The Death of the Centre of Mass Theorem.

[Rogue Physicist]

Testing upload feature...
Okay. I have to admit the innuendo here is low key enough not to make a big fuss about it. We haven't strayed into insults or name-calling, for which I am grateful. Having prefaced my next remark with that, it still is a bit annoying to have people constantly imply that either I don't know how to integrate a simple summation equation, or that if only I would do this, miraculously all the theoretical problems with Newtonian gravitational theory would vanish, like an error term in an approximation.

For the purposes of establishing this small but annoying point once and for all, I have taken the time to convert several pages of one of my books on the flaws of Newtonian gravitational theory into an html file (the word file was too large). The graphics for the equations are a bit primitive (I did this 5 years ago) but I hope you will satisfy yourselves that I can both do the integration and I know what it means when I look at the result. Again I repeat that the integration results mathematically were never in question for the Sphere theorem. If someone can help me convert this into a .pdf document, I can attach it to this post.

 

[Hurkyl]

Arg, I edited out one of the main questions I had wanted to ask in my last post.


When I had said that Newton's law of gravitation was:

<br /> \vec{F} = \int \int \frac{G f(\vec{x}) g(\vec{y})}{|\vec{y} - \vec{x}|^3} (\vec{y} - \vec{x}) d\vec{x}^3 d\vec{y}^3<br />

but doing it verbally rather than formulaically, you responded with:

(1) Hurkyl, you obviously believe that F=Gmm/d^2 is only an approximation of an integral formulation of gravitational potential fields. I am sorry to say, (if mass is treated as a continuum as it is in the integration), it is not an approximation but an exact result identical to the integral calculation.

I had wanted to ask if you were asserting that the above integral was not Newton's law of gravitation.

But now that I've reread the quoted section, it sounds like you are asserting the following equality:

<br /> \frac{G m_1 m_2}{|\vec{r}|^3} \vec{r} = \int \int \frac{G f(\vec{x}) g(\vec{y})}{|\vec{y} - \vec{x}|^3} (\vec{y} - \vec{x}) d\vec{x}^3 d\vec{y}^3<br />

Where \vec{r} is the displacement vector between the center of mass of the two bodies in question, and m_1, m_2 are their masses. Is my interpretation of that quote correct?

 

[Doc Al]

There's no point in getting into semantics. It's more than just a 'concept' and less than a 'theory'. It is best described by the usual grey-zone in the middle handle, a 'theorem'. A fairly short statement of some definitions, an axiom or two, some hidden assumptions, and an optionally syllogistic assertion. It has been described well enough here and elsewhere, as well as in physics textbooks for the last 50 years.

And yet you seem unable to give a simple, clear statement of this "theorem".

The most I can figure from your posts is that you are saying that Newton's law of gravity, written as:
F = \frac{Gm_1m_2}{r^2}
is only approximately true applied to macroscopic objects. Well... of course! That law applies to point masses only. For extended bodies one must integrate.

If you are saying somehow that that law is meant to apply exactly to macroscopic objects where r is the distance between the center of masses... well, no, that's not generally true.

Again, I'll reiterate, but it is clear from the previous post. The Centre of Mass theorem has nothing to do with uniform gravitational fields. You are confusing it with the Centre of Gravity concept.

You are the one talking about "center of mass" and gravity together. Please explain the connection.

 

[Dr.Brain]

(It does *not* claim the sphere is capable of shielding a test-mass inside from other forces.)

It does claimly directly but its understandable.This phenomena is known as "Internal Shielding" , it says that a chrge enclosed inside a conductor is shielded from all other outside electrical influences.

(2) The Centre of Mass Theorem. Another practical method of simplifying the calculation of gravity between two massive objects. The centre of mass itself is carefully defined as the arithmetic mean average of all the point-mass positions weighted by their individual masses in three dimensions. This is just the definition and calculation of the location itself.

The theorem states that any 'system' of point-masses (massive object) can be treated as if it were a point-mass of equivalent mass located at this point. It was known to be an 'approximation' even when Newton originally proposed it. But it was assumed to be a good approximation, with only a minor error factor at closer ranges.

As per my information , this error is larger in non-symmetrical objects and negligible in round objects .These approximations don't violate anything , they were made to make problem-solving and motion-analysing a bit simple.

(3) The Centre of Mass Concept Itself. This has been a useful idea with many practical applications not just in gravity problems, but also in the application of forces, inertia, and also circular motion.

Two simple cases for rigid bodies: (a) To move an object efficiently without rotating it (wasting energy), you apply force in a straight line through the Centre of Mass. (b) If you rotate an object in the absence of other forces, it will naturally rotate on an axis passing through the Centre of Mass.

In both of these concepts, application to rigid bodies is preferred since plastic or flexible bodies complicate both the exchange of energy and can cause changes in the position of the Centre of Mass itself.

The assumption that if force is applied at a point randomly then the object will rotate around the centre of mass is not correct. Tests have shown that there are intermediate-points around which the whole rod/rigid body rotates in between the rotations.


QUOTE]Briefly, the Centre of Mass of bodies rotating and/or flying through the air or through space will indeed trace a parabola or move in a straight line, implying that the object's motion as a system is strictly obeying Newton's laws of force and gravity. This is the only theory which mentions or requires a uniform gravitational field over the area or volume of interest.

[/QUOTE]

I think you can refer to the example of the parabolic-projectile motion which follows the rule that c.m keep son moving the parabolic path even if there is a clast within the ball which is going along the parabolic path.Yes, its true that projectiles being fired near the Earth , are analysed assuming the force on it does not change for that period of time, which is an approximation and results in minor errors.

I should say you pretty much summed up the setbacks in centre of mass concept but as per me , I feel it was more for a problem solving and motion-analysing theory and it didnt ensure that it will give correct results.This does not mean end of the theory .It is a theory of approximations accurate enough to give us near about good results.

BJ.

 

[arun_mid]

Hi, I'm new to this forum, and just out of high school, and I didn't even know that the Center of Mass Theorem was an approximation (is this because gravitation is not included as one of the internal forces acting in a system during its derivation? this could have been countered by Newton's third law, but that was proved wrong-I don't know whether later derivations included the gravitational force or not), but coming back to RP's reasoning with the sphere, if his reasoning were correct, I could apply the same reasoning to a ring of uniform mass M and radius R; then the field at a point on its axis at a distance 'd' from its center would experience a field E=GM/(d^2) (because its CM is at its center) while the actual answer is E=GM/(d^2+R^2). I think this point has been implied by many time and again, in even plainer words. Can RP explain this before going on to the complex aspects of it?

 

[Stingray]

Rogue, please look up what a multipole expansion is. I agree that the force between two objects in Newtonian gravity (or electrostatics) is not given by the inverse square law between their center's of mass. This is hardly a great conspiracy, though. It is just the monopole approximation to the infinite series that actually describes the forces between two extended bodies. It is an approximation valid only when the distance between the center's of mass is much greater than the radii of either object. Here's a quick lesson:

Hurkyl gave the exact expression for the force, but I'm going to put it into a more convenient form. The potential due to body A is equal to (using slightly different notation)
<br /> V_{A}(x) = G \int \frac{\rho_{A}(x&#039;)}{|x - x&#039;|} d^{3} x&#039;<br />

Then the force on body B is
<br /> F_{AB} = - \int d^{3} x \rho_{B}(x) \nabla V_{A}(x)<br />

Looking at the potential in some more detail, we can use the identity
<br /> \frac{1}{|x-x&#039;|} = 4 \pi \sum_{l,m} \frac{1}{2 l +1} \left( \frac{ |x|_{&lt;}^{l} }{ |x|_{&gt;}^{l+1} } \right) Y_{lm}^{*} ( \theta, \phi ) Y_{lm} (\theta&#039;,\phi&#039;),<br />
where the Y_{lm}'s are spherical harmonics, and |x|_{&lt;} is the smaller of |x| and |x&#039;|.

For simplicity, center the coordinate system on body A's center of mass. Also assume that every point in body B is at a greater radius than any point in body A. Then
<br /> V_{A}(x) = 4\pi G \sum_{l,m} \frac{ M_{lm} }{2 l +1} \frac{ Y^{*}_{lm} (\theta, \phi) }{ |x|^{l+1} }<br />
where M_{lm} are the multipole moments of body A:
<br /> M_{lm} := \int d^{3} x&#039; \rho_{A}(x&#039;) |x&#039;|^{l} Y_{lm} (\theta&#039;, \phi&#039;)<br />
In spherical symmetry, the only moment that survives is the monopole, and we find that the potential is given by the standard point mass formula. In general, though, this does not occur. I'll leave it as an exercise to figure out what the full force equation is, why the point mass (monopole) approximation is usually very good.

By the way, a theorem is not something of "intermediate" certainty. It is something which is proven with complete rigor from a set of axioms (whether the axioms are useful ones or not is irrelevant). If something is a theorem, it has a precise proof.

 

[arun_mid]

whoops, sorry, the field due to the ring at the point on the axis is
E=GMd/(R^2+d^2)^3/2; it is however not E=GM/d^2, but in the sphere problem(suppose it has mass M), RP has split the sphere into two hemispheres and taken the fields due to the hemispheres at an axial point to be equal to that due to mass M/2 present at the CM of the hemispheres. This is an incorrect approach. Nowhere does the CM theorem justify this. It cannot be applied like this.

 

[Rogue Physicist]

First I'd like to thank everyone for their excellent replies. This has been an interesting thread for me, and hopefully to others too. Now I would like to clarify my position as is my obligation in response to your efforts and questions:

And yet you seem unable to give a simple, clear statement of this "theorem".

I probably deserve this, on the following analogy: Just as one cannot claim not to have stepped on someone's foot simply because you didn't know you did (or might have), but you have to take into consideration the experience of the person who feels the pain of a squashed toe, I have not been as clear as I could have. We can judge this partly by the results.

But now that I've reread the quoted section, it sounds like you are asserting the following equality:
<br /> \frac{G m_1 m_2}{|\vec{r}|^3} \vec{r} = \int \int \frac{G f(\vec{x}) g(\vec{y})}{|\vec{y} - \vec{x}|^3} (\vec{y} - \vec{x}) d\vec{x}^3 d\vec{y}^3<br />

I am going to try to reply to this more clearly:
What I was doing was applying the premise of the approximation known as the Centre of Mass "method" (we will avoid calling it a 'theorem' so as not to disturb mathematicians who like precision of descriptive terms). I applied it in a way so as to show that it is not just a simple approximation with a predictable 'error' term, but one that produces arbitrary and self-contradictory results when applied in different ways. The purpose was educational, not an attempt to use the Centre of Mass 'method' to realistically calculate the forces. If I had the choice with the solid sphere example, I'd prefer the Sphere Theorem because it will naturally be more accurate and consistent at least with a large mass that itself approximates a continuum distribution of mass. The Sphere Theorem fails for different reasons when applied to discrete distributions of mass in cases like the ones discussed in my other thread.

Dr. Brain was correct that in the Electrostatic case, it is possible for charges to 'shield' one another or counter effects. This is not the case for gravitational forces (at least according to Newtonian Gravitational Theory as normally proposed).

The assumption that if force is applied at a point randomly then the object will rotate around the centre of mass is not correct. Tests have shown that there are intermediate-points around which the whole rod/rigid body rotates in between the rotations.

This is quite interesting: Can you articulate this in more detail? I am not sure exactly what effect is being discussed here.

I could apply the same reasoning to a ring of uniform mass M and radius R; ...Can RP explain this before going on to the complex aspects of it?

In fact, with a ring the force experienced by a particle moving along the axis of symmetry (make that the x-axis) is actually interesting: there is an angle to the ring edge at which the force is maximum, then it fades to zero as the particle moves even closer to the Geometric Centre, then it gradually increases again (still attractive) as the particle continues its path, (the force pulling in the opposite direction) and finally fading out like a 1/d^2 law. I will post the equation and the graph shortly for you.
Again the Centre of Mass 'approximation' will be in error. This time it will be a quantifiable error, due to symmetry, even if the mass is arbitrarily divided up and summed.

Stingray's response is excellent, however I have one fault with it, and that is, it is impenetrable to people like arun_mid who have no advanced engineering methods at their fingertips. Perhaps you could break it down into more steps with some real discussion for ordinary humans: otherwise it will appear to be more like 'showing off' than illuminating a physics problem. You've got DEL operators, spherical harmonics, unidentified angles (or more exotic objects) and no diagrams. It seems a bit unfair to just say 'go find a book on dipole analysis techniques'.

 

[Hurkyl]

The error term is predictable. It's exactly:

<br /> \int \int \frac{G f(\vec{x}) g(\vec{y})}{|\vec{y} - \vec{x}|^3} (\vec{y} - \vec{x}) d\vec{x}^3 d\vec{y}^3<br /> -<br /> \frac{G m_1 m_2}{|\vec{r}|^3} \vec{r}<br />




I'm not sure what you think is so revolutionary about discovering that two different approximations of the same thing aren't equal -- one should only expect them to be approximately equal. (The size of the error there is at most the sum of the sizes of the errors of the two individual approximations)


BTW, I've caught some mistakes in post #5 that I can't believe I missed:

Each half will have its own centre of mass, located 3/8 radians down the axis of symmetry.

"radian" is an angle measure, yet you're using it to talk about a length.

Gravity is an inverse exponential force.

No, gravity is an inverse square force. Inverse exponential looks like: f(x) = K/e^(ax)

 

[Stingray]

What I was doing was applying the premise of the approximation known as the Centre of Mass "method" (we will avoid calling it a 'theorem' so as not to disturb mathematicians who like precision of descriptive terms). I applied it in a way so as to show that it is not just a simple approximation with a predictable 'error' term, but one that produces arbitrary and self-contradictory results when applied in different ways. The purpose was educational, not an attempt to use the Centre of Mass 'method' to realistically calculate the forces.

If your intention in this thread was purely educational, then I think you need to work on how you word things a bit. It sounded more like you were trying to claim that Newtonian gravity contradicted itself. Anyway, I agree that your main point is often a source of confusion for beginners, and it is important to correct them.

This might just be another instance of imprecise wording, but I disagree with your claim that the "center of mass method" does not have a predictable error term. This is not correct. You can estimate errors from the expansion I gave before. You'll also find that those errors reach 100% (or more) if you try to look at center of mass separations comparable to the radii in the problem. These are exactly the cases where you're pointing out that the method fails.

Stingray's response is excellent, however I have one fault with it, and that is, it is impenetrable to people like arun_mid who have no advanced engineering methods at their fingertips. Perhaps you could break it down into more steps with some real discussion for ordinary humans: otherwise it will appear to be more like 'showing off' than illuminating a physics problem. You've got DEL operators, spherical harmonics, unidentified angles (or more exotic objects) and no diagrams. It seems a bit unfair to just say 'go find a book on dipole analysis techniques'.

I apologize for being terse, but I cannot possibly explain all of that in an internet posting. I think some people followed it. The rest can (and should) go learn about these things at some point. For more reference, these types of multipole moments are commonly explained in E&M books. Any applied mathematics book discussing the Poisson equation in detail will do the same.

The basic conclusion is explainable, though. The gravitational potential generated by anybody can be expanded in an infinite series. The first term is the same as the point-mass potential. Each higher term falls off faster and faster with radius (they're also not spherically symmetric). The point mass potential is therefore a good approximation only when very far away from the source.

More specifically, let the source's approximate radius be equal to r (it's not necessarily a sphere, so 'radius' is being used roughly), and say that we want the potential a distance x away from the source's center of mass. Then the nth term in the potential is going to be at most a factor of (r/x)^n times the point-mass potential. The "center of mass method" is therefore only reasonable when r/x is extremely small.

All I've done is given a way of writing down the full integral expression of the force (as given by Hurkyl) in a more convenient way for talking about this problem.

 

[arun_mid]

ok, this is what I meant:
for your argument about the sphere, you divided the sphere(kept at the origin), suppose of mass m and radius R, into two equal halves, and considered the CMs of the two hemispheres. You assumed that the sphere could be considered as the system of these two point masses at equal distances 3R/8 from the center of the sphere, at opposite ends. Now you kept a point test mass m at a distance 'd' from the center of the sphere. Suppose the frame is such that the sphere was cut into two halves by the plane x=0, and that this test mass is on the line y=0,z=0. Then you calculated the grav. force as follows:
F = (GMm/(d-3R/8)^2)+GMm/(d+3R/8)^2)
If this is what you meant, then I could use the same approach to calculate the field due to a ring of uniform mass M and radius R at a point on its axis at distance 'd' from its center. The field thus calculated would be
E = GM/d^2
as the CM of the ring is at its center. This formula is not correct except at very large distances. So this approach is wrong. If you mean that this contradicts CM theorem, I disagree as CM theorem was never designed for gravitational force. it was assumed that each differential mass element of the body did not interact; that mass was just that-mass, with no gravitational field --a valid assumption, given that the gravitational force is negligibly small between such particles, and this is true, even the gravitational force between two atoms can't bind them in a molecule.

 

[Rogue Physicist]

First to respond to Hurkl:

No integral or algebraic expression properly explains the basic fact that in my examples, it is critically important to know *how* one has bisected the sphere, that is, what the orientation of the cut was, in order to predict both the direction of the error as a geometrical vector term in 3-space, and its magnitude. Your equation is fine, but completely un-descriptive for humans trying to understand the effect of orientation.

Something cannot be classified as a 'mistake' if it is done with purpose, successfully meeting the desired goal.

I chose radians deliberately to both normalize the equations and descriptions, and legitimately talk of radians according to their original definition and essential meaning in Euclidean 2 or 3-space: The distance along the perimeter of a unit circle in the axis units of distance, the distance of a radius unit. Radians are abstracted into angles as a consequence, but were fundamentally defined as distances and remain true in Euclidean space (Newtonian Space). To me this reminds the reader of our 'roots', and it remains perfectly legitimate to speak of and use radians as distances in Euclidean space, handily tying the units of measure to the geometrical scale of relevance. Plainly the effects discussed directly depend upon the 'radius' of the sphere and its ratio to the test-mass distance, which should naturally be in this unit.



'exponential' is a natural adjective formed from 'exponent', and quite a general term, used both historically, and still widely, to mean any and all expressions and polynomials that take a non-trivial exponent. Of course it has also taken on a specific reference to that special number e, always itself expressed with an exponent term. In order to be more precise, not less, I always use the capital 'Exponential', often as a noun or substantive phrase when speaking of e. This is less ambiguous and retains the ordinary adjective 'exponential' for its natural meaning according to English usage as 'of or having the qualities of an exponent'. Again, hardly a 'mistake' but rather a useful educational procedure for human beings whose first language is English.

 

[Rogue Physicist]

To respond to Arun Mid:

...I disagree as CM theorem was never designed for gravitational force. it was assumed that each differential mass element of the body did not interact; that mass was just that-mass, with no gravitational field --a valid assumption, given that the gravitational force is negligibly small between such particles, and this is true, even the gravitational force between two atoms can't bind them in a molecule.

In the first part here you appear to think that the CM theorem was never associated with the calculation of gravitational forces. Of course this is teleologically unsound, as even when used to handle problems involving translational energy of projectiles and balancing forces, the forces are of course gravitational. Perhaps you have only come across the concept in a discussion of inertia or angular momentum. I can assure you that this is not the whole story, nor does it reflect the history of ideas in Newtonian Mechanics and gravity.

Next you refer to what appears to be a discussion of independant (free-floating) systems of particles, again a common Newtonian application of both the CM concept and the Newton's 3rd Law of action/reaction, in which typically only the kinetic energy of collisions is usually discussed, and the (external) gravitational field is assumed uniform. In your (1st year?) textbook these subjects are usually limited to a half-page or so, avoiding all the problems we discuss in this thread.

Another way of looking at things might be to take you as having the General Relativity idea of a *real* gravitational field that can carry mass and energy in the back of your mind as non-applicable to Newtonian gravity, and you would be right there also (two different theories).

But I think the basic problem for you is that your working textbooks are not adequately covering the concepts in the depth necessary to really understand them, and this is not your fault, but actually one of my complaints about University Introductory Physics texts.

 

[James Jackson]

What are you talking about? The radian is definied as the ANGLE subtended at the centre of a circle by an arc of equal length to the radius. Are you getting confused as it is a dimensionless quantity?

 

[Rogue Physicist]

...an arc of equal length to the radius. Are you getting confused as it is a dimensionless quantity?

I don't know why you are having trouble with the fact that the radian is defined as a length in physical space describing a distance along an arc or circle's perimeter. You yourself were forced to use the word length to describe it.

Yes it is dimensionless, and that is why it is the ideal measure to describe a dimensionless effect, the failure of the CM 'method', which fails in circumstances having to do with the ratio or proportion of the distance to the radius. Describing the CM 'method' in terms of units of the radius is the best and only way to generalize and concisely convey the effect discussed.

 

[James Jackson]

No, the radian is the ratio of two lengths and henceforth itself is not a length, it is a dimensionless quantity. Dimensions used to 'create' a property and performing dimensional analysis on that quantity are separate entities.

 

[Hurkyl]

Describing the CM 'method' in terms of units of the radius is the best and only way to generalize and concisely convey the effect discussed.

And you could say that -- you could say something like "3/8ths of the way along the radius", or "at a distance of 3r/8 from the center (where r is the radius)". I've even seen "radius" used directly as a unit of measure, as in "3/8 radii from the center". But using radians as a unit of measure is patently incorrect.

Of course it has also taken on a specific reference to that special number e

No, it has not. It is merely customary to write an exponential function in terms of e, but any base can be used. Specifically, for any nonzero a and b, any function of the form f(t) = a^rt can be written in the form f(t) = b^st.

(edit: I think, now, you were referring to the fact exp(x) is called the exponential function, as opposed to talking about exponential functions)

used both historically, and still widely

The graphite in your pencil is still widely referred to as "lead", but it's still wrong to say that you're writing with the element Pb. Whether or not your claim that the use of "exponential" is widely used to describe polynomial growth is correct, that usage is still incorrect.

 

[arun_mid]

definitely you can apply the 'fact' that the body acts as if its mass were concentrated at it's CM for a uniform gravitational field, so I did apply CM to gravity there without much problem, although with large objects you would consider CG instead because the grav field is not uniform anymore.
But this loose definition of CM does NOT hold everywhere. However the fact that mass times the acceleration of the CM equals the net external force is UNIVERSAL. So if you had a free uniform hemispherical mass you could calculate the net grav force on it by a test mass on its axis you can say that the mass times the acceleration of CM equals the net grav force. but you CANNOT say that the entire mass is concentrated at that point, so apply the grav. force law for particles at the CM.

 

[arun_mid]

This definition of CM theorem is true universally:
mass*acceleration of CM = Net external force on system

Try to prove THAT wrong. If you are simply saying that the previous loose definition was wrong, then you are right, and if this is the only definition in textbooks, that is confusing--yet I'm sure that's not the case, I didn't study it that way.

 

[Rogue Physicist]

I guess all good threads must eventually deteriorate into semantic quibbling, so this should be no exception. I'd like to thank everyone for participating, and say how much I appreciate the theoretical contributions to the discussion.

Whether or not your claim that the use of "exponential" is widely used to describe polynomial growth is correct, that usage is still incorrect.

I think we can all chuckle a little bit and lighten up after this line of reasoning.

Since word definitions are based upon usage (at least in real languages like English), if there is any argument to be had for one definition over another it would be based on the convention of statistical usage. The purpose of language is communication, and its success is determined by convention. Six million Welshmen can be stupid, but they can't be 'incorrect' about their own usage of a word.

Correct me if I'm incorrect, but I believe I am probably just stupid. And I fear I'll go on referring to 'pencil lead' to the end of my days, and have to suffer being understood perfectly by my peers, and by 'graphite' men too.

I guess we're both right: its a candymint and a breathmint. Longer *and* milder. Tastes good, really works. You say tomatoe I say tomatoe with a different accent.

Henceforth radian is not a length!
It shall be defined as the portion of my breast exposed during the intermission of a football game.

 

[James Jackson]

So I take it you accept that the concept of the radian as a unit of length is a completely imaginary construction.

 

[krab]

(2) Naturally, the CM for the other half will be the same distance away from the geometric centre of the sphere in the opposite direction away from the test-particle. The total force will be the vector addition of the two halves. But by inspection this is impossible. The increase in force for one half the mass now located closer cannot balance the decrease for the other half, because while the distances are equal, the forces have changed by an unequal amount. Gravity is an inverse exponential force.

Hurkyl gave a poor example with the pencil thing. In the context here, namely quantities and formulas, "exponential" is unambiguously incorrect. It's as bad as saying that a constant speed requires a constant force. A notion that was common knowledge before Newton. But the fact that it was common usage did not make it correct. "Exponential" means that is you increment the independent variable, the effect grows by a constant factor. Hence, without deaths, populations grow exponentially; under constant interest, your savings grow exponentially; etc. Gravity neither grows nor diminishes in this fashion.

 

[quantumdude]

I don't know why you are having trouble with the fact that the radian is defined as a length in physical space describing a distance along an arc or circle's perimeter.

He's not having trouble with the idea of radians. You are. A radian is not a distance. And if you think it is, then please tell us how long a radian is.

You yourself were forced to use the word length to describe it.

Good grief. Just because I have to use the word "distance" to define average speed as the total distance traveled divided by the total elapsed time, it doesn't imply that a speed is a distance.

How you expect to disprove any theorem without having the basics down, I'll never know.

 

[Rogue Physicist]

So I take it you accept that the concept of the radian as a unit of length is a completely imaginary construction.

This is the only reply worth responding to.
Of course radian is essentially a measurement of angle.
It is formally defined by the length along a unit circle in Euclidean space which is proportionate to the angle of concern. In the days of post-Euclidean geometry, the fact that the metric is involved is even more important, and I have no hesitation using radians as the unit of measurement for lengths in problems which are essentially tied to proportions of a circle's relative size. It may have been extreme to actually call the distances 'radians' rather than 'units of radius' or 'radii'.

If we are done nitpicking, we could move on to admitting that it is not trivial that the two main methods (Centre of Mass and Sphere Theorem) of generalization and simplification of gravitational calculations in Newtonian gravitational theory are inaccurate except when d >> r for discrete mass distributions.

In many cases, when the shape of a rigid body, its density distribution and its orientation is known, you can use calculus to find the contribution to the gravitational field of a given massive body. This is very limiting.

When you move to non-rigid bodies (systems of particles), you will I hope agree that the Conservation of Momentum (which is claimed to be even more fundamental than Newton's gravity theory) requires one to again rely upon the axioms and hidden assumptions of the Centre of Mass 'Concept' as defined for general systems of particles. In this case the requirement of 'rigidity' is not required or even assumed.

And this is why all this is important and where the facts lead: The Centre of Mass concept is necessary for the Conservation of Momentum, but in Newtonian Mechanics, the Conservation of Momentum is a circular and poorly defined concept based upon misguided and sloppy descriptions.

The Classical definition and description of the Conservation of Momentum is wrong because it is based upon the misunderstanding and misuse of vectors and their mis-application in Euclidean space.

 

[Doc Al]

What nonsense!