Robittybob1 Posted February 12, 2015 Author Share Posted February 12, 2015 Yes they can (at least in inertial frames). That's the point. You said the barycenter was at rest with respect to the solar system, and then that "solar system" meant barycenter. That's a tautology. It's true by definition. Nothing revealing about it. Everything is moving with respect to the barycenter. Also not particularly revealing, since everything is orbiting it. I was really taken by surprise as to what you think I said, but if it is the case I have obviously not expressed my ideas clearly enough, yet Whiskers in general understood and discussed the idea fairly well. I have proposed the barycenter (SSB) orbits the Sun and the Sun is also a movable object wobbling about the "RB Spot". Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted February 12, 2015 Share Posted February 12, 2015 What is the definition of "orbit" that we are comfortable using here? The planets are not drawn gravitationally toward the SSB. At any one time one might be by coincidence, but for the most part the gravitational force vector does not favour that point. Link to comment Share on other sites More sharing options...
Robittybob1 Posted February 12, 2015 Author Share Posted February 12, 2015 (edited) What is the definition of "orbit" that we are comfortable using here? The planets are not drawn gravitationally toward the SSB. At any one time one might be by coincidence, but for the most part the gravitational force vector does not favour that point. All the sites will tell you the planets are orbiting the SSB which is so close to the Sun that in the purposes of general discussion one would say they orbit the Sun but on a more accurate level they orbit the SSB. So why would you doubt that? How could you change the Gravitational force vector away from the SSB? I was just testing whether it was possible to make the greater mass (99% of the Solar System) start orbiting for that to me would need a fairly constant orbital energy. When the SSB and the center of mass of the Sun coincide does the Sun in fact have any orbital energy then? Yet some years later it will be found to be "orbiting" at a distance to the SSB, so where did this orbital energy come from? Edited February 12, 2015 by Robittybob1 Link to comment Share on other sites More sharing options...
swansont Posted February 12, 2015 Share Posted February 12, 2015 All the sites will tell you the planets are orbiting the SSB which is so close to the Sun that in the purposes of general discussion one would say they orbit the Sun but on a more accurate level they orbit the SSB. So why would you doubt that? How could you change the Gravitational force vector away from the SSB? I was just testing whether it was possible to make the greater mass (99% of the Solar System) start orbiting for that to me would need a fairly constant orbital energy. Pick a coordinate system. Lets say the SSB is at 0º and located right on the surface of the sun. Any planet whose current position is at ±90º is not attracted to the SSB, it's attracted to the sun's center. The angle may be small, especially for distant planets, but it's not zero. When the SSB and the center of mass of the Sun coincide does the Sun in fact have any orbital energy then? Yet some years later it will be found to be "orbiting" at a distance to the SSB, so where did this orbital energy come from? The energy already present in the solar system. PE and KE. If something sped up, distances changed and/or something else slowed down; that could include rotational kinetic energy as well as orbital. (e.g. the earth's rotation is slowing as the moon recedes from it) Link to comment Share on other sites More sharing options...
Acme Posted February 12, 2015 Share Posted February 12, 2015 ... Now there is something called gravitational tidal locking where the gravity of one body can influence the spin of another object. Mercury is totally locked by the sun for example. ... To clarify, Mercury is not tidally locked with the Sun, rather it has a spin-orbit resonance. Were Mercury tidally locked with the Sun it would always present the same face sunward. Link to comment Share on other sites More sharing options...
Mordred Posted February 12, 2015 Share Posted February 12, 2015 Oh cool guess I need to get more up to date astronomy textbooks lol they used to consider it tidally locked. Thanks for the info Link to comment Share on other sites More sharing options...
Robittybob1 Posted February 13, 2015 Author Share Posted February 13, 2015 (edited) Pick a coordinate system. Lets say the SSB is at 0º and located right on the surface of the sun. Any planet whose current position is at ±90º is not attracted to the SSB, it's attracted to the sun's center. The angle may be small, especially for distant planets, but it's not zero. The energy already present in the solar system. PE and KE. If something sped up, distances changed and/or something else slowed down; that could include rotational kinetic energy as well as orbital. (e.g. the earth's rotation is slowing as the moon recedes from it) To me what you have just said sounds wrong, and I know it will come as a shock but I do read a lot of your posts and you are usually right but this time I'm sorry, I think you're wrong. Now, why I think you are wrong is that taking the top situation if the angle is any amount between 0 and 90 degrees what is it then? Are you saying it depends on the angle? In the second situation there is a known mechanism for the transfer of momentum and energy between the Earth and the Moon (gravitational tidal acceleration) but what is the mechanism for transferring the momentum to and from the Sun? Note: in my idea the Sun wobbles so even I add and subtract energy to and from the Sun but that is simply by applying gravitational attraction, F=ma i.e the combined gravitational attraction of the remainder of the SS will apply a gravitational force to the Sun and hence it will accelerate, the ever changing strength and direction of this force wobbles the Sun's position. [i'd be willing to reconsider for at what point is the SSB not involved? If you move an object on Earth the gravitational pull from the Sun or the SSB is rather insignificant, but even then it could still be calculated for accuracy.] Edited February 13, 2015 by Robittybob1 Link to comment Share on other sites More sharing options...
astromark Posted February 13, 2015 Share Posted February 13, 2015 Answering questions from children at a public viewing at a local observatory.. Has tought me not to over answer the questions asked or implied. The mass of the Salar System is in motion. It has been sinse before the Sun ignited it's fusion core..and swept the system clear.. The center of mass is not the same thing as the barycenter.. We all seem to understand that the motion and orbital paths are constantly adjusting the barycenter of each object of mass.. This little side issue of the ISS and of what it's orbiting.. I will go out on the point.. Planet Earths center of mass.. which is constantly in motion. Link to comment Share on other sites More sharing options...
Robittybob1 Posted February 13, 2015 Author Share Posted February 13, 2015 (edited) Answering questions from children at a public viewing at a local observatory.. Has tought me not to over answer the questions asked or implied. The mass of the Salar System is in motion. It has been sinse before the Sun ignited it's fusion core..and swept the system clear.. The center of mass is not the same thing as the barycenter.. We all seem to understand that the motion and orbital paths are constantly adjusting the barycenter of each object of mass.. This little side issue of the ISS and of what it's orbiting.. I will go out on the point.. Planet Earths center of mass.. which is constantly in motion. I can see your point. The barycenter between the Earth and the moon (EMB) is always shifting as the Moon orbits the Earth. OK the ISS doesn't immediately orbit the EMB but each time it passes between the Earth and the Moon is the ISS orbit being given a nudge toward a different orbit? The effect could what is called "Third Body Perturbation" http://ocw.upm.es/ingenieria-aeroespacial/modeling-the-space-environment/contenidos/material-de-clase/mse05_3rdbody.pdf It isn't the easiest subject but Wikipedia has this to say Third-body perturbations[edit]Gravitational forces from third bodies can cause perturbations to an orbit. For example, the Sun and Moon cause perturbations to Orbits around the Earth.[13] These forces are modeled in the same way that gravity is modeled for the primary body by means of Direct gravitational N-body simulations. So this might be one of the many reasons satellites fall back to Earth. Edited February 13, 2015 by Robittybob1 Link to comment Share on other sites More sharing options...
swansont Posted February 13, 2015 Share Posted February 13, 2015 To me what you have just said sounds wrong, and I know it will come as a shock but I do read a lot of your posts and you are usually right but this time I'm sorry, I think you're wrong. Now, why I think you are wrong is that taking the top situation if the angle is any amount between 0 and 90 degrees what is it then? Are you saying it depends on the angle? I gave you the example that shows the maximum effect. What I'm saying is that Newton's Law of gravitation applies. I thought at some point you indicated that you agreed that that was the case. In the second situation there is a known mechanism for the transfer of momentum and energy between the Earth and the Moon (gravitational tidal acceleration) but what is the mechanism for transferring the momentum to and from the Sun? Note: in my idea the Sun wobbles so even I add and subtract energy to and from the Sun but that is simply by applying gravitational attraction, F=ma i.e the combined gravitational attraction of the remainder of the SS will apply a gravitational force to the Sun and hence it will accelerate, the ever changing strength and direction of this force wobbles the Sun's position. Why does simply applying gravitational attraction not account for the sun's motion? Link to comment Share on other sites More sharing options...
whiskers Posted February 13, 2015 Share Posted February 13, 2015 Answering questions from children at a public viewing at a local observatory.. Has tought me not to over answer the questions asked or implied. The mass of the Salar System is in motion. It has been sinse before the Sun ignited it's fusion core..and swept the system clear.. The center of mass is not the same thing as the barycenter.. We all seem to understand that the motion and orbital paths are constantly adjusting the barycenter of each object of mass.. This little side issue of the ISS and of what it's orbiting.. I will go out on the point.. Planet Earths center of mass.. which is constantly in motion. Can you explain the difference between the SS CoM and the SSB ? [i'd be willing to reconsider for at what point is the SSB not involved? If you move an object on Earth the gravitational pull from the Sun or the SSB is rather insignificant, but even then it could still be calculated for accuracy.] There is no gravitational pull from the SSB, since It has no mass. Motion is relative, so there is no objective definition of "fixed position" vs "moving" without a frame of reference. Can you affirm that you understand these propositions? Link to comment Share on other sites More sharing options...
Robittybob1 Posted February 13, 2015 Author Share Posted February 13, 2015 (edited) ....There is no gravitational pull from the SSB, since It has no mass. Motion is relative, so there is no objective definition of "fixed position" vs "moving" without a frame of reference. Can you affirm that you understand these propositions? I know the CoM or the SSB or any barycenter doesn't have all the mass. When using the Newtonian gravitational force formula one says all the mass is at the CoM and that is at r distance between them, now if the shape of the object is irregular and you calculate the CoM and it just happens to be in empty space that is still the spot from which the combined gravity force will be appearing to come from. Yes I understand that, and that is why I am proposing my frame of reference is the "RB spot", which is equivalent to the spot central to all the wobble of the Sun. Is that a valid FoR? Edited February 13, 2015 by Robittybob1 Link to comment Share on other sites More sharing options...
whiskers Posted February 13, 2015 Share Posted February 13, 2015 (edited) I know the CoM or the SSB or any barycenter doesn't have all the mass. Any. the SSB doesn't have *any* mass. "When using the Newtonian gravitational force formula one says all the mass is at the CoM and that is at r distance between them," For the SSB this only applies to gravitational interactions between this solar system and other stars. Inside the solar system, Newtonian gravitation is calculated between bodies and other bodies - the SSB is not used for Newtonian forces. Edited February 13, 2015 by whiskers Link to comment Share on other sites More sharing options...
Robittybob1 Posted February 13, 2015 Author Share Posted February 13, 2015 I gave you the example that shows the maximum effect. What I'm saying is that Newton's Law of gravitation applies. I thought at some point you indicated that you agreed that that was the case. Why does simply applying gravitational attraction not account for the sun's motion? OK so what you have made me think about is what bearing has the barycenter got when approached from multiple angles. They often use the example of a hammer for it has a heavy head end and a lighter handle, but when anyone works out it CoM that is the same CoM no matter from which angle you look at the hammer, and I suppose that too includes the situation of an ant walking across the head. I'm struggling to accept this but is it wrong? Now when you ask "Why does simply applying gravitational attraction not account for the sun's motion?" That is exactly what I'm proposing but just had my doubts that those forces are enough to make the Sun orbit the SSB if you make the SSB the FoR. Any. the SSB doesn't have *any* mass. "When using the Newtonian gravitational force formula one says all the mass is at the CoM and that is at r distance between them," For the SSB this only applies to gravitational interactions between this solar system and other stars. Inside the solar system, Newtonian gravitation is calculated between bodies and other bodies - the SSB is not used for Newtonian forces. I was covering for the occasional situation where the SSB was at the center of mass of the Sun. Who is the authority to determine this "For the SSB this only applies to gravitational interactions between this solar system and other stars. Inside the solar system, Newtonian gravitation is calculated between bodies and other bodies - the SSB is not used for Newtonian forces."? I'm just reading general information but have you got a reference of authority about this? Link to comment Share on other sites More sharing options...
whiskers Posted February 13, 2015 Share Posted February 13, 2015 OK so what you have made me think about is what bearing has the barycenter got when approached from multiple angles. They often use the example of a hammer for it has a heavy head end and a lighter handle, but when anyone works out it CoM that is the same CoM no matter from which angle you look at the hammer, and I suppose that too includes the situation of an ant walking across the head. I'm struggling to accept this but is it wrong? Now when you ask "Why does simply applying gravitational attraction not account for the sun's motion?" That is exactly what I'm proposing but just had my doubts that those forces are enough to make the Sun orbit the SSB if you make the SSB the FoR. I was covering for the occasional situation where the SSB was at the center of mass of the Sun. Who is the authority to determine this "For the SSB this only applies to gravitational interactions between this solar system and other stars. Inside the solar system, Newtonian gravitation is calculated between bodies and other bodies - the SSB is not used for Newtonian forces."? I'm just reading general information but have you got a reference of authority about this? it's all about context. When you are on the surface of the earth, there are plenty of little variances in the gravitational field - trees and rocks exert subtle gravitational forces on each other and on you.. The further you get away from the earth, the more the gravitational field can be treated as a single vector pointing toward the center of the Earth. so it is in the solar system, only so much more so. If you are near Jupiter, the gravitational attraction of the Sun will take a back seat to Jupiter's. So it only makes sense to see the SSB as "exerting gravitational attraction" if you are far away from the solar system. Link to comment Share on other sites More sharing options...
Robittybob1 Posted February 13, 2015 Author Share Posted February 13, 2015 it's all about context. When you are on the surface of the earth, there are plenty of little variances in the gravitational field - trees and rocks exert subtle gravitational forces on each other and on you.. The further you get away from the earth, the more the gravitational field can be treated as a single vector pointing toward the center of the Earth. so it is in the solar system, only so much more so. If you are near Jupiter, the gravitational attraction of the Sun will take a back seat to Jupiter's. So it only makes sense to see the SSB as "exerting gravitational attraction" if you are far away from the solar system. I understand where you are coming from , but go back to the ant on the hammer head and he is trying to calculate gravity strength does he only consider the CoM of the head or should he use the CoM of the hammer? Link to comment Share on other sites More sharing options...
whiskers Posted February 13, 2015 Share Posted February 13, 2015 (edited) I understand where you are coming from , but go back to the ant on the hammer head and he is trying to calculate gravity strength does he only consider the CoM of the head or should he use the CoM of the hammer? What is gravity strength? The force of gravity as experienced? To get absolute and total accuracy would require summing up the gravitational forces from everything in the universe - understood? For an ant on a hammer - treating the hammer as only the CoM will yield a very poor approximation. If this hammer were perfectly symmetrically spherical then u could use the CoM. Edited February 13, 2015 by whiskers Link to comment Share on other sites More sharing options...
Robittybob1 Posted February 13, 2015 Author Share Posted February 13, 2015 (edited) I have no particular point of view on all this, but I'm wanting to understand it and you've definitely helped. If it was possible to fire a rocket off the "surface" of Jupiter heading directly toward the SSB it does seem stupid to think that Jupiter's mass contributes to the pull of gravity ahead and behind the rocket. Edited February 13, 2015 by Robittybob1 Link to comment Share on other sites More sharing options...
whiskers Posted February 13, 2015 Share Posted February 13, 2015 (edited) I have no particular point of view on all this, but I'm wanting to understand it and you've definitely helped. If it was possible to fire a rocket off the "surface" of Jupiter heading directly toward the SSB it does seem stupid to think that Jupiter's mass contributes to the pull of gravity ahead and behind the rocket. Nope. Behind only. There's no "there there" at the SSB. In terms of movements of bodeis inside the solar system, it is a very practical origin for coordinates. In terms of gravitational pull of the SS for a point beyond the SS, the SSB is slightly more accurate single direction of the gravity vector than the Sun itself. Edited February 13, 2015 by whiskers Link to comment Share on other sites More sharing options...
Robittybob1 Posted February 13, 2015 Author Share Posted February 13, 2015 (edited) Nope. Behind only. There's no "there there" at the SSB. In terms of movements of bodeis inside the solar system, it is a very practical origin for coordinates. In terms of gravitational pull of the SS for a point beyond the SS, the SSB is slightly more accurate single direction of the gravity vector than the Sun itself. So are you saying none of the SS planets orbit the SSB, and all those other people who say they do are incorrect? Wikipedia http://en.wikipedia.org/wiki/Barycentric_coordinates_%28astronomy%29 "The Sun orbits a barycenter just above its surface." I assume they are talking about the SSB. So maybe contrary to my OP thoughts the Sun is the only body orbiting the SSB! another: http://zidbits.com/2011/09/the-earth-doesnt-actually-orbit-the-sun/ This center of mass is called the barycenter. The Earth, the sun and everything else in our solar system actually orbit this barycenter – not the sun. Another one: http://homepages.wmich.edu/~korista/solarsystem_barycenter.pdf Although it is convenient to think of the Sun as the stationary anchor of our solarsystem, it actually moves as the planets tug on it, causing it to orbit the solar system's barycenter. Edited February 13, 2015 by Robittybob1 Link to comment Share on other sites More sharing options...
Strange Posted February 13, 2015 Share Posted February 13, 2015 So are you saying none of the SS planets orbit the SSB He said almost exactly the opposite. Link to comment Share on other sites More sharing options...
Robittybob1 Posted February 14, 2015 Author Share Posted February 14, 2015 (edited) This one has some connection to NASA: http://spaceplace.nasa.gov/barycenter/en/ In the case of the Earth and the Sun, both bodies orbit around the very center of the mass (similar to center of gravity) between them. This point is called the "barycenter."Earth and the Sun are "connected" by the gravity pulling them together. It's just like the light end and heavy end of the sledge hammer. Compared to the size of the Sun, though, Earth is about like a flea on a cat! So the center of mass between the Earth and the Sun is almost—but not quite—the very center of the Sun.Jupiter, on the other hand, is 318 times as massive as Earth. Therefore, the barycenter of Jupiter and the Sun is a bit further from the Sun's center. So, as Jupiter revolves around the Sun, the Sun itself is actually revolving around this slightly off-center point, located just outside its surface. He said almost exactly the opposite. Well Whiskers clear this misunderstanding up please. Do all/any/some of the planets in your view orbit the SSB? And Strange clearly state your view too please. Edited February 14, 2015 by Robittybob1 Link to comment Share on other sites More sharing options...
whiskers Posted February 14, 2015 Share Posted February 14, 2015 This one has some connection to NASA: http://spaceplace.nasa.gov/barycenter/en/ Well Whiskers clear this misunderstanding up please. Do all/any/some of the planets in your view orbit the SSB? The big problem here is "what is an orbit" - it is used in 2 different ways: 1) the center of the orbit is something fixed, and other things move relative to it This is the most general definition which works for the Barycenter, or approximately for the Sun, and earlier the earth was seen to occupy this place 2) the massive thing that attracts everything else to it so that they orbit around it This fits the Sun more or less, but it doesn't fit the barycenter in the slightest Link to comment Share on other sites More sharing options...
Robittybob1 Posted February 14, 2015 Author Share Posted February 14, 2015 (edited) The big problem here is "what is an orbit" - it is used in 2 different ways: 1) the center of the orbit is something fixed, and other things move relative to it This is the most general definition which works for the Barycenter, or approximately for the Sun, and earlier the earth was seen to occupy this place 2) the massive thing that attracts everything else to it so that they orbit around it This fits the Sun more or less, but it doesn't fit the barycenter in the slightest I still can't tell if that answers my question, sorry. But the SSB is the center of mass of the SS so that is going to be the largest CoM around in our region. Edited February 14, 2015 by Robittybob1 Link to comment Share on other sites More sharing options...
whiskers Posted February 14, 2015 Share Posted February 14, 2015 (edited) Here's an interesting thing. Imagine that the Earth is a hollow sphere, the mass of which adds up to the true earth mass. If you were outside of it, you would feel that attraction "down" towards the center of the earth. But! If you were *inside* of it, you would feel no gravitational attraction at all! http://en.wikipedia.org/wiki/Shell_theorem Nasa and all these other places describe the bodies going around the barycenter, but this is the kind of language which is more like saying that the Sun is setting - it is a casual description which does the job, but it can lead to misinterpretations. Edited February 14, 2015 by whiskers Link to comment Share on other sites More sharing options...
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