# Some help needed in interpreting a problem - Keplerian velocity

1. Mar 8, 2013

### Sefrez

Some help needed in interpreting a problem -- Keplerian velocity

1. The problem statement, all variables and given/known data
Code (Text):

Consider a galaxy made by a uniform thin disk of surface density ρ.
The total mass of the galaxy is M = 1040 kg and its radius r = 1020 m.
Integrate numerically the gravitational force as a function of the distance
from the center (for distances up to the galaxy’s radius) and plot the
Keplerian velocity as a function of the distance from the center.

I formulated an integral at which I attempted to calculate numerically, but the solutions were all over the place (as if the integral did not exist / did not converge.) In this problem, when it asks for the gravitational force, does it mean on any arbitrary mass within the disk/galaxy? And am I supposed to model this as a sum of forces given by small mass elements within the galaxy? (i.e. dF = Gm/r^2 dM, where G is the gravitational constant, m is the mass of an arbitrary body, dM is a mass element of the galaxy, and r is the distance between these)

If this is correct, I came up with the following integral:
F = ∫Gmρ(x-r0)/((x - r0)2 + y2)3/2da

where I have chosen a Cartesian coordinate system in x and y, and so this integral is taken over (x,y) in a disk of radius r = 1020 m. r0 is the distance from the center of the galaxy.

I realize now I should have chose a polar coordinate system due to the symmetry of the problem, but nonetheless, you can see with the integrand, it goes to infinity when (x, y) -> (r0, 0). Have I done something incorrectly or mis-interpretted the problem?

2. Mar 8, 2013

### CompuChip

It looks like you're overcomplicating it, leading to expressions like $(x - r_0)^2 + y^2$ (is that even correct)?

Try setting the center of the disk to (0, 0) and switching to polar coordinates, as you suggested.

3. Mar 8, 2013

### haruspex

Yes, this integral won't converge. The problem is that it is unrealistic close to the target location. In reality, a galactic disc has some thickness. An easy way around this is to note that everything must cancel within a disc centred at the target location and contained entirely within the galactic disc. So omit that region from your summation. That will still break down as you approach the edge of the galactic disc, though, since the omitted region will shrink to nothing.
(Compuchip, I believe $(x-r_0)^2+y^2$ is correct. Sefrez is taking an element at (x, y) relative to galactic centre, and calculating the acceleration it produces in the x direction at (r0, 0).)

4. Mar 8, 2013

### Sefrez

Some help needed in interpreting a problem.

Ok, so what I should do is simply neglect contributions to the sum when they are close to the target? That is, perhaps, neglect contributions for which sqrt((x-r0)2+y2) < r is true? I guess the chosen r would depend on the accuracy of my summation.

5. Mar 8, 2013

### haruspex

I meant specifically neglecting contributions where (x-r0)2+y2 < (r-r0)2, where r is the radius of the galaxy. But as I said, you'll need some other approach as r0 → r. E.g. if you assume some thickness 2h << r then for (x-r0)2+y2 < h2 consider it as a 3-dimensional region.

6. Mar 8, 2013

### Sefrez

Some help needed in interpreting a problem.

Oh, I see. Essentially it is the same as what I said but maximizing the radius within the disk. This is justified by the fact that contributions within the radius are canceled, as you said before. Given the large extent of the galaxy, r = 10^40, I may just neglect the edge all together, assuming I can get still a good plot. Of course I need to figure out where that limit should be made though.

7. Mar 8, 2013

### Sefrez

Well it seems this integral is proving hard to approximate. I have calculated a few in the [10^19, 10^20] range, but I am honestly not sure if this is correct. See the attachment for a smooth line plot I created quickly in Excel with the few data I had. On the vertical axis is the Keplerian velocity and on the horizontal is the distance from the center. Should it be closely linear at first with a rapid increase toward the end of the disk?

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8. Mar 9, 2013

### haruspex

That general shape is what I'd expect.

9. Mar 9, 2013

### Sefrez

Ok, thank you haruspex.

Would we be correct in saying that the rate of increase in velocity nearer the edge is due to the finite extent of the galaxy disk? That is, as the edge is approached, there is lesser and lesser mass to cancel with that on the other side.

10. Mar 9, 2013

### haruspex

Yes. As noted, if you take it as literally two dimensional then it tends to infinity at the edge. That's because the zone of cancellation shrinks to zero.