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Computing orbital speed in a cloud of particles

Posted on 2014-02-09
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We have a random 2D distribution of many particles (a cloud) within distance "R" from the origin (forming a disk shape of points with the origin in the middle).  Total mass of the system is “M”.

QUESTION:  Considering that the total mass of the system is distributed evenly throughout this volume, what is the linear orbital speed of a particle "P", with mass “m”, at distance “r” from the origin?

Givens:
Assume r < R
Assume M>>m

Requests:
Newtonian gravity only please (F=GmM/rr), though using calculus fine.
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Question by:Korbus
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Author Comment

My thoughts on this so far; quite possibly wrong, but I want to show you how far I got, and where I’m stuck:

Let’s say, for easy of discussion, the particle in question is on the x axis, at positive position “r”.  We will arbitrarily have it orbit counter-clockwise.

The direction of orbital velocity, at this moment, will be straight up, parallel to the y axis.   The magnitude of this velocity is what we need to determine.

We can cut the disk along this velocity direction line into two sections; the larger section (containing the center) will exert a net force towards the center “Fleft”.  The smaller section will exert a net force directly away from the center, “Fright”.

We can get the total force towards the center Fc = Fleft-Fright  (since Fleft contains the center, we know Fleft>Fright)

Once we have Fc, we can compute the orbital velocity.

Fleft will consist of a sum of all the forces for every point in that section of the disk.
Fright will consist of a sum of all the forces for every point in that section of the disk.

I suspect that computing the sum of forces for those points will involve integration, but I’m just not sure how to proceed.
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Expert Comment

If the mass is distributed evenly in a 3-d volume, and we assume that the orbits all circular, the orbital speed of each particle would depend only on the radius of its orbit and the total mass contained within that radius.
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Author Comment

Hi Ozo,

So, it sounds like orbital speed computation inside the cloud would be no different than orbiting outside the cloud, given the same Mass is contained within that orbit.   I was not expecting that result!

When orbiting inside:  Why wont the mass outside of that orbit (to the "right") have the effect of pulling the particle outward?  Or, what is cancelling it out?  Is it the mass outside the orbit but to the "left"?

Perhaps this will be easier for me to figure out the math for.  Or, can you easily demonstrate this mathematically? (or link)
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Expert Comment

>>  Or, what is cancelling it out?
Here is a link that shows the cancellation effects of a spherical shell of uniform density. Once you see the math, you can try to apply the principles to a disk. In fact, the figure shows a ring. Using their derivation, move the mass to any point within that ring and based on ozo's comment, you should see a net force of 0 on the particle from the ring.
http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/sphshell2.html
The net gravitational force on a point mass inside a spherical shell of mass is identically zero! Physically, this is a very important result because any spherically symmetric mass distribution outside the position of the test mass m can be build up as a series of such shells. This proves that the force from any spherically symmetric mass distribution on a mass inside its radius is zero. If a given mass m is inside a spherically symmetric distribution of mass, that part of the mass outside its radius does not contribute to the net force on it.
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Expert Comment

Note that this works for spherical shells, not circular disks, so if the distribution is really 2D, it will be more complicated.  But you also said "distributed evenly throughout this volume" instead of "throughout this area", so I wasn't sure whether we were dealing with volume or a plane.
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Expert Comment

In OP:
>> random 2D distribution of many particles (a cloud) within distance "R" from the origin (forming a disk shape...
>> distributed evenly throughout this volume

A bit ambiguous. Are you referring to a large disk radius having a very small height? That is how I was interpreting your OP. If so, then ozo has clarified that the cancellation will not result in a zero force.
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Expert Comment

Another ambiguity is that a random distribution is not the same as distributed evenly.
And should we assume that the orbits are all circular, or should we consider the velocities to be randomly distributed?
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Expert Comment

I understand that you are asking what is the speed of a particle at a given radius. This almost suggests that you are expecting circular orbits for every particle. Is that correct?

Are you first trying to find out whether there exists a set of initial conditions in a system of evenly distributed particles in a 2d plane such that each particle's speed does not change?

Let's assume the occurrence of the unlikely event where a 2d plane would fill up in an even distribution of particles in the form of a 2d circle. I would guess that there would be instabilities to the orbits due to small perturbations, and that the resultant motions would be extremely difficult to predict even if the particles were point masses and the chance of collisions were close to zero.
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Accepted Solution

ozo earned 400 total points
I'm getting elliptic integrals trying to solve the 2d case.
Would you want to just do a numerical solution?

I agree that the configuration seems unstable.
I'd expect that interactions would tend to circularize the orbits,
but that you'd also get spreading beyond the R boundary, while other particles
migrate toward the center.
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Author Comment

Hi Guys,

I am indeed trying to solve for a theoretical DISK, rather than sphere.   Sorry I said volume,  that was an error, I should have said area.
To clarify on random distribution, I did indeed mean an “even” distribution.

I would like to solve for what would, initially at least, be a circular orbit yes.
Regarding the evolution of the system: variations and clumping, I’m sure this will happen, and become more and more exaggerated over time.
But do we even need to consider that at this point?  We just need to determine an initial circular orbital velocity of a single particle.

Unfortunately I do need a formula for this, not just a particular solution.

What integrations are you doing Ozo?
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Assisted Solution

phoffric earned 100 total points
The mass of the entire system is M spread out evenly over an area of pi R*R. That gives you a density (kg/m^2).

For a particle at point p1 which is r1 distance from the center, you can compute the mass of the inner circle
mc_r1 = M (pi r1*r1)/(pi R*R) = M (r1/R)^2

mc_r1 can be treated as a point mass at the center of the circle.
The gravitational force due to mc_r1 on a particle at r1 is radial towards the center.

For the donut region with R > r > r1, you can break the donut into an infinite number of concentric rings of width dr. You know the delta area, dA, by subtracting the area of the inner circle from the outer circle.

Divide the dA into sections by delta theta. The area of this delta theta section is proportional to the 2pi around the circle. That area*density -> delta mass, dM. From symmetry, you don't have to worry about non-radial forces due to that delta mass.

Draw a radial line from the center through p1 to the rim at radius R.
Draw a line perpendicular to the radial of the element at p1. This perpendicualar line divides the concentric circle into two arcs.

For all of the delta mass in the smaller arc (i.e., above the line), you get a radial force away from the center of the circle. For all the delta mass below the line, you get a radial force towards the center of the circle. You can integrate half of an arc (keeping only the radial force) and multiply by 2. I think that if you get clever, you can compute the center of mass of that delta ring and put the dM at that point. Now you have dF due to one arc. Integrate these delta F's from r1 to R to get the outward force from r1 to R.

Now do the same type of computation for the inward force.
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Expert Comment

> mc_r1 can be treated as a point mass at the center of the circle
Unfortunately, that only works for a spherical distribution, or as an approximation when when r1>>R

When I tried integrating the radial component around theta, I got elliptic integrals,
and didn't see an obvious way to simplify it.
This paper got a slightly simpler equation solving for the potential instead of the force, but still had to use elliptic integrals:
http://www.tp4.rub.de/~jk/science/gravity/gravpot-disk.pdf
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Author Comment

Hi guys,

Stayed up a bit too late last night working on this.   I got to the differential equation, but no way I can solve it. (I'm trying to use Maxima to solve it now, but having troubles.)

Is this the same differential equation you got Ozo?

I am putting up my math, and diagrams as an image, so it's easier to read & post.
(by the way:  where I say "bit" in the equation descriptions  I mean an infinitesimally small section of the area.  Also, I have a wrong sign in one place on the diagram, too lazy to fix.)

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Expert Comment

Essentially the same integral that I got and that author of the gravpot-disk paper got.
There is no closed form solution for the dtheta integral,
but gravpot-disk paper shows a solution in terms of the complete elliptic integral of the first kind.
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Author Comment

Thanks for the quick reply Ozo!
I don't really understand much after the first line... but it doesn't sound good.

>>"There is no closed form solution for the dtheta integral"
Regarding closed form: Can we expression the solution in terms of say... a summation?  Sometimes we can stop iterating through the summation when we have achieved desired precision .. like.. Sin(x).  Or, perhaps we go the numerical solution you suggested, where the integration is expressed/approximated as a summation of finitely sized bits (as opposed to infinitesimal, like dR dTheta)?  I guess if I need to write a program and make a couple of loops to get an approximate solution, that would be better than nothing, though not ideal.

>>"...a solution in terms of the complete elliptic integral of the first kind"
I don't really understand what this means;  Can we somehow use this like a formula to get a final numerical solution, given the starting values?
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Expert Comment

Yes, there are summations to approximate elliptic integrals.  For example:
http://en.wikipedia.org/wiki/Elliptic_integral#Asymptotic_expressions
http://functions.wolfram.com/EllipticIntegrals/EllipticK/introductions/CompleteEllipticIntegrals/05/
http://functions.wolfram.com/EllipticIntegrals/EllipticK/06/01/01/01/
Which might be sufficiently complicated, with enough opportunities to apply the wrong elliptic integral that it may be easier to just sum the forces from the bits.
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Author Comment

Had a thought, but not enough math skills to see if it will actually help:

We can intuit that solving the integral for the Y component, however difficult that may be,  yields ZERO.

The only difference between the two integrals would be the multiplier:
(R-cos(t))    (for x component)
vs
sin(t)          (for y component)

Again, not quite sure how, or if, we can use that fact.
Going to read that stuff now, though I suspect it will be beyond me.
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Author Closing Comment

Unexpectedly, we found the problem not really possible to solve in terms of an exact formula.
Many thanks for all your help guys.
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Expert Comment

For the y component, sin(t) always cancels sin(2pi-t) when you do the integral from 0 to 2pi.
No such cancellation occurs for (R-cos(t))
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