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eburacum45
2004-Jun-11, 09:21 AM
Okjay, I've done a bit of Googling, and I have come up with a wide range of answers;


but this is the question;
what is the population density of the galaxy in stars per cubic light year, particularly in the region of the Sun?

Including the dimmest of red dwarfs, if possible - but perhaps figures won't be available for those objects, seeing as we are still discovering them relatively near Sol.

Manchurian Taikonaut
2004-Jun-11, 03:24 PM
as far as I'm aware we are still not sure about the density, we find new suns every day in the sky. Radius of the sun = 70000 km ? Number of Suns in Andromeda 3* 10^"..andromeda is very like our galaxy. Think about the stars near us
R = 10 light years radius.
checking shows that there are in that volume roughly 37 stars in this radius = R
Do you know how many stars there are in a distance 10,000 light yrs from out Sun?


we have the Galactic Bulge radius of 6,000 light years
Composed of old stars (population II) and young stars it is also thought there are many black hole stars in the middle. Many astronomers believe the mass equivalent to about 3,000,000 suns has collapsed to the center of the MW.
Then as matter continues to be pulled in, large amounts of radiation will be emitted. Mass of our galaxy in M0 = 2*10^11 ? ( not unlike Andromeda 3 * 10^" )


The Galactic Disk a radius of 60,000 light years
old stars (population II) and young stars (population I), gas, and dust spinning. Radius of the disc of our galaxy = 60000 ly

Average Thickness of the disc = 2000 ly. Then the Galactic Halo
Radius of 65,000 light years
Includes globular clusters

and finally Dark Halo
Radius of 200,000 or 300,000 light years
Contains 80%-95% of the mass of the galaxy (dark matter)
What is this Dark matter?

Volgal = volume of our galaxy = pi * rgal * rgal * dgal

densgal = Mgal / Volgal Calculate the number of stars in this volume


number of stars in this volume = nsun1 = vol1 * densgal

this is how you could answer this problem of knowing the stars and how many are in the density of our galaxy

Ricimer
2004-Jun-11, 06:09 PM
around the sun? I think its ~1 star every 4 light years (that's the radius of a sphere)

eburacum45
2004-Jun-11, 06:59 PM
Grant Hutchinson on the Celestia board has given me a very good answer;
0.004 stars per cubic light year.
from
Allan's Astrophysical Quantities apparently. That is about what I expected.

Brady Yoon
2004-Jun-12, 06:40 PM
I think a cubic light year is too small a space when finding stellar densities in places like the sun. It seems like a huge area, but it's really not.

Ricimer
2004-Jun-12, 09:00 PM
I found an interesting way of looking at things.

The stellar "Mass surface densitity" of the area round the sun is 50 solar masses per square parsec.

Thats right, square parsec.

The idea is if you look down on the galactic disk from above, and section off a square parsec of space, at a distance from the core ~where the sun is..you'll get 50 solar masses worth of stars.

5 solar masses of HI and 1.3 solar masses of H2.

It's like a column density if you're familiar with that concept.

stu
2004-Jun-13, 06:13 AM
Grant Hutchinson on the Celestia board has given me a very good answer;
0.004 stars per cubic light year.
from
Allan's Astrophysical Quantities apparently. That is about what I expected.


I actually had to calculate this as part of an astronomy homework set the Spring. I used the Hipparcos data set out to a radius of 10 pc so that I could be fairly certain of a complete mass function. I found the stellar density to be approximately 0.043 stars per cubic pc.

[Edit] Note that this is just within the local neighborhood, and that stellar density varries greatly in different areas of the Galaxy.

Tobin Dax
2004-Jun-13, 07:35 AM
I found an interesting way of looking at things.

The stellar "Mass surface densitity" of the area round the sun is 50 solar masses per square parsec.

Thats right, square parsec.

The idea is if you look down on the galactic disk from above, and section off a square parsec of space, at a distance from the core ~where the sun is..you'll get 50 solar masses worth of stars.

5 solar masses of HI and 1.3 solar masses of H2.

It's like a column density if you're familiar with that concept.

So, if we assume that the disc is about 500 parsecs thick (reasonable, plus nice and easy to work with) and the average stellar mass in the local region is one solar mass (see previous comment), then that gives us an average density of 0.1 stars per cubic parsec, or about 0.029 stars per cubic light year. This is pretty close to stu's answer (to within astronomical accuracy, ansyway ;) ).

Ricimer
2004-Jun-14, 02:51 AM
::Nod:: It should add up.

It's from the first 3 chapters of a book my current supervisor is writing. He specailizes in High Latitude Molecular clouds, so galaxy wide star, dust, and gas distributions are his schtick.

eburacum45
2004-Jun-15, 06:41 PM
Cubic light years per cubic parsec (http://www.google.com/search?hl=en&ie=UTF-8&q=cubic+light+years+per+cubic+parsec&btnG=Google +Search) =34.6981707

Stu's answer = 0.043 stars per cubic parsec

= 0.001239 per cubic light year.

I understand the Solar system is in a slightly less dense part of the local neighbourhood, which might account for the differences.

Ricimer
2004-Jun-15, 06:47 PM
Disclaimers for the "surface density" I pointed out:

1) It doesn't account for star size. You could have several small, or a single big, since all it does is look at mass.

2) Specific number density requires one to know the depth of the disk, and distribution within the disk (thining near the edge increases density near the center for example) since it isn't uniform.

But it is a neat alternative way to look at densities.

Tobin Dax
2004-Jun-15, 07:37 PM
In addition to ricimer's disclaimers, my calculation was just back-of-the-envelope. It was a rough calculation, with a few assumptions in it, and it is more than twice stu's answer (now that I look at it again). That's a barely acceptable agreement, IMO, and it's probably mostly from my assumptions. However, it does help to show that there is a wide range of acceptable answers to this quetion, since there are a lot of details about our galaxy that we don't know.

Ricimer
2004-Jun-16, 02:57 PM
actually, in the astronomical community, with the conditions it faces data wise, a factor of two is pretty acceptable, especially for "back of the envelope" calculations.

It's when you start hitting orders of magnitude that you need to scratch your head more.

stu
2004-Jun-16, 11:03 PM
In addition to ricimer's disclaimers, my calculation was just back-of-the-envelope. It was a rough calculation, with a few assumptions in it, and it is more than twice stu's answer (now that I look at it again). That's a barely acceptable agreement, IMO, and it's probably mostly from my assumptions. However, it does help to show that there is a wide range of acceptable answers to this quetion, since there are a lot of details about our galaxy that we don't know.

I'm not really arguing for my calculations, but they do agree with a few websites that I'd found when I initially did this. Also, keep in mind that my calculation was within the VERY local neighborhood (what exactly is considered the "local neighborhood?"), and only involved 182 stars.