Paradoxes and Dilemmas

tvanflandern
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21 years 10 months ago #3993 by tvanflandern

Replied by tvanflandern on topic Reply from Tom Van Flandern

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>When they take the acceleration vector in polar cordinates and equate its two componenets to the forces present, the following results: M x [r-dotdot - r x omega ^2] = F (GxMxMs/r^2)<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

This makes no sense in a couple of ways. For near-circular orbits, the changes in distance are minor compared with the changes in direction. So you need vector equations, not scalar equations. And you should be using expressions for acceleration, not force, because accelerations are what we observe, whereas forces are unobservable (much like barycenters) and have no role to play in Newtonian celestial mechanics. Remember the special property of gravity (unlike any other force of nature) that gravitational acceleration is independent of the mass of the target body.

In other words, the correct equation is
r-dot-dot=-(GM/r^3)r
where bold indicates a vector. This equation must be applied independently to every active source mass M in the problem, for example to the Sun and to the Earth separately. In a trivial way with vector equations, the sum of the two accelerations (Earth's caused by the Sun and the Sun's caused by Earth) simply add, giving the same net vector acceleration as if M in the above equation were simply the mass of (Sun+Earth).

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>So you already assume there is a radial acceleration. (if it were not, that would be a perfectly circular orbit and no need to solve any equations). Then, the other term in the acceleration vector must be zero. That is: [M/r]x[d(r^2xomega)/dt) = 0
Integrating wrt to t, we get angular momentum:
Mxr^2xomega = J, angular momentum is constant.
But r is already assumed to be a function of time itself. Then, for J to be constant, omega must change (M does not change, hopefully) to keep J constant. Therefore, there is omega-dot. What you say is true only in circular orbits, where r and omega are both constant. But we explained about that already. There is angular acceleration contrary to your claims.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

I cannot follow this argument. (This is not the right medium for an equation-based argument.) But I know that, if the vector equations are handled properly, everything works out just as expected. I've been through that development in fine detail myself, as has every celestial mechanics student. You need to switch to vector equations to get the fundamentals right.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>If the force on the sun were to be a REACTIO, it would be centrifugal and the sun must be fixed to maintain the system stability. But since it is not a centifugal but a centripetal, as you also claim, sun must also scribe a trajectory. So, what is this trajectory?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

If the Sun is adopted as an origin, the Earth's relative ellipse becomes a bit larger.

If the Earth is adopted as origin, the Sun orbits the Earth in a mirror ellipse.

If the barycenter is adopted as origin, The Sun makes a small ellipse while Earth makes a large ellipse, both locked in phase.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>I maintain that if the model would be true (Newton), the sun would kiss the earth. Unless, the sun has independent motion, caused by something else that takes care of the situation.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

Of course the Sun has independent motion. Earth's gravity cannot be neglected. But when the directional components are considered, approaching the Earth is clearly impossible.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>Think about it in another common sense way: For the system to stay attached, there must be a restoring force to any force applied. That's what gives marginal stability to a dynamic system (periodic stable motion. Unless one of the two bodies makes an independent motion to simulate a restoring action, the system cannot have stability.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

Think of Newton's third law as inapplicable to gravity because the whole concept of force is inapplicable. There are definitely no "reaction forces" in gravitation. One can argue there are no direct forces either. For example, consider the gravitational orbit of a massless particle around the Sun. It is the same as that os a massive particle. Yet, although its acceleration is normal, the inferred force (mass times acceleration) on a massless particle is zero. What sense does that make?

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>I am trying to communicate to you what you know already and resisting to admit. There are limitations in communications, so I appreciate your patience and hope you do likewise.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

You have only succeeded in communicating that you do not understand the fundamentals of celestial mechanics (orbital dynamics). Many textbooks are confused about the basics, such as the points mentioned above. Many authors try to do gravity with forces and are led to absurd conclusions. (There are forces in gravity, but only at the graviton level.) Find a better author to learn from.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>Newton's laws apply only to non-inertial reference frames.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

Even if you meant "inertial" when you said "non-inertial", the claim is false. Newton's laws hold even when the accelerating Sun is taken as the origin.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>If the sun is used as an origin, it is a non-inertial frame since there is r-dot dot. Then, you must place a fecticious D'Alembert force on the sun to make it an inertial frame.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

No, you only need to put the negative of the Sun's acceleration into the other body or bodies, and the equations still describe the relative motion perfectly.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>How is it possible that two books use different fundamental assumptions (dealing with degrees of freedom), i.e. one uses a fixed sun and the other a moving, but both get the same result, i.e. the earth orbit is a conic section?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

As just explained, one can always throw the acceleration of the origin into all the other bodies and still get the correct relative motion.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>If you would like we terminate the discussion here.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote><font face="Verdana, Arial, Helvetica"

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Jim
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21 years 10 months ago #3961 by Jim

Replied by Jim on topic Reply from

If you can determine the exact solar income minute by minute the exact orbit can be MEASURED and that is a part of this thread-what is the exact orbit of the Earth and does the REAL sun move or do tides rise to absorb the gravity effect the Earth has on the sun. I.m not talking about models as you are but in real motion.

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tvanflandern
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21 years 10 months ago #4032 by tvanflandern

Replied by tvanflandern on topic Reply from Tom Van Flandern

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>If you can determine the exact solar income minute by minute, the exact orbit can be MEASURED and that is a part of this thread-what is the exact orbit of the Earth and does the REAL sun move or do tides rise to absorb the gravity effect the Earth has on the sun. I'm not talking about models as you are but in real motion.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

The "models", as you call them, are successful attempts to predict the real, observed motions.

Why use an indirect, low-accuracy indicator such as the solar luminosity when we can directly observe the Sun's motion relative to the star/quasar background? Transit circles observe the Sun and bright stars in daylight every clear day. Radio telescopes measure the quasars day or night also.

So there is no question in any astronomer's mind what the exact orbit of the Earth or the exact motion of the Sun is. Modern data has achieved a precision of just meters for the orbit -- better than a part in ten billion.

Not only does the Sun move, but it moves so much that it sometimes leaves the non-movable point (the barycenter) outside the entire physical body of the Sun; i.e., the radius of the Sun's motion is more than a million kilometers! -|Tom|-

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Jim
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21 years 10 months ago #4547 by Jim

Replied by Jim on topic Reply from

You got it right when you say your thinking is a tad off. The real problem is the math is not applied to both bodies in the same way. There is no barycenter and if there was and it worked as the model says you can see(if you open your eyes) the small body pushing the big body and that is really a dumb model.

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tvanflandern
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21 years 10 months ago #3995 by tvanflandern

Replied by tvanflandern on topic Reply from Tom Van Flandern

I promised you one more round. We should probably leave it here unless others express some interest.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>[Makis]: If you set the angular momentum constant and you assume omega-dot is zero then r-dot must also be zero.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

Of course. That simply means that a circular orbit does not change its distance from the Sun. (BTW, angular momentum is constant at every point even for an elliptical orbit because it consists of the combination of speed and potential.)

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>Then how can r-dotdot be equal to something?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

That is indeed the key to where you need to go next to increase your understanding. r-dotdot is of course zero if r-dot is zero (meaning that, for a circular orbit, the distance neither changes nor accelerates).

However, r-dot-dot (a vector) is the total acceleration toward the Sun, which is large and everywhere non-zero. For a circular orbit, all this acceleration is going into changes of direction, not changes of distance.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>Then, some place, somewhow, angular momentun is inceased or decreased. Magic! Any suggestions? Or we will go by this too?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

Angular momentum is conserved at every point, even for an elliptical or parabolic or hyperbolic orbit. But the vector radial acceleration does change angular speed if the orbit is not a circle. But faster angular speed is compensated by deeper (negative) gravitational potential, with the sum remaining constant.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>You cannot add accelerations in non-inertial reference frames just by that, without making appropriate reference frame corrections. If you don't, you will think each body moves relative to the other.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

Of course each body moves relative to the other. There is no such thing as "absolute motion" in this universe. And there is no problem adding vector accelerations. A little experience will put your mind at ease about the legitimacy of this.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>I give up because I see no grounds of advancing the current state of circular thinking propagated by prevailing methodologies.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

If you have not always kept in mind that your own understanding might be incomplete, it will be very difficult for you to learn new things in this or any area of life where your initial thinking is incorrect. Many people have this problem. I do not envy the task that faces you to overcome that type of obstacle. I suggest studying what earns the respect of peers and the effect that acknowledging error has on that respect.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>Ego cogito ego sum.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

If you meant "ergo", you might look up the meanings of "ego" and "irony". -|Tom|-

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tvanflandern
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21 years 10 months ago #3996 by tvanflandern

Replied by tvanflandern on topic Reply from Tom Van Flandern

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>[Jim]: You got it right when you say your thinking is a tad off. The real problem is the math is not applied to both bodies in the same way. There is no barycenter and if there was and it worked as the model says you can see(if you open your eyes) the small body pushing the big body and that is really a dumb model.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

Your understanding of dynamics is extremely limited. Get and read a book on celestial mechanics. Take the math classes you need to follow the equations and their meanings. The Earth pulls on the Sun. (It does not push!). Masses of all size respond equally, even when they are as big as the Sun. Galileo showed that in his Tower of Pisa experiment. -|Tom|-

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