31
$\begingroup$

Here's an intriguing thought - I think so anyway. Let's suppose for a minute that there's an Earth-like planet with sentient life orbiting a star near the edge of the Sagittarius Dwarf Spheroidal Galaxy. Specifically, it would need to be on the side nearest the Milky Way.

When the inhabitants of this world look up at the sky at night, if they're looking in the direction of their own galaxy, I'd assume that they'd see a similar view to what we do (with the naked eye) - a milky, light shape in the sky.

But if they were facing the Milky Way, would they see one huge, well-defined spiral galaxy sitting in the sky?

I've chosen this galaxy specifically because I suspect it might be close enough (~50,000 light years from the core of the Milky Way, with the Milky Way ~100+ light years in diameter) that this is what you would see. What I'm not sure of is if this would really be the case in the real world (hence the hard-science tag)

For the purposes of this question, assume that the Sagittarius Dwarf Spheroidal Galaxy is at a point in its orbit where it is passing directly over the Milky Way's galactic pole.

edit: I've changed the tag from "hard science" to "science based" after having looked at the descriptions of these two tags

$\endgroup$
  • 4
    $\begingroup$ Absolutely brilliant question. I regularly have "what would this look like questions" but never thought about this one. Thanks. $\endgroup$ – Draco18s May 3 '17 at 13:40
  • 1
    $\begingroup$ This came into my memory i.stack.imgur.com/jA0H0.jpg $\endgroup$ – The One May 3 '17 at 16:22
  • $\begingroup$ Can you just go there and look in Celestia ? $\endgroup$ – JDługosz May 5 '17 at 7:05
  • $\begingroup$ @JDługosz strange you should mention - I actually installed it and tried this last night. It looks like the Milky Way would indeed fill up a massive amount of the night sky (not surprising given the answer below), but it's a bit unclear how bright it would look in from a planet's surface $\endgroup$ – danl May 5 '17 at 7:33
31
$\begingroup$

Solution by simple trigonometry

The Sagittarius Dwarf Spheroidal Galaxy (Sgr dSph) is about 10,000 ly in diameter, but has an apparent arc of 450' at a distance of 70,000 ly away from us. This can be solved simply with trigonometry; the tangent of a right triangle is the opposite leg over the adjacent. In this case, the opposite leg is 10,000 ly, while the adjacent is 70,000 ly. We find that the inverse tangent of this ratio $$\arctan\left(\frac{10000}{70000}\right) = 0.1433 \text{ radians} = 488 \text{ minutes}$$ is just about what we are given as a visible arc, especially given the uncertainty in the actual diameter of Sgr dSph. So this is a reasonable way to approximate the visible arc of the Milky Way.

From Sagittarius, the Milky Way (which is 160,000 ly in diameter) is about 50,000 ly away. This gives us $$\arctan\left(\frac{160000}{50000}\right) = 1.27 \text{ radians} = 73 \text{ degrees}. $$

So the Milky way would be perfectly sized to fill the sky of your world, its diameter extended almost halfway across a clear night sky. It is small enough that the whole thing could be visible at once, but large enough to fill a (human) field of vision on a dark night.

The reason this is a simple solution is that I am not making calculations about brightness. Apparent magnitude is easy to calculate for a point source, but not at all for an object smeared across half the sky. There would also be important factors such as how close the planet in question would be to the edge of Sgr dSph. If the planet was on the far side of the dwarf galaxy, obviously the Milky Way would be largely obscured; if it was on the near side, there may be almost nothing between the Milky Way and the viewer.

I will add one simple note on brightness. The Andromeda Galaxy (twice as massive a the Milky Way) has an apparent magnitude of 3.44 at a distance of 2540000 ly. If Andromeda were at 50000 ly, we would expect its apparent magnitude to increase by the square of the ratio of distances, or a factor of 2580. This is good for an apparent magnitude of -5; a little brighter than Venus at its brightest. We don't really know the Milky Way's absolute magnitude, since we can't look at it, but it we assume that it is less than half as bright as Andromeda, we still get an apparent magnitude of -4 from Sgr dSph. So the Milky Way as a whole will definitely be bright, and visible, but how exactly it would look is complicated.

$\endgroup$
  • 1
    $\begingroup$ Thanks for the hard-science answer! I took the hard-sci tag off as I wasn't sure if this would be a bit offputting - also wasn't sure if my question fit the tag. (I've not actually asked a question on here before so was a bit unfamiliar with the tag details) This is a great answer regardless though - I was thinking near-side by the way, and very close to the edge, I assumed the Milky Way would otherwise be obscured. $\endgroup$ – danl May 3 '17 at 12:47
  • $\begingroup$ @danl Hard science is usually the kind of thing where you want people to go look up scientific papers, so not really necessary for this question, I don't think. Great question though! Welcome to Worldbuilding. $\endgroup$ – kingledion May 3 '17 at 12:49
  • 3
    $\begingroup$ Good answer. I got a slightly different value for the first calculation using the same formula, although it comes out to a difference of less than one arcminute, so it's insignificant. If you use the formula for angular diameter, you get something slightly larger. $\endgroup$ – HDE 226868 May 3 '17 at 13:38
  • 4
    $\begingroup$ I think the brightness question is paramount, though. Andromeda is close enough that if you could see the whole thing, it would be significantly larger than our moon; but all we can see with the naked eye is the densest part of its core, which looks like a fuzzy star: waitbutwhy.com/2014/06/andromeda-brighter-youd-see.html $\endgroup$ – yshavit May 3 '17 at 15:01
  • 2
    $\begingroup$ @yshavit Yeah, my concern would be that maybe the view of the Milky Way from this planet would be such that people would only discern the core and wisps about it, and it might not actually look like a spiral galaxy at all due to certain parts of it being too dim. $\endgroup$ – Michael May 3 '17 at 17:29
0
$\begingroup$

So a solar arm or the bulge of the milky way would be about the same area as a 10 cm by 10 cm shape held at a distance of 1 meter. (Bulge is about 2 kpc in diameters; spiral arm

What would you see?

There are about 10 to 20 billion stars in an arm of the Milky Way. The core has about 100 billion stars.

The Orion–Cygnus Arm is roughly the same area. The other arms are a few times larger.

The entire galaxy is about as big as a 4x3 meter object held at a distance of 1 meter; quite large.

The bulge has 5E9 solar lumosities, ignoring dust blocking the view. At 50,000 ly this is 5E-10 times as bright as the Sun is from Earth, or 0.02% as bright as the Moon, and a larger cross-section.

The Moon is about as large as a 4 mm by 4 mm object held at a distance of 1 meter. The galactic Bulge is is 600 times larger, and 5000 times dimmer, making it 3 million times less bright per unit area. And that is before accounting for dust.

That would make the galactic core as bright per unit area as a magnitude ~4 star the size of the moon.

Spiral arms are 1x-6x larger and each have about 5 times fewer stars as the Bulge does, so they are up to 100 times less bright per unit area than the core (again, ignoring dust). That makes the arms as bright per unit area as a magnitude 9 stars the size of the moon (but much larger).

You'd be able to see the galactic bulge, and maybe some wisps around it (especially if you blocked it out), but I suspect it would be less impressive than our milky way view. Which makes sense, as what we see contains stars that are much closer than 50,000 ly.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.