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Philistine
2004-Jun-27, 03:06 AM
Uranus's axis of rotation is nearly perpendicular to the plane of the solar system, unlike any other planet.

The explanation for this that I used to believe is that Uranus was "knocked on its side" by an impactor. My problem with this is that Uranus's rings and the orbits of its moons are in line with the equator and are therefore also "tipped over" relative to the plane of the solar system. An impactor couldn't have done that!

The obvious answer is that Uranus and its moons originally formed in their current orientation, but that just takes the question one step further.

Kaptain K
2004-Jun-27, 03:38 AM
1) The impactor hit during the formation period (4.7 billion years ago).
2 ) Most of the moons of Uranus are captured debris (KBO's and Oort cloud objects).
3) Rings are transient phenomena. They are less than 100 million years old (probably a lot less).
4) Due to torque from the equatorial bulge (oblateness caused by planetary rotation), moons tend to be drawn into equatorial orbits over billions of years*.

* Note that the moons of Mars are certainly captured asteroids, yet their orbits are close to equatorial.
Also, the Earth's moon is now thought to be the result of such an impact, yet its orbit is nearly equatorial.

stu
2004-Jun-27, 04:50 AM
3) Rings are transient phenomena. They are less than 100 million years old (probably a lot less).
4) Due to torque from the equatorial bulge (oblateness caused by planetary rotation), moons tend to be drawn into equatorial orbits over billions of years.

The same point raised in (4) is also applicable to (3), even though the rings aren't around for billions of years.

Kaptain K
2004-Jun-27, 04:58 AM
3) Rings are transient phenomena. They are less than 100 million years old (probably a lot less).
4) Due to torque from the equatorial bulge (oblateness caused by planetary rotation), moons tend to be drawn into equatorial orbits over billions of years.

The same point raised in (4) is also applicable to (3), even though the rings aren't around for billions of years.
True! However, the rings were formed from a moon that I assumed was already in an equatorial orbit - either from a collision or because it ventured inside Roche's limit.

milli360
2004-Jun-27, 05:03 AM
Kaptain K:
Also, the Earth's moon is now thought to be the result of such an impact, yet its orbit is nearly equatorial.
The moon's orbit can be as much as 30 degrees away from the equator. Seems like a lot to be called "nearly equatorial"

Kaptain K
2004-Jun-27, 05:15 AM
Kaptain K:
Also, the Earth's moon is now thought to be the result of such an impact, yet its orbit is nearly equatorial.
The moon's orbit can be as much as 30 degrees away from the equator. Seems like a lot to be called "nearly equatorial"
True. Actually, it is closer to being in the ecliptic, which I doubt is a coincidence.

Candy
2004-Jun-27, 05:19 AM
Actually, it is closer to being in the ecliptic, which I doubt is a coincidence. I really like this kind of stuff, so don't think I am a retard but what is an ecliptic?

Brady Yoon
2004-Jun-27, 05:25 AM
I really like this kind of stuff, so don't think I am a retard but what is an ecliptic?

Good question! :D The ecliptic is the apparent path of the sun across the sky over the course of the year. Of course, this is only relative motion, caused by the revolution of the Earth around the sun.

Candy
2004-Jun-27, 05:32 AM
The ecliptic is the apparent path of the sun across the sky over the course of the year. Of course, this is only relative motion, caused by the revolution of the Earth around the sun. Earth's sky? Why is that?

Kaptain K
2004-Jun-27, 05:38 AM
The Earth orbits the Sun. As it does so, the Sun appears to move against the background stars. The path the Sun takes is known as the ecliptic. It is a projection of the Earth's orbit (as seen from the Sun) onto the sky.

Candy
2004-Jun-27, 05:48 AM
I hate to sound like a Dumas, but shouldn't we be taking the calculations based on a more perfect ecliptic world from the Sun? Earth doesn't seem to be that stable. :-k

Tobin Dax
2004-Jun-27, 06:22 AM
I hate to sound like a Dumas, but shouldn't we be taking the calculations based on a more perfect ecliptic world from the Sun? Earth doesn't seem to be that stable. :-k

You hate to sound like a French author? :wink:

Anyway, which calculations are you referring to? If you mean the inclination (angle of the spin axis) of Uranus, then that is taken from the ecliptic (well, from the line perpendicular to it), as are the inclinations of all the planets.

In general, coordinates of an object are taken from the most convient reference object. That can be the earth, the solar system, the galaxy, or even the "fixed stars."

Oh, and one final note is that while the earth's properties do change (i.e. precession), most happen very slowly compared to our lifetimes, so using Earth-based coordinates isn't a huge problem. That these properties change is known, though, and known well enough to compensate if need be. Just one of the perks of being an astronomer. :wink:

Candy
2004-Jun-27, 06:33 AM
In general, coordinates of an object are taken from the most convient reference object. That can be the earth, the solar system, the galaxy, or even the "fixed stars." I understand that, but wouldn't a better calculation come from a 'fixed star' to formulate everything? This would be so much better to comprehend from a future generations point of view? Just steppin' outside the box.

Kaptain K
2004-Jun-27, 06:54 AM
The ecliptic is the plane of the Earth's orbit by definition.

And besides, there are no fixed stars. Everything in the universe is in motion.

Candy
2004-Jun-27, 06:59 AM
The ecliptic is the plane of the Earth's orbit by definition.

And besides, there are no fixed stars. Everything in the universe is in motion. Are you sure? 8-[

Candy
2004-Jun-27, 07:01 AM
What about the Sun? :-k

Kaptain K
2004-Jun-27, 07:05 AM
The ecliptic is the plane of the Earth's orbit by definition.

And besides, there are no fixed stars. Everything in the universe is in motion. Are you sure? 8-[
Yes, the ecliptic is defined as the plane of the Earth's orbit.
and
Yes, I'm sure there are no "fixed" stars. The Black hole at the center of the Milky Way "appears" to be immobile with respect to the mass of the galaxy as a whole, with all the stars, gas and dust revolving around it. But, it still moves with respect to the other galaxies.

PhantomWolf
2004-Jun-27, 07:37 AM
Actually my big problems with the "massive impactor knocked it over" idea, is that the stats of Uranus' Orbit don't seem match.

Uranus' eccentricity is the 4th smallest at 0.0485 (only behind Venus, Neptune, and Earth) meaning that it's orbit hasn't been greatly perturbed by the impact. Surely an impact large enough to knock a planet over would have caused orbital irregularities.

The second problem is its inclination to the eliptic. At 0.77 degrees it is all by equal to Earth's. Compared to the other planets it is nearly half that of the next closest, Jupiter to 1.30 degrees. Again I would have expected that such a massive collision would have thrown the inclination off by far more than than is shown.

Philistine
2004-Jun-27, 09:00 AM
:o =D> Really fast response. Cool! Thanks everyone!
I should have checked back sooner.


1) The impactor hit during the formation period (4.7 billion years ago).
2 ) Most of the moons of Uranus are captured debris (KBO's and Oort cloud objects).
3) Rings are transient phenomena. They are less than 100 million years old (probably a lot less).
4) Due to torque from the equatorial bulge (oblateness caused by planetary rotation), moons tend to be drawn into equatorial orbits over billions of years*.


Some thoughts on this:
1) That the impactor hit during the formation period shouldn't matter much. If Uranus was only partially formed at the time, material added later would have at least partially cancelled out the impactor's effect by transferring angular momentum at different angles to the ecliptic. Uranus's current axial tilt is just about the most extreme angle possible. If it had been knocked to a greater angle, other infalling material would have slowed down the rotation, and we would be asking questions about that, as well as about why it's turning backwards. ( :-k That brings Venus to mind...)
2) Moons like Titania and Oberon are way too big and regular to be captured objects. Plus, if they were captured objects, they would have wound up in orbits at more normal angles to the ecliptic.
3) I already knew that. I shouldn't have mentioned them. :oops: My assumption was that the rings came from one of the moons that "mysteriously" ended up orbiting at that weird angle.
4) That's a concept that never occurred to me. I'm a bit skeptical that equatorial bulge torque could get all of the moons to where they are now, but, in the absence of other plausible explanations, I'll take it.

Philistine
2004-Jun-27, 09:13 AM
Uranus' eccentricity is the 4th smallest at 0.0485 (only behind Venus, Neptune, and Earth) meaning that it's orbit hasn't been greatly perturbed by the impact. Surely an impact large enough to knock a planet over would have caused orbital irregularities.

I think that the momentum needed to knock over a gas giant (mini-giant?) is miniscule compared to the momentum that a significant orbital change requires. I bet that Uranus could be hit by an earth-sized world and remain in a nearly unchanged orbit.



The second problem is its inclination to the eliptic. At 0.77 degrees it is all by equal to Earth's. Compared to the other planets it is nearly half that of the next closest, Jupiter to 1.30 degrees. Again I would have expected that such a massive collision would have thrown the inclination off by far more than than is shown.

The ecliptic is defined as the plane of Earth's orbit. That is arbitrary. Uranus could be close to it by sheer random chance.

PhantomWolf
2004-Jun-27, 10:16 AM
Uranus' eccentricity is the 4th smallest at 0.0485 (only behind Venus, Neptune, and Earth) meaning that it's orbit hasn't been greatly perturbed by the impact. Surely an impact large enough to knock a planet over would have caused orbital irregularities.

I think that the momentum needed to knock over a gas giant (mini-giant?) is miniscule compared to the momentum that a significant orbital change requires. I bet that Uranus could be hit by an earth-sized world and remain in a nearly unchanged orbit.

Actually you don't knock it over. ;) The vector for the rotation is added to the vector caused by the torque of the Impacter and the result is the new rotation vector which defines the direction of rotation. To change that angle by 90 degrees requires a huge hit, especially when we have a large mass, this is because before it can have an effect, the torque needs to overcome both the inertia of the planet and the gyroscopic effects of the planet's rotation. (Try spinning a bicycle wheel and then turn it from directly up to 90 degrees.) Subsquent large hits would have to reinforce the angle, or continue to rotate it. The trouble is that should an impacter hit and stick, then the rules of physics tell us that the orbit of the combined mass MUST be the same as orbit of the centre of mass prior to the collision. Unless both objects have very similar orbits (which wouldn't cause the 90-degree rotational vector change.) That is why the lack of eccentricity is strange to say the least.




The second problem is its inclination to the eliptic. At 0.77 degrees it is all by equal to Earth's. Compared to the other planets it is nearly half that of the next closest, Jupiter to 1.30 degrees. Again I would have expected that such a massive collision would have thrown the inclination off by far more than than is shown.

The ecliptic is defined as the plane of Earth's orbit. That is arbitrary. Uranus could be close to it by sheer random chance.

Again, an impactor should stand out here. Assuming a plane of debris, all planets should be very close to each other, and in fact are, only a degree of two off for most of them. However, just as above, a major impactor with Uranus would have had to come from either above or below the eliptic of the math just doesn't work. With that occuring the COM would have to also move off the eliptic meaning that for it to have an angle of less than 1 degree would seem very strange to me.

milli360
2004-Jun-27, 05:49 PM
Philistine:
The ecliptic is defined as the plane of Earth's orbit. That is arbitrary. Uranus could be close to it by sheer random chance.
Doesn't matter. As I've mentioned before, the ecliptic is also very close to the orbit of Jupiter, and lines up pretty well with the plane of the the Sun's own rotation. In other words, if we were to use the Sun instead of the Earth as the reference, we'd end up with nearly the same system. That may be a coincidence--and it may explain why we haven't "changed over." The only planets that are more than a couple degrees off this "standard" are Mercury, Venus, and Mars. And of course Pluto. But none of the gas giants.

Hmm. Another way of saying that is Earth and the gas giants match up, the others don't.

Brady Yoon
2004-Jun-27, 07:22 PM
Can a gas giant planet be tipped over by a giant impact, even though it is covered in gases? Or did it happen before the gases were accreted?

milli360
2004-Jun-27, 07:28 PM
Can a gas giant planet be tipped over by a giant impact, even though it is covered in gases? Or did it happen before the gases were accreted?
I think I see what you're getting at, but these gas giants are pretty dense, and as Earth's thin atmosphere constantly shows, it can play havoc with pretty substantial meteoroids. A lot of energy gets transferred.

That would have been something, eh, to see a part of Shoemaker-Levy pop into Jupiter and pop out the other side. :)

Brady Yoon
2004-Jun-27, 07:35 PM
I understand how energy can be transferred, but can the planet itself be tipped over? For example, if you throw a rock at a solid object, the object will be tipped over or damaged, but if you throw a rock at gas...

milli360
2004-Jun-27, 07:48 PM
I understand how energy can be transferred, but can the planet itself be tipped over? For example, if you throw a rock at a solid object, the object will be tipped over or damaged, but if you throw a rock at gas...
If the rock and gas remain together, and continue on as a single body, the original momentum has to be accounted for, in the final value.

Tobin Dax
2004-Jun-27, 09:04 PM
I understand how energy can be transferred, but can the planet itself be tipped over? For example, if you throw a rock at a solid object, the object will be tipped over or damaged, but if you throw a rock at gas...

But if you throw the rock at an independent mass of gas in a vacuum, you don't need contact with the gas to affect it. A "gravitational slingshot" is a good example of this--you can steal rotational energy from, say, Jupiter, giving to a "rock" passing by without the rock needing to physically collide with the planet.

The same thing holds for tipping Uranus: if a large object passing by pulls (gravitationally) on the equatorial bulge in the right way, it will "tip" the planet.

Philistine
2004-Jun-28, 03:21 AM
:-k I've been chewing over the "equatorial bulge torque" idea for a while now, and I have a problem with it.

In order for a torque to be exerted there would have to be some sort of major irregularity causing a distortion in the gravitational field that moves with the rotation of the planet. Such a distortion would cause a gentle tide, dragging other objects with it. A uniform planet-girdling bulge would not create that kind of moving distortion. It would just make the moons' orbits less round.

I hope that made sense. It's very hard to describe what I'm thinking without drawing a picture!

Candy
2004-Jun-28, 05:27 AM
I understand how energy can be transferred, but can the planet itself be tipped over? For example, if you throw a rock at a solid object, the object will be tipped over or damaged, but if you throw a rock at gas... Being the physics dummy here, wouldn't the effect, if an object hit gas, be based on the speed that the object is traveling? I would think there would be different outcomes from kaplunk to kaboom. :lol:

MrObvious
2004-Jun-28, 06:28 AM
It's all dependant on the energy transfer. Which means it's dependant on: velocity (speed & direction) , mass of objects , density of objects , point of impact etc etc.

Think of a bullet hitting an object, if the bullet is too hard it transfers little energy to the object- instead passing through and leaving a drill hole in it. If the bullet is softer it deforms and transfers more energy. Imagine the object is a ball suspended by a wire, then the position of impact can impart rotation onto it or not depending on where it hit the ball. If the ball had rotation to begin with it can add or reduce the rotational speed. Transfer this view into a full 3d one (ball on string has one stable plane of rotation) and all manner of rotations can be possible.

Remember too, a gas planet also has a core, this too can either add or remove rotation as observed from a distance.

eburacum45
2004-Jun-28, 09:57 AM
One theory of gas giant formation is 'disk instability';
http://www.aas.org/publications/baas/v32n3/dps2000/46.htm
http://www.aas.org/publications/baas/v35n3/aas202/19.htm

perhaps if this theory is true (to a greater or lesser extent) it might explain Uranus' tilt as a feature of the original instability-

without any need for an impactor.

Candy
2004-Jun-28, 05:02 PM
It's all dependant on the energy transfer. Which means it's dependant on: velocity (speed & direction) , mass of objects , density of objects , point of impact etc etc. That's what I meant to say. :wink:

Doodler
2004-Jun-28, 05:25 PM
Here's a tangent question.

Is the fact that some kind of impact shifted the rotation of a gas giant sufficient foundation for the arguement that gas giants form with solid cores rather than coalesced gases from distortion nodes in the original dust disk?

After all, if it were just gas, an impactor of any sufficient mass would just pass through or be crushed by internal pressures while only causing massive disruption to the structure of the planet.

Bottom line, if its really an impact that toppled it, does it not imply by default that there's something in there to hit?

Brady Yoon
2004-Jun-28, 08:53 PM
Bottom line, if its really an impact that toppled it, does it not imply by default that there's something in there to hit?

The gases in gas giant planets become solid and liquid deep inside because of the enormous pressure. If you mean does only this piece of evidence prove that gas giants have rocky cores, I would say no.

milli360
2004-Jun-28, 08:59 PM
Doodler:
Is the fact that some kind of impact shifted the rotation of a gas giant
Is that a fact?

Doodler
2004-Jun-28, 09:32 PM
Doodler:
Is the fact that some kind of impact shifted the rotation of a gas giant
Is that a fact?

No, but it appears to be a frontrunner idea. Perhaps I should have phrased it.

"If its believed that an impact..."

Candy
2004-Jun-29, 05:00 AM
I've noticed you can alter the direction of a bubble by blowing on it. Meaning a bubble of gas might alter it's direction, too, if something blew on or by it hard enough. Let's say a huge celestial object. :-k

Duane
2004-Jun-29, 05:28 AM
Interesting conversation =D>

I think a couple of things are being overlooked here. First of all the "gas" of a gas giant is compressed to the point where it acts more like a solid than a gas. It's not like looking at a bubble of gas in your bath tub :wink:

Secondly, as was mentioned by someone earlier (how the heck do you bring up the whole topic?) it is a question of mass. When two bodies interact the results of the interaction are dependant on their combined mass. While you may think of such an impact as a huge explosion, when you are talking about bodies as massive as these (the proto-Uranus and the earth-sized object that hit it) there is no explosion per se, the objects actually merge. The angle of the tilt, the rotational period, the rotational direction, the orbit and the final mass of the combined object are all dependant on the mass of the objects, their velocity and the angle of impact.

There could have been enough debris thrown up to have gone into orbit around Uranus and formed its moons after the event. The direction of the impactor could be the result of interacting with Jupiter or any of the other gas giants, so it could have come in from almost any direction, including head on. (doubtful given its orbital speed).

Candy
2004-Jun-29, 05:03 PM
When two bodies interact the results of the interaction are dependant on their combined mass. While you may think of such an impact as a huge explosion, when you are talking about bodies as massive as these (the proto-Uranus and the earth-sized object that hit it) there is no explosion per se, the objects actually merge. The angle of the tilt, the rotational period, the rotational direction, the orbit and the final mass of the combined object are all dependant on the mass of the objects, their velocity and the angle of impact.Hey, I like my bubble theory better. :lol: You know, two bubbles can merge when they collide. A bubble can also merge with a solid object. They don't always explode upon impact. I know Uranus is not a bubble. I am just trying to put it into a vision I can comprehend, since I don't know the definitions to all the terminology used. 8-[

milli360
2004-Jun-29, 05:13 PM
Doodler:
Is the fact that some kind of impact shifted the rotation of a gas giant
Is that a fact?

No, but it appears to be a frontrunner idea. Perhaps I should have phrased it.

"If its believed that an impact..."
OK, I think we're clear then.

Doodler (http://www.badastronomy.com/phpBB/viewtopic.php?p=287946#287946):
Is the fact that some kind of impact shifted the rotation of a gas giant sufficient foundation for the arguement that gas giants form with solid cores rather than coalesced gases from distortion nodes in the original dust disk?

After all, if it were just gas, an impactor of any sufficient mass would just pass through or be crushed by internal pressures while only causing massive disruption to the structure of the planet.

Bottom line, if its really an impact that toppled it, does it not imply by default that there's something in there to hit?
It doesn't seem to me that having a solid core now implies that it had a solid core to begin with, regardless of whether a solid core is needed to "tip it over."

Anonymous
2004-Jul-01, 04:40 PM
I see no problem with a near miss tipping a planet over provided that my initial assumptions are fairly accurate.

First, I take as a given that the core of Uranus formed in the Jupiter-Saturn region and then migrated outward to its current position. Next, I assume that Uranus acquired its tilt early in the accretion era and therefore had a lower mass. (8-10 Earth masses)

Next, I assume that the early outer system was more densely populated than today and qualified as a ‘target rich environment’.

Next, I assume that in such an environment, a near-miss is almost as likely as a direct hit.

If these assumptions are correct, then I would expect all of the usual suspects to be hotter and to possess more bloated-distended atmospheres. (due to the heat of accretion)

The remaining questions are: how massive is the interloper and how near is the ‘near-miss’?

Too close (within the roche limit) and the encounter may be catastrophic. (Yoinking an isosphere of material from one or both objects)

Too far (several dozen planetary radii) and all you get is a couple of egg-looking objects. (at least temporarily)

I guess it comes down to the right distance with the right mass at the right approach velocity. We’d have to move enough mass in the outer layers to drag the core ‘over’.

No, I can’t prove any of this, but I suspect that extreme tidal alteration may explain a few of the ‘odd-ball’ objects in our solar system.

Brady Yoon
2004-Jul-01, 05:57 PM
No, I can’t prove any of this, but I suspect that extreme tidal alteration may explain a few of the ‘odd-ball’ objects in our solar system.

How is this hypothesis better than the impact one?

Anonymous
2004-Jul-02, 12:22 AM
Brady Yoon wrote:

“How is this hypothesis better than the impact one?”

I favor the ‘near miss’ over the ‘direct hit’ because we already have an abundance of evidence that the process of accretion is characterized by an unknown, yet enormous number of direct hits which transformed all the proto-planets and embryos into the planets we see today.

So, just what could have been different about the direct hit on Uranus which changed its obliquity so severely yet leaves the remainder of the planets in the solar system with lesser obliquities?

Personally, I rule out the ‘severe angle of attack’ notion because I think that the entire proplyd (proto-planetary disk) underwent a chaotic phase immediately following ignition and concurrent with the ‘clearing’ of the disk. Specifically, I suspect that virtually all planetary embryos were subjected to impacts at a variety of angles.

Which brings me to my next problem. If near-misses were more likely in the early accretion era and potentially most objects were processed by near-misses, why is only one (Uranus) knocked on its side? I’m not sure yet. Mercury appears to be a good candidate for a ‘rocheing’. (I simply could not accept that collisional stripping would be so…uniformly efficient) Pluto-Charon looks sufficiently suspicious. And there’s something that still bugs me about Terra-Luna…..’big yoink’ instead of ‘big splat’? Oh, what I would give for some crunch time on one of DARPA’s non-existent computers. Better simulations and more of them.

milli 360.

You appear to hold a similar position yet have not explicitly said so. Am I alone out here on the end of a long creaky limb?

milli360
2004-Jul-02, 12:54 AM
Heathen:
milli 360.

You appear to hold a similar position yet have not explicitly said so. Am I alone out here on the end of a long creaky limb?
I am just a little dubious about any theory for which it is said that there is direct evidence, because I don't think any impact is necessary, although it could happen I know.

If the planets started as an undifferentiated disk of material revolving around the Sun, and accreted into the separate planets, it would seem that the angular momentum of the constituent parts would equal that of the result, but that includes the angular momentum around the Sun, as well as its rotation. Depending upon where the part(icle)s coalesced, in solar orbit, that might determine a whole range of rotation aspects. I'm not sure...let me do some calculating.

milli360
2004-Jul-02, 02:41 PM
milli360:
...let me do some calculating.
OK, the Earth's angular momentum due to its orbit around the Sun is 2.7 x 10^40 kg m^2/s, but its angular momentum from rotating is 0.2 millionth of that. Because of conservation of angular momentum, the sum of the two is equal to the combined angular momentum of the masses that came together to make the Earth. Of course, the AM is changing even now, due to tidal friction, and impacts, but this is a back-of-the-envelope sort of calculation.

By chance, a few pieces accumulated, and then were able to grow faster, because of their size. Had the pieces first accumulated somewhere else, the result might have been different--the orbit of the Earth might have been a little farther out. If so, then the orbit AM would be been greater, which means that the rotational AM would have to be less.

How much and how far?

Unless my Excel spreadsheet has precision problems, it looks like this: if the Earth had coalesced 30 kilometers farther out, it would have an additional amount of angular momentum equal to about twice its AM due to its rotation. In order for the sum to remain the same, the rotation of the Earth would have to be the same as it is now, but reversed.

That would have been the result of many many impacts, but no single one would have "tipped over" the Earth.

Eroica
2004-Jul-03, 04:39 PM
I have a problem with the impactor theory. If a large body impacted Uranus, causing it to topple over, why would the planet stop when its axial inclination reached a certain value? Would it not continue to rotate about a second axis indefinitely? :-k

umop ap!sdn
2004-Jul-04, 07:34 AM
Moons like Titania and Oberon..., if they were captured objects, they would have wound up in orbits at more normal angles to the ecliptic.

Actually, it depends on how far out the moons orbit. Close orbits tend to precess about the planet's equator; farther out orbits like our Moon tend to precess about the ecliptic (Kaptain K is right), and in-between moons have their own Laplace planes (http://scienceworld.wolfram.com/physics/LaplacePlane.html).


I have a problem with the impactor theory. If a large body impacted Uranus, causing it to topple over, why would the planet stop when its axial inclination reached a certain value? Would it not continue to rotate about a second axis indefinitely? :-k

Depends on whether the impact managed to slow its rotation down enough. As fast as it is spinning now, there must be a major gyroscopic action taking place, so no it wouldn't.

Jens
2005-Jul-13, 05:43 AM
4) Due to torque from the equatorial bulge (oblateness caused by planetary rotation), moons tend to be drawn into equatorial orbits over billions of years*.


I was wondering how Uranus' moons happened to be aligned along with its axial tilt, and came across this thread. I suspected something like this might be at work, because the other alternative, that the axial tilt happened while the moons were forming, doesn't seem to make sense.

It got me wondering, though. Would this work for the solar system as a whole? I recall that the sun doesn't have very much oblateness, so maybe it wouldn't. But if this effect does work, wouldn't it, over a long period of time, encourage the planets to orbit in the sun's equatorial orbit?

Jens
2005-Jul-13, 08:57 AM
There's another thing I forgot about, that I meant to ask. I'm reading a pretty informative book, though a bit old, called Asimov's New Guide to Science. He mentions that the equatorial regions of Uranus were found to be warmer (a relative term in this case!) than the polar regions, even though the planet is tilted on its side so that the poles essentially point to the sun. He says there is no explanation for this, but the book is about 20 years old. Is there an accepted explanation now?

Fram
2005-Jul-13, 10:18 AM
There's another thing I forgot about, that I meant to ask. I'm reading a pretty informative book, though a bit old, called Asimov's New Guide to Science. He mentions that the equatorial regions of Uranus were found to be warmer (a relative term in this case!) than the polar regions, even though the planet is tilted on its side so that the poles essentially point to the sun. He says there is no explanation for this, but the book is about 20 years old. Is there an accepted explanation now?

I have no explanation, but only one of the poles will be pointing at the Sun, the other one is pointing away from it, so that one would be considerably cooler, I suppose.

NEOWatcher
2005-Jul-13, 12:25 PM
You're all wrong. It's rolling on the kupier belt. :D

Grey
2005-Jul-13, 12:50 PM
I have no explanation, but only one of the poles will be pointing at the Sun, the other one is pointing away from it, so that one would be considerably cooler, I suppose.
And of course the axis of rotation will stay fixed throughout the orbit, so each pole takes turns being the one aimed at the Sun, with a "spring" and "fall" in between, when the equator faces the Sun and the axis is more or less parallel to the orbital path.

Doing a bit of a search, I see that when Voyager 2 passed by, one pole was more or less pointing straight toward the Sun, but that in 2007 the Sun will be over the equator. I can see the references to the equatorial regions being warmer, and no clear explanation known, but I also don't know when those measurements were taken. Right now, I'd certainly expect the equator to be warmer, but back when Voyager 2 was passing it would seem that the facing pole should have been warmer.

Still, it's perhaps not as weird as it might first seem. Maybe just the fact that each pole has a chance to get really cold during its turn facing directly away from the Sun accounts for it. Because of the rotation, the equator always has at least an oblique angle to the Sun throughout the year (if it were below the horizon for part of the day, it would be above the horizon for the other part of the day), while each pole spends much of the orbit (out of an 84 year period) in a long cold night.

Fram
2005-Jul-13, 12:54 PM
That's what I said! :^o :lol: Thanks Grey, that's a much better explanation. Mine was a bit simplistic...

NEOWatcher
2005-Jul-13, 01:04 PM
You're all wrong. It's rolling on the kupier belt. :D Like This (http://www.amazon.com/exec/obidos/ASIN/B0006IGX8O/ref=pd_sxp_elt_l1/102-3666891-2478560)

Grey
2005-Jul-13, 01:12 PM
That's what I said! :^o :lol: Thanks Grey, that's a much better explanation. Mine was a bit simplistic...
Hey, I'm just making it up as I go along! :D I was just realizing that it's not quite as implausible as it might first seem on first look. I didn't find what the actual temperature difference was. Anyone have better data?

Stardate
2005-Jul-13, 02:54 PM
You're all wrong. It's rolling on the kupier belt. :D Like This (http://www.amazon.com/exec/obidos/ASIN/B0006IGX8O/ref=pd_sxp_elt_l1/102-3666891-2478560)

Kuiper Belt :)

gopher65
2005-Jul-13, 04:02 PM
Kaptain K:
Also, the Earth's moon is now thought to be the result of such an impact, yet its orbit is nearly equatorial.
The moon's orbit can be as much as 30 degrees away from the equator. Seems like a lot to be called "nearly equatorial"

IIRC Luna doesn't orbit Earth, it orbits Sol. Therefore its orbit should be lined up with Sol's Rotation, not Earth's.

Klausnh
2005-Jul-13, 08:59 PM
Could Uranus be a planet not of this solar system that was captured by the sun's gravity?

Grey
2005-Jul-13, 09:24 PM
Could Uranus be a planet not of this solar system that was captured by the sun's gravity?
Anything is possible. :) But it doesn't seem very likely, actually. Although it's axis of rotation is pretty strange, it's inclination to the ecliptic is only 0.772 degrees, and a nearly circular orbit with an eccentricity of only 0.0457. There would be no particular reason for a captured object to orbit in the same plane as all the other planets. Also, to be captured it needs to have some kind of interaction with another massive object, and there's no particular reason why that would favor a nearly circular orbit.

publiusr
2005-Jul-13, 09:49 PM
The ecliptic is the plane of the Earth's orbit by definition.

And besides, there are no fixed stars. Everything in the universe is in motion. Are you sure? 8-[
Yes, the ecliptic is defined as the plane of the Earth's orbit.
and
Yes, I'm sure there are no "fixed" stars. The Black hole at the center of the Milky Way "appears" to be immobile with respect to the mass of the galaxy as a whole, with all the stars, gas and dust revolving around it. But, it still moves with respect to the other galaxies.

I thought the ecliptic plane was defined as extending from the sun's equator out into the solar system.

Grey
2005-Jul-13, 10:34 PM
I thought the ecliptic plane was defined as extending from the sun's equator out into the solar system.
No, it's the plane of the Earth's orbit (http://en.wikipedia.org/wiki/Ecliptic).

Klausnh
2005-Jul-14, 01:59 PM
Could Uranus be a planet not of this solar system that was captured by the sun's gravity?
Anything is possible. :) But it doesn't seem very likely, actually. Although it's axis of rotation is pretty strange, it's inclination to the ecliptic is only 0.772 degrees, and a nearly circular orbit with an eccentricity of only 0.0457. There would be no particular reason for a captured object to orbit in the same plane as all the other planets. Also, to be captured it needs to have some kind of interaction with another massive object, and there's no particular reason why that would favor a nearly circular orbit.
Aren't all captured moons in the same plane? How many captured moons don't have a nearly circular orbit?
As I understand it, the earth-moon was formed when a large object collided with the earth. Why would the earth not have been tilted as much as Uranus after that collision? Where is the large object that hypothetically collided with Uranus?

Grey
2005-Jul-14, 03:21 PM
Aren't all captured moons in the same plane?
More or less, but remember that those captured moons are all assumed to have originated within the solar system. So, like everything else around here, they were near the plane of the ecliptic before they were captured.


How many captured moons don't have a nearly circular orbit?
That's a valid point. I don't know offhand, but I'm not sure that there are very many. Circular orbits are less likely to run close to the other objects and be disturbed by them, so I'll agree there's a tenedency to favor them.


As I understand it, the earth-moon was formed when a large object collided with the earth. Why would the earth not have been tilted as much as Uranus after that collision?
Well, why would it have been, necessarily? It seems to me the resulting tilt would presumably be some fairly random value, which could be large or small. I think 23 degrees seems both fairly large and pretty random.


Where is the large object that hypothetically collided with Uranus?
No idea, though it doesn't have to be a separate object. If something collided, it's entirely possible that it was just subsumed into Uranus. In fact, that's a pretty good way to transfer rotational angular momentum.

Klausnh
2005-Jul-15, 11:08 AM
Aren't all captured moons in the same plane?
More or less, but remember that those captured moons are all assumed to have originated within the solar system. So, like everything else around here, they were near the plane of the ecliptic before they were captured.
I originally thought that the plane of the solar ecliptic is the same as the plane of the galactic ecliptic and that is why a captured planet would be in the same plane as the solar system. However, after doing some reading, I discovered the plane of the galactic ecliptic actually matches the tilt of Uranus. So an interstellar planet orbiting in the plane of the ecliptic of the Milky Way would have the same tilt as Uranus.



Where is the large object that hypothetically collided with Uranus?
No idea, though it doesn't have to be a separate object. If something collided, it's entirely possible that it was just subsumed into Uranus. In fact, that's a pretty good way to transfer rotational angular momentum.
With such a large collision, shouldn't there be more of a difference between the rotational angular momentum of Uranus and the other large planets?
Could there ever be evidence found that Uranus is a captured planet?

Jens
2005-Jul-15, 11:20 AM
However, after doing some reading, I discovered the plane of the galactic ecliptic actually matches the tilt of Uranus. So an interstellar planet orbiting in the plane of the ecliptic of the Milky Way would have the same tilt as Uranus.


To me, that's an intriguing thing to learn. I don't really know how to find the data myself, but just out of curiosity, how close is it? It would be fascinating to find that Uranus was captured from the interstellar space.

Grey
2005-Jul-15, 12:58 PM
I originally thought that the plane of the solar ecliptic is the same as the plane of the galactic ecliptic and that is why a captured planet would be in the same plane as the solar system. However, after doing some reading, I discovered the plane of the galactic ecliptic actually matches the tilt of Uranus. So an interstellar planet orbiting in the plane of the ecliptic of the Milky Way would have the same tilt as Uranus.
That's not correct, though. The galactic plane is inclined about 63 degrees to the ecliptic, so Uranus is still off by around 30 degrees.


With such a large collision, shouldn't there be more of a difference between the rotational angular momentum of Uranus and the other large planets?
There is a big difference, in direction if not magnitude. Angular momentum is a vector quantity, and changing its direction is just as profound a difference as changing its magnitude. I'd imagine that it might be fun to run simulations of such a collision to see if you could get it to work. Heck, someone may have already done so. Besides, remember that to capture an extrasolar planet, you'd actually need a more energetic collision or encounter to kill the planet's velocity and allow capture than is necessary just to change the rotation. So with either of these ideas, we still have to assume that Uranus had a pretty serious encounter.


Could there ever be evidence found that Uranus is a captured planet?
I have no idea. You'd probably want to be able to show that its overall composition was significantly different from the rest of the solar system, but to do that, you'd probably need to be able to get an accurate measurement of the material of gas giants at the core as well as the upper portion fo the atmosphere, and that sounds really hard.

dgavin
2005-Jul-15, 02:39 PM
I seem to remember some modeling done on an old Nova program, which attributed Uranus's orbit shift to a massive object (a star of some kind) having passed through the outskirts (Just outside of Krupier belt) of our solar system.

I also think this was the same model that proposed that the astroid belt was once a planet (of which Mars was a moon of) that had been shattered by comeing within the roche limit of one of the passing star's planets.

Mars was devestated by a huge chunk of that, (hellas impact crater)

They had called the asteroid belt former planet, Titanica...heh

I think the last half of this model was thought to be incorrect later as the Asteriod belt seems to be made of rock of approimately 3 different composistions? (although it could be possible Titanica had multiple moons and only Mars survived i guess)

Klausnh
2005-Jul-15, 03:20 PM
I originally thought that the plane of the solar ecliptic is the same as the plane of the galactic ecliptic and that is why a captured planet would be in the same plane as the solar system. However, after doing some reading, I discovered the plane of the galactic ecliptic actually matches the tilt of Uranus. So an interstellar planet orbiting in the plane of the ecliptic of the Milky Way would have the same tilt as Uranus.
That's not correct, though. The galactic plane is inclined about 63 degrees to the ecliptic, so Uranus is still off by around 30 degrees.

from www.seds.org (http://www.seds.org/~spider/spider/ScholarX/coords.html)

The inclination of the galactic equator to Earth's equator is thus 62.9 deg.
As I understand it ecliptic and plane of the earth's equator differ by 23 degrees.



With such a large collision, shouldn't there be more of a difference between the rotational angular momentum of Uranus and the other large planets?
There is a big difference, in direction if not magnitude. Angular momentum is a vector quantity, and changing its direction is just as profound a difference as changing its magnitude. I'd imagine that it might be fun to run simulations of such a collision to see if you could get it to work. Heck, someone may have already done so. Besides, remember that to capture an extrasolar planet, you'd actually need a more energetic collision or encounter to kill the planet's velocity and allow capture than is necessary just to change the rotation. So with either of these ideas, we still have to assume that Uranus had a pretty serious encounter.

I don't know enough about orbital mechanics to continue this part of the discussion. Thanks for the info.

Grey
2005-Jul-15, 04:05 PM
from www.seds.org (http://www.seds.org/~spider/spider/ScholarX/coords.html)

The inclination of the galactic equator to Earth's equator is thus 62.9 deg.
As I understand it ecliptic and plane of the earth's equator differ by 23 degrees.
These are both correct statements, and it looks like the 63 degrees I found was indeed the inclination between the galactic plane and the celestial equator. I'd remembered that it was about 60 degrees, but looked quickly for a more precise figure, and it looks like my source was wrong there, so thanks for correcting me on that. However, that would only add up to about 90 degrees if both tilts were aligned in the same direction, and they aren't. The two really are aligned at about 60 degrees, as you can see in this (http://coolcosmos.ipac.caltech.edu/image_galleries/IRAS/allsky.html) nice picture from IRAS. The bright red-orange horizontal band is the plane of the galaxy, and the broad diagonal blue band is from dust in the plane of the solar system.

If you prefer hard numbers, though, here (http://asc.harvard.edu/toolkit/precess.jsp)'s a really nice little Java applet that can let you convert between galactic, equatorial, and ecliptic coordinates, so you can put in 90 degrees latitude galactic, and see that in ecliptic coordinates, that comes out to about 29.8 degrees.


I don't know enough about orbital mechanics to continue this part of the discussion. Thanks for the info.
I was only doing rough estimates, myself. However, if I've done the math right, the total energy from rotation for Uranus is about 1.2 x 10^32 J (that's assuming a uniform sphere, so the real number would be less than this, since it's presumably denser toward the center), so if you applied twice that much energy, you could pretty much stop its rotation entirely and then start it up rotating at a comparable rate in whatever direction you'd like. On the other hand, as an object comes into the solar system from outside, it picks up speed as it falls toward the Sun. Unless you get rid of that excess kinetic energy, it will send it right out again. For an object the mass of Uranus at its present distance, that works out to about 4 x 10^33 J, and you need to get rid of half of that to drop into a stable circular orbit, so that's about an order of magnitude more energy that you'd need to transfer.

[edited to correct math]

Disinfo Agent
2005-Jul-15, 05:56 PM
I seem to remember some modeling done on an old Nova program, which attributed Uranus's orbit shift to a massive object (a star of some kind) having passed through the outskirts (Just outside of Krupier belt) of our solar system.
Strange that it had so little effect on Neptune, no?


I also think this was the same model that proposed that the astroid belt was once a planet (of which Mars was a moon of) that had been shattered by comeing within the roche limit of one of the passing star's planets.
There isn't enough mass in the Asteroid Belt for such a large planet to have existed. (http://seds.lpl.arizona.edu/nineplanets/nineplanets/asteroids.html)

aurora
2005-Jul-15, 08:06 PM
As I understand it, the earth-moon was formed when a large object collided with the earth. Why would the earth not have been tilted as much as Uranus after that collision?
Well, why would it have been, necessarily? It seems to me the resulting tilt would presumably be some fairly random value, which could be large or small. I think 23 degrees seems both fairly large and pretty random.


And the moon tends to keep the Earth's tilt fairly constant, unlike Mars (for example).

so perhaps the Moon tended to pull the Earth back slightly if it was tilted more than 23 degrees following the collision.

Klausnh
2005-Jul-16, 01:38 PM
from www.seds.org (http://www.seds.org/~spider/spider/ScholarX/coords.html)

The inclination of the galactic equator to Earth's equator is thus 62.9 deg.
As I understand it ecliptic and plane of the earth's equator differ by 23 degrees.
These are both correct statements, and it looks like the 63 degrees I found was indeed the inclination between the galactic plane and the celestial equator. I'd remembered that it was about 60 degrees, but looked quickly for a more precise figure, and it looks like my source was wrong there, so thanks for correcting me on that. However, that would only add up to about 90 degrees if both tilts were aligned in the same direction, and they aren't. The two really are aligned at about 60 degrees, as you can see in this (http://coolcosmos.ipac.caltech.edu/image_galleries/IRAS/allsky.html) nice picture from IRAS. The bright red-orange horizontal band is the plane of the galaxy, and the broad diagonal blue band is from dust in the plane of the solar system.

If you prefer hard numbers, though, here (http://asc.harvard.edu/toolkit/precess.jsp)'s a really nice little Java applet that can let you convert between galactic, equatorial, and ecliptic coordinates, so you can put in 90 degrees latitude galactic, and see that in ecliptic coordinates, that comes out to about 29.8 degrees.
Thanks fpr the links and info. It looks my sources were wrong on the angle of the ecliptic to the palne of the MW.



I don't know enough about orbital mechanics to continue this part of the discussion. Thanks for the info.
On the other hand, as an object comes into the solar system from outside, it picks up speed as it falls toward the Sun. Unless you get rid of that excess kinetic energy, it will send it right out again. For an object the mass of Uranus at its present distance, that works out to about 4 x 10^33 J, and you need to get rid of half of that to drop into a stable circular orbit, so that's about an order of magnitude more energy that you'd need to transfer.

[edited to correct math]
After thinking about this all night, I do not understand. Wouldn't the amount of kinetic energy just need to be re-directed? Depending on the initial velocity, I thought there were there 3 scenarios on what happens to an object that comes within the gravitational field of the sun:
1) the object falls into the sun when its velocity is too small
2) the object continues on an altered course when its velocity is large
3) the object goes into orbit around the sun when its velocity is just right.

Grey
2005-Jul-16, 04:47 PM
Thanks fpr the links and info. It looks my sources were wrong on the angle of the ecliptic to the palne of the MW.
No trouble. It meant I got to track down the IRAS picture, which I thought was really stunning.


After thinking about this all night, I do not understand. Wouldn't the amount of kinetic energy just need to be re-directed? Depending on the initial velocity, I thought there were there 3 scenarios on what happens to an object that comes within the gravitational field of the sun:
1) the object falls into the sun when its velocity is too small
2) the object continues on an altered course when its velocity is large
3) the object goes into orbit around the sun when its velocity is just right.
This is perhaps pretty tricky, but unless you kill the excess kinetic energy, it will always end up being option two. Why is that? Well, remember that when something wanders into the solar neighborhood, it isn't just passing through unaffected. As it gets steadily closer to the Sun, it's moving deeper into the gravity well, and so potential energy is turned into kinetic energy. In essense, it's "falling" toward the Sun and picking up speed as it does so. If you think about conservation of energy, it shouldn't be too surprising that the object converts potential energy to kinetic energy as it gets closer, and that the amount of kinetic energy it now has would be just enough, if converted back into potential energy, to push it exactly as far away as it started. Since the minimum velocity you need to escape is exactly the velocity that would give you this amount of kinetic energy, the actual velocity of an object coming in is always at least equal to the escape velocity from the Sun at that point (it probably started with some velocity toward the Sun to start with).

Well, since it can start with an initial velocity, though, what if it started out with some velocity that would counterbalance what it picks up from the Sun? Well, since the acceleration is toward the Sun, that initial velocity would have to be away from the Sun, and if it starts out moving away from the Sun, it will never get close at all! As for option one, it could hit the Sun if it happened to be headed right for it, but as it turns out from orbital mechanics, if it isn't already aimed pretty much straight on, it won't actually hit. The Sun is actually one of the hardest places in the solar system to get to.

So how can you capture something at all? It has to come in, and then have some kind of encounter (collison, near passage to some other object, or something) that will get rid of some of the excess kinetic energy. It works out that the kinetic energy an object needs for a circular orbit at any particular distance is exactly half the kinetic energy it would have if it fell in to that distance from infinitely far away starting at zero velocity, which is a pretty good approximation if you assume the object starts from light years away. Technically, of course, it doesn't have to lose all of that energy in a single encounter (it could lose just enough in an intial encounter to leave it in a highly elongated elliptical orbit, and then lose the rest in a series of later encounters), but whether through a single encounter or through many separate ones, that energy has to be dissipated somehow to get to a circular orbit.

Does that help, or have I just confused the issue further?

Klausnh
2005-Jul-16, 09:49 PM
Thanks, Grey. As always very helpful and informative. Give me some time to digest your post and if I have any more questions, I'll get back to you.
On a slightly different, but related vein: When I was reading about the ecliptic, I noticed that the Sun has an angle of about 7 degrees (if my sources are correct) in the plane of the ecliptic. I would think that angle should be zero, as the sun formed at the center of the rotating gas/debris? Any theories or explanation why the sun is not rotating at 0 degrees in the plane of ecliptic?

Arneb
2005-Jul-16, 10:37 PM
On a slightly different, but related vein: When I was reading about the ecliptic, I noticed that the Sun has an angle of about 7 degrees (if my sources are correct) in the plane of the ecliptic. I would think that angle should be zero, as the sun formed at the center of the rotating gas/debris? Any theories or explanation why the sun is not rotating at 0 degrees in the plane of ecliptic?

Oh, not at all - The ecliptic is defined by Earth's rotation around the sun, not by any property of the Sun itself:



I thought the ecliptic plane was defined as extending from the sun's equator out into the solar system.

No, it's the plane of the Earth's orbit. (http://en.wikipedia.org/wiki/Ecliptic)

You could also say that Earth's plane of rotation is 7° off the rotational plane of the Sun. Every planet has its own "ecliptic", as the orientations of their respective orbital planes are slightly tilted (=inclined) with respect to each other (Pluto being the odd man out with its large inclination). The primordial cloud from which the planets formed was not a razor sharp ring rotating exactly over the equator of its star, and later events (collisions, etc.) worked to spread the planes of the planetary orbits somewhat.

Faultline
2005-Jul-25, 03:28 AM
By straining too hard over the toilet.

:wink:


'scuse me, couldn't resist that one.

Faultline

01101001
2006-Apr-30, 06:53 AM
Here's a new one:

Planetary Society Weblog: How Uranus got its tilt (http://www.planetary.org/blog/article/00000553/)


Argentinian scientist Adrián Brunini published a paper in Nature proposing a novel mechanism by which the planet Uranus could have acquired the nearly horizontal tilt of its rotation axis.
[...]
He says you don't need impacts to make the giant planets tilt. "The present obliquities of the giant planets were probably achieved when Jupiter and Saturn crossed the 1:2 orbital resonance," he proposes.
[...]
During all of this orbital dancing, Brunini says, the planets exchange a great deal of angular momentum. In particular, the very close approaches of Uranus to Saturn causes them to exchange momentum, which, over time, changes their axial tilts. Now, this process is relatively fast, happening over a few hundred thousand years, but it is much slower than a single humongous impact tipping over a planet. The slower pace of the process that Brunini proposes means that as the planets slowly tilt, their satellite systems can actually follow the change in tilt.

neilzero
2006-Apr-30, 12:50 PM
Do tide forces move moon orbits gradually to the plane of the planet's equator? If not, why would we expect moons to follow the axial tilt of their planet? Neil

jlhredshift
2006-May-03, 07:02 PM
After reading all the posts I have several thoughts:

AGN and their associated accreation discs are an oversized model of what Grey explained about incoming objects to the sun.

The data we have on Uranus's magnetic field comes from Voyager (if there is new data please refer me to it), but I would think that it's differential angle to that of rotation for a world as cold as it is would also need to be explained.

Venus was mentioned in one of the posts and I would think that the same thoughts could be applied to it as well for an explanation of axial tilt, with main difference being the average density being higher and the total mass being lower; with obvious difference in total angular momentum and depth in the suns gravitational well. Brunini's theory probably would not work for Venus.