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Jim

1607 Posts

Posted - 03 Jan 2003 : 11:51:47

I want to get away from all the distractions at the paradox forum and focus on the issue of gravity effects as TVF, Makis and I were chatting about. TVF says the current model correctly applies math to the acceleration of both the sun and Earth. And Makis and I have a different opinion here mainly because acceleration is not focce and it not equal for both bodies. To be cont,

Jim

1607 Posts

Posted - 03 Jan 2003 : 12:27:33

Picking up on the remarks by TVF on the dynamics of the Earth/sun orbit-I see several perplexing issues. The acceleration of the Earth by the sun is not equal to the acceleration of the sun by the Earth so the fact that two bodies fall as the same rate seems to be applied in different ways for both bodies. The force of gravity is greater on the sun than on Earth so this is also applied different for both bodies. There are different ways the math is applied to the two bodies. I think you(TVF) say the sun and Earth orbit a barycenter and also say this is not real. This is still another way math is applied in different ways. It is my opinion that the mass of the sun is not being pushed around a barycenter and the model requires this action. Another example of misuse of math to suit the end result.

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tvanflandern

USA
2793 Posts

Posted - 03 Jan 2003 : 13:11:59

quote:
The acceleration of the Earth by the sun is not equal to the acceleration of the sun by the Earth so the fact that two bodies fall as the same rate seems to be applied in different ways for both bodies.

No, it is applied in identical ways to the two bodies. Only the amounts differ, just as you say.

quote:
The force of gravity is greater on the sun than on Earth so this is also applied different for both bodies.

Is that a typo? In any case, in gravitation, forces are not observable, and are therefore a theoretical distraction. Only accelerations matter for determining orbits. And obviously, the Sun's acceleration due to Earth is much smaller than Earth's acceleration due to the Sun.

quote:
There are different ways the math is applied to the two bodies.

No, the math is applied identically. Only the amounts differ.

quote:
I think you (TVF) say the sun and Earth orbit a barycenter and also say this is not real.

A barycenter is also a mathematical abstraction, and is not observable. It is useful for some purposes, but is just a distraction in this discussion. Forget barycenters here.

quote:
This is still another way math is applied in different ways.

I have no idea what the basis for this claim might be, or even what it means.

quote:
It is my opinion that the mass of the sun is not being pushed around a barycenter and the model requires this action. Another example of misuse of math to suit the end result.

I would say "another example of your not understanding dynamics", because you continue to state that dynamics makes absurd claims. Claiming something absurd, then knocking the studding out of it, is called a "straw man argument".

Barycenters have no forces or accelerations, and nothing gets "pushed around a barycenter". Eliminate barycenters from your thinking. You have to understand the basics first. Those basics consist of Newton's acceleration law: acc. = GM/r^2, applied to both bodies in the system. (M = Sun's mass to get Earth's acceleration, or Earth's mass to get Sun's acceleration.)

What problems do you have with this basic picture? -|Tom|-

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Jim

1607 Posts

Posted - 03 Jan 2003 : 20:00:13

There is data at JPL Horizons that clearly shows the barycenter model in operation. It is the Earth/Moon orbit and the Earth is nearer the sun at Full Moon than at New Moon. This can be drawn on paper also. This is the one of the problems and I think the generator of the data- that is the barycenter model needs to push or how else do the results occur? I'm quite willing to forget this model for this forum, however. So onward-The force is as you say F=ma. Doing the math here I get a greater force exerted on the Earth from the sun than the Earth exerts on the sun. The force can be calculated and I suspect observed if you look in the right place. F=ma for the Earth is very small compared to the force exerted by the sun. Is this OK so far? If so-I think the sun is not moved by the Earth. The force is too small and tidal effects absorb the force. I know you tell me the motion is measured so I am wrong here. Thats fine if the data exists that indicates the motion then I'll rethink my idea. I'm very open to anything that makes sense and rebel against unproven theory. I also know Newton designed most of this stuff and I'm aware of LaPlace, Einstein and others.

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tvanflandern

USA
2793 Posts

Posted - 03 Jan 2003 : 21:09:33

quote:
There is data at JPL Horizons that clearly shows the barycenter model in operation. It is the Earth/Moon orbit and the Earth is nearer the sun at Full Moon than at New Moon. This can be drawn on paper also. This is the one of the problems and I think the generator of the data- that is the barycenter model needs to push or how else do the results occur?

Forget the barycenter and just look at the accelerations actually occuring. During two weeks of each month, the Moon pulls Earth in the same direction as the Sun. That eventually results in the Earth getting closer to the Sun than it would if only the Sun were pulling it. Right? Two forces are stronger than one, right?

Of course, for the other two weeks, the Moon pulls opposite to the Sun, which eventually results in the Earth getting farther away from the Sun than it would be if there were no Moon. Right?

So without any barycenter or any pushes or centrifugal forces or anything at all expect simple Newtonian gravity, we get the observed behavior shown in the models.

quote:
I'm quite willing to forget this model for this forum, however. So onward-The force is as you say F=ma. Doing the math here I get a greater force exerted on the Earth from the sun than the Earth exerts on the sun. The force can be calculated and I suspect observed if you look in the right place. F=ma for the Earth is very small compared to the force exerted by the sun. Is this OK so far?

Solar system gravitational forces are not observable and cannot be observed, even in principle. So let's stick to observable accelerations or we will not be able to explain the basics. Forces and barycenters are unnecessary, advanced concepts introduced for special purposes where they are useful. They cannot help us understand the fundamentals.

The acceleration law is not F = ma. (Forget you ever learned that. It has no value for gravitation.) The acceleration law for any target body in a gravitational field is acc. = GM/r^2, where M is the mass of a source of gravity and r is the distance of a target body from the source mass. Because every mass in a system can be a source mass, every body in the system must take its turn as a target body, and that acceleration formula must be applied to it. So everything moves, no matter how big or small it is.

quote:
If so-I think the sun is not moved by the Earth. The force is too small and tidal effects absorb the force. I know you tell me the motion is measured so I am wrong here. Thats fine if the data exists that indicates the motion then I'll rethink my idea. I'm very open to anything that makes sense and rebel against unproven theory.

We all rebel against unproven theory. But most of us accept theory that has proved itself.

** 300 years ago, the acceleration law was new, and it was amazing to see that the planets and their moons all obeyed it so well.

** 200 years ago, comets and asteroids could be added to the list of successful predictions, even though observations were an order of magnitude better.

** By 100 years ago, the acceleration law was known to apply to many double stars and to certain laboratory experiments; and observations had become another two orders of magnitude more accurate. The acceleration law had been used to predict Neptune, leading to its discovery in 1845. Remaining anomalies indicated that at least one additional planet remained to be found.

** By 50 years ago, the acceleration law was verified to eight decimal places of accuracy. But is was found inaccurate (finally) in going from the eigth to the ninth decimal. This led to Einstein's correction to explain (among other things) why Mercury's elliptical orbit had a slight precession that the acceleration law did not predict. The acceleration law was extended to stellar clusters and even galaxy dynamics.

** In the last 50 years, we have added another four orders of magnitude to the accuracy of the observations. GPS, VLBI, laser ranging, and spacecraft data are accurate to meters, or even centimeters in some cases, even over distances of billions of meters. Yet the acceleration law (except for Einstein's small corrections) has still lived up to its name: the universal law of gravitation.

Instead of just thinking about these matters, why not get a few books and read up on the state of the field today? That is really the only way to get up to speed. If you want to make a contribution, you will first have to see where everyone else is and how they got there. Then you can lead the pack in some new direction. But you cannot just go off in a new direction and expect anyone in the pack to follow. The route to where the pack is now is straightforward, easy to follow, and known to be safe because it was forged by geniuses. To the pack, those who go off in other directions before they get to the safe place where the pack is appear "mad" because all other directions have been explored many times before, and no one who went that way has every returned.

So either enjoy your adventure -- alone -- or learn the basics so you can communicate your original ideas to others. -|Tom|-

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Jim

1607 Posts

Posted - 04 Jan 2003 : 13:22:34

The issue is still the same and here at least there a a tiny window that is open to the issue. Lets not bring the baggage from the other forum here- that is why I left the paradox forum and started this one. I want to know dynamics-not dogma-not how egnorant "they" are. Do that at the other forum. So, TVF-you say F=ma is not an issue in orbital dynamics? I ask you,"is that a typo"?

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tvanflandern

USA
2793 Posts

Posted - 04 Jan 2003 : 13:38:26

quote:
[Jim]: you say F=ma is not an issue in orbital dynamics? I ask you, "is that a typo"?

No, it is not a typo, but I'm glad you asked.

In astronomy, all of dynamics can be done with the acceleration law I quoted. Forces are never needed and cannot be observed, so it can be quite confusing to invoke them. In the case in point, your understanding of concentional dynamics would be greatly enhanced if you eliminated all thought of forces and Newton's 2nd and 3rd laws of motion from your thinking. That throws out the equivalence principle too, in case that matters to you.

What is left is the pure motions of bodies -- their accelerations -- and the single, simple rule that governs them. Once you understand what these motions are, and how the rule governing them has been verified (that list of observation techniques I gave in my last message), you will understand the basics of conventional dynamics. And once you understand conventional dynamics, you will be in a position to criticize it if there is something not right. -|Tom|-

Jim

1607 Posts

Posted - 04 Jan 2003 : 15:24:42

Then Gm=rv2 must be the right road. Is this true? And if so how is this rule equally applied to large and small bodies when the moon is only moving the Earth according to its mass ratio of 81:1? The equivalance of m and v2 is evident since the other factors are exactly the same quantities.

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tvanflandern

USA
2793 Posts

Posted - 04 Jan 2003 : 15:42:36

quote:
[Jim]: Then Gm=rv2 must be the right road. Is this true?

No, that is another red herring. It is true under special circumstances only, such as circular orbits in the 2-body problem. But that is an "advanced topic", and should be forgotten for now until we get the basics right.

The only law that matters for 99% of all gravitational phenomena is acceleration = GM/r^2, where M is the mass of the source of gravity and r is the distance from source mass to target body.

quote:
And if so how is this rule equally applied to large and small bodies when the moon is only moving the Earth according to its mass ratio of 81:1?

When the acceleration law is applied to source mass Earth and target body Moon, we get the Moon's normal orbit around Earth. When it is applied to source mass Moon and target body Earth, we get the Earth's much smaller (by a factor of 81) orbital motion around the Moon. Earth keeps trying to fall toward the Moon, just as the Moon tries to fall toward the Earth. The difference is that the Earth doesn't get very far before the Moon moves on and starts pulling the other way.

quote:
The equivalance of m and v2 is evident since the other factors are exactly the same quantities.

GM/r = v^2 only for circular orbits, and only when there are just two bodies involved. Forget these complicated advanced matters until we get the basic motions straight. Nothing in this so-called "virial theorem" will help you understand the basics. -|Tom|-

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Jim

1607 Posts

Posted - 04 Jan 2003 : 17:08:10

OK I'll do this with a=Gm/r2. a for the moon is 1/81 if Earth a=1. The r2 is the same for both bodies. Gm=ar2. a=v2/r for circular orbits if that is ok with you? so v2/r=Gm/r2 to v2=Gm/r to Gm=rv2. It looks like the same idea to me. (I hope my math is right)

tvanflandern

USA
2793 Posts

Posted - 04 Jan 2003 : 19:05:21

quote:
OK I'll do this with a=Gm/r2. a for the moon is 1/81 if Earth a=1. The r2 is the same for both bodies. Gm=ar2.

Okay so far.

quote:
a=v2/r for circular orbits if that is ok with you?

The relation is correct for 2-body circular orbits. But I advise against going there. Stick with formulas that tell you something about the motion. It is too easy to get into paradoxes when you switch from motion equations to energy equations, which is where you appear to be headed.

quote:
so v2/r=Gm/r2 to v2=Gm/r to Gm=rv2. It looks like the same idea to me. (I hope my math is right)

Yes, the math is right, and so is the relationship, with the previous caveats still in effect. Your second-last form is an energy equation, and no longer contains information about how any single body moves. Instead, it connects the kinetic and potential energies for one body in a circular orbit. The last formula tells you that the product rv^2 is constant for all the planets. But it would be a different constant for the Moon and Earth satellites because they are orbiting a different source mass.

So where do you plan to go with this formula? -|Tom|-

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Jim

1607 Posts

Posted - 04 Jan 2003 : 19:50:02

I want to resolve the energy effect of mass in orbital motion it that makes sense. You lost me on the last post when you divided the math relationships into two types-Energy and Motion. I have never run into that before and will need to ponder it some. I guess tidal effects are one or the other and energy is needed to perform either type. Back to the topic-I just did a bit of math with acceleration and mass since the mass of the sun moon and Earth are known and the acceleration is proportional to mass. Now, I know you are not going to agree but I get a very large difference in the force of these three bodies have on each other. And I am now more sure the moon does not move Earth and Earth does not move the sun. Also tides are for sure a gravity effect that is currently modeled wrongly. I say this because you asked where I want to get to with this link not to belabor my point. I'm looking for truth and find some once in a while.

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tvanflandern

USA
2793 Posts

Posted - 04 Jan 2003 : 21:28:18

quote:
I want to resolve the energy effect of mass in orbital motion if that makes sense.

The sentence is not self-explanatory.

quote:
You lost me on the last post when you divided the math relationships into two types-Energy and Motion. I have never run into that before and will need to ponder it some. I guess tidal effects are one or the other and energy is needed to perform either type.

Tidal effects are a type of "non-gravitational force", even though gravity is involved. But we can learn little or nothing about the nature of gravity from tides because they disrupt normal orbital motion.

And I don't know what your interest in energy is. It is an effect of gravity, not a cause.

quote:
Back to the topic-I just did a bit of math with acceleration and mass since the mass of the sun moon and Earth are known and the acceleration is proportional to mass. Now, I know you are not going to agree but I get a very large difference in the force of these three bodies have on each other.

We agreed to leave forces out of the discussion because they are mathematical fictions. Stick with accelerations -- that is what drives orbital motion. There certainly is a large difference in the accelerations produced by Sun, Earth, and Moon on one another. Why wouldn't I agree? Only the accelerations of Earth and Moon produced by the Sun are nearly equal. All other accelerations are very different. The acceleration formula is easy to apply and easily gives those results.

Examples:
If acceleration of Sun caused by Earth is 1,
then acceleration of Earth caused by Moon is 1900,
and acceleration of Moon caused by Earth is 150,000,
and acceleration of Earth or Moon caused by Sun is 330,000.

quote:
And I am now more sure the moon does not move Earth and Earth does not move the sun.

Regarding logical consistency, you just jumped off a cliff.

If each body applies an acceleration to each other body, that means all the bodies move by definition of the word "acceleration". How did our discussion derail so badly, out of the blue?

quote:
Also tides are for sure a gravity effect that is currently modeled wrongly.

From your descriptions, I have to doubt that you know what a "tidal effect" really is. But let's get gravity straight first, and leave more complicated matters for chapter two. -|Tom|-

Jim

1607 Posts

Posted - 05 Jan 2003 : 16:29:27

There are no cliffs here-you asked me where I want this to go and I told you. My views and yours are quite different and it is not a problem for me. The model as you have so far described is good as are the several other models I'm aware of. This one assumes acceleration is a primary factor I think you say and that is ok by me. I look forward to the day when force, energy and the tidal effect are factored in. It is good to learn that tides are a gravity effect and that must mean energy is part of the overall modeling. So once I have the other part understood I'll be ready for that.