Thread: galaxy rotation speed and mass distribution

1. Well, this is astonishing. In response to Cougar's post, I changed the dimensions of the disk in the program from 100,000 light years across and 12,000 light years thick for the Milky Way to 100,000 light years across and 1,000 light years thick, which accounts only for the stars themselves and not the dust and gas which extends out further, to see what that would give for a lesser thickness. Well, the force still comes out to almost twice as great as that of a sphere and the energy 1.2 times as great. I even found for twice as many points along each axis, for eight times as many points in all, and it comes out the same. If that continues with smaller thicknesses, then the gravity of a disk plane is finite. I'm not sure what that means yet, unless the integration must somehow be performed in a much different than the ordinary way or something. It might also have something to do with the natural "clumping" that is done when considering the mass over a finite number of points in the program instead of an even spreading out of the mass throughout the disk, as would be the case when performing integrations for it. I will also recheck the program. This sort of clumping does definitely seem to be the way to go with this.

2. Originally Posted by grav
Well, this is astonishing. In response to Cougar's post, I changed the dimensions of the disk in the program from 100,000 light years across and 12,000 light years thick for the Milky Way to 100,000 light years across and 1,000 light years thick, which accounts only for the stars themselves and not the dust and gas which extends out further, to see what that would give for a lesser thickness. Well, the force still comes out to almost twice as great as that of a sphere and the energy 1.2 times as great. I even found for twice as many points along each axis, for eight times as many points in all, and it comes out the same. If that continues with smaller thicknesses, then the gravity of a disk plane is finite. I'm not sure what that means yet, unless the integration must somehow be performed in a much different than the ordinary way or something. It might also have something to do with the natural "clumping" that is done when considering the mass over a finite number of points in the program instead of an even spreading out of the mass throughout the disk, as would be the case when performing integrations for it. I will also recheck the program. This sort of clumping does definitely seem to be the way to go with this.
Whoops. I rechecked the program and it appears I changed the numbers of points found for along the z axis, but not the actual thickness of the galaxy itself, so it was still finding for the same thickness, but just using a lesser number of points along it. Sorry. So for a thickness that is 3/25 that of the diameter of the galaxy, the force at the rim is about 1.9 times greater than that of a sphere of the same mass and the energy is 1.2 times greater. For a thickness that is 1/12 as great, or 1/100 of the diameter, the force at the rim is 2.75 times as great as that of a sphere of the same mass and the energy is 1.25 times as great. I'm still going to try working with a "clumping" variable in the integrations, though.

3. Originally Posted by grav
Originally Posted by Cougar
Then there's the dark matter effect, and its distribution....
Well, dark matter was actually introduced to account for this effect, that of the rotation speed curves, but it may just simply lie within the geometry and mass distribution itself. But some relatively accurate level of the geometry and mass distribution must be known, or at least theorized according to the results of the integrations themselves. As well as this, the correct integrations must also be performed. This does not appear to be the case with some of the earlier models for galaxies, incorporating spheres for simplicity and rings used in the same way as spheres, which led to the hypothesis of dark matter to begin with. But that is what I am working on now.
By the way, to be fair, this is not saying that a galaxy is regarded in the same way as a sphere. If that were the case, the force would climb in direct proportion to the distance all the way to the rim, and the rotational speed would be constant, so that the entire galaxy should rotate together with the same period of rotation for all points. The rotational speed at the rim, figuring for the total mass and radius of the Milky Way galaxy would then be about twice as great as is observed, which would actually require some form of "negative" dark matter, or lesser matter to account for the lesser speed observed at the rim. Also, if regarded as just a thin disk of uniform density, then the rotational speed would be expected to climb very rapidly at the rim, so a leveling off of the rotational speed would also require some form of "negative" dark matter. As it is, though, the rotational speed is predicted to climb in the same way it is observed to do up to some peak and then drop back down, but instead it just levels off at the peak, as shown in this link. The difference between the predicted curve and that of a geometry incorporating a uniform density would be that of the predicted mass distribution for the galaxy. I guess my next step, then, is to find the mass distribution that would give a curve that matches that for the predicted curve for a galaxy and go from there.

4. Cougar,

You might want to check my post #30 again. I accidently erased it , so I redid it, but I know I responded differently to your second paragraph after rereading it than I did the first time, not that I remember my response to that anyway, but I think I got the other two responses the same.

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Originally Posted by grav
I guess my next step, then, is to find the mass distribution that would
give a curve that matches that for the predicted curve for a galaxy and
go from there.
It sounds to me like you still have to get the math right before the
simulations can have any value. The fundamental problem causing
infinities must be a common one -- I'm sure it has a common solution.

-- Jeff, in Minneapolis

6. Originally Posted by Jeff Root
It sounds to me like you still have to get the math right before the
simulations can have any value. The fundamental problem causing
infinities must be a common one -- I'm sure it has a common solution.

-- Jeff, in Minneapolis
The math is right, although I am still trying to find out the predicted or observed mass distribution of galaxies, yes. The infinity is for the edge of a flat disk plane of uniform density, not one with some thickness, but as far as I know, the exact solution for a thick disk is not known either, only approximated, so that is why I am just using the program unless I can manage to work it out through the integrations. But even then, the program does the same thing as the integration, just point by point, and I can vary the mass distribution at will instead of reperforming the integrations accordingly, and it's much quicker to boot.

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Originally Posted by grav
Here is a link I found a little while ago that summarizes some methods. It looks like I have already tried most of the ones mentioned. Is this pretty much all that has been explored of the subject so far?
Hi grav,
I think you are on the right track in referring to the above paper. Nicholson's calculation was discussed extensively in this forum here:

http://www.bautforum.com/questions-a...lculation.html.

Nicholson's solution avoided the dreaded elliptical integral solution by using a numerical interation method in a computer program he used while working on the Space Shuttle Program. Dont stop reading the thread after the first parts, as it takes a turn later, and Nicholson joins the discussion.

TomT

8. Thanks, Tom. Good to hear from you again. I remember you helped out with this when I first went through it. Hh and publius have since given me a two week crash course for finding integrations and I have learned a lot from them. I have been trying to use it for this, but you say it's not necessary, at least not with a computer program for it, or by using iterations, eh? I have read through some of Nicholson's papers before. Maybe this time I can follow it better. I'll read through it again and try to use any calculations he provides to see what it comes up with.

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Originally Posted by grav
The math is right,
I don't think it can be right if it produces infinities, even if only for
the unrealistic case of zero thickness. I suspect that the added
calculations for points out of the plane are partially masking the
error which is apparent at zero thickness.

Originally Posted by grav
The infinity is for the edge of a flat disk plane of uniform density, not
one with some thickness,
If it happens at the edge, it should happen everywhere. I suspect
that in the interior, the error in one direction is masked by the error
in the opposite direction.

Does your program actually calculate the attraction between a mass
at one point and the same mass at the same point? Or does it
calculate attraction between masses at two points that are very
close together? Or what?

-- Jeff, in Minneapolis

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Originally Posted by grav
They would still both lead to an infinite acceleration of gravity and therefore an infinite orbital speed for the point at the center of the mass that is right at the disk edge for a straight forward calculation of the disk plane. The calculation just blows up there.
I strongly disagree, for one simple reason - you have two stars, one on the edge of the circular field, the other positioned such that half the mass is inside it's orbit, the other half outside it's orbit. That mass is finite, and so is the gravitational attraction towards the center. Nothing goes to zero. Nothing approaches infinity.

We're dealing with finite entities, here, and the model is patterned after, and close to, our own Milky way. Clumping doesn't enter into the picture, nor do the spiral arms, because it's at such a distance, and there are so many clumps and arms, that while these might cause pertubations in our orbit, those are minimal unless we get fairly close to a clump. If that happens, then, yes - we might get tossed outwards (slingshot) or decayed inwards (orbital braking).

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Originally Posted by grav
The thing about the gravity at the edge of a disk plane, though, is that since it comes out to 4 G (M/A) [atanh(1) - 1], the gravity is just right on the "edge" of infinity.
You're really missing something, here, grav...

Instead of stars, let's use steel ball bearings, 1" in diameter, arranged in a circular plane. Let's use the same numbers as in the first problem I stated, and set things in motion such that all initial orbits are circular given the initial conditions. Let's assume we're in the center of that supermassive void they discovered so the local gravitational gradient from even the closest object is zero.

Forget the math you're using for just one second and answer this question: Is gravity "just right on the 'edge' of infinity?"

I'll answer it for you: Hardly.

In fact, it'll be so low that the outermost ball bearing will have an orbital velocity measured in micrometers per second, if not hour.

If we blow this back up to the model I proposed, the proportions increase along with it, but nothing, especially the gravitational tug at the edge of the disk, get's anywhere close to infinity, any more than it does for our own Milky way.

Again, the model is patterned after our Milky Way, and for all intents and purposes, when you get beyond about 1% of the Milky Way's radius, things can be approximated by using a thin disk model.

Thus, I conclude that since the only substantial difference between my model and the Milky Way is local clumping, I think you're tripping over your calculations, somewhere...[/QUOTE]

Thanks, Jeff Root, and Cougar, for posts 27 and 28.

12. To answer these last three posts at once, hopefully, the integration for a disk plane does indeed come out to a result of infinite acceleration at the very edge for a point particle. A larger sphere at the edge, half in and half out should experience the same thing, at least for the point at its center, or considering the sphere a point mass at the center. In other words, if I were to perform separate integrations for every point in the sphere toward the disk, the point in the center would still blow up regardless of the rest of the points, and other points in the sphere shouldn't make much difference adding or subtracting with their finite contributions, then, I wouldn't think.

If we took a bunch of spheres and arranged them in a disk, then they would have thickness, and the acceleration toward the disk would be finite.

The computer program doesn't blow up at the edge, since it is calculating over clumped points within the disk instead of a mass spread evenly throughout the disk, as with an integration. However, as I make the thickness of the disk smaller and smaller, the gravity at the edge climbs higher and higher.

I know it doesn't seem to make much sense that the gravity at the edge of a finite mass disk should be infinite, but I have two theories about this. The first is with the thickness. With zero thickness, the particle does not lie against a surface as it would with a sphere or thick disk. Since the force varies with the square of the distance, we might need two dimensions to cancel out the infinities. That can be tested by finding for the potential energy, which only varies with the inverse of the distance, should become finite against the one dimension of the disk's edge. The other is with the clumping. I'm thinking stars, even gas and dust, are not spread evenly throughout a geometry, so the minimum possible distance between points must be taken into account.

I am wondering about the precise relationship between the resulting force and energy in all of this.

13. I'm not sure what "integration" you're talking about. All you need is the distance from your test particle to each object in the galaxy and the approximate mass of each object. I think you could simplify it to a single central black hole 5% to 10% of the total mass, and the rest solar-sized masses. Let the computer add up the resultant accelerations....

This "infinite gravity" is obviously an error.

14. Originally Posted by Cougar
I'm not sure what "integration" you're talking about. All you need is the distance from your test particle to each object in the galaxy and the approximate mass of each object. I think you could simplify it to a single central black hole 5% to 10% of the total mass, and the rest solar-sized masses. Let the computer add up the resultant accelerations....

This "infinite gravity" is obviously an error.
For a central black hole, we would still perform the same integration for the disk, and then just add the gravity of a point mass for the central body, which wouldn't add much considering the mass of the rest of the galaxy. In any case, here's the integration for a particle on the rim of a flat disk plane from post #14.

If we integrate for the acceleration of gravity of a disk plane on a particle, we have Int G (M/A) dx dy (R-x) / d^3, where M/A is the mass per area which is uniform, R is the distance of the particle from the center, d is the distance from the particle to each point in the disk, and (x,y) are the coordinates of a point using the center of the disk as the origin. From this, we get

Int G (M/A) dx dy (R-x) / sqrt[(R-x)^2 +y^2]^3

= Int G (M/A) dx (R-x) * [2 * sqrt(r^2 - x^2) / (R-x)^2 / sqrt[(R-x)^2 + (r^2 - x^2)] ] for y = -sqrt(r^2 - x^2) to sqrt(r^2 - x^2)

but the last integration for dx does lead to complicated elliptic integrals, three different kinds, in fact, so let's avoid that by finding a solution for a specific R, such as R=r, for acceleration of gravity acting on a particle right at the edge of the disk. Then we get

Int 2 G (M/A) dx sqrt(r^2 - x^2) / (r-x) / sqrt[(r-x)^2 - r^2 -x^2]

= Int 2 G (M/A) dx sqrt(r+x) sqrt(r-x) / (r-x) / sqrt[2r * (r-x)]

= Int 2 G (M/A) dx sqrt(r+x) / (r-x) / sqrt(2r)

= 2 G (M/A) / sqrt(2r) * [ 2 sqrt(2r) atanh[sqrt(r+x) / sqrt(2r)] - 2 sqrt(r+x)]

from x = -r to r, so

= 4 G (M/A) [ atanh(1) - 1]

Here, since x = -r to r, it reduces to just zero for the whole thing for -r, but for +r, the part in atanh becomes sqrt(r+x)/sqrt(2r) = 1, and atanh(q) = (1/2)[ln(1+q) - ln(1-q)], so if q=1, then - ln(1-q) = -ln(0) = - (-infinity) = infinity. We have nothing else that might cancel out the infinity here, so that is the result for a disk plane. The gravity is infinite at the edge of the disk plane.

Now, I am thinking that finite masses cannot result in infinite force, so the only thing I see that might cause this is due to considering any and all point masses that lie directly in front of the particle, even within zero distance, like singularities. I am thinking that a better method would be to allow some distance between the particle and the closest possible point mass. For the gravity of something like a planet or the sun, this could possibly be as small as the distance between atoms, so wouldn't make much difference in the integration for those, at least that of spheres. With galaxies, however, especially being disks, I'm thinking an average or minimum possible distance between stars must be accounted for, at the very least being that of the diameter of a typical star.

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grav,

I'm going to reply to a couple of things you just said in a separate post in a
few minutes, but right now I want to suggest a test I just thought of.

Can you have your program calculate the force on a plane shape other than
a disk? Like, the force on a point at the middle of an edge of a square? Or
on the edge of a U-shape? Like this:

Code:
```  _____       _____
|    |     |    |
|    |     |    |
|    |__.__|    |
|               |
|_______________|```
The period indicates the location of the test point.

From your descriptions, I expect that your program will give a result of
infinite force toward the bottom of the U, and the force from the upper
part of the U, while calculated, will have no effect on the result. If it
does have an effect, that could tell you something about the problem.

You might want to use an L-shape instead of a U. The lopsidedness
could provide more diagnostic data.

-- Jeff, in Minneapolis

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Originally Posted by grav
The computer program doesn't blow up at the edge, since it is
calculating over clumped points within the disk instead of a mass
spread evenly throughout the disk, as with an integration. However,
as I make the thickness of the disk smaller and smaller, the gravity
at the edge climbs higher and higher.
That should make the problem much easier to resolve than if it only
showed up when it reaches zero thickness. This way, you will get
a definite indication that you have found the problem and fixed it,
rather than applying some arbitrary restriction which makes the
infinities go away but doesn't necessarily give correct results.

Originally Posted by grav
With zero thickness, the particle does not lie against a surface as it
would with a sphere or thick disk.
I don't understand the geometry you are describing when you say
"...lie against a surface...".

Originally Posted by grav
Since the force varies with the square of the distance, we might need
two dimensions to cancel out the infinities.
I think you shouldn't be producing infinities in the first place. Cancelling
them out sounds like a hack, and a particularly dangerous one.

-- Jeff, in Minneapolis

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Cougar,

I know very little about how gravity simulations are done in general, or how
grav is doing it in particular, but I suspect that it isn't practical to simulate
individual stars. There are estimated to be something like 300 billion stars
in the Milky Way, and most of the mass is supposedly in dark matter which
is spread around rather evenly. So my guess is that grav is dividing his
simulated galaxy up into a few hundred or maybe a few thousand smallish
chunks, and treating each chunk as a point mass.

Did I guess right, grav?

-- Jeff, in Minneapolis

18. Originally Posted by Jeff Root
Cougar,

I know very little about how gravity simulations are done in general, or how
grav is doing it in particular, but I suspect that it isn't practical to simulate
individual stars. There are estimated to be something like 300 billion stars
in the Milky Way, and most of the mass is supposedly in dark matter which
is spread around rather evenly. So my guess is that grav is dividing his
simulated galaxy up into a few hundred or maybe a few thousand smallish
chunks, and treating each chunk as a point mass.

Did I guess right, grav?

-- Jeff, in Minneapolis
Yup. You're right about how the simulation is done, anyway. I'm using about 378,000 points most of the time. I chose that many because it gives a nice balance between the preciseness of the result and my patience for how long it takes the program to run. But if I want to be more precise, I just add more points. I could add as many points as there are stars in the galaxy if I wanted to, but it would take very, very, uh, very long time to run. But it wouldn't make that much difference as far as the result goes if I did anyway, except for where the numbers run extraordinarily high, like on the rim of a flat plane galaxy. That's why I had to perform the integration for that, to gain a better preciseness for right at the edge, and see just how high the numbers would run, only to find it to be infinite. For a galaxy with some reasonable thickness, though, the simulation works just fine all the way from center to rim.

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It appears there may be something more to galaxy formation that dark matter. They've found one that, so far, shows no evidence of DM formation & behaves more like you'd expect from Newtonian effects.
http://space.newscientist.com/articl...tronomers.html
Originally Posted by New Scientist
What do you call an absence of darkness? Dark matter is supposed to be spread throughout the universe, but a new study reports a spiral galaxy that seems to be empty of the stuff, and astrophysicists cannot easily explain why...

...In the spiral galaxy NGC 4736, however, the rotation slows down as you move farther out from the crowded inner reaches of the galaxy. At first glance, that declining rotation curve is just what you would expect if there is no extended halo of dark matter, and no modification to gravity...

20. Originally Posted by Jeff Root
grav,

I'm going to reply to a couple of things you just said in a separate post in a
few minutes, but right now I want to suggest a test I just thought of.

Can you have your program calculate the force on a plane shape other than
a disk? Like, the force on a point at the middle of an edge of a square?
That's an interesting idea, and you've got me curious. I could use the program to find the gravity for a square just as easily as that of a circular plane, yes (although not right on the edge, actually, because the numbers are too high), but working out the integration is much better if one can get a workable solution, so I'll do that. The integration is the same as before, but with a different shape, so if we integrate for a particle centered in a side of length b of a rectangle, the other side being length a, then

Int G (M/A) (R-x) dx dy / d^3, where d = sqrt[(R-x)^2 + y^2]

= Int G (M/A) (R-x) dx * [b / [(R-x)^2 sqrt[(R-x)^2 + b^2]]]

= Int G (M/A) dx b / [(R-x) sqrt[(R-x)^2 + b^2]]

= G (M/A) b * [(log[(2b + 2 sqrt[(R - a/2)^2 + b^2]) / (R - a/2)] - log[(2b + 2 sqrt[(R + a/2)^2 + b^2]) / (R + a/2)]) / b]

= G M / (a b) * [log(2) + log(b + sqrt[b^2 + (R - a/2)^2]) - log(R - a/2) - log(2) - log(b + sqrt[b^2 + (R + a/2)^2]) + log(R + a/2)]

= G M / (a b) * [log(1 + sqrt[1 + (R - a/2)^2 / b^2]) - log(1 - a/2R) - log(1 + sqrt[1 + (R + a/2)^2 / b^2) + log(1 + a/2R)]

This works out fine to find the result for the acceleration of gravity at any distance R (where R>r, though, I believe. I'll have to compare it to the program) for the particle from the center of the rectangle, perpendicular to side 'b' and with the other side 'a'. But as you can see, if we put the particle right on the edge of side b, then R = a/2, and the -log(1 - a/2R) part of the solution becomes -log(0) = - (-infininity) = infinity, and has nothing to cancel out with, so the infinity remains.

21. Sorry. I missed this post somehow.

Originally Posted by Jeff Root
That should make the problem much easier to resolve than if it only
showed up when it reaches zero thickness. This way, you will get
a definite indication that you have found the problem and fixed it,
rather than applying some arbitrary restriction which makes the
infinities go away but doesn't necessarily give correct results.
The infinity only shows up for a particle right at the rim and only for a zero thickness disk. For any other position at zero thickness and/or for any thickness other than precisely zero, there are no infinities.

I don't understand the geometry you are describing when you say
"...lie against a surface...".
I mean that a particle at the surface of a sphere, say, is lying against two dimensions. At the edge of a disk, there is only one, linearly along the circumference. I think that makes the difference with the force since it is inversely proportional to the square of the distance, or inverse to two dimensions of space, so requires two dimensions to operate against, otherwise a zero dimension length will still exists in the denominator of the result, producing an infinity. I think this is reasonable, especially considering that any real object must have greater than zero thickness to begin with.

I think you shouldn't be producing infinities in the first place. Cancelling
them out sounds like a hack, and a particularly dangerous one.
Well, I agree. I use a greater than zero thickness galaxy in the program. I mainly just showed the infinite gravity of a zero thickness disk plane 1) because it is one of the easiest things to integrate 2) to demonstrate how the gravity of a thin disk is very much different than that of a sphere of the same mass, and 3) to show that the gravity at the edge of a disk can be very large indeed, potentially with no limit, depending on how thin we can squeeze the matter in the disk.

22. Originally Posted by Acolyte
It appears there may be something more to galaxy formation that dark matter. They've found one that, so far, shows no evidence of DM formation & behaves more like you'd expect from Newtonian effects.
http://space.newscientist.com/articl...tronomers.html
Thanks, Acolyte. Welcome to the forum.

23. Mmmm - that diagram at Wikipedia is for a "typical spirial galaxy", not the Milky Way. And one thing that I have not seen so far is mention that The Milky Way is a barred spiral. This has been confirmed since 2005 (but I only learned it this year!).

I don't understand barred spirals at all! Could there be a concentration at the end of each end of the bar holding the stars in place? Or a cosmic string? Very weird.
Last edited by Vanamonde; 2008-Apr-28 at 08:06 AM. Reason: Deleted inaccurate dark matter questioning statement

24. I believe I just thought of a really easy way to think about the infinities.

For a point particle lying directly against a zero-dimensional point mass, the gravity varies with the square of the distance and so becomes infinite upon direct contact, even if the mass itself is finite, and so it follows the same reasoning when considering a finite mass with infinite gravity.

For a point particle lying directly against a line, the gravity is infinite. The force in this case varies inversely with the distance instead of the square of the distance, since one of the dimensions is cancelled out, but it still results in infinite gravity.

For a particle lying against a one-dimensional curved line, such as a ring, the gravity is infinite.

For a particle lying against the one-dimensional outer rim of a disk or other two-dimensional shape, the gravity is infinite.

For a particle lying against the two-dimensional surface of a three-dimensional gravitating mass, the two inverse dimensions for the force are cancelled out, and the gravity is finite.

For a particle lying against or at any distance from the surface of a two-dimensional infinite plane, the gravity is finite and constant.

That last one might play against intuition also, being an infinite mass that produces only a finite gravity, but when one thinks about how the dimensions work in accordance with each other, it starts to sound reasonably sensible, I think.

{By the way, just to point out real quick, I actually don't believe a mass could produce an infinite gravity anyway for other reasons as well, though, being a proponent of push gravity theory and all, where the force of gravity would be limited by the total energy density of the gravitational field, and could never be greater than that full pressure would allow, but that's another story. For this, I am just following the classical Newtonian calculations.}

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Could you calculate the force on a point on the edge of a disk one millimetre
thick and two metres in diameter, with a mass of 10 kilograms? What is the
force when the point is at the center of the millimetre-wide edge? What is
the force when the point is on the edge of that edge?

-- Jeff, in Minneapolis

26. Originally Posted by Jeff Root
Could you calculate the force on a point on the edge of a disk one millimetre
thick and two metres in diameter, with a mass of 10 kilograms? What is the
force when the point is at the center of the millimetre-wide edge?
I could try. It would be finite, but I would have to find a way to bypass the elliptic integrals. I found a solution for a thin rectangular mass, but it involves eighteen logs, half a dozen atans, and a few complex numbers thrown in just to mix it up a bit. I'm too lazy to work through all of that right now. I was going to then use it to find for a disk, by lining up a succession of thin rectangles with lengths 'b' that would fit within the disk when placed in a row with sides with small width 'a'. I might eventually get that if I reduce the result down well enough.

What is the force when the point is on the edge of that edge?
Interesting. That may be easier to find than for the center of the edge, and not too much different with a finite gravity, probably. I'll see what I can do.

27. Well, it looks like the integration for the edge of an edge of a disk is the same as before with the ellipticals, but with +4h^2 added with some other stuff in the denominator instead of just +h^2. Since the disk has some thickness, I'll use the program for it, and just run it over a larger and larger number of points until the accuracy of the result converges to some limit.

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Originally Posted by grav
I found a solution for a thin rectangular mass, but it involves eighteen
logs, half a dozen atans, and a few complex numbers
Yow!!! My mind is truly boggled. Even if you've made a mistake
somewhere and all the calculations are wrong, I'm still impressed.
You're doing calculations in less time than it would take me just
to look them up.

-- Jeff, in Minneapolis

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I just thought of another geometry that might be a useful diagnostic.
Either of the following should work, so you can choose whichever is
easiest to handle: The force on a point at the apex of a cone (maybe
2 or 3 cones with different angles), or at the apex of a pyramid (2 or 3
pyramids with different angles). See how the force varies with angle.

-- Jeff, in Minneapolis

30. Originally Posted by Jeff Root
Yow!!! My mind is truly boggled. Even if you've made a mistake
somewhere and all the calculations are wrong, I'm still impressed.
You're doing calculations in less time than it would take me just
to look them up.

-- Jeff, in Minneapolis
Thanks, Jeff. You wouldn't believe how many variations of the integrations I've run through trying to get this.

I put the numbers into the computer program for a disk that is 2000 times as wide as it is thick for that 1 millimeter disk you had. The actual mass and scale of size don't matter in the program, though, only the ratio of the diameter to the thickness, and the result will be a ratio of the gravity of a disk to that of a sphere of the same mass and radius. For the first run, I used a bare minimum of five points along each column of the z axis. It is ironic that the thinner we make the disk, the more points I must use to get a decent result, since if I am using the minimum number of points along z, I must supply more along the x and y axes to make the disk wider and thereby thinner. So for a ratio of 2000 to 1 for the diameter to thickness ratio, I ended up having to use a minimum of about 37.7 million points, which took two and a half hours to run. That gave a very rough estimate of about 5 times the gravity of a sphere with the same mass and radius. So for a ten kilogram mass with a disk radius of one meter, the acceleration of gravity on a particle at the rim would be 5GM/r^2 = 3.337 * 10^(-9) m/s^2.

I didn't have the patience to run that any more precisely the way I was doing it, but here's the good news. I finally remembered what I had written at the end of post #16. I already have the integrations along the z and y axes worked out in the integrations with absolute preciseness for a disk with any thickness greater than zero. So all I need to do now is to apply the result for those axes along the x axis only in a program. So instead of the time I just wasted running it the way I did, for all three axes, I could have had a much better preciseness in, oh, about 1/20000 of the running time. Oh, well. Hindsight's 20/20, I guess. More than that, though, I'll bet that most of the gravity that a particle at the rim experiences for a thin disk is from that part of the disk that lies right in front of the particle, so I should only have to run the program for a distance along the x axis that is just a few times greater than the thickness of the disk and still get a very precise result. If that is the case, then I will be able to run the program for any thickness whatsoever with the same running time. I will get to writing that program right away, and with any luck, we will be able to find the gravity very quickly and precisely for any thickness we desire.

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