Jump to content

Gravity wells determine orbital plane?


DeckerdSmeckerd
 Share

Recommended Posts

For example, if you drew a line along Uranus's rings and equator and did that through one entire revolution of Uranus around the Sun, where is that intersection? (Do I need to explain the orientation of the line in respect to the Sun. I can try to but I don't have the terms.) Maybe it points to something with mass? It is further from the Sun than Saturn is. Saturn's are oriented towards the Sun, I read. (Or perhaps there is something undetected that is closer to Uranus).

Edited by DeckerdSmeckerd
Link to comment
Share on other sites

On 11/20/2021 at 12:12 PM, DeckerdSmeckerd said:

From what I was able to determine, even though the solar system's orbital plane is 60 degrees tilted from the galactic plane, the planetary bodies fall toward and rise away from the center of the galaxy. 

I think of a gravity well as a direction down (just for the sake of describing my thought) since that is how we think of it while on earth. Up is away from the gravity and down is towards it. If the center of the galaxy exerts even the slightest of gravity on our solar system then the direction towards it is down. (Just for the sake of describing my thought process). That means everything wants to fall toward it unless there is some force that counteracts it. Even though it might be too weak to move the ocean, I don't see why the planets would not fall toward that distant gravity well because no matter how weak it is, it is still the natural direction for things to fall. I have checked our solar system and they do fall around our Sun in this way.

 

The point is that this "down" is the same for everything in the solar system and they all react equally to it.  While the planets do "fall" towards the center of the galaxy, they do not tend to do so any more than the Sun does. The solar system is so small compared to the distance to the center of the galaxy, that any difference in gravity acting over it caused by the galaxy's mass is insignificant. ( a quick calculation indicates that some nearby stars would have a larger effect.)

Secondly.  Even if there was a measurable differential across the solar system, it would not cause the orbital plane to align with it. The solar system is a rotating structure, and when you apply a force in order to try to change the axial orientation of a rotating structure you get a precession instead. (the axis "wobbles")

Now, since the rate of this precession depends on a number of factors, including the angular velocity of the rotating structure, and each planet orbits the Sun at different angular velocities, each planet's orbit acts as a singular rotating structure with its own independent precession rate.  Thus, over time, you would end up with planets orbiting at all kinds of orientations to each other and not nearly in the same plane as we have now.  The fact that all the major planets, for the most part, share the same orbital plane shows that there has been no significant force acting on it in this way.

16 hours ago, DeckerdSmeckerd said:

For example, if you drew a line along Uranus's rings and equator and did that through one entire revolution of Uranus around the Sun, where is that intersection? (Do I need to explain the orientation of the line in respect to the Sun. I can try to but I don't have the terms.) Maybe it points to something with mass? It is further from the Sun than Saturn is. Saturn's are oriented towards the Sun, I read. (Or perhaps there is something undetected that is closer to Uranus).

If Saturn's rings were oriented toward the Sun, then we would not see them as well as we can from Earth (since we'd always be looking at them nearly edge on.)

Sure, twice an orbit, the ascending or descending nodes of the rings align with the Sun, But since the ring orientation remains nearly fixed relative to the stars as it orbits, this has to happen no matter how they are oriented.

Link to comment
Share on other sites

2 hours ago, Janus said:

The point is that this "down" is the same for everything in the solar system and they all react equally to it.  While the planets do "fall" towards the center of the galaxy, they do not tend to do so any more than the Sun does. The solar system is so small compared to the distance to the center of the galaxy, that any difference in gravity acting over it caused by the galaxy's mass is insignificant. ( a quick calculation indicates that some nearby stars would have a larger effect.)

Secondly.  Even if there was a measurable differential across the solar system, it would not cause the orbital plane to align with it. The solar system is a rotating structure, and when you apply a force in order to try to change the axial orientation of a rotating structure you get a precession instead. (the axis "wobbles")

Now, since the rate of this precession depends on a number of factors, including the angular velocity of the rotating structure, and each planet orbits the Sun at different angular velocities, each planet's orbit acts as a singular rotating structure with its own independent precession rate.  Thus, over time, you would end up with planets orbiting at all kinds of orientations to each other and not nearly in the same plane as we have now.  The fact that all the major planets, for the most part, share the same orbital plane shows that there has been no significant force acting on it in this way.

If Saturn's rings were oriented toward the Sun, then we would not see them as well as we can from Earth (since we'd always be looking at them nearly edge on.)

Sure, twice an orbit, the ascending or descending nodes of the rings align with the Sun, But since the ring orientation remains nearly fixed relative to the stars as it orbits, this has to happen no matter how they are oriented.

Thanks Janus. I really appreciate everyone's help with this even though there is likely nothing much to gain from this conversation.

Maybe I could understand this better if I put forward an example and if you tell me if that causes precession.

Suppose I have a star system like ours. Imagine a 2d isosceles right triangle. If the star system was at the right angled corner and there were two equal gravity wells at the angles adjacent to the hypotenuse, close enough to a have a real effect on the star system, would "down" in that case be toward the center point along the hypotenuse and would that cause precession in the star system?

Imagine a triangular pyramid where each side is identical. At the top of the pyramid is the star system. The other 3 vertexes at the base each have an equal gravity well. Would "down" be the direction of the center point of the base triangle? Does that cause precession?

Link to comment
Share on other sites

8 minutes ago, DeckerdSmeckerd said:

Thanks Janus. I really appreciate everyone's help with this even though there is likely nothing much to gain from this conversation.

Is there a point to this?  You ask a question, a member takes the time to answer it and then you say essentially, "Yeah, what ever".  Very odd behavior it seems to me.

Maybe we aren't answering the question you think you are asking.  Your questions are not very clear, for instance:

11 minutes ago, DeckerdSmeckerd said:

Suppose I have a star system like ours. Imagine a 2d isosceles right triangle. If the star system was at the right angled corner and there were two equal gravity wells

First this is clearly this is not a star system like ours.  The other problem is that your isosceles right triangle will only exist for a moment in time since the objects will be orbiting a common center of gravity.

17 minutes ago, DeckerdSmeckerd said:

If the star system was at the right angled corner and there were two equal gravity wells at the angles adjacent to the hypotenuse, close enough to a have a real effect on the star system, would "down" in that case be toward the center point along the hypotenuse and would that cause precession in the star system?

Where is the observer located?  By "down" do you mean the barycenter?

18 minutes ago, DeckerdSmeckerd said:

Imagine a triangular pyramid where each side is identical. At the top of the pyramid is the star system. The other 3 vertexes at the base each have an equal gravity well. Would "down" be the direction of the center point of the base triangle?

No, it is not possible to answer without defining down, specifying the observers location and specifying the size of the 'gravity wells'.  Of course this pyramid arrangement would also only occur for a moment in time.

Link to comment
Share on other sites

I found a nice thread regarding nodal precession (that shifting of the rotational plane of the lightest object...) for a special, well studied case example we're witnessing from the inside - the Sun-Earth-Moon-three-body-system:

https://www.physicsforums.com/threads/what-is-the-cause-of-lunar-nodal-and-apsidal-precession.851208/

It refers to textbook opinion on the relative influences of sun, earth's shape, and the other planets - and states that the Sun's gravitational influence is beating the others by five orders of magnitude. After all, gravity weakens by d² (d being the distance).

 

As to your tetraedric (concise term for the equilateral triangular pyramid) model system:

This depends.

First and simplest case: They all start out with velocity zero. Then the whole shebang will simply collapse in the center.

Which brings us back to momentum:

In order to keep anything stably separate when gravity is involved, there has to be some velocity, which then leads to momentum, and from there to angular momentum.

Long story short: These three+-body-systems are always chaotic systems, where you can extrapolate the future to quite some degree of precision, but without knowing the exact starting parameters for that prediction, not much can be said...

Edited by Godot
typo correction
Link to comment
Share on other sites

25 minutes ago, Bufofrog said:

Is there a point to this?  You ask a question, a member takes the time to answer it and then you say essentially, "Yeah, what ever".  Very odd behavior it seems to me.

Let me respond to this part first. I just mean, I acknowledge I am probably not figuring out anything new, so not much to gain from it for those that already know these answers. I am gaining from it more than you, for example. 

29 minutes ago, Bufofrog said:

First this is clearly this is not a star system like ours.  The other problem is that your isosceles right triangle will only exist for a moment in time since the objects will be orbiting a common center of gravity.

Where is the observer located?  By "down" do you mean the barycenter?

No, it is not possible to answer without defining down, specifying the observers location and specifying the size of the 'gravity wells'.  Of course this pyramid arrangement would also only occur for a moment in time.

Yes, this would only be a hypothetical configuration to understand how the forces work. If I am understanding barycenter, it would mean the gravity wells are orbiting each other. In these hypotheticals, I was not implying there were other forces at work on the gravity wells, but the star system is still operating normally. From what you say is needed, it makes me wonder, is the degree of precession correlated to the strength of the gravity wells? So, if the gravity wells were very distant and the gravity weaker, it might affect the degree of precession?

37 minutes ago, Godot said:

I found a nice thread regarding nodal precession (that shifting of the rotational plane of the lightest object...) for a special, well studied case example we're witnessing from the inside - the Sun-Earth-Moon-three-body-system:

https://www.physicsforums.com/threads/what-is-the-cause-of-lunar-nodal-and-apsidal-precession.851208/

It refers to textbook opinion on the relative influences of sun, earth's shape, and the other planets - and states that the Sun's gravitational influence is beating the others by five orders of magnitude. After all, gravity weakens by d² (d being the distance).

 

As to your tetraedric (concise term for the equilateral triangular pyramid) model system:

This depends.

First and simplest case: They all start out with velocity zero. Then the whole shebang will simply collapse in the center.

Which brings us back to momentum:

In order to keep anything stably separate when gravity is involved, there has to be some velocity, which then leads to momentum, and from there to angular momentum.

Long story short: These three+-body-systems are always chaotic systems, where you can extrapolate the future to quite some degree of precision, but without knowing the exact starting parameters for that prediction, not much can be said...

Well, I have to read some more about precession. If I have the general idea, the wobble is caused because the body in question gets nearer or farther away from a second gravity well. So it gets pulled to a greater or lesser degree by the second well but this affects the axis as well. If we don't observe the axis shifting then we can presume there isn't a second gravity well. Did I say that correctly?

Link to comment
Share on other sites

I have learned that there is a difference in the way astronomers and physicists use the word "precession", though I don't quite have my mind wrapped around it in any context. However, since I was more concerned about forces that might not be observable but still be real, maybe it will just be too confusing to talk about this concept.

In my hypothetical example, I was just interested in the forces 2-3 stationery gravity wells would have on a normal solar system. Not in any way a realistic scenario though, but Bufofrog said it was important to know where the observer is. It took me a while to understand that this might be the difference I read about. 

Quote

"Precession" and "procession" are both terms that relate to motion. "Precession" is derived from the Latin praecedere ("to precede, to come before or earlier"), while "procession" is derived from the Latin procedere ("to march forward, to advance"). Generally the term "procession" is used to describe a group of objects moving forward. The stars viewed from Earth are seen to proceed from east to west daily, due to the Earth's diurnal motion, and yearly, due to the Earth's revolution around the Sun. At the same time the stars can be observed to anticipate slightly such motion, at the rate of approximately 50 arc seconds per year, a phenomenon known as the "precession of the equinoxes".

In describing this motion astronomers generally have shortened the term to simply "precession". In describing the cause of the motion physicists have also used the term "precession", which has led to some confusion between the observable phenomenon and its cause, which matters because in astronomy, some precessions are real and others are apparent. This issue is further obfuscated by the fact that many astronomers are physicists or astrophysicists.

The term "precession" used in astronomy generally describes the observable precession of the equinox (the stars moving retrograde across the sky), whereas the term "precession" as used in physics, generally describes a mechanical process.

https://en.wikipedia.org/wiki/Axial_precession

 

edit: I didn't understand what Bufofrog meant, and I was about to say I observe the pyramid as if I was sitting on the floor and it was about 6 inches tall with a tiny little solar system at the top with planets swirling around getting closer and then farther from the different gravity wells at the other vertexes. I was trying to understand why that affects their axis, or if it does in that scenario. I need a simple example because a real solar system with all it's moving parts is hard to comprehend all at once, for me. 

Edited by DeckerdSmeckerd
Link to comment
Share on other sites

1 hour ago, DeckerdSmeckerd said:

Let me respond to this part first. I just mean, I acknowledge I am probably not figuring out anything new, so not much to gain from it for those that already know these answers. I am gaining from it more than you, for example.

Don't worry, the point of the forum to learn.  There is almost no chance anyone on any forum will come up with anything new, so don't sweat that!

Link to comment
Share on other sites

1 hour ago, DeckerdSmeckerd said:

as if I was sitting on the floor and it was about 6 inches tall with a tiny little solar system at the top with planets swirling around getting closer and then farther from the different gravity wells at the other vertexes.

Not sure I understand but here is a quick simulation I created using an online tool, It's purpose is to briefly illustrate the dynamics mentioned by other members. There are three stars of identical mass; the top one having two planets to illustrate a "solar system" and the bottom two stars are acting as "gravity wells". Initially the planets are moving while the three stars are at rest relative to one another. Note that the system is not stable. Also note that this is not an attempt at presenting a realistic or existing configuration of celestial bodies. 

In this specific case one planet is ejected from the system while the other is beginning to orbit around one of the gravity wells. A small change in initial setting would have impact on the system's development.

 

Edited by Ghideon
clarification
Link to comment
Share on other sites

24 minutes ago, Ghideon said:

Not sure I understand but here is a quick simulation I created using an online tool, It's purpose is to briefly illustrate the dynamics mentioned by other members. There are three stars of identical mass; the top one having two planets to illustrate a "solar system" and the bottom two stars are acting as "gravity wells". Initially the planets are moving while the three stars are at rest relative to one another. Note that the system is not stable. Also note that this is not an attempt at presenting a realistic or existing configuration of celestial bodies. 

 

In this specific case one planet is ejected from the system while the other is beginning to orbit around one of the gravity wells. A small change in initial setting would have impact on the system's development.

 

Nice. What tool is that. I would have to anchor the suns and see the effect on the planets that the two other suns have.

Link to comment
Share on other sites

8 minutes ago, DeckerdSmeckerd said:

What tool is that.

I used Gravity Simulator available at testtubegames .com*, it did not require me to download anything or do a registration. Google for "gravitation simulation online" or "Gravity simulator sandbox online" for various alternatives.

16 minutes ago, DeckerdSmeckerd said:

I would have to anchor the suns and see the effect on the planets that the two other suns have.

That is not possible in these kind of simulations as far as I know since celestial bodies will always mutually attract one another. It sounds like you wish to try something else than gravity; if so you may require a custom model.  

 

*) Disclaimer: I have not done any check of the quality of this site or used any other content than the simulator.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
 Share

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.