The point of the question is that gravitationally bound objects, (the atoms in the planet earth, or the cloud of gas that collapses to form a star, or the stars in a galaxy) behave as if the net force of gravity is coming from the center of mass.
Any galaxy, even a hypothetical galaxy with uniform density, will have a center of mass, and there will be an attraction to it. The condition of uniform density will be unstable.
Total nonsense. Are you just trying to be obstinate? If you are somewhere in the galaxy, a star somewhere on the galactic arm, there is stuff on the side of the arm towards the centre of mass of the entire galaxy that pulls you towards the centre of the galaxy. The amount of pull is determined by how much stuff there is on that side of the arm and the centre of mass of the stuff on that side of the arm which is in a different place than the centre of mass of the galaxy. That is straightforward. There is also stuff on the arm which is away from the centre of mass of the galaxy on the other side of you, going out from the galaxy on the spiral arm. That section of the galactic arm also has its own centre of mass. That section of the arm pulls you away from the centre. The two forces combine and since there is more stuff towards the centre, you are pulled towards the centre. Both forces are attractive forces, one towards the centre of mass towards the centre of the galaxy and one towards the centre of mass of stuff away from the galaxy. Add both forces and factor out the 1/r^2. The grand total of mass both fore and aft which you subtract from each other is proportional to r. Combining we get a force proportional to 1/r which is still gravity and is still an attractive force towards the centre of the galaxy. The force must equal m(v^2)/r. The r's cancel, the m's cancel and the resultant orbital velocity is equal to a constant. I believe this is the fourth time I have explained this to you. This is simple high school physics. The forces are gravitational through mutual gravitational attraction and the force acting on each member of the galaxy is proportional to 1/r in a linear orientation of matter having constant density. r is the distance to the centre of mass of the galaxy. The r's cancel according to the force required to keep the member substances of the galaxy in circular orbit about the centre. The resultant force is constantly directed towards the centre. The forces are gravitational and are attractive forces. Please don't say silly things like forces outside the galaxy which have not been mentioned. Also, please read the paper I have posted before making any more silly comments. Furthermore, you have stated many times that a uniform density will be unstable. It will not. That is because all the stars orbit at the same rotational velocity. If you wish anyone to believe you in saying the condition is unstable then provide the mathematical steps to reach your conclusion because all you have done so far is wave your arms. Your concluding statement "The condition of uniform density will be unstable." is hogwash. Prove it.
Total nonsense. Are you just trying to be obstinate? Your concluding statement "The condition of uniform density will be unstable." is hogwash. Prove it.
Heh. Actually the reason we know there is no dark matter is because, if you look veeeery closely at veeeeery good digital photographs of NGC 3198 very, very closely and squint real hard, you can distinctly see the Flying Spaghetti Monster and its tendrils reaching out to all the stars in the galaxy helping them in their orbits. There is no dark matter; it's the Flying Spaghetti Monster!
Ha! An exceptionally good discussion. Thank you.
Great stuff! Here's my Patzer's elementary math question (you will have a lot of fun with this) 
From your paper: "Also, with the distance and rotational velocity measurements reported here and an angular size of 8.3 arc minutes, or 126,000 ly, we calculate a linear density of 8.7× 1020 kg/m giving a mass of 1.04× 1042 kg or 5.22× 1011 solar masses and angular momentum of 2.5 × 1047 J-s."
At a distance of 7.2 million parsecs, or 23.5 million LY, wouldn't 8.3 arc minutes correspond to about 57,000 ly? Wouldn't you need about 18.5 arc minutes for 126,000 ly?
Interestingly, +/- 18 arc minutes is the figure usually given in catalogs for this galaxy, although in most photos in moderate instruments it appears much smaller. I assume that's a combination of the oblique angle of the galaxy and the fact that the outer regions are just too faint to register on most instruments.
You are correct. I rechecked everything and found I had entered 8.3 arc minutes instead of the value I had measured which is 18.6 arc minutes. There is a local dust cloud obscuring the view of a section of the galaxy. I honked up the magnitude, measured half and doubled it. I got 18.6 arc minutes. I did calculations and all was well. I wrote paper and put in the correct calculations and measurements except for the arc length of the galaxy, which I mistakenly entered as 8.3 rather than 18.6.
It is very difficult to do this type of work without peer review. I am very grateful for you finding this glitch. I believe the other numbers are correct. Thank you.
ps. a corrected paper has been submitted.
Just to let you know, an updated paper with the typo corrected is now in place. Thanks again and a major tip of the hat to Patzer. One more paper to go.
Actually it's finished but I have to wait for the rejection from Physical Review. It usually only takes a couple of days and then I can post to Vixra. I send it in to get peer review but so far no editor has sent it on and just rejects it out of hand. Therefore, the meticulous criticism you have given me and the wonderful discussion that we had, which makes the model easier to understand, is invaluable to say the least. I am very appreciative of any and all criticism and review.
That didn't take long. Here is the third and final paper of the series:
http://vixra.org/abs/0911.0023
Have fun and looking forward to your comments.
I meant to ask this earlier -- How do you derive the masses of the galaxies?
linear density times length. Forget the spiral. Think stick. Mass is a scalar and independent of reference frame.
Sorry. I guess I should have asked, How do you derive the linear density?
I do recall a series of equations in the first paper that wound up with a formula M(r)=kr for the equivalent center of mass, but I never quit got how you convert that to linear density, thence to total mass.
Apologies, but you have to go slow on this part 
Hey, no problem. It's the gravitational n-body relativistic problem. You asked me to do it as a student home-work assignment. You now have to mark it. Heh. The linear density is (v^2)/G. \rho * r = M(r). Then \rho = (v^2)/G. Scary.
Let me see if I can get this to work...
in cgs V^2 is in cm/s ?
G is 
?
Are those units correct?
If you are a star in the galaxy, the stuff farther out pulls you away from the centre, the stuff farther in, pulls you towards the centre. Simple physics.
And what pulls you away if you are the outer star? Are you proposing an infinite galaxy?