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utesfan100
2012-Aug-30, 05:53 AM
The mass distribution of the Milky Way is given in http://arxiv.org/pdf/astro-ph/9612059v2.pdf.

As a simplification, consider that 1/4 of the mass of the Milky Way is in a spherically symmetric central region with a radius of 10,000 LY, and 3/4 of the mass is in a uniform disk-like region with a radius of 40,000 LY. A factor or 17/16 by making the disk occupy the entire region is not significant to this order of magnitude estimate of force.

For R<<10,000 most of the mass is due to the central mass. Only M=\frac{1}{4}\left(\frac{r}{10,000}\right)^3 impacts the dynamic. This gives us \frac{GM}{r^2}=\frac{G r}{4\times10^{12}}=\frac{v^2}{r}, using the equation of circular motion. This gives us that velocity is proportional to radius within the spherically symmetric mass, which is observed in galactic rotation curves.

For R>10,000, the internal mass is M=\frac{1}{4}+\frac{3}{4}\left(\frac{r}{40,000}\ri ght)^2.
This gives us \frac{G}{4r^2}+\frac{3G}{64\times10^8}=v^2/r. Finally we get the velocity distribution v^2=\frac{G}{4r}+\frac{3Gr}{64\times10^8}

Taking the derivative relative to r gives us 2V\frac{dV}{dr}=\frac{-G}{4r^2}+\frac{3G}{64\times10^8}, which is 0 at roughly 20,000 LY. Thus for the region 20,000<r<40,000 we should expect the orbital velocity to be increasing as we move further out. This produces a profile very similar to that shown at wikipedia.

http://en.wikipedia.org/wiki/Milky_Way#Galactic_rotation

How to galactic orbital velocities differ from this classical model?

Jens
2012-Aug-30, 08:54 AM
I can't give you an answer, but it seems to me that if you know enough to be able to do those calculations, you should be able to understand the papers that discuss actual observations of galactic rotation curves.

StupendousMan
2012-Aug-30, 12:21 PM
Please read these notes

http://spiff.rit.edu/classes/phys443/lectures/gal_dark/gal_dark.html

and pay special attention to the graphs of observed rotation curves.

Read some of the papers from which those curves were taken.

Cougar
2012-Aug-30, 01:08 PM
...and 3/4 of the mass is in a uniform disk-like region with a radius of 40,000 LY...

Shouldn't that be "...with radii between 10,000 and 50,000 LY"? Er, wait. Your simplification is very unclear....

onomatomanic
2012-Aug-30, 01:41 PM
The assumption that a symmetric mass distribution can be treated as if it were a point-mass at the centre of that distribution does not hold for a disk.

Proof: If the assumption were true, then a disk D1 of radius R and surface density s would attract a test-mass at a point P on its circumference with gravitational acceleration a = G M/R^2 = G s (pi R^2) / R^2 = pi G s. Now consider the disk in two parts, as a disk D2 of radius R/2 whose centre is halfway between P and the centre of D1, and the remainder R of D1 after D2 has been taken away. D2 would attract the test-mass with a = G s (pi (R/2)^2) / (R/2)^2 = pi G s. Therefore, R would attract the test-mass with a = 0. This is obviously impossible, if one considers the basic geometry of the problem. By contradiction, the assumption must have been false. QED.

I'm guessing that, paradoxically as it sounds, you have to adopt a more complicated approach than you did to recreate the less complicated result shown as the red curve in the graph you linked to.

utesfan100
2012-Aug-30, 04:49 PM
I can't give you an answer, but it seems to me that if you know enough to be able to do those calculations, you should be able to understand the papers that discuss actual observations of galactic rotation curves.
I was unclear whether and how Dark Matter was included in the calculations of that paper.

utesfan100
2012-Aug-30, 04:59 PM
The assumption that a symmetric mass distribution can be treated as if it were a point-mass at the centre of that distribution does not hold for a disk.

Proof: If the assumption were true, then a disk D1 of radius R and surface density s would attract a test-mass at a point P on its circumference with gravitational acceleration a = G M/R^2 = G s (pi R^2) / R^2 = pi G s. Now consider the disk in two parts, as a disk D2 of radius R/2 whose centre is halfway between P and the centre of D1, and the remainder R of D1 after D2 has been taken away. D2 would attract the test-mass with a = G s (pi (R/2)^2) / (R/2)^2 = pi G s. Therefore, R would attract the test-mass with a = 0. This is obviously impossible, if one considers the basic geometry of the problem. By contradiction, the assumption must have been false. QED.

I'm guessing that, paradoxically as it sounds, you have to adopt a more complicated approach than you did to recreate the less complicated result shown as the red curve in the graph you linked to.
The equation was presented as an order of magnitude estimate. Your proof shows that it can not be correct, and actually underestimates the acceleration due to the disk. It seems reasonable to expect that the acceleration from region R would would be much less than the acceleration due to region D2. Thus D2 accounts for most of the disk acceleration.

If we are on the interior of the disk, the acceleration due to the disk area outside our radius is also not 0, but slightly outward.

I would expect the fact that the disk becomes less dense with distance to be a much more significant factor, but this function is hard to find in literature. The simplification of constant disk density is likely to be the biggest issue. If the density falls off with 1/r we will get a limiting constant velocity curve. If we parametrize the density as r^{\gamma} we get increasing velocity for \gamma<-1 and decreasing velocity for \gamma>-1.

onomatomanic
2012-Aug-30, 06:31 PM
If we are on the interior of the disk, the acceleration due to the disk area outside our radius is also not 0, but slightly outward.


Yes, that occurred to me as well. As far as I can remember, whenever rotation curves are discussed, one of the first assumptions made is that only mass within the orbit need be considered. Doesn't really make sense, does it...



The simplification of constant disk density is likely to be the biggest issue.


Agreed.