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flawedprefect
2010-Jun-08, 05:04 AM
Hi guys,

I am wondering if anyone knows the answer. Say debris falls into a planet's orbit; a small moon falls within the planet's roche limit, or whatever, where does the ring end up and why?

01101001
2010-Jun-08, 05:19 AM
Cassini Mission FAQ: Why do Saturn's rings all lie in the same plane? Why are planetary rings always found in their equatorial planes and not sometimes crossing their poles? (http://saturn.jpl.nasa.gov/faq/FAQSaturn/index.cfm#q14)


If Saturn captures particles coming in from other directions (along the polar plane, for example), they will tend to be pulled toward the equatorial plane, too. Saturn's rapid rotation creates a centrifugal effect that produces a bulge around its equator. With more mass around its equator than at its poles, Saturn's gravity is stronger around its middle, so incoming particles would tend to be drawn there. Once in the area, they'd be likely to collide with some of the many other particles orbiting in that plane, which would rob them of their initial momentum and encourage them to join the throng moving along the equatorial plane.

Bearded One
2010-Jun-08, 11:59 PM
Hmm, so what if you have a planet that's perfectly spherical and of uniform density? Say a loose ball bearing from a truly giganormous machine?

Spoons
2010-Jun-09, 12:02 AM
I don't think that will generally occur anyway, since rotation will change the shape of the body, effectively squashing it at the poles. If there's any rotation at all the orbits would tend towards the equator, wouldn't they?

Bearded One
2010-Jun-09, 12:55 AM
The engineers made it so it keeps it's shape. It's actually a thought experiment, IF such a body existed would rings still end up around the equator? I'm thinking there would still be a tendency for them to end up at the equator due to frame dragging but with small planet sized bodies I suspect it would take a very long time to happen.

Spoons
2010-Jun-09, 01:25 AM
Yes, I'd expect in the fullness of time there would be an eventual tendency for the orbiting object toward the equator, though if it were made to perfectly retain its shape I expect it'd be a very slow process. If it were to have any degree of retained or collected atmosphere that would exacerbate the rate it moves towards the equator, wouldn't it?

RalofTyr
2010-Jun-09, 01:55 AM
I suppose if you had an object that didn't rotate...at all, then there'd be no tidal-bulge to pull the particles towards the equator.

cjameshuff
2010-Jun-09, 03:10 AM
The engineers made it so it keeps it's shape. It's actually a thought experiment, IF such a body existed would rings still end up around the equator? I'm thinking there would still be a tendency for them to end up at the equator due to frame dragging but with small planet sized bodies I suspect it would take a very long time to happen.

They would still form rings, simply due to momentum perpendicular to the plane of the rings tending to cancel out as particles collide and interact gravitationally with each other...there would be little tendency for those rings to align to the giant ball bearing's rotation, though. I'm pretty sure the ring system would decay long before frame dragging effects became relevant. Any other orbiting bodies might cause the rings to align to their orbits.

novaderrik
2010-Jun-09, 05:07 AM
I suppose if you had an object that didn't rotate...at all, then there'd be no tidal-bulge to pull the particles towards the equator.

what if it was designed so that when it rotates, it becomes a perfectly spherical body with the mass distributed evenly?

Megatarius
2010-Jun-14, 05:18 PM
If it had a perfectly spherical shape or not, it wouldn't matter.

An object of any shape, turning around one axis, is still going to be heavier at the furthest out from the axis.

Get a bunch of people together, holding hands in a line, have them all line up coming out from a single point, and have them run around that point. The person standing at that point will simply turn in place. The person on the other end will be running to keep up. That person has further to run.

And we all know the faster something moves, the more mass it gains. And the heavier it gets.

Thus, the equator of most planets rotates along with it, not frame-dragging. It has to go faster than the poles to keep up its solid (or semi solid) shape. It becomes heavier, and attracts all the debris into line with it.


Right?

chornedsnorkack
2010-Jun-14, 05:55 PM
An orbiting body can be perturbed by the equatorial bulge of the primary, or by remoter objects. For example, the orbit of Moon is more perturbed by Sun than by equator of Earth. So if a planet is slowly rotating, it will have a small equatorial bulge and the rings will be perturbed mainly by other objects not the bulge.

What kind of orbits will ring particles follow around a body that is seriously nonaxisymmetric?

Hungry4info
2010-Jun-14, 11:04 PM
Providing a simple answer,

Assuming the planet has a perfectly, spherically symmetric, Fg is the same on all points of the surface gravitational field, then the ring inclination will depend only on how it was formed.

Regarding the planet engineered to rotate as a perfect sphere, the increased relativistic effects can be cancelled out by having less matter in the equatorial region (or less dense), thus lowering the relativistic mass to equal the rest mass at the poles.


What kind of orbits will ring particles follow around a body that is seriously nonaxisymmetric?

Unstable ones. Recall the masons at the moon and the probes deployed by the Apollo spacecrafts. This is an appropriate enough representation of a nonaxisymmetric body as far as the gravitational field is concerned.

grant hutchison
2010-Jun-14, 11:21 PM
I'd imagine that a ring around a planet with a perfectly spherical mass distribution (however that unlikely configuration is achieved) would settle into the ecliptic plane. In the absence of any oblateness in the primary, that would be the invariant plane for all the ring particles. They would precess in that plane at different rates, exchange momentum, and settle into the invariant plane, by the same process that drives the formation of equatorial rings around a conventionally oblate planet.

Grant Hutchison

chornedsnorkack
2010-Jun-15, 07:35 PM
I'd imagine that a ring around a planet with a perfectly spherical mass distribution (however that unlikely configuration is achieved)
How close is the very slowly rotating Venus?

would settle into the ecliptic plane. In the absence of any oblateness in the primary, that would be the invariant plane for all the ring particles. They would precess in that plane at different rates, exchange momentum, and settle into the invariant plane, by the same process that drives the formation of equatorial rings around a conventionally oblate planet.

Grant Hutchison

But if the planet does rotate, there is no invariant "plane". For the inner rings, their precession would be closer to the planetīs equator; outer rings would be closer to ecliptic, but not at the ecliptic.

At which distance (if any) from Venus does the attraction of Venusī equatorial bulge (small as it is) equal the attraction of Sun?

grant hutchison
2010-Jun-15, 07:58 PM
But if the planet does rotate, there is no invariant "plane".I presume you're talking about a planet which is oblate because of its rotation, rather than the hypothetical "ball bearing" we've been discussing?
In the case of an oblate planet, there is still an identifiable invariant plane for each orbital radius: the plane just varies smoothly from equatorial to ecliptic with increasing orbital radius.

Grant Hutchison

chornedsnorkack
2010-Jun-15, 08:21 PM
They would precess in that plane at different rates, exchange momentum,

Which particles can exchange momentum?

If two neighbouring circular rings could exchange momentum, then rings could not exist - the inner ring would lose momentum and shrink, the outer one would expand. Therefore the neighbouring rings cannot exchange momentum.

Now elliptical rings cannot remain next to each other, because their apsides would precess at a different speed and they would therefore run into each other.

But if we have a circular ring away from equator/eclipse, then all the particles of the ring have their nodes precessing at the same speed. What prevents the ring from continuing to precess and causes it to settle in a single plane?

grant hutchison
2010-Jun-15, 08:27 PM
Which particles can exchange momentum?All particles can exchange momentum.


If two neighbouring circular rings could exchange momentum, then rings could not exist - the inner ring would lose momentum and shrink, the outer one would expand.That's exactly what happens. That's why rings have limited life spans, which can be prolonged by the existence of relatively massive "shepherd" moons at their edges.


But if we have a circular ring away from equator/eclipse, then all the particles of the ring have their nodes precessing at the same speed. What prevents the ring from continuing to precess and causes it to settle in a single plane?The nodes do not precess at the same rate, since the precession rate is a function of orbital radius. An out-of-plane ring of any significant width therefore smears into a thick band, the particles exchange momentum by gravitational interaction, and the ring settles into the local invariant plane.

Grant Hutchison

Nereid
2010-Jun-15, 10:09 PM
Providing a simple answer,

Assuming the planet has a perfectly, spherically symmetric, Fg is the same on all points of the surface gravitational field, then the ring inclination will depend only on how it was formed.

Regarding the planet engineered to rotate as a perfect sphere, the increased relativistic effects can be cancelled out by having less matter in the equatorial region (or less dense), thus lowering the relativistic mass to equal the rest mass at the poles.



Unstable ones. Recall the masons at the moon and the probes deployed by the Apollo spacecrafts. This is an appropriate enough representation of a nonaxisymmetric body as far as the gravitational field is concerned.(bold added)
I think you meant mascons (http://www.universetoday.com/2006/06/03/huge-asteroid-crater-in-antarctica/), as in mass-concentrations.

Nereid
2010-Jun-15, 10:13 PM
The small moon/test particles have mass.

They will raise tides on the planet.

The planet therefore cannot retain its perfectly spherical shape. And so ...

A planet - unless it's a rogue planet - will orbit a star.

The star will affect the planet, via many physical mechanisms, such as tides, winds, magnetic fields, differential heating, and many more.

The planet will not, as a result, remain perfectly spherical.

Also, it will not continue to have zero angular momentum.

And so on.

Hungry4info
2010-Jun-16, 01:47 AM
(bold added)
I think you meant mascons (http://www.universetoday.com/2006/06/03/huge-asteroid-crater-in-antarctica/), as in mass-concentrations.

Yep, thanks.
I need to proof-read for typos before hitting that submit button.

Noclevername
2010-Jun-24, 01:57 AM
Here's a thought-experiment: An ovoid moon formed by tidal stretching, which gets knocked out of orbit of its primary and winds up as a planet of its own, spinning on its long axis. It's core is too cool and solid to slump at the equator. Then it gets a moonlet breaking up around it-- ring or no ring? And, how stable would such a body be in its rotation? Would it instantly wobble into a Tumbling Pigeon, or stay "up" on its long axis like a top, and if so, for how long?

Spoons
2010-Jun-24, 04:16 PM
I would think that it's an "easier", more efficient mode of rotation than any other, so it would keep that rotation, though I don't know that I can back that up with any valid physics. Would that be correct? If so, I'd expect the rings to evolve around the axial centre of gravity, but be more chaotic or at least spread out.

antoniseb
2010-Jun-24, 04:37 PM
As a practical matter, Saturn's recently discovered giant ring is not in the equatorial plane... now because it is so far from the planet it hasn't had time to settle, but it is also being shepherded (I think) by an outer moon which is also orbiting far from the equatorial plane. Perhaps this can serve as a counter example for the OP.

Noclevername
2010-Jun-28, 06:03 PM
I would think that it's an "easier", more efficient mode of rotation than any other, so it would keep that rotation, though I don't know that I can back that up with any valid physics. Would that be correct? If so, I'd expect the rings to evolve around the axial centre of gravity, but be more chaotic or at least spread out.

But would it be dynamically unstable? With mass distributed towards the poles, would there be positive feedback that would throw it out of balance before a ring could form?

cjl
2010-Jun-28, 06:27 PM
Here's a thought-experiment: An ovoid moon formed by tidal stretching, which gets knocked out of orbit of its primary and winds up as a planet of its own, spinning on its long axis. It's core is too cool and solid to slump at the equator. Then it gets a moonlet breaking up around it-- ring or no ring? And, how stable would such a body be in its rotation? Would it instantly wobble into a Tumbling Pigeon, or stay "up" on its long axis like a top, and if so, for how long?

Rotation around the least axis of inertia is a stable configuration, but it is also the maximum energy state. If there is any energy loss at all, from flexing, or sloshing (if it happens to have liquid or an atmosphere), or similar, then the body will eventually depart from this maximum energy state, going through an unstable period and eventually ending up with a spin about the axis of greatest inertia. For long term spin stability, an object must be rotating about the axis of greatest inertia. The only exception is if the object is perfectly rigid, in which case both the axis of greatest and the axis of least inertia are stable (the intermediate axis is always unstable).

chornedsnorkack
2010-Jun-28, 06:47 PM
Rotation around the least axis of inertia is a stable configuration, but it is also the maximum energy state. If there is any energy loss at all, from flexing, or sloshing (if it happens to have liquid or an atmosphere), or similar, then the body will eventually depart from this maximum energy state, going through an unstable period and eventually ending up with a spin about the axis of greatest inertia. For long term spin stability, an object must be rotating about the axis of greatest inertia. The only exception is if the object is perfectly rigid, in which case both the axis of greatest and the axis of least inertia are stable (the intermediate axis is always unstable).

So what does a perfectly rigid object do if it possesses a spin not aligned with any axis of inertia? Does it have any way to stop free precession?

cjl
2010-Jun-28, 07:45 PM
If it has a spin not aligned with any axis of inertia, and it is perfectly rigid (I.E. zero energy dissipation), then it will freely precess unless there is some force acting on it.

Nereid
2010-Jun-28, 09:10 PM
As a practical matter, Saturn's recently discovered giant ring is not in the equatorial plane... now because it is so far from the planet it hasn't had time to settle, but it is also being shepherded (I think) by an outer moon which is also orbiting far from the equatorial plane. Perhaps this can serve as a counter example for the OP.
As I understand it, it's not only a transient ring, but is constantly being replenished by Phoebe (much of the ring material is lost by collision with Iapetus). Further, it's a very different kind of ring than I think is being asked about in the OP; for example, the Phoebe ring's thickness is ~40 times Saturn's radius.

Tangent
2010-Jun-29, 01:12 AM
Hmm, the question is. You have a loose ball bearing in space with out rotation. Nothing else acting on it. Where will the rings form?
They wouldn't. There would be no rings. Might have a deep space comet on your hands.

mugaliens
2010-Jun-29, 06:34 AM
Cassini Mission FAQ: Why do Saturn's rings all lie in the same plane? Why are planetary rings always found in their equatorial planes and not sometimes crossing their poles? (http://saturn.jpl.nasa.gov/faq/FAQSaturn/index.cfm#q14)

Isn't there a frame-dragging (general relativity) effect in operation here, as well? While with Saturn it'd take millions if not trillions of orbits, the Lense-Thirring effect is what causes a black hole's accretion disk (http://en.wikipedia.org/wiki/File:Galaxies_AGN_Inner-Structure-of.jpg)to be a disk rather than a collapsing sphere, and it does it in less than one rotation.

Gravity Probe B (http://en.wikipedia.org/wiki/Gravity_Probe_B)measured rotational frame dragging around Earth. Surely if we can measure it around Earth, then Saturn's far greater mass would have an appreciable effect after a few million to trillion orbits of its debris.

Jeff Root
2010-Jun-29, 11:10 AM
The collisions referred to in the Cassini Mission FAQ are the primary
cause of particulate matter ending up in a single plane, making rings.
Even a perfect, nonrotating ball bearing would acquire rings. Saturn's
equatorial bulge, its moons, and the Sun just help determine where
the plane of the rings ends up. Without those influences, the plane
will be determined by the original orbits of the particles, and how
they collide with one another.

-- Jeff, in Minneapolis

Noclevername
2010-Jun-29, 06:40 PM
Rotation around the least axis of inertia is a stable configuration, but it is also the maximum energy state. If there is any energy loss at all, from flexing, or sloshing (if it happens to have liquid or an atmosphere), or similar, then the body will eventually depart from this maximum energy state, going through an unstable period and eventually ending up with a spin about the axis of greatest inertia. For long term spin stability, an object must be rotating about the axis of greatest inertia. The only exception is if the object is perfectly rigid, in which case both the axis of greatest and the axis of least inertia are stable (the intermediate axis is always unstable).

Not tracking so well after the seizure, so dumb it down for me please-- does this mean there will be a ring, or no ring?

(For the record, my example has no atmosphere and a solid core.)

cjl
2010-Jun-30, 03:41 AM
Not tracking so well after the seizure, so dumb it down for me please-- does this mean there will be a ring, or no ring?

(For the record, my example has no atmosphere and a solid core.)

That wasn't actually referring to rings at all. It was purely talking about the stability of a spinning object's rotation.

Noclevername
2010-Jun-30, 06:37 PM
That wasn't actually referring to rings at all. It was purely talking about the stability of a spinning object's rotation.

And what was it saying about it?

Noclevername
2010-Jul-02, 01:29 PM
Also, would there be a ring? Or would everything go into a polar orbit?

John Mendenhall
2010-Jul-02, 04:20 PM
Seems to me that all objects rotate on the grand scale. You, me, Earth, the solar system. etc.

cjl
2010-Jul-03, 08:50 PM
And what was it saying about it?

Simply that unless the object is perfectly rigid, it will eventually settle down into a spin around the axis about which it has the greatest moment of inertia - the axis where as much of the object as possible is as far from the axis of rotation as possible. If the object is perfectly rigid, there's a second stable spin condition, in which the object is spinning about the axis about which it has the least moment of inertia (in other words, the axis where as much of the object as possible is as close to the axis as possible). This spin state is only stable though if the object has no way of internally dissipating energy, such as through a fluid sloshing or the object flexing.

(was that clearer?)

chornedsnorkack
2010-Jul-03, 09:04 PM
So, suppose that we have a perfectly rigid triaxial body, with no external torques and with no internal way to dissipate energy. And it happens to have a spin whose direction does not coincide with any axis of inertia.

What will it do? Undergo free precession forever?

cjl
2010-Jul-04, 02:27 AM
So, suppose that we have a perfectly rigid triaxial body, with no external torques and with no internal way to dissipate energy. And it happens to have a spin whose direction does not coincide with any axis of inertia.

What will it do? Undergo free precession forever?

Yep.

chornedsnorkack
2010-Jul-04, 05:33 AM
Then let us consider the possibilities.

For a completely rigid body, with no internal dissipation, what will happen if it has spin directed at:

A random direction far from any axis?

Close to, but not exactly, the shortest axis?

Exactly the middle axis?

Close to, but not exactly, the middle axis?

Close to, but not exactly, the longest axis?

Exactly the longest axis?

bunker9603
2010-Jul-04, 12:07 PM
I am confused, if our moon broke up could it/would it form a ring around the earth? If it would how would the ring appear to us, would it reflect the sunlight as the moon does now?

cjl
2010-Jul-04, 08:49 PM
Then let us consider the possibilities.

For a completely rigid body, with no internal dissipation, what will happen if it has spin directed at:

A random direction far from any axis?

Free precession.



Close to, but not exactly, the shortest axis?

Precession about the short axis, with the spin staying fairly close to the same direction



Exactly the middle axis?

It will stay spinning about the middle axis, no precession



Close to, but not exactly, the middle axis?

It will precess in an unstable fashion, sometimes spinning almost stably close to the mid axis, sometimes flipping all the way over.



Close to, but not exactly, the longest axis?

It will precess about the longest axis, with the spin staying fairly close to the axis\



Exactly the longest axis?
No precession, just spin about the axis.

Nereid
2010-Jul-04, 10:48 PM
I am confused, if our moon broke up could it/would it form a ring around the earth?
Depends, at least to some extent, on how it broke up, and where.

In general though, yes, it would form a ring around the Earth.


If it would how would the ring appear to us, would it reflect the sunlight as the moon does now?
Again, it depends on how and where it broke up. From the equator, any ring would likely be very hard to see; from high latitudes, north or south, much easier (but it depends a great deal on many details of the nature of ring(s)).

How reflective the ring(s) would be depends heavily on the composition and location of the rings; however, from afar (NOT the surface of the Earth!) any rings would very likely be much brighter than the Moon is today, even if the material had the same albedo as the Moon does - can you work out why?

chornedsnorkack
2010-Jul-05, 02:23 PM
I think I am getting the logic then.

If there is no damping, then there is no way to change momentum of inertia. The body can precess relative to its angular momentum (which remains unchangeable relative to fixed stars, damping or no damping) between orientations that have same momentum of inertia.

If the body is close to minimum or maximum momentum of inertia, it can only change its orientation in the same vicinity. Whereas momenta of inertia close to middle axis of inertia can be found both near the middle axis and far from the middle axis somewhere quite else between the short and long axis.

Correct?

cjl
2010-Jul-06, 12:25 AM
I think I am getting the logic then.

If there is no damping, then there is no way to change momentum of inertia. The body can precess relative to its angular momentum (which remains unchangeable relative to fixed stars, damping or no damping) between orientations that have same momentum of inertia.

If the body is close to minimum or maximum momentum of inertia, it can only change its orientation in the same vicinity. Whereas momenta of inertia close to middle axis of inertia can be found both near the middle axis and far from the middle axis somewhere quite else between the short and long axis.

Correct?

That seems to be a reasonable way to put it, yes.

Noclevername
2010-Jul-06, 11:32 PM
So what I'm getting is, my spinning-egg world could be stable, but it's a trillion-to-one shot that it will be. So that gets back to the question of, assuming that it is stable, with the mass distributed towards the poles rather than the equator, would/could it develop a ring?

cjl
2010-Jul-07, 10:50 PM
So what I'm getting is, my spinning-egg world could be stable, but it's a trillion-to-one shot that it will be. So that gets back to the question of, assuming that it is stable, with the mass distributed towards the poles rather than the equator, would/could it develop a ring?

Well, it could be stable if and only if it were perfectly rigid. At the size scale of a planet, no matter what the planet is made of, it will tend to assume a surface at equal potential, which for a spinning world, means that it will bulge at the equator. Rock simply doesn't have the structural strength needed to hold a different shape, and even if it did, rock (or metal) flexes enough that it would tend to dissipate energy through flexing, causing the spin to go unstable.

Jeff Root
2010-Jul-08, 03:21 AM
cjl,

Going even farther off-topic, I don't see your last point. Would a prolate
planet rotating on its long axis really flex if there are no outside forces?

-- Jeff, in Minneapolis

cjl
2010-Jul-08, 05:11 AM
cjl,

Going even farther off-topic, I don't see your last point. Would a prolate
planet rotating on its long axis really flex if there are no outside forces?

-- Jeff, in Minneapolis

No, as long as it's perfectly balanced (in which case it would be stable in its rotation). As I said though, I don't think any material is strong enough to maintain any shape significantly deviating from an equipotential surface at planetary scales.

chornedsnorkack
2010-Jul-08, 05:30 AM
No, as long as it's perfectly balanced (in which case it would be stable in its rotation). As I said though, I don't think any material is strong enough to maintain any shape significantly deviating from an equipotential surface at planetary scales.

Not at planetary scales. But at slightly smaller scales, rock and metal asteroids can and will maintain irregular shapes over long periods of time.

Suppose that an asteroid of irregular shape does, as a result of a collision, acquire angular momentum which coincidentally is close to its long axis. If it is exactly the long axis, it will be perfectly balanced. If it is exactly the middle axis, it will be perfectly balanced too.

But if it is close to the long axis, it will be imperfectly balanced. It will precess freely, close to the long axis, flex and get ever further from the long axis, as a result of which the flexing will grow.

Where is the rotation axis of Hyperion?

Noclevername
2010-Jul-08, 05:22 PM
Well, it could be stable if and only if it were perfectly rigid. At the size scale of a planet, no matter what the planet is made of, it will tend to assume a surface at equal potential, which for a spinning world, means that it will bulge at the equator. Rock simply doesn't have the structural strength needed to hold a different shape, and even if it did, rock (or metal) flexes enough that it would tend to dissipate energy through flexing, causing the spin to go unstable.

It's a former moon, so probably it's what the kids call these days a "dwarf planet". But yes, for purposes of this question, it's solid enough to be stable for a very long time.

Noclevername
2010-Jul-13, 04:16 PM
So should I start a new thread for the ring question? Because I've asked several times and gotten no answer.

Nereid
2010-Jul-13, 04:19 PM
So should I start a new thread for the ring question?
That would seem to be a good idea ...


Because I've asked several times and gotten no answer.
Usually a good signal that a new thread is needed ... :p

Spoons
2010-Jul-14, 02:06 AM
If you start a new thread for that, Noclevername, would you mind linking to it here (just in case I don't see it there)? I'm interested in the answer to that.

Noclevername
2010-Jul-15, 07:36 AM
Started. (http://www.bautforum.com/showthread.php/105884-Ring-around-the-Ovoid)