Paradoxes and Dilemmas

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21 years 10 months ago #4033 by Jeremy
Replied by Jeremy on topic Reply from
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
I promised you one more round. We should probably leave it here unless others express some interest.
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Well I for one am willing to trust the orbital mechanicians with basic planetary motion. Since we are getting the free lesson here I have to admit I am unfamiliar with why r^3 appears in Newton's formula when I thought it was supposed to be r^2. Where is the extra r coming from?

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21 years 10 months ago #4034 by Quantum_Gravity
what formula of newtons does R^3 come from and has anybody tried to argue r^2, fill me in here k

The intuitive mind

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21 years 10 months ago #4612 by tvanflandern
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>I am unfamiliar with why r^3 appears in Newton's formula when I thought it was supposed to be r^2. Where is the extra r coming from?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Newton's universal law of gravitation, generalized to vector form to retain directional information, is:
<font color=orange><b>r</b></font id=orange>-dot-dot = -(GM/r^3) <font color=orange><b>r</b></font id=orange>
where bold orange coloring indicates a vector, "dot" indicates a rate of change with respect to time, and "dot-dot" indicates acceleration.

"Vector form" means that distances such as r become vectors (arrows) with components in the x- and y-directions. For example, let <font color=orange><b>i</b></font id=orange> be a unit vector in the x-axis direction, and <font color=orange><b>j</b></font id=orange> be a unit vector in the y-axis direction. Then:
<font color=orange><b>r</b></font id=orange> = r cos(theta) <font color=orange><b>i</b></font id=orange> & r sin(theta) <font color=orange><b>j</b></font id=orange>,
where "&" indicates a plus sign (which doesn't show up on this message board for some reason), and theta is the angle at the origin (e.g., the Sun) between <font color=orange><b>r</b></font id=orange> and the x-axis. Note the difference between the simple distance r and the vector distance with direction <font color=orange><b>r</b></font id=orange>.

Once the notation is understood, the answer to the question asked becomes easy to see. The right-hand-side of Newton's law contains a divisor of r^3, and a factor of <font color=orange><b>r</b></font id=orange>. The latter contains a factor of r in each of its components. So if we substitute the equation defining <font color=orange><b>r</b></font id=orange> into Newton's law, we get:
<font color=orange><b>r</b></font id=orange>-dot-dot = -(GM/r^3) [r cos(theta) <font color=orange><b>i</b></font id=orange> & r sin(theta) <font color=orange><b>j</b></font id=orange>]
We can now cancel a factor of r, and we see that the law is simply:
<font color=orange><b>r</b></font id=orange>-dot-dot = -(GM/r^2) [cos(theta) <font color=orange><b>i</b></font id=orange> & sin(theta) <font color=orange><b>j</b></font id=orange>]

In this final form, we see that the generalized Newtonian law simply says that gravitational acceleration is a vector with a magnitude of (GM/r^2), and with a direction toward (because of the minus sign) the Sun. That magnitude is factored by cos(theta) to get the component in the x-direction, and by sin(theta) to get its component in the y-direction.

Those who make the effort to follow the above will get some insight into the power of vectors in dynamics. -|Tom|-

[I am delighted to see that others are following this discussion closely enough to have asked this very good question.]


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21 years 10 months ago #3998 by Jim
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Why is my observation about a model and comments here about the facts of the model dynamics used by TVF to show how ignorant I am? The model says this and I'm just a messager-so don't put the dumbness in my lap look at the model.

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21 years 10 months ago #3999 by Jim
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It is my opinion that the sun is not being moved by the planets because they are too small. A very large mass would be required to move the sun(~.05solar mass or more). This can be proven too with observations of the true motion of bodies rather than just hoping models are correct and dismissing anything that does not fit well with whatever theory is being touted at the moment. The simple fact is the observations that have been made are trimmed to suit theories that are well established and work well. It is very poor science to accept this as it is.

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21 years 10 months ago #4613 by Larry Burford
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>[makis]
Your whole thinking is based on a fixed sun Tom. Attached with big bolts on a certain origin. Non-movable and if it is, magically powered to counter act any forces from the earth.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

??? It is hard to follow you sometimes, Makis, so I may be trying to solve the wrong problem. But...

Step One - fix the Sun
The Sun is not bolted to the coordintate origin; rather, the origin is bolted to the Sun. As the Sun continues to move in whatever directions and with whatever accelerations it actually has (see note below), this coordinate system moves with it. We do this to make analysis of the Earth's motions <b>relative to the Sun</b> easier (because we no longer need to explicitly subtract the Sun's movements from the Earth's movements).

Step two - fix the Earth
Having done this (the first of several steps), we must now unbolt the origin of our coordinate system from the Sun and bolt it to the Earth. Now another analysis is performed, which determines the movement of the Sun <b>relative to the Earth</b>.

I think this might be what you are calling "magic", but of course it isn't.

NOTE - other masses can and do move the Sun. But they also move the Earth, and in general they are so far from both Earth and Sun that these additional motions are the same and cancel out. Jupiter is about the only significant mass that gets close enough (~4 AU at closest approach to Earth) to violate this conditon. But even so it causes only very minor deviations. (This does not seem to be playing a role in your disagreement.) In particular, Jupiter will accelerate the Earth slightly more than it accelerates the Sun when Jupiter is near closest approach to Earth.

Step Three - combine the partial results
The final step is to combine the results of the previous steps to get the whole answer, which is a predictive description of the moving Earth and the moving Sun relative to each other (or to their center-of-mass). This combination of results from the first two steps shows that the Sun makes a small orbit and the Earth makes a large orbit around their mutual center-of-mass; that each orbit takes one year; and that a line drawn between the centers-of-mass of each will always pass through their combined center-of-mass.


<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>[makis]
Disagreeing with someone does not mean I disrespect him. For me, TVF is a rare personality of immense intelligence. Very few people with his knowledge about cosmology and other subject matters can be found.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

It is not necessary to make an explicit disclaimer like this. We can tell from the tone of your voice and the words you choose what emotions you are expressing. At least most of the time - translation difficulties can sometimes make it hard to be sure. When in doubt we generally do NOT assume disrespect.

Having said this, I do still appreciate the gesture. I suspect others do, too.

Regards,
LB

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