27-April-2008, 09:31 PM
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Senior Member
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Join Date: May 2006
Posts: 2,317
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Quote:
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.
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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.
Quote:
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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