Post by: David LaPierre on 04/03/2013 20:15:18
I'm having trouble extending this to a thin circle, a disc, oriented the same way. Can anyone help me out?
Cheers :-)
Post by: lightarrow 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
Post by: David LaPierre on 04/03/2013 23:18:34
Post by: David LaPierre on 05/03/2013 10:37:31
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
Post by: lightarrow 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.
Post by: lightarrow 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
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.
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
Post by: lightarrow on 05/03/2013 18:40:57
Post by: lightarrow 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 ?
Post by: David LaPierre 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.
Post by: Pmb on 05/03/2013 22:53:08
Post by: David LaPierre on 06/03/2013 12:25:52
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.
Post by: lightarrow 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 [???]
Post by: flr on 07/03/2013 02:03:33
Post by: Pmb on 07/03/2013 02:42:09
I meant that I love this thread. A lot of technical cool mathy physics. :)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 [???]
Post by: imatfaal on 07/03/2013 11:43:58
the tags it uses are [_tex][_/tex]
remove the underscores obviously for beautiful latex (or in fact cramped oldyworldy latex)
Post by: lightarrow 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
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.
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.
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).
Post by: David LaPierre on 08/03/2013 02:05:03
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?
Post by: imatfaal on 08/03/2013 09:27:36
Post by: David LaPierre on 08/03/2013 12:29:48
Post by: imatfaal 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.
Post by: JP on 08/03/2013 17:32:12
.The magnitude of force due to a point on the disk that is located a distance
from the axis and at an angle 
measured going "around" the disk is:
,where the term
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
from 0 to R, where R is the disk radius, and from 0 to 
.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
and angle , 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:
This can be integrated to get you (assuming I didn't drop any factors)
Post by: JP on 08/03/2013 18:36:52
from that. The entire sphere will be many disks displaced by 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 from the center of the sphere. From the formula for a sphere
.The next trick is to figure out what z is in terms of
. 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 from the sphere's center, so 
.Using all this in the above expression for the force due to a disk, we get:
,where
can now be interpreted as mass density per unit volume. So the trick is taking the integral of this over from -r to r. There are a few tricks to do this*, but if you do it, you come out with 
,which is the usual formula for a spherical mass.
*: The trick I used was to realize that I can express
and then integrating by changing variables to 
. You could also use integration by parts, which takes more steps.
Post by: lightarrow on 08/03/2013 21:55:10
This can be integrated to get you (assuming I didn't drop any factors)Yes, but that is the force per unit mass of the point particle (on which the disk acts). If it has mass m1:
==
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.
Post by: JP on 09/03/2013 00:46:56
Post by: lightarrow 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...