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Gravitational potential of a ring at an off axis point

  • Thread starter NruJaC
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  • #1
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Homework Statement



Find the potential at off-axis points due to a thin circular ring of radius a and mass M. Let R be the distance from the center of the ring to the field point and [tex]\theta[/tex] be the angle between the line connection the center of the ring with the field point and the axis of the ring. Assume R>>a so that terms of order (a/R)3 and higher may be neglected.

Homework Equations



[tex]\Phi[/tex]=[tex]\int[/tex][tex]\frac{-Gdm}{r}[/tex]

The Attempt at a Solution


The main problem is to express r as a function of a, R, and [tex]\theta[/tex]. I approached it by first looking at the potential due to an arbitrary point on the ring, and letting the field point lie in the yz plane (the ring is in the xy plane). r can then be decomposed via the pythagorean theorem into two parts: the distance in the xy plane from the y intercept of the ring to the point on the ring, and the distance from the y-intercept up to the field point. I'll call the former l and the latter k. k can be found via the law of cosines as k2=a2+R2-2aRcos(90-[tex]\theta[/tex]). lf I then let [tex]\phi[/tex] be the angle from the y-axis to the radial line connecting the point on the ring with the center of the ring, l is the third side of an equilateral triangle and so sin([tex]\frac{\phi}{2}[/tex])=[tex]\frac{l}{2a}[/tex]. At this point, dm can be set up. dm=[tex]\rho[/tex]*dx where dx is the portion of the ring that we are looking at. Since the segment of the ring we are looking at is marked off by a d[tex]\phi[/tex], and for a small change like that, l makes a good estimate, dx=dl, and so dm=[tex]\rho[/tex]dl. dl of course can be put in terms of d[tex]\phi[/tex]=a*cos([tex]\frac{\phi}{2}[/tex])*d[tex]\phi[/tex]. Putting all of this together yields the following integral:
[tex]\frac{-GM}{4*\pi*a}[/tex][tex]\int[/tex][tex]\frac{2*a*cos(\frac{\phi}{2})d\phi}{\sqrt{4*a^{2}sin^{2}(\frac{\phi}{2})+k^{2}}}[/tex]
This really complex looking integral simplies (way too nicely) under a change of variables to:
[tex]\frac{-GM}{4*\pi*a}[/tex]*[tex]\int[/tex]sec([tex]\varphi[/tex])d[tex]\varphi[/tex] from 0 to 2[tex]\pi[/tex].
This yields 0. If you can spot what I did wrong, that'd be great. If you can think of an easier method, I'd appreciate that as well. Thanks for any insight you guys can provide!

Arjun
 

Answers and Replies

  • #2
288
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Since R>>a use a binomial expansion to the given order. Note also that due to symmetry the potential is independent of phi. The binomial expansion should be performed before integration so the integral is easy to evaluate. Use the expression

[tex]U_{g}=\int\frac{Gdm}{|\overrightarrow{r'}-\overrightarrow{r}|}[/tex]

where

[tex]\overrightarrow{r'}[/tex]

is the vector from the center of the ring to the ring element and

[tex]\overrightarrow{r}[/tex]

is the vector from the center of the ring to the point in space. The magnitude of the difference of these two vectors is the the distance from the ring element to the point in space.
 
  • #3
23
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Since R>>a use a binomial expansion to the given order. Note also that due to symmetry the potential is independent of phi.
I'm not sure I see what you mean. It can't be independent of [tex]\phi[/tex] because as [tex]\phi[/tex] changes, the distance from the new point on the ring to the field point has changed.

In your formulation, what would I use dm as in order to integrate? dr'? In that case, I'd set it up as a polar derivative, bringing [tex]\phi[/tex] right back into the picture.
 
  • #4
288
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Phi is the angle that lies in the plane of the ring so this is axial symmetry. Your expression will have the angle phi but when evaluating the integral the phi dependence will drop out. The value dm can be expressed in terms of the linear density, sigma, the radius of the ring, and the differential angle d(phi). Namely,

[tex]dm=\sigma ad\phi[/tex]
 
  • #5
23
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Thanks, I think I see what I did wrong. Had the wrong dm->d[tex]\phi[/tex] relationship. Having a little trouble with the taylor expansion, but I'm gonna keep trying to work it out. I just wish I was get a nice series of [tex](\frac{a}{R})^{n}[/tex], but I probably need to tweak it until I see it.
 
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  • #6
288
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The binomial expansion is

[tex](1+b)^n=1+nb+\frac{n(n+1)}{2!}b^2+\frac{n(n-1)(n-2)}{3!}b^3+...[/tex]

where b<<1.

Your expression for the distance should be

[tex](R^2+a^2-2aRsin\theta cos\phi)^{\frac{-1}{2}[/tex]

Factor out R2 from the distance expression then use the expansion to the order indicated. This will put it in a form that can be integrated with respect to phi.
 
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  • #7
23
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Oh! I see. I was trying to do a taylor expansion as f(x)=f(0)+f'(0)*x/2!... etc., and trying to figure out why it was going to hell, and why nothing looked the way it was supposed to. Thanks so much, I think I can take it from there. Hehe, looks like I really need to go review my series expansion stuff, it appears I've forgotten it.


Thanks,
Arjun
 
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  • #8
23
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Wait, I don't see where you got the expression for the distance. It appears that you're using a different angle phi than I am. I have phi as the horizontal angle that I plan to integrate around the ring. Am I missing some part of the geometry?

Thanks,
Arjun
 
  • #9
23
0
Got it! I had my diagram wrong. I was connecting k and l to the ring for some reason, and by dropping that requirement, I quickly get the result you got.

EDIT: My final answer came out to

[tex]-\frac{GM}{2R^{3}}[2R^{2}-a^{2}(1+\frac{1}{4}sin^{2}(\theta}))][/tex].

Look right?

Thanks for all your help,
Arjun
 
Last edited:
  • #10
288
0
Arjun,

My answer is a little different. Looks like a bit of algebra tweaks. Open my website below to see how I solved it. You are very close and put in good effort so I think it's ok to show you how it can be solved.

http://www.allenisd.org/facstaff2.nsf/Pages/1BC035BA0B2BE6BF86257512005D5DE9/$FILE/Symon%206-15.PDF [Broken]

chrisk
 
Last edited by a moderator:
  • #11
23
0
Oh wow, I'm retarded, I was using (n+1) rather than (n-1). Had the same thing as you other than that.

Thanks so much, I couldn't have done it without your help.

Arjun
 

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