# Gravity due to a ring and gravity due to a disc

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#1
Hi everyone! Just wanted to clarify something when I was deriving expressions for the gravitational field strength due to a single ring, and due to a single disc (separately) . for the ring I had to include the cos(theta) whereas for the disc i didn't? Is this correct? and any reasoning? 0
3 years ago
#2
(Original post by marinacalder)
Hi everyone! Want to clarify something when I was deriving expressions for grav field strength due to a ring and disc. For the ring I had to include [email protected], for the disc I didn't, is this correct? I believe you mean grav field strength at point P on an axis through centre of ring, perpendic to ring's plane. P is at distance x from ring's centre.

At P the vert components of grav field strength vectors all cancel one another in diametric pairs. So you're left with their horiz components (hence using [email protected]), all point towards ring (accelerates P towards ring). Sum g's from all dm's by integrating from L = 0 to L = 2πR, via dm = (M/2πR)dL.

This helps you find grav field strength at P for a disc; same radius R, on the same plane, sharing same centre. Just sum the g's of an infinite no. of rings, varying radius of 0 to R. Note dm = (2πrdr)(M*/πR^2) here, integrate from r = 0 to r = R. 0
#3
(Original post by Physics Enemy)
I believe you mean the grav field strength at some point P on the axis running through the centre of the ring, perpendicular to ring's plane, at any distance x.

At any such P, the vert components of the grav field strength vectors all cancel one another in diametric pairs. So you're left with their horiz components (hence using [email protected]), all pointing towards the ring (accelerates P towards the ring). Sum the g's from all dm's by integrating from L = 0 to L = 2πR, via dm = (M/2πR)dL.

This enables you to calc grav field strength at point P for a disc instead; same radius R, on the same plane, sharing same centre. All you do is sum g's of an infinite number of rings, varying in radius from 0 to R. Just need to note dm = (2πrdr)(M*/πR^2) here, and you're integrating from r = 0 to r = R instead. so it's correct to be leaving out the cos(theta) for the disk? 0
3 years ago
#4
(Original post by marinacalder)
Correct to leave out [email protected] for the disk? Yes, as you've already used it for the ring. You then integrate over an infinite no. of rings, radius 0 to R. So the issue of cancelling vector components and summing g_x's only, doesn't repeat itself. 0
#5
(Original post by Physics Enemy)
Well yes as you've already used it for the ring - you're then just integrating over an infinite number of rings, varying in radius from 0 to R. So the issue of cancelling vector components doesn't repeat itself. ahh ok, so first you find the ring, with the cos (theta) and then you use that value?

You don't start again separately for the disk and not use cos(theta)?

Sorry I'm having a blond moment 1
#6
(Original post by Physics Enemy)
Yeah, the most common or only way to do it. I think it was the way the question was worded that confused me, Starting from scratch for a disc... I'd consider one ring first, using the cos(theta), and then I'd integrate this expression for all the radii that make up a disk? 0
3 years ago
#7
(Original post by marinacalder)
...
Kinda. Use general result for a ring, this is the most common way. Then integrate to sum infinite no. of rings, radius 0 to R. But note expression for dm is different, we integrate radially instead of along a path. May be worth googling derivations. 0
#8
(Original post by Physics Enemy)
Kinda, use test case for a ring then integrate to sum an infinite no. rings, radius 0 to R. But note expression for dm is different and we integrate radially rather than along a path. May be worth googling derivations. ok, awesome!
One more thing, I'm guessing I would ignore the resolving for deriving the potential due to either the ring or disc since its a scalar?
0
3 years ago
#9
(Original post by marinacalder)
Guess I'd ignore resolving to derive potential due to ring/disc, as it's scalar?
Yes, on a given ring each dm contributes the same potential at P. Then go through steps for ring/disc derivations similar to those of grav field strength.

Btw it's possible to do such problems using polar coordinates (r, θ), then do double integrals (dr dθ), but it requires more knowledge:

Ring: https://youtu.be/q5EWLNdv_pE
Disc: https://youtu.be/Xdk8PR_SZSI
0
3 years ago
#10
Hey, I've attached a diagram to help, this is how I've derived them: marinacalder

Ring: To derive g_p we need to sum g_x's from all dm's around the ring; g_x is the same for each dm (@, g_dm are the same). Distance between P and any dm is √(x^2 + R^2). So [email protected] = x/√(x^2 + R^2). So g_x = [Gdm/(x^2 + R^2)]x/√(x^2 + R^2) = Gxdm/(x^2 + R^2)^3/2.

Sum all of these around the ring to get g_p = Int Gx/(x^2 + R^2)^3/2 dm. But dm = (M/2πR)dL, so g_p = Int (0 to 2πR) [Gx/(x^2 + R^2)^3/2](M/2πR)dL = [Gx/(x^2 + R^2)^3/2](M/2πR) Int (0 to 2πR) dL = [Gx/(x^2 + R^2)^3/2](M/2πR)(2πR)
=> g_p = GMx/(x^2 + R^2)^3/2.

If P is far away from C, x >> R, x^2 + R^2 -> x^2, so g_p -> GM/x^2; this is Newton's grav law for point masses, so we can consider ring to be point mass M at C. If P is close to C, R >> x, x^2 + R^2 -> R^2, so g_p -> GMx/R^3 i.e) linear dec for smaller x. As x -> 0 (P -> C), g_p -> 0 as expected (net F = 0 at C).

Re: g_p (max), we can use calculus: d(g_p)/dx = GM[1/(x^2 + R^2)^3/2 - 3x^2/(x^2 + R^2)^5/2] = 0 => 1/(x^2 + R^2)^3/2 = 3x^2/(x^2 + R^2)^5/2 => x^2 + R^2 = 3x^2 => x = (R√2)/2. So g_p (max) = GMR/{√2[(3/2)R^2]^3/2} = 2GM/(3√3.R^2).

Disk: All we do is sum an infinite no. of rings of varying radius 0 to R. Can use g_p = GMx/(x^2 + R^2)^3/2, but adapted for any ring of mass dm and radius r, so g_r = Gxdm/(x^2 + r^2)^3/2. Then g_p* = Int Gxdm/(x^2 + r^2)^3/2. For any such ring of thickness dr: dm = (M*/πR^2)dA = (M*/πR^2)(2πrdr) = (2M*/R^2)rdr.

So g_p* = Int (0 to R) Gx/(x^2 + r^2)^3/2 dm = (2M*Gx/R^2) Int (0 to R) r.(r^2 + x^2)^(-3/2) dr = (GM*x/R^2) Int (0 to R) 2r.(r^2 + x^2)^(-3/2) dr = (GM*x/R^2).(-2)(r^2 + x^2)^(-1/2) | (0 to R) = (-2GM*x/R^2)[1/√(R^2 + x^2) - 1/x]
=> g_p* = (2GM*/R^2)[1 - x/√(x^2 + R^2)] > 0.

So if P is far away from C, x >> R, x^2 + R^2 -> x^2, so g_p* -> 0. If P is close to C, R >> x, x^2 + R^2 -> R^2, so g_p* -> (2GM*/R^2)(1 - x/R). i.e) linear inc for smaller x. As x -> 0 (P -> C), g_p* -> 2GM*/R^2, so max g_p* occurs at C.

Also worth re-visiting the other (edited) posts, helps it all sink in. 0
#11
(Original post by Physics Enemy)
Hey, I've attached a diagram to help, this is how I've derived them: marinacalder

Ring: To find g_p we need to sum g_x's from all dm's around the ring; g_x is the same for each dm as @, g_dm are the same. Distance between P and any dm is √(x^2 + R^2). So [email protected] = x/√(x^2 + R^2). So g_x = Gdm/(x^2 + R^2) . x/√(x^2 + R^2) = Gxdm/(x^2 + R^2)^3/2.

Sum all of these around the ring to get g_p = Int Gx/(x^2 + R^2)^3/2 dm. But dm = (M/2πR)dL, so g_p = Int (0 to 2πR) Gx/(x^2 + R^2)^3/2 . (M/2πR) dL = Gx/(x^2 + R^2)^3/2 . (M/2πR) Int (0 to 2πR) 1 dL = Gx/(x^2 + R^2)^3/2 . (M/2πR)(2πR) => g_p = GMx/(x^2 + R^2)^3/2.

If P is far away from C, x >> R, x^2 + R^2 -> x^2, so g_p -> GM/x^2; Newton's grav law for 2 point masses, so we can consider ring to be point mass M at C. If P is close to C, R >> x, x^2 + R^2 -> R^2, so g_p -> GMx/R^3 i.e) linear dec for smaller x. As x -> 0 (P -> C), g_p -> 0 as expected i.e) net F is 0 at C => net g is 0 too.

Re: g_p (max), we can use calculus: d(g_p)/dx = GM[1/(x^2 + R^2)^3/2 - 3x^2/(x^2 + R^2)^5/2] = 0 => 1/(x^2 + R^2)^3/2 = 3x^2/(x^2 + R^2)^5/2 => x^2 + R^2 = 3x^2 => x = (R√2)/2. So g_p (max) = GMR/{√2[(3/2)R^2]^3/2} = 2GM/(3√3.R^2).

Disk: All we do is sum an infinite no. of rings of varying radius 0 to R. So we can use g_p = GMx/(x^2 + R^2)^3/2, but adapted for any ring of mass dm and radius r, so g_r = Gxdm/(x^2 + r^2)^3/2. Then g_p* = Int Gxdm/(x^2 + r^2)^3/2. For any such ring of thickness dr: dm = (M*/πR^2)dA = (M*/πR^2)(2πrdr) = (2M*/R^2)rdr.

So g_p* = Int (0 to R) Gx/(x^2 + r^2)^3/2 dm = (2M*Gx/R^2) Int (0 to R) r.(r^2 + x^2)^(-3/2) dr = (GM*x/R^2) Int (0 to R) 2r.(r^2 + x^2)^(-3/2) dr = (GM*x/R^2).(-2)(r^2 + x^2)^(-1/2) | (0 to R) = -2GM*x/R^2[1/√(R^2 + x^2) - 1/x] => g_p* = (2GM*/R^2)[1 - x/√(x^2 + R^2)] > 0.

So if P is far away from C, x >> R, x^2 + R^2 -> x^2, so g_p* -> 0. If P is close to C, R >> x, x^2 + R^2 -> R^2, so g_p* -> (2GM*/R^2)(1 - x/R). i.e) linear inc for smaller x. As x -> 0 (P -> C), g_p* -> 2GM*/R^2, so max g_p* occurs at C.

Also worth re-visiting the other (edited) posts, helps it all sink in. *virtual hug* 1
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