## Prof. Strassler and Kepler's Rule

- Prof. Strassler writes about Kepler's Rule in the context of "dimensional analysis."

- Strassler assumes that the words "dimensions" and "units" are synonymes and can bu used interchangeably.
- So I skip to where he mentions Kepler's Rule

[Kepler's Rule] relates the radius of a planet's orbit R to its orbital period T, specifically, that \(R^3 \propto T^2\)

So Prof. Strassler admits, clearly, that Kepler's Rule explains orbits perfectly well only with R and T. Kepler's Rule which explains orbits perfectly well, does not include a term for Newtonian force of attraction. So Kepler's Rule explains orbits perfectly well without using any terms of Newtonian force. This can only mean that orbits are independent of Newton's force of attraction.

- From the undeniable truth that Kepler's Rule perfectly explains orbits, a scientist, someone who uses scientific reasoning would conclude that orbits are independeno of the Newtonian force of attraction. Not physicists! People who call themselves physicists are Newtonian faitful and see Newtonian force F in Kepler's Rule. We don't see a force term in \(R^3 \propto T^2\) but physicists do. At least they will put F and G and M and say their prayers to Newton and then they cancel or hide Newton's branded terms and do their orbital cmoputations with Kepler's Rule.
- Let's continue with Strassler:

[Kepler's Rule] is true for objects that orbit the sun.

So, Kepler's Rule which does not contain a force term is true for objects that orbit the sun. Kepler's Rule is also true for objects that orbit the earth.

If Kepler's Rule is so true why do physicists brand it with Newtonian ideological terms? Is it because academic physics is Newtonian scholasticism.

- Strassler:

Can we find a formula which is true both for the sun and the earth?

Kepler's Rule is true for the Sun's satellites and it is true for the sun's satellite but with "details that were different."

What are these details? I don't know.

Can we find a formula which is true both for the sun and the earth – one that explains the difference?

Strassler says that if we assume that Newton's gravity is involved somehow we can find such a formula.

- According to physicists, Kepler's Rule explains orbits perfectly well but it is not reallp perfect because it does not include terms branded as Newton's own. So physicists will try to make Kepler's Rule like a Newtonian equation.
- Prof. Strassler is talking like a Newtonian physicist:

First, if [Newtonian] gravity is at work, an experienced [Newtonian] physicist knows that Newton's constant G

alwaysappears, because this constant charaterizes the overall strength of gravity.

What is meant by "overall strength" of gravity? Does he mean "unit" strength of gravity?

The dimensions [or the units] of G have to be consistent with Newton's gravitational force equation, \(F=GMm/r^2\) which gives the force of gravity between two objects of mass \(M\) and \(m\) that are separated by a distance \(r\).

I don't agree with this fundamental physics incantation. This is not a practical or effective formula. No orbit can be computed with this formala the way is written above. You need to eliminate the little \(m\).

- Then Strassler solves for G, and writes

- As a physicist, Strassler cannot start any derivation anywhere else, he must write down $F=GMm/r
^{2}" and repeat the usual incantations to Newton: "This is the force of gravity between two modies \(M\) and \(m\), \(r\) distance apart." - Without \(G\) this expression does not make sense beause its units do not match:

But \(kg^2/L^2\) is not the units of force.

- He then solves for \(G\),

I don't know if this makes sense, force \(F\) is a dummy term.

- Force has dimensions

It's acceleration with the addition of kg which must be cancelled in the case of orbital motion.

- So, from

we find the dimensions of G:

\begin{equation*} G=\frac{L^3}{T^2}\cdot \frac{1}{kg} \end{equation*}So this is Kepler's Rule \(\times\) 1/kg.

Here e switched to orbits.

\begin{equation*} F=\frac{GMm}{r^2} \end{equation*}is meaningless for orbits. You have to eliminate \(m\) to do orbital calculations.

Equations,

\begin{equation*} F=\frac{GMm}{r^2} \end{equation*}and

\begin{equation*} F=\frac{GM}{r^2} \end{equation*}are not the same equation.

- "Assume that gravity is at work" says Strassler.

So now to Kepler's law: Might there be an equation that relates gravity's ever-present constant G, the mass of central body \(M\), the period \(T\) of an object orbiting that central body and the radius R of that orbit?"

[Where is \(m\)? Why doesn't he mention the orbiting body?]

\begin{equation*} G M_{\text{Sun}}= \text{#}\; \frac{R^3}{T^2} \end{equation*}He finds that \(\text{#}=4\pi^2\)

But I'm not interested in his dimensional analysis, I'm interested to his analysis of Kepler(s Rule.

- He writes some equations, then takes theri ratio an G cancels. G disappears. Just before starting to compute dynamical orbits, orbits that obey Newton's force of gravity, the symbol G that represents Newton's force of gravity cancels and orbits are computed as if they were kinematical not dynamical. But physicists cancel the symbol which gives equations their Newtonian force power but still pretend that it is there and orbits are dynamical. But G is not there anymore. Orbits cannot be Newtonian. If the existence of G is the sign of Newton's gravity, the lack of G is the sign that no Newtonian gravity enters orbits.
- If G indicates gravity, lack of G indicates lack of gravity.
- I was trying to prove

- In order to save Newton's authority they put it into Kepler's Rule which does not have a force term and explains orbits without force.
- \begin{equation*}

Density = \frac{Mass}{Unit Volume} \end{equation*}

\begin{equation*} Mass = Density \times Unit Volume \end{equation*}- G is the unit term in Kepler's Rule.
- So he has,

- We must not forget that

is simply the half of Kepler's Rule.

- Strassler started from

he wrote it as

\begin{equation*} G = \frac{Fr^2}{Mm} \end{equation*}and from this he defined the units of G. But the problem is that litlle \(m\) is not in this expression, it is written as a placeholder or dummy term and it will be cancelled. So on the left hand side of this equation F too is a placeholder and it will be replaced by two terms and one of the terms is \(m\). So that ghost term, invisible while physicists repeats his incantations to Newton, in fact, \(m\) is wrtten on both sides of this equation, so it does not exist, anytihng written on both sides of an equation does not exist and has no effect on the workings of the equation.

- So to account for this non-existent \(m\) physicists add a \(1/kg^2\) which later they struggle to eliminate by writing G as GM.
- Units are byproduct of physics because physics is a consistent system of units. Equations must be written with stahdard physics units to make sense. That's why in physics there exists by-products of unitful equations, like G. I mean G is a byproduct of physics equation and it must be cancelled when you need to make orbit calculations.
- If you work with proportions units lose their importance. When you work with proportionalities constants or sanctified constants such as G disappear.
- He gets G from half of Kepler's Rule.
- They try desperately to Newtonize Kepler's Rule.
- From dimensional analysis he switches to calculations with proportions.
- He just applies Kepler's Rule written with G\(\times\) Mass
_{Sun}and G\(\times\) Mass_{Earth} - He takes the ratio and G disappears! The
**universal**constant of gravity ceases to be**universal**and disappears. - He ends up with

- We need to show that his Mass terms are alsa part of Kepler's rule, they are not Newtonian terms. He wrote the Newtonian term, the sanctified universal constant G, he said his prayers to Newton then he cancelled it! He is doing orbit calculations with Kepler's Rule. So why did he write all that Newtonian paraphelnalia? To pray to Newton. Because physics is schlasticism.
- When he is ready to do orbit computations he cancels the sacred constant of gravitation which gives orbits their Newtonian flavor. So from this we must conclude that orbits are not Newtonian.
- And later he transitions to total relative computations skipping standard physics units all together.