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3

You can absolutely see it in rural areas. I have a gorgeous view of it on clear nights, living thirty miles west of Lincoln.


4

The stars in the Galactic bulge are predominantly metal-rich (by that I mean have a metallicity similar to the Sun or even a little higher). Even though these stars are predominantly old, the bulge is thought to have formed extremely quickly and the interstellar medium from which the stars were formed would have been enriched with metals very quickly. Here ...


4

Just to clarify a concept here: there are no pictures of the Milky Way other than the ones taken from within the Solar system, and they all look somewhat like this: Everything else you see are artists' depictions of the Milky Way based on the latest research. This is the currently most accepted shape: i.e.: a spiral galaxy with four main arms. There ...


1

The ones in the halo have blue colors, which is the "old" part of the spectrum. Each type of metallicity is usually measured using a form of logarithmic scale, since the flat percentage of metals (especially iron peak elements) of any star will be very small. In this case, we have $$[Fe/H] = \log(N_{\text{Fe}}/N_H)_{\text{star}} - ...


0

Red is higher blue lower metalicity, so the halo has low metalicity and the disk higer.


0

In a simpler way than the preceding answers, think probability. The probability of a galaxy colliding with another is small, but it affecting a single star greatly is very tiny, especially 3/4 from the center. The most affect we would get is a wonderful show... https://www.youtube.com/watch?v=qnYCpQyRp-4 The stars in our (and all other) galaxies are ...


0

I've seen some articles, which I've referenced in a recent answer of mine concerning this collision, which make a claim to the tune of "the Earth and Sun will probably not be significantly affected" (other than that the Sun is dead by this point, and Earth possibly consumed by it). This was based on N-body simulations of the merger, themselves based in part ...


0

I've always found that somewhat strange when they say no stars will hit out of billions, but I also trust the scientists that they know their stuff. Lets look closer at your numbers. If the Oort Cloud has a diameter of 2 light years (I think it's a bit less than that, but numbers vary), that's a radius of 1 light year, and the Solar System to the Oort Cloud ...


3

Outside my field, but collecting information from the Wiki article, I can make an attempt nevertheless: You're right that we will probably pass through Oort-like clouds of other stars. But the largest of the planetesimals of the Oort Cloud is measured in kilometres, whereas the distance between these objects is measured in AU, so the total cross section of ...


-2

If you look at a recent picture of our galaxy (top view), then you'll be able to see that the stars are distributed in not so much a spiral in the middle, but in the outer arms. The middle is shaped like a bar, which then extends into the two spiraling outer arms. It is because of this particular arrangement that we say that it is a barred spiral galaxy.


1

It may make more sense to convert the exponential into the product of the individual exponential terms: $$ \exp(-\frac{R}{L}) \exp(-\frac{Z}{H}) \exp(\frac{Z_{\odot}}{H}) $$ This makes it clearer that the $Z_{\odot}$ term is a constant, which is just part of the overall normalization (similar to the $\exp(R_{\odot}/L)$ term earlier in the original equation). ...



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