lgl
Level 2 Rank
Posts: 93
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Post by lgl on Dec 8, 2009 10:21:31 GMT
"I see no reason to doubt their calculations." Neither do I, but AM being constant does not mean that there is not some extra AM 'transfer' as result of inter-planet perturbations. You are asking me to calculate the AM without the perturbations to compare with real world but I have no idea how to do that. But no one has been able to find such a transfer! Carlson did the calculation in that prior thread and from my understanding of what he did, he did it correctly. It showed no such transfer. I think you mean Carsten Arnholm. His calculation showed a lot of 'transfer' in the sense when the total planetary AM increases the solar AM decreases equally so that the total for the system is constant, but like I said perturbations may increase the planetary AM even more, which the Sun will have to counter. The calcs Carsten did will not show that so no reason to redo them. In fact we know that the Sun moves because of perturbations. When Ea and Ve is accelerated by Ju they will move away from the Sun (or the Sun away from them). Jupiter will slow down a little so move closer to the Sun, which will also be closer to the SSBC because it is always somewhere between Sun-Ju, so the total motion of the Sun will probably be closer to the barycenter. Then the question is: How can the Sun be in free fall when it's motion is not equal to the sum of adding all the individual two-body systems Su&Me+Su&Ve+Su&Ea+...Su&n ? In a two-body system, which I agree is a true free fall scenario, the Sun will accelerate when the planet is getting closer increasing the gravitation. But in a three-body system, when the planets accelerate each other, the one that gets its speed increased will move away from the Sun, decreasing the Sun-planet gravity, so this is the total opposite to free fall. Now the Sun will accelerate while gravity decreases.
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Post by Graeme on Dec 8, 2009 20:34:01 GMT
I think you mean Carsten Arnholm. Sorry, yes. I was writing from memory since that thread is no longer available. His calculation showed a lot of 'transfer' in the sense when the total planetary AM increases the solar AM decreases equally so that the total for the system is constant, but like I said perturbations may increase the planetary AM even more, which the Sun will have to counter. The calcs Carsten did will not show that so no reason to redo them. I don't understand. His calculations showed that the total angular momentum was constant. What is it that they didn't show? And, as an aside, the Sun doesn't have to 'counter' anything. It just has to obey the well known laws of motion and gravity. Nothing more. Nothing less. In fact we know that the Sun moves because of perturbations. When Ea and Ve is accelerated by Ju they will move away from the Sun (or the Sun away from them). Jupiter will slow down a little so move closer to the Sun, which will also be closer to the SSBC because it is always somewhere between Sun-Ju, so the total motion of the Sun will probably be closer to the barycenter. Accepted. It is, of course, a matter of perspective. The observations you are talking about can only be observed from outside the solar system, or from the point of view of the barycenter (the mathematical 'center of gravity' of the solar system). Then the question is: How can the Sun be in free fall when it's motion is not equal to the sum of adding all the individual two-body systems Su&Me+Su&Ve+Su&Ea+...Su&n ? You've lost me again. Where has it been shown that the Sun's motion around the barycenter is not obeying the laws of gravity? If there's a paper anywhere that shows that, it should be headline news! I am assuming that by two-body systems you're talking about the gravitational interaction of two bodies, as per Newton's formula. If that's the case, then I believe that the net force on any single object (eg. the Sun) is the vector sum of the gravitation forces from all the other objects. In a two-body system, which I agree is a true free fall scenario, the Sun will accelerate when the planet is getting closer increasing the gravitation. But in a three-body system, when the planets accelerate each other, the one that gets its speed increased will move away from the Sun, decreasing the Sun-planet gravity, so this is the total opposite to free fall. Now the Sun will accelerate while gravity decreases. Again, you've got a completely different definition of free fall to me. It seems to me that by your definition, free fall can only exist if there are only two objects in the universe. Otherwise there are gravitation effects from other objects and you say that that means that things are not in free fall. I don't understand why you're saying that. All the objects are still interacting via gravitational forces, and no other forces, so why aren't they in free fall?
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Post by squire on Dec 9, 2009 2:05:39 GMT
Hi esarver, I found your question interesting. I first came into this idea from Fairbridge and Richard Mackey who describes the solar orbit as an epitrochoid, a series of large and small loops as you say a spirograph. Complete cycle of ~178 years, in three ~59 year cycles. These cycles are evident in the oscillation of Earth temperatures. Each 59 year cycle is roughly devided into a large loop and a small loop. In the large loop the Sun moves at double the speed as it does in the small loops as it nears the s/s barycenter. In the 80's and 90's the Sun was in a large loop. It is now into the beginning of a small loop and is slowing down. Activity is reducing and will continue to do so for another 20 something years according to the theory. This is why many feel the Sun will continue to be less active. Other factors will most certainly amplify this or not. Oh, and by the way, Charvatova, that is HER research. And smart she is! If this means it is in free fall, raise your hands. Sorry but I wasn't following the discussion. My loss not yours. Please continue. Sorry to intrrupt. I've never been good at threads. Cheers!
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lgl
Level 2 Rank
Posts: 93
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Post by lgl on Dec 9, 2009 10:05:06 GMT
"Where has it been shown that the Sun's motion around the barycenter is not obeying the laws of gravity?"
I don't think I am saying that. In a two-body system the Sun will accelerate when it is getting closer to the planet. It's losing height losing potentional energy and gaining kinetic energy, so perfect. But in the three-body system one object will accelerate while gaining height and the Sun will have to accelerate to counter. I can't understand how these totally different situations both can be free fall. The motion of the second planet will not counter the motion of the first because there's an angle between their motions. (sorry if that was bad english)
"His calculations showed that the total angular momentum was constant. What is it that they didn't show?"
You can accelerate any or all planets wildly and the Sun has to counter because the SSBC can't move. Carstens calcs would again show a constant (but higher) AM after you stopped applying the force(s). In real world there is no external force but that doesn't mean that the planets can't 'borrow' some AM from the Sun and give it back as result of perturbations. Carstens calcs will not show.
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Post by Graeme on Dec 10, 2009 0:38:13 GMT
But in the three-body system one object will accelerate while gaining height and the Sun will have to accelerate to counter. I can't understand how these totally different situations both can be free fall. The motion of the second planet will not counter the motion of the first because there's an angle between their motions. (sorry if that was bad english) Let's go back to 'what is free fall?'. Do you agree that free fall is when a body is reacting freely to gravitational forces? I think that's essentially what we agreed earlier. Well, in a three body system the net gravitational forces can be quite complex, but for any given situation they are easy to calculate. NASA does this to allow them to predict where objects in the solar system will be in the future. So far, those predictions match the observations, so that's a pretty good indication that their solution works. As I understand it, the only thing they are doing is applying Newton's laws of gravity (with a slight adjustment for Einstein for Mercury's orbit). Given that they can predict the movement of the bodies in the solar system just using the laws of gravity, what other force is being applied that makes you think that the objects will no longer be in free fall? Just because the behavior isn't intuitive, that doesn't mean another force is involved. Everything can be explained with just the laws of gravity. "His calculations showed that the total angular momentum was constant. What is it that they didn't show?" You can accelerate any or all planets wildly and the Sun has to counter because the SSBC can't move. Carstens calcs would again show a constant (but higher) AM after you stopped applying the force(s). In real world there is no external force but that doesn't mean that the planets can't 'borrow' some AM from the Sun and give it back as result of perturbations. Carstens calcs will not show. I thought he calculated the AM on a daily basis, and found that it always summed to the same value. There was no 'borrowing' that he could find. If you believe there is, it would therefore have to be at a frequency of less than a day. Again, it'll be up to you to show that, or find someone who is willing to redo Carsten's exercise with data at a frequency greater than a day so you can see for yourself if there is any 'borrowing' of AM between the Sun and the planets.
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lgl
Level 2 Rank
Posts: 93
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Post by lgl on Dec 10, 2009 9:42:38 GMT
And let's look at a simple system, Su-Ve-Ea. Su-Ve alone and Su-Ea alone are perfect free fall systems, then I'm saying the Su-Ve-Ea system should give the Sun the same motion as the motions of Su-Ve and Su-Ea added for the Sun to still be in free fall. You are saying it's ok to add the motion caused by the Ve-Ea interaction. This is where we disagree. The Sun moves because the planets moved altering the net gravitation so there is only gravitation involved, no problem there, but I see a difference when the planets move 'on their own' and not by gravity from the Sun. Like I said, we will never agree on this. (and again, Carstens calcs do not separate what I call the free fall component from the perturbation component so any extra transfer will not be visible)
Edit: I have to moderate myself a little. Sun&Ve&Ea=Su&Ve+Su&Ea is a bit 'over simplified'. The point is the inter-planetary perturbations should not be included.
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