Gravitational force between a thin circle and a particle?

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Offline David LaPierre

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I was able to calculate the force between a thin ring and a particle somewhere on the line perpendicular to the plane of the ring as G*r*m1*m2/(r^2+d^2)^1.5 where r is the radius of the ring and d is the distance from the particle to the center of the ring.

I'm having trouble extending this to a thin circle, a disc, oriented the same way. Can anyone help me out?

Cheers :-)

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Offline lightarrow

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Re: Gravitational force between a thin circle and a particle?
« Reply #1 on: 04/03/2013 20:51:03 »
I was able to calculate the force between a thin ring and a particle somewhere on the line perpendicular to the plane of the ring as G*r*m1*m2/(r^2+d^2)^1.5 where r is the radius of the ring and d is the distance from the particle to the center of the ring.

I'm having trouble extending this to a thin circle, a disc, oriented the same way. Can anyone help me out?

Cheers :-)

If m2 is the ring's mass, you have to imagine it as infinitely thin, if you want to compute its contribute in a disk of radius R. It means its infinitesimal mass dm is: dm = k*a*da and k is such that the total mass of the disk M is the integral of dm:
M = Integral[a=0;a=R] dm = Integral[a=0;a=R] k*a*da = k*R2/2
so k = 2M/R2.

Then you sum the gravitational contributes of every infinitesimal ring of radius a:

Integral[a=0;a=R] {G*a*m1/(a2+d2)3/2}  k*a*da =

= 2M*G*m1/R2 * Integral[a=0;a=R] {a2/(a2+d2)3/2} * da
« Last Edit: 04/03/2013 20:52:36 by lightarrow »

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Offline David LaPierre

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Re: Gravitational force between a thin circle and a particle?
« Reply #2 on: 04/03/2013 23:18:34 »
Thanks for the reply! Would you mind explaining the terms used in your equations? I'm having a hard time trying to figure it out still.
« Last Edit: 05/03/2013 00:24:24 by David LaPierre »

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Offline David LaPierre

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Re: Gravitational force between a thin circle and a particle?
« Reply #3 on: 05/03/2013 10:37:31 »
So I need to integrate G*r*m1*m2/(r^2+d^2)^1.5 in terms of r, from 0 to the radius of the circle. But to do that I need to represent the ring mass, m2, as determined by r?
So if M is the disk mass, and R is the disc radius...
m2/M=(pi(r+dr)^2-pir^2)/piR^2=(2rdr+dr*dr)/R^2
So m2=M(2rdr+dr*dr)/R^2
Can I disregard the dr*dr?

So now I have int G*r*m1*M*2rdr/(R^2*(r^2+d^2)^1.5) from 0 to R

Which works out to
2G*M*m1*d/R^2 * (1/d-1/(R^2+d^2)^.5)

Is this correct? My biggest concern is what to do with that dr*dr

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Offline lightarrow

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Re: Gravitational force between a thin circle and a particle?
« Reply #4 on: 05/03/2013 18:04:19 »
Thanks for the reply! Would you mind explaining the terms used in your equations? I'm having a hard time trying to figure it out still.
Integral[a=0;a=R] {a2/(a2+d2)3/2} * da

means that you have to integrate in da between 0 and R the function of a:

a2/(a2+d2)3/2

If you prefer, you can use x instead of a:

Integral[x=0;x=R] {x2/(x2+d2)3/2} * dx

Since you were able to compute the gravitational force between the point and the ring, I assumed you were able to compute integrals...

Anyway, if I computed correctly, the result shoud be:

2M*G*m1/R2 * {log [sqrt(R2/d2 + 1) + R/d] - (R/d) / sqrt(R2/d2 + 1)  }.

Assuming your initial result is correct.
« Last Edit: 05/03/2013 18:36:38 by lightarrow »

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Offline lightarrow

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Re: Gravitational force between a thin circle and a particle?
« Reply #5 on: 05/03/2013 18:13:53 »
So I need to integrate G*r*m1*m2/(r^2+d^2)^1.5 in terms of r, from 0 to the radius of the circle. But to do that I need to represent the ring mass, m2, as determined by r?
Yes, and this time is not a finite mass anylonger, is an infinitesimal mass: you "cut" the disk in infinitely many rings of infinitesimal tickness dr, you compute the infinitesimal gravitational force on the point made by these rings, then you "sum" all these forces. Since the forces are infinite in number, varying continuously (as r varies continuously from 0 to R) you make an "integration" instead of a sum.
Quote

So if M is the disk mass, and R is the disc radius...
m2/M=(pi(r+dr)^2-pir^2)/piR^2=(2rdr+dr*dr)/R^2
So m2=M(2rdr+dr*dr)/R^2
Can I disregard the dr*dr?

So now I have int G*r*m1*M*2rdr/(R^2*(r^2+d^2)^1.5) from 0 to R

Which works out to
2G*M*m1*d/R^2 * (1/d-1/(R^2+d^2)^.5)

Is this correct? My biggest concern is what to do with that dr*dr
Sorry but I didn't understand what you have done (I'm quite tired in this moment). I suggest you to follow the method I have used.

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Offline lightarrow

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Re: Gravitational force between a thin circle and a particle?
« Reply #6 on: 05/03/2013 18:40:57 »
I have a doubt: the point particle is on the disk's axis or not?

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Offline lightarrow

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Re: Gravitational force between a thin circle and a particle?
« Reply #7 on: 05/03/2013 19:42:22 »
I was able to calculate the force between a thin ring and a particle somewhere on the line perpendicular to the plane of the ring as G*r*m1*m2/(r^2+d^2)^1.5 where r is the radius of the ring and d is the distance from the particle to the center of the ring.
Are you sure it's not, instead, G*d*m1*m2/(r^2+d^2)^1.5 ?

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Offline David LaPierre

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Re: Gravitational force between a thin circle and a particle?
« Reply #8 on: 05/03/2013 20:57:06 »
I was able to calculate the force between a thin ring and a particle somewhere on the line perpendicular to the plane of the ring as G*r*m1*m2/(r^2+d^2)^1.5 where r is the radius of the ring and d is the distance from the particle to the center of the ring.
Are you sure it's not, instead, G*d*m1*m2/(r^2+d^2)^1.5 ?

Whoops, typo, you are correct  :)
And I have done a bit with integrals, although that was long long ago haha. I only needed some basic trig to calculate the force of the ring. I've been brushing up on my calculus, I'll post again with some clarification once I get it figured a bit clearer.

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Offline Pmb

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Re: Gravitational force between a thin circle and a particle?
« Reply #9 on: 05/03/2013 22:53:08 »
Ah! I love the smell of physics in the morning. Smells like ... victory! :)

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Offline David LaPierre

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Re: Gravitational force between a thin circle and a particle?
« Reply #10 on: 06/03/2013 12:25:52 »
Is it correct to say that the ratio of the ring mass to the disc mass is equal to the ratio of the ring's surface area to the disc's surface area? (surface area only in regard to the face nearest the particle)

So, dm/M = (pi*(r+dr)^2 - pi*r^2) / (pi*R^2)

Isolating dm and canceling pi gives

dm = M * ((r+dr)^2 - r^2) / R^2
Or
dm = M * (2rdr + (dr)^2) / R^2

If that is all good, what do I do with the squared dr? Those sorts of terms never cease to confuse me..

*edit*
It think it can just be disregarded. Conceptually, it seems as if integrating a function with a term like dr is almost like adding an infinite number of infinitely small objects. Infinitely small because one of their dimensions gets indefinitely close to 0. Whereas with the squared dr term, it falls short by 2 'dimensions,' and is therefore practically meaningless.

That explanation will probably make a lot of people cringe haha. I'd love a more precise explanation if the concept is at all accurate.
« Last Edit: 06/03/2013 12:37:11 by David LaPierre »

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Offline lightarrow

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Re: Gravitational force between a thin circle and a particle?
« Reply #11 on: 06/03/2013 15:43:35 »
Ah! I love the smell of physics in the morning. Smells like ... victory! :)
Probably it's because I have flu, but I can't grasp what you mean [???]

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Offline flr

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Re: Gravitational force between a thin circle and a particle?
« Reply #12 on: 07/03/2013 02:03:33 »
Does this forum has the option to use latex for equations? Or anything else (mathml?) that could help with mathematical formulas.

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Offline Pmb

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Re: Gravitational force between a thin circle and a particle?
« Reply #13 on: 07/03/2013 02:42:09 »
Ah! I love the smell of physics in the morning. Smells like ... victory! :)
Probably it's because I have flu, but I can't grasp what you mean [???]
I meant that I love this thread. A lot of technical cool mathy physics. :)

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Offline imatfaal

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Re: Gravitational force between a thin circle and a particle?
« Reply #14 on: 07/03/2013 11:43:58 »
yeah it does have latex (although the rendering sucks)

the tags it uses are [_tex][_/tex]

remove the underscores obviously for beautiful latex (or in fact cramped oldyworldy latex)

[tex]\frac{top}{bottom}= x^2+3x-1[/tex]
Thereís no sense in being precise when you donít even know what youíre talking about.  John Von Neumann

At the surface, we may appear as intellects, helpful people, friendly staff or protectors of the interwebs. Deep down inside, we're all trolls. CaptainPanic @ sf.n

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Offline lightarrow

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Re: Gravitational force between a thin circle and a particle?
« Reply #15 on: 07/03/2013 13:51:30 »
Is it correct to say that the ratio of the ring mass to the disc mass is equal to the ratio of the ring's surface area to the disc's surface area? (surface area only in regard to the face nearest the particle)
Yes.
Quote

So, dm/M = (pi*(r+dr)^2 - pi*r^2) / (pi*R^2)

Isolating dm and canceling pi gives

dm = M * ((r+dr)^2 - r^2) / R^2
Or
dm = M * (2rdr + (dr)^2) / R^2

If that is all good, what do I do with the squared dr? Those sorts of terms never cease to confuse me..

*edit*
It think it can just be disregarded. Conceptually, it seems as if integrating a function with a term like dr is almost like adding an infinite number of infinitely small objects. Infinitely small because one of their dimensions gets indefinitely close to 0. Whereas with the squared dr term, it falls short by 2 'dimensions,' and is therefore practically meaningless.

That explanation will probably make a lot of people cringe haha. I'd love a more precise explanation if the concept is at all accurate.
It's ok. You could simply say that dr2 is an infinitesim of higher order than dr, so you can neglet dr2 in a sum with dr.
At the beginning I didn't understand what you did because for me it was very simple: the element dA of the ring's area is 2(pi)r*dr (circumference multiplicated radius variation, immediately).
« Last Edit: 07/03/2013 13:54:44 by lightarrow »

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Offline David LaPierre

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Re: Gravitational force between a thin circle and a particle?
« Reply #16 on: 08/03/2013 02:05:03 »
Ok, so a disk of radius 'a' and mass 'M' with a center point with a distance of 'x' from a particle of mass 'm' will have a gravitational force of 2GMm/a^2 * (1-1/(a^2/x^2+1)^.5)

My next question is, how can you apply that equation to calculate the force between a sphere and the particle?
Let's call the sphere radius 'r', the sphere mass 'M', the disc mass 'dM', and the distance from the closest point of the sphere to the particle 'y'.

I would think we could integrate the disk equation for x from x=y to y+2r
'a' could be expressed as (r^2-(r+y-x)^2)^.5
dM is a little more tricky, I believe it can be expressed as M/(4r^3) * (4r^3-(x-y-dx)^2(3r-x+y+dx)-(2r+y-x)^2(r+x-y))

Resulting in this...
F= integration of 2*G*m/(r^2-(r+y-x)^2)*(1-1/((r^2-(r+y-x)^2)/x^2+1)^.5)*(M*(4r^3-(x-y-dx)^2(3r-x+y+dx)-(2r+y-x)^2(r+x-y))/(4r^3)) from x=y to y+2r

Integrating this is a bit beyond me. There are two dx terms embedded in the equation, and I'm not sure how to deal with that haha.

Any suggestions?

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Offline imatfaal

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Re: Gravitational force between a thin circle and a particle?
« Reply #17 on: 08/03/2013 09:27:36 »
David - one problem you might find in trying to check your work is that nobody bothers to work out the gravity from a sphere like that.  We know from Newton that we can treat the gravity of a sphere as if all the mass is located in the centre
Thereís no sense in being precise when you donít even know what youíre talking about.  John Von Neumann

At the surface, we may appear as intellects, helpful people, friendly staff or protectors of the interwebs. Deep down inside, we're all trolls. CaptainPanic @ sf.n

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Offline David LaPierre

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Re: Gravitational force between a thin circle and a particle?
« Reply #18 on: 08/03/2013 12:29:48 »
That's actually what I was thinking about. I wanted to see how GMm/r^2 compared to whatever resulted from this :-)

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Offline imatfaal

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Re: Gravitational force between a thin circle and a particle?
« Reply #19 on: 08/03/2013 16:14:46 »
That's actually what I was thinking about. I wanted to see how GMm/r^2 compared to whatever resulted from this :-)

I am not sure I have ever seen the proof  - but surely it follows (in a hand-wavy way?) from the shell theorem - ie either Gauss or Newton depending on choice. 
Thereís no sense in being precise when you donít even know what youíre talking about.  John Von Neumann

At the surface, we may appear as intellects, helpful people, friendly staff or protectors of the interwebs. Deep down inside, we're all trolls. CaptainPanic @ sf.n

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Offline JP

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Re: Gravitational force between a thin circle and a particle?
« Reply #20 on: 08/03/2013 17:32:12 »
Here's an easier way (with tex), if you're willing to use cylindrical coordinates.  The disk is located a distance z from your observation point, which also lies along an axis through the disk.  Let the disk have constant mass density per unit area of [tex]\sigma[/tex].

The magnitude of force due to a point on the disk that is located a distance [tex]\rho[/tex] from the axis and at an angle [tex]\phi[/tex] measured going "around" the disk is:
[tex]F=\frac{G\sigma}{z^2+\rho^2}\rho\mathrm{d}\rho\mathrm{d}\phi[/tex],
where the term [tex]\rho\mathrm{d}\rho\mathrm{d}\phi[/tex] accounts for the tiny area subtended by a small change in rho and a small change in phi.

To consider the whole disk, you want to integrate [tex]\rho[/tex] from 0 to R, where R is the disk radius, and [tex]\phi[/tex] from 0 to [tex]2\pi[/tex].

BUT there is one more thing to do!  Force is a vector, so you have to add up directions of each bit of force when integrating.  Fortunately, you can make symmetry arguments to remove this.  For every force due to an element at position [tex]\rho[/tex] and angle [tex]\phi[/tex], there is a force on the opposite side of the axis through the disk.  When you add these vectors, only components pointing along the axis remain.  To account for this, you need an additional factor to use only the axial component of force in the above equation:
[tex]F_{\mathrm{ax}}=\frac{G\sigma z}{(z^2+\rho^2)^{3/2}}\rho\mathrm{d}\rho\mathrm{d}\phi[/tex]

This can be integrated to get you (assuming I didn't drop any factors)
[tex]F=2\pi G\sigma[1-\frac{z}{\sqrt{z^2+R^2}}][/tex]
« Last Edit: 08/03/2013 17:43:16 by JP »

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Offline JP

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Re: Gravitational force between a thin circle and a particle?
« Reply #21 on: 08/03/2013 18:36:52 »
Now, to go to a sphere is trickier, but not too bad with a few tricks.  The above equation holds for the force at position z along the axis of a disk, where the disk is assumed to be at z=0.  So first off, we have to consider the sphere centered at z=0 and a disk an amount [tex]\xi[/tex] from that.  The entire sphere will be many disks displaced by [tex]\xi[/tex] from -r to r, where r is the sphere's radius.  The next trick is to ask what the radius of a disk making up the sphere is when it's located at a distance [tex]\xi[/tex] from the center of the sphere.  From the formula for a sphere
[tex]R(\xi)=\sqrt{r^2-\xi^2}[/tex].
The next trick is to figure out what z is in terms of [tex]\xi[/tex].  This is fairly easy, since we can take Z to be the distance from the observation point to the sphere's center, and the disk is located a distance [tex]\xi[/tex] from the sphere's center, so
[tex]z(\xi)=Z-\xi[/tex].

Using all this in the above expression for the force due to a disk, we get:
[tex]F=2\pi G\sigma[1-\frac{Z-\xi}{\sqrt{Z^2+r^2-2Z\xi}}][/tex],
where [tex]\sigma[/tex] can now be interpreted as mass density per unit volume.  So the trick is taking the integral of this over [tex]\xi[/tex] from -r to r.  There are a few tricks to do this*, but if you do it, you come out with
[tex]F=\frac{4\pi G\sigma R^3}{3 Z^2}[/tex],
which is the usual formula for a spherical mass.


*: The trick I used was to realize that I can express [tex]\frac{Z-\xi}{\sqrt{Z^2+r^2-2Z\xi}}=\frac{\partial}{\partial Z}\sqrt{Z^2+r^2-2Z\xi}[/tex] and then integrating by changing variables to [tex]\sqrt{Z^2+r^2-2Z\xi}=x[/tex].  You could also use integration by parts, which takes more steps.
« Last Edit: 08/03/2013 18:40:13 by JP »

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Offline lightarrow

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Re: Gravitational force between a thin circle and a particle?
« Reply #22 on: 08/03/2013 21:55:10 »
This can be integrated to get you (assuming I didn't drop any factors)
[tex]F=2\pi G\sigma[1-\frac{z}{\sqrt{z^2+R^2}}][/tex]
Yes, but that is the force per unit mass of the point particle (on which the disk acts). If it has mass m1:

[tex]F=2\pi Gm_1\sigma[1-\frac{z}{\sqrt{z^2+R^2}}][/tex] =

= [tex]\frac{2 Gm_1M}{z^2+R^2+z\sqrt{z^2+R^2}}[/tex]

where M is the total disk' mass.

From this last equation is easier to compute, for example, the limit of the force when z or R goes to zero or to infinity.

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Offline JP

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Re: Gravitational force between a thin circle and a particle?
« Reply #23 on: 09/03/2013 00:46:56 »
Yes, you're absolutely right.  I neglected the test mass.

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Offline lightarrow

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Re: Gravitational force between a thin circle and a particle?
« Reply #24 on: 09/03/2013 19:58:59 »
Yes, you're absolutely right.  I neglected the test mass.
I wrote it just for the OP, if you had answered to me I would have read "field" and it would have been the same...