Physics derivation of Kepler’s rule from Newton’s laws at Freedom of Science

Physics derivation of Kepler’s rule from Newton’s laws

Published by admin April 25th, 2008 in Physics, Doctors of Philosophy, Newton, Kepler, Force, Newtonism

1. F=ma

Physicists start their derivation with

$F=ma$

This statement may be meaningful to physicists but the truth is that all three terms, F, m and a, disappear during the derivation. None of these Newtonian terms enter the formula used to compute orbits. From this observation we must conclude that F, m and a are not quantities that play any role in orbital motion:

Orbital motion is independent of force, mass and acceleration caused by a supposed tangential motion.

This is not surprising because force is occult and does not exist in nature. Physicists write F, m and a because it is a professional tradition and also to save Newton’s authority. But authority terms do not have a place in orbital motion. Below I demonstrate that physicists’ derivation is nothing more than writing Kepler’s rule with a conventional unit.

2. Acceleration

Let’s use the following figure to look at physicists’ derivation in more detail:

In F=ma we can expand a

$a = \frac{BC}{t^2}$

Acceleration a refers to the supposed tangential motion of point B.¹ To make explicit that BC is tangential and not circular, let’s define BC = d and write a as

$a=\frac{BC}{t^2}=\frac{d}{t} \frac{1}{t}$

where

$v_\mathrm{tan}=\frac{d}{t}$

So F = ma is really

$F=m\frac{d}{t}\frac{1}{t}$

But we know that m will cancel and it’s silly to carry a term we know we will cancel, so, for unit mass,

$F=\frac{d}{t}\frac{1}{t}$

2. Geometry of circular motion and Newtonian gravity

From the geometry of circular motion

$\textrm{Acceleration}=\frac{v_\mathrm{tan}^2}{r}$

$\textrm{Acceleration}=\frac{d^2}{t^2}\frac{1}{r}$

Placing this acceleration into F = ma

$F=\frac{d^2}{t^2} \frac{1}{r}$

Let’s invoke the other Newtonian F

$F=\frac{GM}{r^2}$

and let both forces be reunited²

$\frac{GM}{r^2}=\frac{d^2}{t^2}\frac{1}{r}$

4. Rotation versus revolution

The velocity on the circle and orbital period T are related by

$v_\mathrm{orb}=\frac{2\pi r}{T}$

Now physicists make the hidden assumption that

$v_\mathrm{tan}=v_\mathrm{orb}$

and that

$\frac{d}{t}=\frac{2\pi r}{T}$

But according to this, for a given period T

$v_\mathrm{tan}\propto r$

but this describes a sling-type rotation where velocity increases with increasing r. But orbits obey a different rule

$v_\mathrm{orb}\propto \frac{1}{r^2}$

Therefore, the assumption that the planet has a tangential motion and that this tangential motion is converted into a dynamical orbit by the Newtonian occult force F = ma is an incorrect assumption that is contradicted by observations. This is why v(tan) cancels below.

But physicists equate v(tan) and v(orb)

$\frac{d}{t}=\frac{2\pi r}{T}$

and obtain

$\frac{GM}{r}=\frac{4\pi^2 r^2}{T^2}$

This eliminates the tangential component v(tan) from the equations. Now all of the Newtonian terms physicists started with are eliminated. Kepler’s rule, the true formula used to compute orbits, has no Newtonian term in it:

$\frac{GM}{4\pi^2}=\frac{r^3}{T^2}$

At this point physicists say that they have derived “Kepler’s laws” from “Newton’s laws.”

Conclusion

This derivation amounts to writing Kepler’s rule R3/T2 = k with a conventional unit

$k =\frac{GM}{4\pi^2}$

This replacement of the Gaussian constant of gravity k³ with the Newtonian G occurred in the 19th century. Later physicists elevated G to the status of “constant of nature.” But constant terms in a proportionality cannot be “constants of nature.” What is constant is the proportionality, what physicists call “proportionality constants” are arbitrary constants no different than arbitrary constants of integration.

Therefore, all of the Newtonian terms physicists wrote when they started their derivation turned out to be physics vaporware. F, m and a exists in physics solely for traditional reasons and to save Newton’s authority. They are not needed to describe orbits. And orbits are independent of the Newtonian occult force and animistic mass.

Physicists draw a picture over the geometric point B to represent the mass m. But this term cancels even though it is represented in the figure. So the mass is an occult, or hidden, scholastic dormitive virtue. [↩]
Here we’ve equated two forces, one proportional to 1/r and the other 1/r2. This seems absurd to me. [↩]
or k2 with proper units [↩]

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2 Responses to “Physics derivation of Kepler’s rule from Newton’s laws”

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1 Doug May 1st, 2008 at 6:13 pm

Hi Pioneer1,

I would like to comment on a modification of first figure [circle] in item #2 Acceleration.

Perhaps you could draw the figure from this description.
Let c be the radius of the crcle.
From orthogonal X and Y axes, with the origin at the center of the circle, draw the chord.
Do this only for the usual Quadrant I [ , ], although this could be done for the other three quadrants.
Reflect the right triangle so that a square is formed.

Compare this to the Gini coefficient diagram.
GOTO the Terence Tao Einstein’s derivation of E=mc^2 discussion.

From my perspective, this may bring economics into relativity:
From the wiki page:
a - the shaded segment labeled gini index may be O(v^3) and
b - the remaining portion of the lower right triangle may be E’ or perhaps bonding energy [?];
From the constructed diagram:
c - may correspond to the upper triangle in Quadrant I,
d - the lower triangle in Quadrant I may correspond to the kinetic energy,
e - the entire square may correspond to potential energy.

Insight - maybe?
Rigor - no.
Geometry in GR - perhaps? - when letting m=1, v=c, a=c^2.

The chord may be the relative velocity of orthogonal particle movement or c*sqrt(2).
The diameter is 2c which may be the relative velocity of two particles traveling 180 degrees from each other, although the vector sum of energy is zero.
2 Pioneer1 May 2nd, 2008 at 8:07 am

Hi Doug,

Thanks for your comment. It will probably take me months to understand Terry Tao’s derivation. I remember that I marked it as interesting when I first saw it but I could never find the time to study it carefully. Also, I consider E=mc2 to be a definition and I’m not sure if one should or could derive definitions. But I may be wrong. Since you mention a geometric interpretation for kinetic and potential energies I thought I comment on that. My understanding is that physicists call Kepler’s rule “conservation of energy” and they write it as (for unit mass)

$\textrm{Kinetic energy}=\frac{R^2}{T^2}=\frac{1}{R}=\textrm{Potential energy}$

So they labeled R2/T2 = kinetic energy and 1/R= potential energy (for unit mass). Since this is Kepler’s rule, it has a simple geometric interpretation. Given two orbits R_0 and R I have this geometry:

AE = R_0
BE = w_0
AD = R
DC = w

Then by Kepler’s rule

$\frac{R_0^2\, \omega_0^2}{R^2\, \omega^2}=\frac{R}{R_0}$

R^2 and R_0^2 are the areas swept by the radii. Please let me know what you think.

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