Also see Is Newton’s force occult
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Chatgpt wrote: Newton defined gravitational force mathematically—as an inverse-square law—

I have different view on this: Newton’s \(1/r^2\) is simply one of the terms in Kepler’s Rule: \(r/t^2 = 1/r^2\)

You’re touching on an interesting point—the close connection between Kepler’s empirical laws and Newton’s theoretical formulation. To clarify:

1. Kepler’s Third Law (Kepler’s Rule): Kepler observed that for the planets, the square of the orbital period (\(T^2\)) is proportional to the cube of the semi-major axis (\(r^3\)). In mathematical terms, \(T^2 \propto r^3\).
   
2. Centripetal Acceleration in Circular Motion: For a circular orbit, the centripetal acceleration needed to keep a body in orbit can be approximated by a \(r/T^2\) (more precisely, \(a = (4\pi^2 r)/T^2\) when you work through the full derivation).
   

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1. « Centripetal Acceleration in Circular Motion »
   
   1. We must distinguish between two types of circular motion. Spin rotation and orbital revolution are not analogous motions. So the circular motion mentioned here is orbital revolutionary motion, it is not spin rotation.
      
2. « For a circular orbit, the centripetal acceleration needed to keep a body in orbit… »
   
   1. This is Newton’s unjustified assumption based on the analogy of spin rotation and orbital revolution. We don’t know if orbital motion produces an acceleration toward the center. In orbits there is no input of energy as in a sling. Newton tries to invent this input of energy with his fairy tale or voodoo forces.
      
   2. Is Newton’s assumption justified?
      
   3. Can we assume that this supernatural force does not exist?
      
   4. Yes. Then what happens? If we assume that Newton’s supernatural cause called force? If so, how can we explain orbits without force? How would such an orbit look like?
      
3. « a = \(r/T^2\) (more precisely, \(a = (4\pi^2 r)/T^2\) »
   
   1. Yes, this can also be written as \(v^2 / r\).
      
   2. We’ll work through the full derivation. Is this derivation independent of Kepler’s Rule?
      

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1. Newton’s Inverse-Square Law: Newton proposed that the gravitational acceleration \(a\) provided by a mass \(M\) is given by \(a = G\cdotM/r^2\). For planetary motion, setting the centripetal acceleration equal to the gravitational acceleration leads to
   \[ \frac{4\pi^2 r}{T^2} = \frac{GM}{r^2}. \]
   Rearranging this equation indeed recovers Kepler’s third law (\(T^2 \propto r^3\)), showing that an inverse-square dependence (\(1/r^2\)) is necessary for the observed relationship between orbital period and radius.
   

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1. « Newton’s Inverse-Square Law… »
   
   1. Where is this law?
      
   2. Is there any laws other then Kepler’s Rule?
      
   3. As far as I know, there is not other law or rule other than Kepler’s Rule.
      
2. « Newton proposed that the gravitational acceleration \(a\) provided by a mass \(M\) is given by \(a = G\cdotM/r^2\) »
   
   1. This is anachronism. Newton never wrote an equation like this: \(a = G\cdotM/r^2\). Newton did not work with equations. He worked with ratios and proportionalities. And Newton did not know this thing called \(G\).
      
3. « Gravitational acceleration… »
   
   1. So now “centripetal acceleration” became “gravitational acceleration”
      
   2. This gravitational acceleration is providid by a mass \(M\) at the center now we are in the Newton’s fairy tale land.
      
   3. We suspend disbelief but we uphold our faith in Newton’s authority and accept Newton’s voodoo orbits explaining as physical explanation.
      
4. « Rearranging this equation… »
   
   1. There is no need rearrange. This is already Kepler’s Rule.
      
5. « Newton’s mathematical statement of force…* »
   
   1. I object to the statement that Newton “defined gravitational force mathematically as an inverse square law.
      
   2. \(1/r^2\) is one of the terms in Kepler’s rule: \[\frac{r}{t^2}=\frac{1}{r^2}\]
      
   3. « Kepler’s empirical laws »
      
   4. « Newton’s theoretical formulation »
      
   5. Kepler’s Third Law is called more appropriately “Kepler’s Rule: \[T^2 \propto r^3\]
      
6. « Kepler’s Third Law or Kepler’s Rule »
   
   1. We prefer to say Kepler’s Rule not Kepler’s Third Law.
      
   2. Not semi-major axis but radius because orbits of all major planets are circular.
      
   3. Newton needs an acceleration directed toward the center of the orbit.
      
   4. His proofs in the Principia is childish.
      
   5. How do we derive it for orbital motion?
      
7. « Centripetal acceleration »
   
   1. They say circular motion has an acceleraiton toward the center but this is an unjustified assumption and generalization.
      
   2. There are two types of circular motion:
      
      1. In slign motion, for instance, energy is put in constanly to sustain uniform circular motion.
         
      2. The energy put in by the hand that turns the string manifests itself as the outward tendency and tension in the string.
         
   3. There is gyroscope motion where there is no input of energy. The disk of the gyroscope rotates with its own inertia. It constantly looses speed because there is no perpetual motion. The speed of the gyroscope cannot be uniform. It’s angular speed is reducing.
      
   4. I’m not sure if there is a centrifugal tendency in gyroscope motion.
      
   5. In the spin rotation motion all points on the radius move with the same angular velocity, so they are forced to move. The point further away from the center had to move a greater arc. The point closer to the center moves a samller arc. So in unit time the point further from from the center moves a greater arc and closer one moves a smaller arc.
      
   6. Then we hava the orbital revolution. In this case angular velocity is not same for various \(r\), the point at greater \(R\) moves more slowly then a point at smaller \(r\).
      
   7. Here there is no input of energy unless we assume Newton’s voodoo force.
      
8. « Kepler’s empirical rule… »
   
   1. No one disputes this.
      
   2. Then we have Newton’s appropriation or rebranding of Kepler’s Rule.
      
   3. But there is a subtlety here. We need to derive this acceleration or tendency for orbits.
      
   4. Again \[\frac{R}{T^2} = \frac{1}{R^2}\]
      
   5. Does \(R/T^2\) make sense on its own?
      
   6. Chatgpt does not make the distinction about circular motions that I mentionde above: “Centripetal acceleraiton in circular motion,” and “for a circular orbit”… So chatgpt ignores my distinction between spin rotation motion and orbital revolution.
      
   7. There must be an acceleration in orbital motion to keep the body in orbit. But this is the dynamical unstable orbital explanation.
      
   8. Where is this acceleration?
      
   9. We need input of energy for this acceleration to exist.
      
   10. Where is that energy coming from?
       
   11. This gravitational acceleration is provided by a “mass” \(M\) at the center. Now we are in Newton’s fairy tale land.
       
   12. We suspend disbelief but we uphold our unwavering faith in Newton’s authority and accept Newton’s voodoo force as the physical cause powering orbits.
       
9. « Newton’s inverse square law (again) »
   \[ \text{cent. acc.} = \frac{4\pi^2 r}{T^2} = \frac{GM}{r^2}s= \text{grav. acc.} \]
   
   1. So, \(GM/r^2\) is called gravitational acceleration and this acceleration is created by this mass \(M\). I guess this supernatural cause Newton calls force residing in mass \(M\) (although Newton denies that attractive force resides in \(M\)) reaches out to the satellite traveling the distance \(r\) magically without time passing, and also calculating without time passing the exact mass of \(m\) and applying the exact attraction as decsribed by “Newtonian mechanics” and bends the rectilinear motion of the mass \(m\) just the right amount and then calculates the exact amount of attraction to keep \(m\) in orbit forever. If you belive this Newtonian fairy tale I have a bridge to sell you.
      
   2. And the rationalization of this fairy tale by physicists is that “we don’t how all these things happen but this is what our great master Newton said so it must happen somehow and hypotheses non fingo! We are great scientists so we don’t try to understand how all these things happen by breaking all known laws of physics. Maybe it is the work of God. So shut up and trust Newton’s authority”.
      
10. « Rearranging this equation indeed recovers Kepler’s third law (\(T^2 \propto r^3\)), showing that an inverse-square dependence (\(1/r^2\)) is necessary for the observed relationship between orbital period and radius. »
    
    1. There is no need to rearrange: \(r/t^2 = GM/r^2\) is already Kepler’s Rule.
       
    2. And I don’t understand what “an inverse-square dependence (\(1/r^2\)) is necessary for the observed relationship between orbital period and radius” means. \(1/r^2\) is a term in Kepler’s Rule. Yes, in this sense it is necessary, without it there is no Kepler’s Rule. But Kepler’s Rule does not say that there is a quantity that varies as \(1/r^2\). The true relationship is \(r/t^2 = GM/r^2\). Without \(r/t^2\) the term \(1/r^2\) makes no sense.
       
    3. \(1/r^2\) is necessary not because it is the “inverse square law of Newton” but because it is a term in Kepler’s Rule.
       
11. « Why are gravitational acceleration and centripetal acceleration equal? »
    
    1. The only reason is to make Newtonian fairy tale orbits orbits powered by his voodoo force work.
       
    2. This is their excuse to re-combine Kepler’s Rule that they divided into two.
       
    3. Newtonians developed an ironclad rhetoric to hide the fact that they are using supernatural causes to explain natural phenomena like orbits. But there are cracks in their argument.
       

12. « Let’s look at “Newton’s inverse square law one more time »
    \[ \frac{4\pi^2 r}{T^2} = \frac{GM}{r^2}. \]
    

    1. They call this “Newton’s inverse square law”.
       

    2. To me this is Kepler’s Rule written with Newtonian sounding labels like \(G\) which is nothing more a fudging factor invented by Newton’s disciples to make the units work.
       \[ A_{cent} = \frac{4\pi^2 r}{T^2} \]
       

       \[A_{grav} = \frac{GM}{r^2}\]
       

    3. What justification do we have that these two expressions are not the two terms of Kepler’s Rule. This is already Kepler’s Rule but labeled as \(A_{cent}\) and \(A_{grav}\). Kepler’s Rule did not change because Newtonians chopped it into two parts and labeled each part with their own jargon.
       
    4. \(A_{grav}\) must be the acceleration caused by this mass \(M\) at the center. How can a mass \(M\) which does not have this acceleration cause an acceleration in a distant object? This would be creating new motion. This violates the axiom that no new motion can be created. Motion can only be tranferred. So, \(M\) must have this exact accelarition it creates in the satellite. But \(M\) does not have such a motion.
       
    5. But the fairy tales goes like this: \(M\) reaches out to \(m\) (which does not exist in these formulas) and pulls it magically towards itself. Here we have the Newton’s voodoo force showing its magic by traveling the distance \(r\) between \(M\) and \(m\) without time passing. But we suspend desbelief and allow this to happen we would be assuming that nature is supernatural. But nature is not supernatural. This is our fundamental axiom that cannot admit an exception. Even if the exception is requested by the greatest scientist eved lived, the Slick Sir Isaac.
       
    6. This Newtonian voodoo force residing in mass \(M\) is also intelligent. We must suspend disbelief again and we must admit that this timeless traveler is also intelligent. Since when we allow matter to be intelligent? We don’t. But Newton’s force must be intelligent because it needs to “weigh” the mass of \(m\) in order to apply the exact amount of attraction. If mass \(M\) fails to apply the exact amount of attraction according to the mass of \(m\) orbits cannot be formed. How does \(M\) compute the exact amount of mass required to pull \(m\) exactly to form an orbit. Well, if it can travel without time passing it can also be intelligent. So Newton’s force is some magical thingy that is both intelligent and timeless traveler. And Newtonian have no difficulty accepting that such a thing exists in this world and creates and powers orbits. But of course hypotheses non fingo. If you repeat this magic Latin word a few times all fairy tale nonesense invented by Newton will be physical and natural.
       
    7. I don’t understand how they don’t see Kepler’s Rule in this expression
       

    \[ \frac{4\pi^2 r}{T^2} = \frac{GM}{r^2}. \]
    

    6. Newton labeled (rather Newtonians labeled, because Newton never wrote such an expression with standard units) each term of Kepler’s Rule “acceleration.”
       
    7. There is a visaul illusion here.
       

    \[ \frac{4\pi^2 R}{T^2} = \frac{GM}{R^2}. \]
    

    8. is nothing more than
       

    \[ \frac{4\pi^2}{GM} = \frac{T^2}{R^3}. \]
    

    9. This is equivalent to
       

    \[ \frac{4\pi^2}{GM} = \frac{T_0^2}{R_0^3} = \frac{T^2}{R^3}. \]
    

    10. Where the constant term of Kepler’s Rule is stated with Newtonian looking symbols.
        
    11. I see Kepler’s Rule written with Newtonian looking terms (\(G\) and \(M\)) but they see Newton’s inverse square law. They exploiting this visual illusion.
        
13. « Gravitational acceleration »
    \[A_{grav} = \frac{GM}{r^2}\]
    
    1. This thing on its own makes no sense. Can we compute orbits with this expression? No. Unless we combine it with \(A_{cent} = \frac{4\pi^2 r}{T^2}\) we cannot compute orbits. \[A_{grav} = \frac{GM}{r^2}\] does not have the period \(T\). This is a lame equation. It is meaningless without the other term of Kepler’s rule.
       
    2. So clearly what Newtonians call \(A_{grav}\) is one of the terms of Kepler’s Rule.
       
    3. Newtonians labeled one of the terms in Kepler’s Rule, \(1/R^2\), as “gravitational acceleration” by fudging its dimensions and units with the fudge factor thel call \(G\). The fact that this fudge factor is called with thi ideological and pompous name of “Newton’s Universal Constant of Gravitation” does not make Kepler’s Rule Newtonian or dynamical \[A_{grav} = \frac{GM}{r^2}\].
       
    4. In this expression I only see \(1/R^2\), a term in Kepler’s Rule.
       
    5. The label \(A_{grav}\) is a placeholder that will be replaced with \(A_{cent}\) or with \(4\pi^2 R/T^2\)
       
    6. So both \(A_{grav}\) and \(A_{cent}\) are labels or placeholders, they are not physical quantities, they disappear from formulas when Kepler’s Rule is recovered. You cannot cancel a physical quantity from formulas.
       
    7. There is a story behind these two species of accelerations. Can there be different species of acceleration. Acceleration is acceleration but it is said that they are caused by two different causes. One is the product of circular motion of the spin rotation type the other magically appears from the central mass.
       
    8. Here enters the so-called equivalence principle.
       
    9. Let’s summarize: Newtonians write Kepler’s Rule with their own jargan, \(G\) and \(M\) and \(4\pi^2 R\) and claim that they transformed kinematic Kepler’s Rule into Newton’s dynamical inverse square law.
       
    10. Yes, this is an ironclad, air-tight rhetorical sophistry but we’ll find the crack and refute this sophistry
        

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In your formulation, you express this connection by writing
\[ \frac{r}{T^2} \sim \frac{1}{r^2}, \]
which—after rearranging—essentially gives you Kepler’s law (\(r^3 \propto T^2\)). This demonstrates that the \(1/r^2\) term is not arbitrary; rather, it’s exactly what’s required to account for the orbital dynamics observed by Kepler.
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I asked chatgpt why he used \(\sim\) notation:

Why did you use the symbol for “is similar” \(\sim\) rather than the usual proportionality symbol \(\propto\)

1. I used the symbol “∼” to emphasize that the relationship is one of similar scaling or order-of-magnitude behavior, rather than an exact proportionality with a fixed constant. In many physics contexts, especially when discussing scaling laws or approximations, “∼” is used informally to indicate that two quantities behave in a similar way as one changes, even if the exact constant of proportionality isn’t specified.
   
2. On the other hand, “\(\propto\)” is reserved for situations where one quantity is exactly proportional to another—meaning there is a constant \(k\) such that \(A = kB\).
   
3. In our discussion, I was emphasizing the idea that the quantities scale similarly (or “are similar in behavior”) without necessarily pinpointing the exact proportionality constant, which is why I chose “\(\sim\)”. However, if the relationship is strictly proportional, then “\(\propto\)” would be the more precise symbol to use.
   

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I don’t exactly understand. In Kepler’s Rule one quantity is exactly proportional to another quantity. So \(\propto\) should be appropriate here. So, let use \(\propto\) \[\frac{r}{T^2} \propto \frac{1}{r^2}\]

« …after rearranging—essentially gives you Kepler’s law (\(r^3 \propto T^2\)) »

This is already Kepler’s Rule. There is no need to rearrange.

It seems that chatgpt considers \(r/T^2 \propto 1/R^2\) and \(R^3 \propto T^2\) two different rules. I don’t think so. These are the same rule written differently.

\[ \frac{R}{T^2} \propto \frac{1}{R^2} \] and
\[ R^3 \propto T^2 \] and
\[ R^{3/2}\propto T \] are all identical and state Kepler’s Rule with different algebraic forms.

For instance, do you think \(v^2/R\) and \(R/T^2\) are two different statements? They are the same. You just make some algebraic transformations to obtain one from the other. The underlying idea remains the same. These are two different ways of saying the same thing. Six of one, half a dozen of the other. Same thing. And this goes for all permutations of Kepler’s Rule.

So it makes no sense to say “after rearranging” \(R/T^2 \propto 1/R^2\) gives you Kepler’s Rule. There is no need to rearrange. Both forms are Kepler’s Rule. The rule is the same.

We can express Kepler’s Rule as \(R^3 \propto T^2\), \(R^{3/2} \propto T\), \(R/T^2 \propto 1/R^2\) or \(R^2/T^2 \propto 1/R\), these are all Kepler’s Rule. Do you agree?

« This demonstrates that the \(1/r^2\) term is not arbitrary; rather, it’s exactly what’s required to account for the orbital dynamics observed by Kepler. »

This is play on words. Of course, \(1/r^2\) is not arbitrary. How can it be? It is one of the terms in Kepler’s Rule. This term on its own has nothing to do with Newton and Newton’s dynamical jargon.

Kepler did not observe “orbital dynamics”. Kepler’s Rule is kinamatical. It contains only two terms \(R\) and \(T\).

So first you write Kepler’s Rule with Newtonian jargon then you eliminate Newtonian jargon and claim to have derived Kepler’s Rule from Newton’s laws. This is nothing less than fraud.

I never said that \(1/R^2\) was arbitrary. I said that \(1/R^2\) is one of the terms of Kepler’s Rule and for this reason \(1/R^2\) makes no sense without \(R/T^2\).

Stated alone, \(1/R^2\) can be anything. It just has no meaning on its own. This is why we need to label it somehow. Newtonians write \(A_{grav}=1/R^2\). But this is meaningless too. You read this as “gravitational acceleration varies as \(1/R^2\)”

If I write two \(1/R^2\)’s can you tell which one is Newton’s inverse square law: (1) \(1/R^2\), (2) \(1/R^2\). I guess not.

So, \(A_{grav}\) is just a label attached to \(1/R^2\), it is a placeholder.

Do you agree that unless we combine \(1/R^2\) with \(R/T^2\), \(1/R^2\) makes no sense?

But Newtonians write \(1/R^2\) term of Kepler’s Rule with Newtonian labels as \(GM/R^2\) to steal Kepler’s Rule and make it look Newtonian and dynamical. But both \(G\) and \(M\) are cosmetic additions that cannot change or transform Kepler’s Rule into Newton’s Laws.

So you read this as “gravitational acceleration”.

If this is gravitational acceleration, can you compute orbits with \(GM/R^2\)? No. Because this thing make no sense on its own. You must unite it with the other part of Kepler’s Rule.
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While it might seem like “Newton’s \(1/r^2\)” is just a term extracted from Kepler’s rule, historically Newton went further by providing a theoretical framework that not only explained why Kepler’s observations held true but also generalized the gravitational interaction to all bodies. In other words, Newton’s insight was to show that the inverse-square law wasn’t merely a coincidental match to planetary orbits but was a universal law governing gravitational attraction.

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Chatgpt makes good points:

(1) It looks like \(1/R^2\) is a term that belongs to Kepler’s Rule.

Well, yes. \(1/R^2\) is one of the terms of Kepler’s Rule. Newton is trying to appropriate it and brand it as Newton’s inverse square law.

(2) Newton provided a theoretical framwork

Not exactly. Newton labeled the terms of Kepler’s Rule with his idieological labels like \(G\) and \(M\) (ignoring for now that this statement is anachronism)

(3) Newton explained why Kepler’s observation [Kepler’s Rule] held true.

So, the story goes that Newton discovered the underlying cause of Kepler’s Rule. And what is that underlying cause that Newton discovered? It is a supernatural cause that does not exist in this world. The Slickest Sophist of All Times our Sir Isaac, names this supernatural cause “force” and gets away with it.

I don’t think Kepler’s Rule has a supernatural cause.

(4) Newton generalized gravitational attraction to all bodies

No. Newton was the first one who recognized the general nature of Kepler’s Rule. What is generalized is not some supernatural cause Newton calls gravitational attraction but Kepler’s Rule. And indeed, Newton’s orbital computations in his Principia are confirmations of Kepler’s Rule, for the moon, for the solar system and for Jupiter’s moons and saturn’s moons. This is wht Newton’s computations amount to: they are confirmations of Kepler’s Rule. Nothing more. Newton proves no dynamical laws with his orbital computations.

(5) Newton showed that the inverse-square law wasn’t merely a coincidental match to planetary orbits but was a universal law governing gravitational attraction.

What Newton calls his inverse square law is just a term in Kepler’s Rule. There is no inverse square law that has dynamcal consequences. There is no inverse law that sets distant bodies in motion instantly without time passing. These are all part of Newtonian fairy tales.

So what is exactly Newton’s inverse square law so that I can understand what this thing means?

I don’t see any gravitational attractions in nature the way it was defined by Newton.
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Your view highlights that the inverse-square dependence arises naturally from the orbital mechanics described by Kepler—something Newton recognized and mathematically justified. It’s a powerful example of how empirical observations (Kepler’s laws) and theoretical reasoning (Newton’s law) can reinforce one another in physics.
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There is no inverse square dependence arising naturally from the orbital mechanics described by Kepler’s empirical rule. This inverse square term simply appears when you apply some simpl algebraic manipulations to Kepler’s Rule. But what appears as inverse square “law” is not an inverse square law because when you cannot that inverse square term with the other term of Kepler’s Rule, namely, \(R/T^2\) the inverse square thing disappears simply because \(R/T^2\) has also an \(R\) in it.

What was Newton’s “mathematical justification”? Newton simply applied some simple algebraic manipulations to Kepler’s Rule. There is nothing more to Newton’s mathematics. The mathematics in the context of orbits is given by Kepler’s Rule. Newton’s contribution to the mathematics of orbital motion is nil. Newton only try to embellish Kepler’s Rule with his own dynamical decorations.

Where is Newton’s theoretical reasoning? There is none. There is Newton’s voodoo force and fairy tale and supernatural and absurd and childish explanation of orbits. Do you call these Newton’s “theoretical reasoning”? There is no “Newton’s law”. Newton only paraphrased Kepler’s Rule. There is only one law describing orbital motion and that is called Kepler’s Rule.
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No, what I meant is that, in R/T² = 1/R, we have R in both sides so the left had side cannot vary as 1/R²

I see what you’re getting at! You’re pointing out that if we start with the proportionality:

\[ \frac{r}{T^2} \propto \frac{1}{r^2} \]

then the left-hand side (which involves \( r \)) cannot simultaneously vary as \( 1/r^2 \), which would seem to contradict Newton’s inverse-square law. This is an interesting observation!
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« …which would seem to contradict Newton’s inverse-square law… »

There is no “Newton’s inverse-square law.” As soon as we strip the constant terms \(G\) and \(M\) we end up with \(1/R^2\). And this can be anything.

Therefore, there is no contradiction, \(1/R^2\) is simply one of the terms of Kepler’s Rule: \(R/T^2 = 1/R^2\). The term \(1/R^2\) on its own without \(R/T^2\) make no sense.

But chatgpt does not address the fact that \(R/T^2\) cannot change as \(1/R^2\).
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In order to show that writing Kepler’s Rule with Newtonian sounding units and ideological terms does not change anything, we can easily show that that \(G\) and \(M\) are parts of the constant term of Kepler’s Rule.

In order to do orbital computations we need to write Kepler’s Rule with a constant term like this \[\frac{R_0^3}{T_0^2} = \frac{R^3}{T^2}\] The constant term \(R_0^3/T_0^2\) is formed by choosing a known \(R\) and a known \(T\) and making them our units. So \(R_0\) and \(T_0\) are our units.

Since you (chatgpt) “derived” Kepler’s Rule and wrote is as \[T^2 = \frac{4\pi^2}{GM}\cdot R^3\] or \[\frac{R^3}{T^2} = \frac{GM}{4\pi^2}\] it is clear that \(GM/4\pi^2\) is the Kepler’b Rule’s constant term expressed in Newtonian looking units: \[\frac{GM}{4\pi^2} = \frac{R_0^3}{T_0^2}\]

So Newtonians wrote Kepler’s Rule with their standard units. But this does not make Kepler’s Rule Newton’s laws!

You are saying that this expression \[\frac{R_0^3}{T_0^2}=\frac{R^3}{T32}\] describes orbital properties while \[\frac{GM}{4\pi^2}=\frac{R^3}{T^2}\] governs the forces per unit mass. In other words, you are saying that the first is kinematical Kepler’s Rule and the latter is dynamical Newton’s laws.

This cannot be true. The only thing that is different between these two expressions is that the first is stated with natural units (we chose a known \(R\) and a known \(T\) as our units) and the second is stated with standard units that are ideologically named to imply that Kepler’s Rule is dynamical. Both are Kepler’s Rule one is stated with natural units the other with standard, named, units.

Do you think the distance between /(A\) and \(B\) will be different if I express it with a unit I defined meyself and if I stated it with a named unit like meters. No. The distance will be the same distance.

The important thing to notice that neither expression contain force terms.

Neither \(G\) nor \(M\) are forces. \(G\) is a fudge factor invented to make units work. Furthermore, \(M\) is not an independent term. In practice, \(GM\) is always used as a single term.

And \(M\) is assumed to be the source of the Newtonian voodoo attraction or Newtonian voodoo force but in fact \(M\) is only a number. Both \(G\) and \(M\) named to sound as dynamical terms. No, they are just units with fancy names.

What chatgpt calls “Newton’s force law” is this \[a = \frac{GM}{r^2}\] But there is no force here. \(G\) is just a fudging factor and a number with fudged dimensions.

\(M\) is called “mass” and we have to supply with our imagination in order to save Newton’s authority that mass is the home or the source of Newton’s voodoo force. But \(a=Gm/R^2\) is \(a \propto 1/R^2\) and says nothing about any force or any force law. We need to supply that force law by using our imagination. The formula does not say this.

Chatgpt thinks that \(R^3 \propto \T^2\) is Kepler’s Rule but \(R/T^2 = 1/R^2\) is not. No. They are both Kepler’s Rule.

Why did you use the symbol for “is similar” \(\sim\) rather than the usual proportionality symbol \(\propto\)

I used the symbol “\(\sim\)” to emphasize that the relationship is one of similar scaling or order-of-magnitude behavior, rather than an exact proportionality with a fixed constant. In many physics contexts, especially when discussing scaling laws or approximations, “\(\sim\)” is used informally to indicate that two quantities behave in a similar way as one changes, even if the exact constant of proportionality isn’t specified.

On the other hand, “\(\propto\)” is reserved for situations where one quantity is exactly proportional to another—meaning there is a constant \(k\) such that \(A = kB\).

In our discussion, I was emphasizing the idea that the quantities scale similarly (or “are similar in behavior”) without necessarily pinpointing the exact proportionality constant, which is why I chose “\(\sim\)”. However, if the relationship is strictly proportional, then “\(\prop\)” would be the more precise symbol to use.

Explain what justification you have to set these two accelerations equal caused by two different causes?

The key justification comes from considering the dynamics of a body in circular motion under gravity. In a stable, circular orbit, the only force acting on the orbiting object is gravity. This gravitational force is what provides the necessary centripetal force to keep the object moving along its curved path. Here’s a step-by-step explanation:

1. Centripetal Acceleration Requirement:
   For an object moving in a circle of radius \( r \) with orbital period \( T \), the centripetal acceleration needed to keep it in that circular path is given by
   \[ a_c = \frac{4\pi^2 r}{T^2}. \]
   This acceleration is not a force in itself but the rate of change of velocity required for circular motion.
   
2. Gravitational Acceleration:
   Newton’s law of gravitation tells us that the gravitational force acting on an object of mass \( m \) due to a mass \( M \) is
   \[ F_g = \frac{GMm}{r^2}. \]
   Dividing by \( m \) gives the gravitational acceleration:
   \[ a_g = \frac{F_g}{m} = \frac{GM}{r^2}. \]
   
3. Equating the Two Accelerations:
   In a stable orbit, the gravitational acceleration \( a_g \) is exactly what causes the centripetal acceleration \( a_c \) required for circular motion. In other words, the gravitational force “supplies” the necessary acceleration to keep the object in orbit. This balance is expressed by setting
   \[ \frac{4\pi^2 r}{T^2} = \frac{GM}{r^2}. \]
   There is no contradiction here because both expressions describe the same acceleration experienced by the orbiting body: one from the perspective of motion (the requirement for circular motion) and one from the force acting on it (gravity).
   
4. Why Equate Accelerations from “Different Causes”:
   Although the expressions arise from different conceptual approaches—one from kinematics (centripetal acceleration) and one from dynamics (gravitational acceleration)—in a circular orbit they are two descriptions of the same physical requirement. The gravitational force is the cause, and the centripetal acceleration is the effect required to maintain the orbit. Since they represent the same acceleration in this context, equating them is justified.
   

In summary, by setting these two accelerations equal, we are applying Newton’s second law in the context of circular motion, where the gravitational force (a dynamical cause) exactly accounts for the centripetal acceleration (a kinematical requirement).

I seem to remember that one of these accelerations is change in direction only. It changes direction but not its speed. Which one is it? And how come an object changes direction but not its speed. This, apparently, would violate conservation laws. Can you comment about these questions?

The acceleration you’re referring to is the centripetal acceleration. This is the acceleration that acts perpendicular to the velocity of an object moving in a circular path. It continuously changes the direction of the velocity vector without changing its magnitude (the speed).

But intuitively, gravitational attraction needs to be greater than the centripetal force or the speed vector so that it can bend it. All these concepts are mixed up in my mind. Can you help me understand?

Your intuition touches on a common confusion. In a stable orbit, the gravitational force is the centripetal force—it doesn’t need to be “greater” than it. They are, in fact, equal in magnitude, which is exactly what keeps the satellite in a circular path.