# Gravitational Potential of a ring, at a point P

1. Nov 28, 2012

1. The problem statement, all variables and given/known data
I am asked to calculate the gravitational potential of the ring at the point Q. I can do this for point P, but Q is killin me...

2. Relevant equations
V = GM/r
M = ρ2∏a

3. The attempt at a solution

Well for the case at point P, it seems pretty straight forward:
dV = GdM/y
where y is the distance between the point and the edge of the ring. It's a fixed distance (so a constant)

Therefore, by integration:
V = G/y ∫dM (from 0 to M)

V = GM/y

However, for the case at point Q, I am baffled...
It seems to me that in this case y is no longer a constant, and therefore requires integration from the near side of the ring to the far side.

But how to express y? A hint would be wonderful...

Last edited: Nov 28, 2012
2. Nov 28, 2012

### rude man

I can only think as follows:

Create new angle ψ subtended by radii between the ring center, the "up" (12 o'clock) position on the ring, and another point along the ring (i.e. ignore θ).

Set up a cartesian coordinate system with the ring center at x=y=z=0, equation of ring being
x2 + z2 = a2 and give position of Q as Q(x,y,z=0) or Q(x=0,y,z), without loss of generality. So y points along the ring axis, x points up and z points to the left, looking at the ring from P.

The distance to be integrated d is now the distance between each ring segment and Q, i.e. a function of ψ, to appear in the denominator of your integrand. Note I have replaced your old y with a new parameter d since y is now part of the xyz coordinate system in which the ring is defined (the ring is at y = 0). The integral would be of the form ρaG∫dψ/d with limits ψ = 0 to ψ = 2π, and ρ is linear ring density, i.e. dM = aρdψ.

The expression for d promises to be messy! But decent tables should help. Hope I am making sense.

3. Nov 29, 2012

### rude man

EDIT: I can only think as follows:
Ignore my previous post.

Create new angle ψ subtended by radii between the ring center, the "up" (12 o'clock) position on the ring, and another point along the ring (i.e. ignore θ).

Set up a cartesian coordinate system with the ring center at x=y=z=0, and give position of Q as Q(x0,y0,z=0), without loss of generality. So y points along the ring axis towards P, x points up and z points to the right, looking at the ring from P.

The distance to be integrated d is now the distance between each ring segment and Q, i.e. a function of ψ, to appear in the denominator of your integrand. Note I have replaced your old y with a new parameter d since y is now part of the xyz coordinate system in which the ring and Q are defined (the ring is at y = 0). The integral would be of the form ρaG∫dψ/d with limits ψ = 0 to ψ = 2π, and ρ is linear ring density, i.e. dM = aρdψ.

d is the distance between a ring element at (acosψ, 0, asinψ) and Q(x0,y0,0) where x0 and y0 are of course constants.

The expression for d promises to be messy! But decent tables should help. Hope I am making sense.

4. Nov 29, 2012

Hmmm that makes more sense. Thank you very much for your insight.

Another quick question: for the case where the distance along OP is >> than the radius of the ring, AND the distance of Q from the line that OP makes: Would the gravitational potential occur at the center of mass of the ring? i.e. would the gravitational potential simply be:
GMm/d where d is the distance from the point mass at Q to the center of mass of the ring?

5. Nov 30, 2012

### rude man

That is correct. You can easily see that that is the case for P if P is far away from the ring's c.g. (i.e. center), since then y ~~ r. By the same token what you say is also correct. (I checked this with a co-worker, a PhD. M.E. !)

6. Nov 30, 2012

What if we looked at this another way?

Let's say the point P lies in the plane of the ring. The gravitational potential at P is:
∫Gdm/s
where m = 2∏Rρ and dm = Rρdθ

So, V = ∫(G/s)Rρdθ
= GRρ ∫dθ/s

Using the law of cosines:
s = (r^2 + R^2 - 2rRcosθ)^1/2

Take out an r^2 (since that is the larger term)
s = r(1 + (R/r)^2 - 2(R/r)cosθ)^1/2

The integral now looks like:
V = GRρ/r ∫ (1 + (R/r)^2 - 2(R/r)cosθ)^-1/2 dθ

Using legendre polynomials, I can rewrite:
V = GRρ/r ∫ ƩPn(cosθ)(R/r)^n dθ

This was all from 0 to 2∏, but lets evaluate from 0 to ∏ and multiply by 2:
V = 2GRρ/r ∫ ƩPn(cosθ)(R/r)^n dθ (from 0 to ∏)

after expanding, integrating, etc, I get:
V = (2GRρ∏/r)(1 - R/r + (R/r)^2 - (R/r)^3 + ....

m = 2∏Rρ, so,

V = (GM/r)(1 - R/r + (R/r)^2 - (R/r)^3 + ....

which looks a LOT like V = GM/r, and you can see that the series terms go away if r >> R.

Now, what about Q? Can you simply say that:

Cos (x) = V/(V at Q)

where x is the polar angle pointing up to Q from the plane of the ring? Or am I crazy?

7. Nov 30, 2012

### D H

Staff Emeritus

You aren't going to be able to solve this using standard integration techniques. Has the class covered special functions? If so, use that knowledge.

If the class hasn't covered special functions, you can always write the potential as an infinite series.

8. Nov 30, 2012

### rude man

Adoniram - I've forgotten all about Legendre polynomials so can't comment on your new idea. Sorry, wish I could.

DH - why can't 'standard integration techniques' be used? Only problem I see is finding a good table of integrals, e.g. Jahnke & Emde.

9. Nov 30, 2012

### rude man

Assuming you worked the Legendre polynomial right, this far looks good. But ..

Problem: then V(Q) = V(P)/cos(x) so what happens at x = π/2? V(P) is then the potential at the ring's center, a finite quantity ... ... is your line construct from Q to the ring's center justifiable except for large distances where we agree the c.m. determines the whole show?

10. Dec 1, 2012

### rude man

Last shot: d of my Nov.29, 3:55 p.m. MST post is not that bad. It simplifies to the form
d = √{c - d*cos(ψ)}, c, d constants (functions of ring radius and position of Q).
.

11. Dec 1, 2012

Yeah, you're right... my idea would be close, but would be better at greater distances, and pretty bad at close distances.

I will try to work it out with your value for d. Thank you again, and everyone, for your help with this one. Sheesh!

12. Dec 2, 2012

### D H

Staff Emeritus
That's right. Using half angle formulae and a bit of manipulation with lead to an elliptic integral of the first kind,
$$F(\theta|k^2) = \int_0^{\theta} \frac{d\phi}{\sqrt{1-k^2\sin^2\phi}}$$
That's the special function I was talking about.

13. Dec 2, 2012

### rude man

OK, you probably noticed I used d twice in that expression for d. Inadvertently! So make the constants b and c instead of c and d. You shouldn't have any trouble identifying b and c with radius a and Q position coordinates x0 and y0, with z0 = 0 as before.

14. Dec 2, 2012