viewing our galaxy edge-wise from the inside, we’re also looking through many tens of thousands of light-years worth of stars. Looking at the Andromeda Galaxy from above, you’d only be looking through a plane of stars on the order of a few thousand light-years thick, which is similar to what we look through when we peer “up” or “down” perpendicular to the plane of our own galaxy (at least it’s on the same order of magnitude). Assuming you weren’t so close that the center of the galaxy filled your field of view (in which case you’d practically be in the galaxy) I think it would look more like a really bright milky way rather than a brilliant fireworks display.

The Andromeda Galaxy is actually pretty large viewed from our vantage point here on Earth; at 4-5 degrees wide, it’s about 10 times wider than a full moon. Like so many objects visible with a small telescope or binoculars, it’s not particularly small so much as it’s just rather faint. Here’s a picture that illustrates this by superimposing an image of the moon onto an appropriately scaled image of the Andromeda Galaxy.

I will be hovering directly above the galactic center, about 80,000 light years away. The total luminosity of about 1 trillion stars, including the immense radiation emanating from the core, will not amount to 1 solar brightness?

I am talking about visual brightness, not integrated luminosity across all wavelengths.

Back of the napkin:

1 solar brightness is observed at 1 AU, or 1.6 x 10^-5 light years of an averageish star. Brightness decreases with the square of distance. At 810^5 light years, the distance will be 210^11 AU, and the brightness will therefore be (210^11)^-2 = 410^-22. Multiply that by one trillion stars (1x10^12) and the approximate brightness of Andromeda will be 4*10^-10 of the sun.

The difference in apparent magnitude between the sun and Andromeda here is 25. The sun has magnitude -26, Andromeda would be -1. From our vantage point on Earth, Andromeda is 3.4, and by comparison the new moon is -2.5. Should be naked-eye visible, but not spectacularly bright.

The galactic core vastly outshines the disk, even from our current perspective looking at the disk on angle. The core is already a very large object in the sky but is pretty faint. I suspect the core would be bright enough to make the disk insignificant to the naked eye in most situations.

Andromeda has a very bright core, and while not a quasar nucleus, the core should be far brighter than the disk of stars, as **FluffyBob **points out. How does that change the visual brightness from 80K LY away? Will my eyes water? Time to wear my ray-ban?

Nope, not even then. If it was concentrated into a single point it would be comparable to the brightest stars or the less bright planets. Even the core will be somewhat spread out, I’d guess with an angular size at least as big as the moon, thus the comparison to the magnitude of the new moon.

With the inverse-square law and the immense distances you’re talking about, you’ll never see anything spectacular with the naked eye at that distance short of a (super)nova.

Think of it this way: all of the naked-eye stars in our sky are within a couple thousand light-years. Beyond that average stars (or even stars millions of times “above average”) don’t appear as points of light, but dense conglomerations of them can appear as diffuse light sources.

Keep in mind that the eye-popping pictures of Andromeda you see are the result of long exposures through telescopes with apertures that are enormous compared to the diameter of your pupil. In addition, we’re not really that far from Andromeda; 2,500,000 ly is only about 15 diameters of Andromeda. At 80,000 ly you’d be about 31 times closer than we are now, so the galactic center (which is all we can see with the naked eye here on Earth anyway) would appear 31^2 or a little less than a thousand times brighter than normal (although this brightness would also be spread out over a larger area).

The human pupil has an area of about 40 mm^2, so looking through a telescope with a diameter of 1.1 m at a magnification of 25x (25401000=1,000,000mm^2, the area of a circle approx 1.1 m in diameter) would give you a pretty good approximation of what you’ll see on your trip. University observatories with scopes that size frequently have public viewing nights where you could get an idea of what’s in store for you. You likely won’t even be able to see it in color.

And it’s getting bigger! :eek: