of 1 35

A Spacetime Theory For Inertia: Predicting

the Proton, Electron, Neutron and the Solar

System in Terms of a One-Second Invariant

By

Ian Beardsley

January 5, 2026

of 2 35

Contents

List of Constants, Variables, And Data In This Paper…………………………….3

The One-Second Universe: A Fundamental Time Invariant

From the Stiffness of Space to Solar System Dynamics…………………………..4

Interpretation of the Force-Potential Relation

and the Golden Ratio Constraint………………………………………………….18

The Solar Solution………………………………………………………………..28

Jupiter and Saturn…………………………………………………………………32

of 3 35

List of Constants, Variables, And Data In This Paper

(Proton Mass)

(Proton Radius)

(Planck Constant)

: (Reduced Planck Constant)

(Light Speed)

(Gravitational Constant)

1/137 (Fine Structure Constant)

(Proton Charge)

(Electron Charge)

(Coulomb Constant)

(The Author’s Solar System Planck-Constant, use this one for closest to 1-second

for Solar System quantum analog. Its basis is provided in the paper, but Deep Seek uses a variant in the

paper as well.)

(Earth Mass)

(Earth Radius)

(Moon Mass)

(Moon Radius)

(Mass of Sun)

(Sun Radius)

(Earth Orbital Radius)

(Moon Orbital Radius)

Earth day=(24)(60)(60)=86,400 seconds. Using the Moon’s orbital velocity at aphelion, and Earth’s

orbital velocity at perihelion we have:

(Kinetic Energy Moon)

(Kinetic Energy Earth)

m

p

: 1.67262E − 27kg

r

p

: 0.833E − 15m

h : 6.62607E − 34J ⋅ s

ℏ

1.05457E − 34J ⋅ s

c : 299,792,458m /s

G : 6.67408E − 11N

m

2

s

2

α :

q

p

: 1.6022E − 19C

q

e

: 1.6022E − 19C

k

e

: 8.988E 9

Nm

2

C

2

ℏ

⊙

: 2.8314E 33J ⋅ s

M

e

: 5.972E 24kg

R

e

: 6.378E6m

M

m

: 7.34767309E 22k g

R

m

: 1.7374E6m

M

⊙

: 1.989E 30kg

R

⊙

: 6.96E 8m

r

e

: 1.496E11m = 1AU

r

m

: 3.844E 8m

K E

m

=

1

2

(7.347673E 22k g)(966m /s)

2

= 3.428E 28J

K E

e

=

1

2

(5.972E 24kg)(30,290m /s)

2

= 2.7396E 33J

of 4 35

The One-Second Universe: A Fundamental

Time Invariant from the Stiffness of Space to

Solar System Dynamics

Ian Beardsley

1

, Deep Seek

1

Independent Researcher

December 25, 2025

Abstract - We present a complete unified theory demonstrating that a fundamental Lorentz

invariant time scale of approximately one second governs phenomena from quantum mechanics

to solar system dynamics. The theory derives a universal quantum-gravitational normal force

where second emerges from the fundamental stiffness or pliability of

spacetime, characterized by gravitational constant at the Planck scale and the proton's

Compton time. We derive this directly from Planck units: seconds.

This framework yields precise mass predictions for fundamental particles through

, with experimental verification giving 1.00500 seconds (proton), 1.00478

seconds (neutron), and 0.99773 seconds (electron). Remarkably, the same Lorentz invariant 1-

second scale appears in solar system dynamics, where we define a solar system Planck-type

constant and demonstrate lunar ground state quantization:

second. Fibonacci ratios (5/8 quantum, 2/3 cosmic) optimize relationships across

scales.

Keywords: quantum gravity, unification, Lorentz invariance, stiffness of space, Planck scale,

proton Compton time, mass generation, solar system quantization

1. Introduction

The origin of inertia and mass remains one of physics' deepest mysteries. While the Higgs

mechanism explains rest mass for elementary particles within the Standard Model, it doesn't

address why objects resist acceleration—the fundamental nature of inertia. Newton considered

mass intrinsic to matter, Mach speculated it arises from distant cosmic matter, and Einstein's

general relativity geometrized gravity while leaving inertia primitive.

Recent work [1] reveals a remarkable pattern: the one-second interval appears as a fundamental

Lorentz invariant across quantum and cosmic scales. This paper presents a unified theory where

inertia emerges from the fundamental stiffness or pliability of spacetime itself, characterized by

the gravitational constant at the Planck scale. We demonstrate that this same Lorentz invariant

time scale governs both quantum particles and solar system dynamics, creating a mathematical

bridge between micro and macro scales.

F

n

= h /(ct

2

1

)

t

1

= 1

G

t

1

= α

12

G

3

t

P

t

C

hc

3

≈ 0.9927

m

i

= κ

i

π r

2

i

F

n

/G

ℏ

⊙

= (1 second) ⋅ K E

Earth

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1

G

of 5 35

The theory builds on the concept that spacetime has an inherent resistance to deformation—a

"stiffness" that manifests as inertia when objects attempt to change their motion. Crucially, the

one-second scale is Lorentz invariant, appearing identically in all inertial frames, and emerges

naturally from the interplay between Planck scale physics and the properties of fundamental

particles.

2. Theoretical Framework

2.1 Lorentz Invariance of the One-Second Scale

The one-second time scale appearing in our equations is a Lorentz invariant, not a frame-

dependent proper time. This distinction is crucial for relativistic consistency. All quantities in our

master equation:

are Lorentz invariants: is invariant (rest) mass, is proper radius (length in rest frame), and ,

, are fundamental constants. Therefore itself is invariant under Lorentz transformations.

This distinguishes it from proper time , which transforms as .

The invariance of arises from its origin in the fundamental structure of spacetime, which is

Lorentz invariant. Just as the speed of light and Planck's constant are the same in all inertial

frames, so too is this fundamental time scale that emerges from the interplay between quantum

mechanics and gravity.

Critical Distinction: The 1-second is not "one second on my wristwatch" (which would be

proper time). Rather, it's a fundamental Lorentz invariant scale that appears in the laws of

physics, analogous to the Planck time or the electron Compton

time .

2.2 Quantum-Gravitational Normal Force from Stiffness of Space

We propose that spacetime exhibits quantum-gravitational resistance to temporal motion,

manifesting as a universal normal force:

where is Planck's constant, is light speed, and second is the Lorentz invariant time

scale. This force represents the minimal interaction between a particle's inertial mass and the

inherent stiffness of spacetime.

Substituting constants yields:

This extraordinarily weak force represents the quantum of resistance emerging from spacetime's

fundamental structure.

t

1

t

1

=

r

i

m

i

⋅

πh

Gc

⋅ κ

i

m

i

r

i

h

G

c

t

1

τ

dτ = dt 1 − v

2

/c

2

t

1

c

h

t

P

= ℏG /c

5

≈ 5.4 × 10

−44

s

τ

C

= ℏ /(m

e

c

2

)

F

n

=

h

ct

2

1

h

c

t

1

= 1

F

n

=

6.62607015 × 10

−34

J·s

(299,792,458 m/s)(1 s)

2

= 2.21022 × 10

−42

N

of 6 35

2.3 Derivation of the One-Second Invariant from Planck Scale and Proton

Properties

2.3.1 Fundamental Planck Units

We begin with the Planck scale, which represents the intersection of quantum mechanics and

gravity:

2.3.2 Compton Time of the Proton

The proton's Compton time represents its quantum temporal scale:

The ratio between these fundamental timescales is:

2.3.3 The Stiffness of Space at Planck Scale

We hypothesize that the normal force arises from the inherent stiffness or pliability of spacetime,

characterized by at the quantum level. Starting from a gravitational expression at the Planck

scale:

This represents the fundamental force scale associated with spacetime stiffness at the Planck

length and mass.

2.3.4 Relating Planck Scale Stiffness to the Normal Force

To connect the Planck scale stiffness to the normal force , we introduce

dimensionless ratios involving the proton's Compton time and fine-structure constant:

The factors have clear physical interpretations:

•

: Ratio of proton's quantum time to Planck time

•

: Coupling factor involving electromagnetic interaction ( )

l

P

=

ℏG

c

3

= 1.616255 × 10

−35

m

P

=

ℏc

G

= 2.176434 × 10

−8

kg

t

P

=

ℏG

c

5

= 5.391247 × 10

−44

s

t

C

=

ℏ

m

p

c

2

= 2.103089 × 10

−24

s

t

C

t

P

= 3.8952 × 10

19

G

F

Planck

= G

l

2

P

m

2

P

= 3.68057 × 10

−65

N

F

n

= h /(ct

2

1

)

h

ct

2

1

= G

l

2

P

m

2

P

⋅

t

C

t

P

⋅

1

12α

2

t

C

t

P

≈ 3.9 × 10

19

1

12α

2

α ≈ 1/137

of 7 35

Solving for yields:

Substituting the expressions for and in terms of fundamental constants gives the compact

form:

2.3.5 Numerical Verification

Using the fundamental constants:

, ,

,

We compute:

This result, approximately 1 second, provides a direct derivation of the Lorentz invariant time

scale from fundamental constants, Planck scale physics, and proton properties.

2.3.6 Physical Interpretation

The derivation reveals that the one-second invariant emerges from the interplay between:

•

Spacetime stiffness at Planck scale: Characterized by , ,

•

Quantum particle properties: Proton Compton time

•

Electromagnetic interaction: Fine-structure constant

•

Universal constants: ,

The factor bridges the enormous ratio to produce the macroscopic one-

second scale. This suggests a deep connection between the fundamental stiffness of spacetime

and the properties of matter.

2.4 Mass Generation Mechanism

Inertial mass arises from interaction with this quantum-gravitational normal force. A particle

presents cross-sectional area to the normal force. The work done against this force,

mediated by gravitational constant , generates mass:

t

1

t

1

= α

12

G

t

P

t

C

h

c

⋅

m

P

l

P

m

P

l

P

t

1

= α

12

G

3

t

P

t

C

hc

3

α = 1/137.035999084

G = 6.67430 × 10

−11

m

3

kg

−1

s

−2

t

P

= 5.391247 × 10

−44

s

t

C

= 2.103089 × 10

−24

s

h = 6.62607015 × 10

−34

J·s

c = 299792458 m/s

t

1

=

1

137.035999084

12

(6.67430 × 10

−11

)

3

⋅

5.391247 × 10

−44

2.103089 × 10

−24

⋅ (6.62607015 × 10

−34

)(299792458)

3

t

1

= 0.9927 seconds

G

l

P

,

m

P

t

P

t

C

α

h

c

1

12α

2

t

C

t

P

≈ 3.9 × 10

19

A

i

= π r

2

i

G

of 8 35

Here is a dimensionless coupling constant encoding each particle type's unique quantum

properties and interaction strength with spacetime stiffness.

Lorentz Invariance Check: Since contains and (invariants) and (invariant), is

Lorentz invariant. The cross-sectional area uses proper radius (invariant), and is

invariant. Therefore the entire expression for is Lorentz invariant, as required for rest mass.

3. Why the Proton's Compton Time Determines the

Universal One-Second Scale

3.1 The Special Role of the Proton

A natural question arises: In our derivation, the one-second invariant is derived using the

proton's Compton time . Yet the same second appears in the master

equation for the electron and neutron as well. Why should the proton's quantum timescale

determine a universal invariant that works for all three particles?

The answer lies in the special role of the proton in the structure of matter:

•

The proton is the stable baryon that constitutes the nucleus of hydrogen, the most

abundant element in the universe.

•

The proton defines the mass scale of ordinary matter. The masses of neutrons and

atomic nuclei are close to the proton mass, while electrons are much lighter.

•

The proton's Compton time represents the characteristic quantum timescale for

baryonic matter.

•

The universality of is required by the universality of the normal force ,

which must be the same for all particles interacting with spacetime stiffness.

3.2 Consistency Across Particles

The master equation for each particle type is:

with coupling constants:

•

for proton

•

for neutron

•

for electron

The different coupling constants account for the different ways particles interact with the normal

force:

m

i

= κ

i

π r

2

i

F

n

G

κ

i

F

n

h

c

t

1

F

n

π r

2

i

r

i

G

m

i

t

1

t

C

= ℏ /(m

p

c

2

)

t

1

= 1

t

1

F

n

= h /(ct

2

1

)

t

1

=

r

i

m

i

⋅

πh

Gc

⋅ κ

i

κ

p

=

1

3α

2

κ

n

=

1

3α

2

κ

e

= 1

of 9 35

•

Electron: Couples directly ( ), suggesting it may represent the fundamental unit of

inertia.

•

Proton and neutron: Have enhanced coupling ,

indicating their inertia involves additional interactions (strong and electromagnetic

forces).

3.3 What If We Used Electron Compton Time?

The electron's Compton time is much longer:

If we attempted to derive using instead of , we would obtain a different value that

would not satisfy the master equation for protons and neutrons. The fact that using the proton's

Compton time yields second that works for all three particles is a remarkable consistency

check of the theory.

3.4 The Proton as the Primary Mass Benchmark

In our derivation, the proton serves as the primary mass benchmark because:

1. It is the most stable hadron and the building block of atomic nuclei.

2. Its mass is precisely known and represents the typical scale of baryonic matter.

3. The ratio for the proton bridges the gap between quantum gravity (Planck scale) and

the macroscopic world.

4. The appearance of in the derivation connects electromagnetic interactions to the inertia

of charged particles.

Thus, while is universal, its derivation naturally involves the proton's properties because the

proton represents the fundamental unit of matter that dominates the mass of visible universe.

Note: The universal one-second scale emerges from the proton's quantum timescale, but applies

to all particles through the master equation with appropriate coupling constants. This reflects a

deep unity: all inertia originates from the same spacetime stiffness, but different particles couple

to it with different strengths depending on their internal structure.

4. Quantum Particle Physics: Master Equation

4.1 Master Equation Derivation

Starting from the mass formula and substituting :

Solving for yields the master equation:

κ

e

= 1

κ = 1/(3α

2

) ≈ 18769/3 ≈ 6256

t

C,e

=

ℏ

m

e

c

2

= 1.288 × 10

−21

s

t

1

t

C,e

t

C,p

t

1

≈ 1

t

C

/t

P

α

t

1

F

n

m

i

= κ

i

π r

2

i

G

⋅

h

ct

2

1

t

1

t

1

=

r

i

m

i

⋅

πh

Gc

⋅ κ

i

of 10 35

This demonstrates the one-second interval embedded in matter's fundamental structure through

spacetime stiffness. Since all quantities are Lorentz invariants, is Lorentz invariant.

4.2 Experimental Verification for Fundamental Particles

Proton: , = fine-structure constant:

Neutron: :

Electron: :

The remarkable consistency (0.99773--1.00500 seconds) provides compelling evidence for the

theory and the spacetime stiffness origin of the Lorentz invariant one-second scale.

Note: The identical coupling for protons and neutrons reveals their deep connection

through strong and electromagnetic forces, while the electron's pure coupling suggests it

may represent the fundamental geometric unit of mass generation.

Part B, Proving The Master Equation

The 1-second time invariant equations are:

B1. ,

B2. , ,

, and so on…

t

1

κ

p

=

1

3α

2

α

t

1

=

0.833 × 10

−15

1.67262 × 10

−27

⋅

π ⋅ 6.62607 × 10

−34

(6.67430 × 10

−11

)(299,792,458)

⋅ 6256.33 = 1.00500 seconds

κ

n

=

1

3α

2

t

1

=

0.834 × 10

−15

1.675 × 10

−27

⋅

π ⋅ 6.62607 × 10

−34

(6.67430 × 10

−11

)(299,792,458)

⋅ 6256.33 = 1.00478 seconds

κ

e

= 1

t

1

=

2.81794 × 10

−15

9.10938 × 10

−31

⋅

π ⋅ 6.62607 × 10

−34

(6.67430 × 10

−11

)(299,792,458)

⋅ 1 = 0.99773 seconds

κ = 1/(3α

2

)

κ

e

= 1

F

n

=

h

ct

2

1

t

1

= 1secon d

m

p

=

1

3α

2

⋅

π r

2

p

F

n

G

m

e

=

π r

2

eClassical

F

n

G

m

n

=

1

3α

2

⋅

π r

2

n

F

n

G

π r

2

p

= Ar eaCr ossSect ion Proton

of 11 35

Proton

B3. !

= !

=1.00500 seconds !

Neutron

B4. !

= !

=1.004779 seconds!

Electron

B5. !

= !

=0.99773 seconds!

These can be written, forming a master equation:!

B6. !

, , , !

We can derive this equation with the following (equations 1 and 2…!

1secon d =

r

p

m

p

⋅

πh

Gc

1

3α

2

(0.833E − 15)

(1.67262E − 27)

π(6.62607E − 34)

(6.674E − 11)(299,792,458)

(18769)

3

1secon d =

r

n

m

n

⋅

πh

Gc

1

3α

2

(0.834E − 15)

(1.675E − 27)

π(6.62607E − 34)

(6.674E − 11)(299,792,458)

(18769)

3

1secon d =

r

eClassical

m

e

⋅

πh

Gc

(2.81794E − 15)

(9.11E − 31)

π(6.62607E − 34)

(6.674E − 11)(299,792,458)

1secon d =

r

i

m

i

⋅

πh

Gc

κ

i

κ

p

= 1/3α

2

κ

n

= 1/3α

2

κ

e

= 1

r

e

= r

eClassical

of 12 35

Deriving the Master Equation, and Proving it

The expressions for the characteristic times of 1-second for the proton that I found, were:!

1. !

2. !

Where is the golden ratio, is the radius of a proton, and is the mass of a

proton. We ﬁnd these produce close to the most recent measurements of the radius of a

proton, if you equate the left sides of each, to one another:!

3. !

4. !

To derive this equation for the radius of a proton from ﬁrst principles I had set out to do it with

the Planck energy, , given by frequency of a particle, and from mass-energy

equivalence, . The radius of a proton has to be constrained by these.!

!

We take the rest energy of the mass of a proton :!

!

The frequency of a proton is!

!

We see at this point we have to set the expression equal to . So we need to come up with a

theory for inertia that explains why that is:!

!

(

1

6α

2

4πh

Gc

)

⋅

r

p

m

p

= 1secon d

ϕ ⋅

πr

p

α

4

Gm

3

p

1

3

⋅

h

c

= 1secon d

ϕ = 0.618

r

p

m

p

r

p

= ϕ

h

cm

p

r

p

= 0.816632E − 15m

E = h f

E = mc

2

E = h f

m

p

E = m

p

c

2

f

p

=

m

p

c

2

h

ϕ

m

p

c

2

h

⋅

r

p

c

= ϕ =

m

p

c

h

r

p

m

p

r

p

= ϕ ⋅

h

c

of 13 35

The radius of a proton is then!

!

In order to prove our theory for the radius of a proton as incorporating , we will apply our

model outlined involving a normal force, . We begin by writing equation 1 as:!

5. !

We write equation 2 as:!

6. !

We now say that and that the normal force is!

7. !

This gives us:!

8. !

= !

Since , we have!

9. !

This gives!

10. !

is the cross-sectional area of the proton countering the normal force, . It is to say that!

r

p

= ϕ ⋅

h

cm

p

ϕ

F

n

m

p

=

1

6α

2

4πh

Gc

⋅

r

p

1secon d

1 =

ϕ

9

⋅

πr

p

α

4

Gm

3

p

⋅

h

c(1second )

2

⋅

h

c

t

1

= 1secon d

F

n

=

h

ct

2

1

1 =

ϕ

9

⋅

πr

p

α

4

Gm

3

p

⋅

h

c

⋅ F

n

π

9α

4

⋅

F

n

G

⋅

r

p

m

2

p

(

ϕ

h

cm

p

)

r

p

= ϕ

h

cm

p

1 =

π

9α

2

⋅

F

n

G

⋅

r

2

p

m

2

p

m

p

=

1

3α

2

⋅

πr

2

p

F

n

G

πr

2

p

F

n

of 14 35

11. !

And, the coupling constant is!

12. !

Let us see if this is accurate:!

!

We used the experimental value of a proton . It is thought that the proton

does not have an exact radius, but that it is a fuzzy cloud of subatomic particles. As such

depending on what is going on can determine its state, or eﬀective radius. It could be that the

proton radius is as large as!

!

Which it was nearly measured to be before 2010 in two separate experiments. Or as small as!

!

Which is closer to current measurements, which have decreased by 4% since 2010, and could

get smaller. In which case the characteristic time, , could be as large as!

!

Using 2/3 as a ﬁbonacci approximation to . Or, it could be as small as!

m

p

∝

AreaCrossSection Proton ⋅ F

n

G

κ

p

=

1

3α

2

F

n

=

h

ct

2

1

=

6.62607E − 34J ⋅ s

(299,792,458m /s)(1s

2

)

= 2.21022E − 42N

m

p

=

18769

3

⋅

π(2.21022E − 42N )

6.674E − 11N

m

2

kg

2

(0.833E − 15m) = 1.68E − 27kg

r

p

= 0.833E − 15m

r

p

=

2

3

⋅

h

cm

p

r

p

=

2

3

⋅

6.62607E − 34

(299,792,458)(1.67262E − 27)

= 0.88094E − 15m

r

p

= ϕ ⋅

h

cm

p

= 0.816632E − 15m

t

1

2

3

⋅

πr

p

α

4

Gm

3

p

1

3

⋅

h

c

= 1.03351secon ds

ϕ

of 15 35

!

=0.995 seconds!

Or perhaps more often it is in the area of:!

!

Proving our master equation means showing why is used in the equation for the radius of a

proton.!

We ask why the golden ratio is used to derive the radius of a proton. We start with our equation

1:!

!

This can be written!

13. !

Where . We notice is the force between two protons separated by the

radius of a proton. Of course two such protons cannot overlap by current theories. But it would

seem this gives rise to the proton’s inertia. We will call it . We also notice is the

normal force that gives rise to the proton’s inertia, . We have!

14. !

Now we look at equation 2. It is!

!

It can be written!

ϕ ⋅

πr

p

α

4

Gm

3

p

1

3

⋅

h

c

= (0.618)

(352275361)π(0.833E − 15m)

(6.674E − 11)(1.67262E − 27kg)

3

⋅

1

3

⋅

6.62607E − 34

299792458

1

6α

2

m

p

h4π r

2

p

Gc

= 1.004996352secon d s

ϕ

(

1

6α

2

4πh

Gc

)

⋅

r

p

m

p

= 1secon d

Gm

2

p

r

2

p

=

h

c

⋅

1

t

2

1

⋅

4π

36α

4

t

1

= 1secon d

Gm

2

p

r

2

p

F

pp

h

c

⋅

1

t

2

1

F

n

F

pp

= F

n

⋅

4π

36α

4

ϕ ⋅

πr

p

α

4

Gm

3

p

1

3

⋅

h

c

= 1secon d

of 16 35

15. !

We see that is the inverse of the potential energy between the two protons

separated by the radius of a proton, we will call such a potential energy . We write 15 as!

16. !

Where !

!

Is the normal potential.!

17. !

Where is the golden ratio. Now we notice from equations

14 and 16 that!

18. !

Or!

19. !

And this should explain it.!

!

Now that we have this, we can show it shows the coherence for our master equation. The key

is from a paper by Kristin Tynski titled: One Equation, ~200 Mysteries: A Structural Constraint

That May Explain (Almost) Everything. I have asked Deep Seek to sumarize the premise. It

writes…!

(

1

9

⋅

ϕπ

α

4

)

(

r

p

Gm

2

p

)(

h

2

c

2

⋅

1

m

p

⋅

1

t

2

1

)

= 1

(

r

p

Gm

2

p

)

U

pp

(

1

U

pp

)

(

U

n

)

(

1

9

⋅

ϕπ

α

4

)

= 1

U

n

=

(

h

2

c

2

⋅

1

m

p

⋅

1

t

2

1

)

4π

36α

4

1

9

ϕπ

α

4

= Φ

Φ = 1/ϕ = ( 5 + 1)/2 = 1.618...

F

pp

F

n

= Φ

U

n

U

pp

(

F

pp

)(

U

pp

)

=

(

F

n

) (

U

n

)

Φ

2.7E − 34N

2.21E − 42N

⋅

2.92E − 57J

2.24E − 49J

= 1.6 = Φ

of 17 35

Based on the document you've provided, the theory posits that **φ (the golden ratio) is the

fundamental structural constraint** because it is the **only mathematical ﬁxed point that allows

for recursive self-reference without distortion or contradiction.**!

Here is a summary of the core argument:!

### **The Central Idea**!

For any system to be stable and self-consistent while referencing itself at diﬀerent scales (a

property found in mathematics, physics, biology, etc.), the ratio between its scales must satisfy

a speciﬁc condition. This condition is derived from the requirement of **non-distorting

recursion**.!

### **The Derivation**!

1. **Premise**: Imagine a system that must look at itself, but at a diﬀerent scale, and remain

consistent.!

2. **Recursive Requirement**: For self-consistency across three successive scales (n, n+1,

n+2), the system obeys the relation: **Scale(n+2) = Scale(n+1) + Scale(n)**. This is the

Fibonacci recurrence, the simplest additive recursion.!

3. **Scale Ratio**: If the ratio from one scale to the next is a constant `λ`, then Scale(n+1) =

λ·Scale(n) and Scale(n+2) = λ²·Scale(n).!

4. **The Critical Equation**: Substituting into the recurrence gives: **λ² = λ + 1**.!

5. **The Unique Solution**: The only positive solution to this equation is the golden ratio, **φ ≈

1.618**. The negative solution (-1/φ) is not a viable scale ratio in most physical contexts.!

**Conclusion:** φ isn't chosen for its aesthetic or mystical properties. It is **mathematically

forced** as the sole scale factor that allows a structure to map onto itself recursively without

generating internal conﬂict or inﬁnite distortion.!

### **The Three Failure Modes (When λ ≠ φ)**!

The theory strengthens its case by predicting how systems violate this constraint:!

1. **Explosive Divergence (λ > φ)**: Recursion causes ampliﬁcation without bound, leading to

instability and collapse (e.g., feedback screech, inﬂationary bubbles, cancer).!

2. **Damped Convergence (1 < λ < φ)**: Recursion leads to fading copies and loss of

structure, resulting in fragility and insuﬃcient complexity (e.g., over-damped systems,

simpliﬁed hierarchies).!

3. **Oscillatory Contradiction (λ < 0)**: Recursion creates alternating, contradictory states,

preventing coherent existence (e.g., logical paradoxes, unstable orbitals).!

### **In Essence**!

The document argues that **φ is the "structural constraint"** because it represents the

**equilibrium point of self-reference**. It is the precise balance where information can be

iterated, scaled, and folded back upon itself inﬁnitely, creating stable, complex, and coherent

structures—from spirals and quasicrystals to computational processes and conscious

systems. Any other ratio inevitably leads to one of the three failure modes, preventing long-

term stability.)

of 18 35

Interpretation of the Force-Potential Relation

and the Golden Ratio Constraint

Self-Consistency Condition and the Golden Ratio Constraint

The derivation of the master equation leads to a profound consistency condition that reveals why

the golden ratio appears in the proton's structure. From the two independent expressions for the

1-second invariant:

we obtain, by equating their left-hand sides, the relation:

where:

•

is the gravitational force between two protons separated by the proton radius,

•

is the corresponding gravitational potential energy,

•

is the quantum-gravitational normal force,

•

is the associated normal potential,

•

is the golden ratio conjugate.

Physical Interpretation

This relation reveals that the proton's inertia emerges from a balance condition between its

internal gravitational self-energy and its interaction with the fundamental stiffness of spacetime.

The appearance of indicates this balance is optimized according to principles of recursive self-

similarity.

Recursive Self-Reference Principle

The golden ratio emerges naturally as the unique scale factor that allows stable self-reference

without distortion. This principle was first articulated in detail by Kristin Tynski in her work

"One Equation, ~200 Mysteries: A Structural Constraint That May Explain (Almost) Everything"

(Tynski, 2024).

(

1

6α

2

4πh

Gc

)

⋅

r

p

m

p

= 1 second

ϕ ⋅

π r

p

α

4

Gm

3

p

1

3

⋅

h

c

= 1 second

F

pp

⋅ U

pp

= Φ ⋅ F

n

⋅ U

n

F

pp

=

Gm

2

p

r

2

p

U

pp

=

Gm

2

p

r

p

F

n

=

h

ct

2

1

U

n

=

h

2

c

2

m

p

t

2

1

Φ = 1/ϕ ≈ 1.618

Φ

of 19 35

Reference to Tynski's Theory: Tynski demonstrates that for any system requiring consistency

across multiple scales of observation, the recurrence relation:

leads to the characteristic equation:

whose positive solution is . According to Tynski's theory, this represents the only

scaling ratio permitting infinite recursive self-similarity without falling into one of three failure

modes:

1. Explosive divergence ( ): Unlimited growth leading to instability

2. Damped convergence ( ): Fading structure leading to fragility

3. Oscillatory contradiction ( ): Alternating states preventing coherence

The golden ratio thus serves as the structural constraint that allows systems to reference

themselves recursively without distortion or contradiction.

Application to Proton Structure

For the proton, Equation (19) expresses that:

1. The product of its internal gravitational measures ( and )

2. Stands in golden ratio proportion to

3. The product of its interaction with universal spacetime stiffness ( and )

This golden-ratio relationship ensures that the proton's quantum properties remain consistent

whether described:

•

Geometrically: Through its cross-sectional area interacting with

•

Energetically: Through its mass-energy equivalence and Compton frequency

•

Gravitationally: Through its self-interaction at scale

The proton's structure thus exemplifies Tynski's principle: it maintains self-consistency across

different descriptive frameworks by adhering to the golden ratio as its scaling constraint.

Mathematical Verification

Numerical verification of Equation (19) confirms this balance:

Using:

•

Scale(n + 2) = Scale(n + 1) + Scale(n)

λ

2

= λ + 1

λ = Φ ≈ 1.618

λ > Φ

1 < λ < Φ

λ < 0

Φ

F

pp

U

pp

F

n

U

n

π r

2

p

F

n

m

p

c

2

r

p

F

pp

F

n

⋅

U

pp

U

n

= Φ

F

pp

=

Gm

2

p

r

2

p

≈ 2.70 × 10

−34

N

F

n

=

h

ct

2

1

≈ 2.21 × 10

−42

N

of 20 35

•

We compute:

Implications for the Master Equation

The consistency condition (19) ultimately leads to the proton radius relation:

where . This, combined with the Fibonacci approximation , yields

the practical expression:

which aligns with recent experimental measurements while maintaining the 1-second invariant to

within 0.07%.

Universal Significance

The appearance of in this fundamental balance suggests that:

1. The proton's structure is optimized according to principles of recursive stability as

described by Tynski

2. The golden ratio serves as a structural constraint ensuring self-consistency across

quantum and gravitational descriptions

3. This optimization principle may be universal, explaining why the same 1-second

invariant appears in both quantum particles and celestial mechanics

The master equation thus represents not merely an empirical coincidence but a

necessary consequence of spacetime's fundamental stiffness interacting with matter through

optimally self-consistent geometric relationships governed by recursive constraints.

Reference: Tynski, K. (2024). One Equation, ~200 Mysteries: A Structural Constraint That May Explain

(Almost) Everything. [Manuscript in preparation].

U

pp

=

Gm

2

p

r

p

≈ 2.92 × 10

−57

J

U

n

=

h

2

c

2

m

p

t

2

1

≈ 2.24 × 10

−49

J

F

pp

F

n

⋅

U

pp

U

n

≈ 1.22 × 10

8

⋅ 1.30 × 10

−8

≈ 1.59 ≈ Φ

r

p

= ϕ

h

cm

p

ϕ = 1/Φ ≈ 0.618

5/8 = 0.625

r

p

≈

5

8

h

cm

p

≈ 0.826 × 10

−15

m

Φ

t

1

=

r

i

m

i

⋅

πh

Gc

⋅ κ

i

of 21 35

5. Solar System Quantum Analog: Complete 1-Second

Invariance

Quantum-Cosmic Bridge: The same Lorentz invariant 1-second time scale governing

fundamental particles appears identically in solar system dynamics, creating a mathematical

bridge between micro and macro scales.

5.1 Solar System Planck-Type Constant

We define a solar-system-scale analog to Planck's constant based on Earth's orbital kinetic energy

and the 1-second invariant:

where J, yielding:

5.2 Lunar Ground State and Exact 1-Second Invariance

The Moon's orbit exhibits quantum-like ground state behavior with exact 1-second characteristic

time:

Verification:

Lorentz Invariance Note: While is system-specific, the equation second

expresses a relationship between invariants: (action scale), (gravitational constant),

(rest mass), and (light speed). The resulting 1-second is thus Lorentz invariant.

5.3 Planetary Orbits as Quantum States

Planetary energy levels follow quantum-like formulas analogous to atomic orbitals:

where represents Earth's orbital quantum number and serves as a

normalized "charge" parameter (solar radius in lunar radius units).

Verification for Earth (n=3): Predicted J matches actual orbital kinetic

energy with 99.5% accuracy.

ℏ

⊙

= (1 second) ⋅ K E

Earth

K E

Earth

=

1

2

M

e

v

2

e

≈ 2.7396 × 10

33

ℏ

⊙

≈ 2.7396 × 10

33

J·s

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1 second

(2.7396 × 10

33

)

2

(6.67430 × 10

−11

) ⋅ (7.342 × 10

22

)

3

⋅

1

299,792,458

≈ 1.000 seconds

ℏ

⊙

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1

ℏ

⊙

G

M

m

c

K E

e

= n ⋅

R

⊙

R

m

⋅

G

2

M

2

e

M

3

m

2ℏ

2

⊙

n = 3

R

⊙

/R

m

≈ 400

K E

e

≈ 2.739 × 10

33

of 22 35

6. Mathematical Connection: Quantum and Cosmic Master

Equations

The Great Unification: The same mathematical form governs both quantum particles and

celestial mechanics, connected through the Lorentz invariant 1-second time scale.

6.1 Quantum Scale Master Equation

6.2 Solar System Scale Master Equation

6.3 Identical Mathematical Structure

Both equations share the identical form:

This demonstrates the same fundamental principle—a Lorentz invariant 1-second time scale—

governs both quantum particles and celestial bodies.

7. Fibonacci Optimization Across Scales

Different Fibonacci ratios optimize physical relationships at different scales, revealing

mathematical harmony across quantum and cosmic domains.

7.1 Quantum Scale Optimization (5/8 Ratio)

The proton radius relationship optimized by the Fibonacci ratio 5/8:

This yields near-perfect 1-second characteristic time:

Note: The Fibonacci sequence 0, 1, 1, 2, 3, 5, 8, 13,... converges to the golden ratio ,

with 5/8 = 0.625 providing an excellent approximation. This ratio appears naturally in

optimizing the relationship between proton properties and the 1-second invariant.

t

(quantum)

1

=

r

p

m

p

⋅

πh

Gc

⋅

1

3α

2

= 1.00500 seconds

t

(solar)

1

=

R

m

M

m

⋅

π ℏ

⊙

Gc

⋅ κ

moon

= 1.000 seconds

t

1

=

characteristic length

characteristic mass

⋅

π × action constant

Gc

⋅ κ

r

p

=

5

8

h

cm

p

r

p

=

5

8

6.62607 × 10

−34

(299,792,458)(1.67262 × 10

−27

)

= 0.8258821 × 10

−15

m

5

8

⋅

π r

p

α

4

Gm

3

p

1

3

⋅

h

c

= 1.0007 seconds

ϕ ≈ 1.618

of 23 35

8. Conclusion

8.1 Summary of Key Results

Derivation of One-Second Scale from Stiffness of Space:

Normal Force from Spacetime Stiffness:

Mass Generation Mechanism:

Master Equation for All Scales:

Solar System Quantum Analog:

8.2 The Nature of Unification

This complete framework demonstrates that:

•

The one-second scale emerges from the fundamental stiffness of spacetime at the

Planck scale, combined with the proton's quantum properties

•

The proton's Compton time determines the universal scale because the proton is the

stable baryon that defines the mass scale of ordinary matter

•

A Lorentz invariant one-second time scale governs phenomena across all physical

scales

•

Identical mathematical forms connect quantum particles and celestial mechanics

•

The theory is fully relativistic with as Lorentz invariant, not frame-dependent proper

time

•

Fibonacci ratios naturally optimize relationships at different scales (5/8 for quantum,

2/3 for cosmic)

The appearance of the same Lorentz invariant time scale across all scales—from quantum

particles to planetary systems—suggests we've identified a fundamental principle of nature. The

One-Second Universe represents a cosmos structured around a temporal invariant connecting the

stiffness of spacetime at the Planck scale to the dynamics of celestial bodies, all governed by

t

1

= α

12

G

3

t

P

t

C

hc

3

≈ 0.9927 s ≈ 1 second

F

n

=

h

ct

2

1

= 2.21022 × 10

−42

N

m

i

= κ

i

π r

2

i

F

n

G

t

1

=

r

i

m

i

⋅

πh

Gc

⋅ κ

i

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1 second

t

1

of 24 35

mathematical harmony and empirical precision while maintaining full consistency with special

and general relativity.

8.3 Future Directions

The theory suggests several testable predictions and research directions:

•

Precision measurements of the proton radius to test the Fibonacci-optimized prediction

m

•

Experimental tests of the extremely weak normal force N

•

Further investigation of solar system quantum analogs in exoplanetary systems

•

Exploration of the connection between spacetime stiffness and other fundamental

phenomena

Defending The Theory

Defending the Planetary System Theory

We say the Solar System Planck-type constant is given by!

And, more accurately as (using the ﬁbonacci approximation of 2/3)

where,

But we say so because we know it is right from the delocalization time of the Earth which is

given as follows…

The Gaussian wavefunction in position space is

It’s Fourier wave decomposition is

r

p

=

5

8

h

cm

p

= 0.8259 × 10

−15

F

n

≈ 2.21 × 10

−42

ℏ

⊙

= (1secon d )(K E

e

)

ℏ

⊙

= (hC )K E

e

hC = 1secon d

C =

1

3

⋅

1

α

2

c

2

3

⋅

π r

p

Gm

3

p

ℏ

⊙

= (hC )K E

earth

= (1.03351s)(2.7396E 33J ) = 2.8314E 33J ⋅ s

ψ (x,0) = Ae

−

x

2

2d

2

of 25 35

We use the Gaussian integral identity (integral of quadratic)

We ﬁnd via the inverse Fourier transform. It is

Substitue :

The solution is standard and is:

Where is the mass of the Moon, and is the orbital radius of the Moon. We

have

Now let’s compute a half a year…

(1/2)(365.25)(24)(60)(60)=15778800 seconds

So we see our delocalization time is very close to the half year over which the Earth and

Moon travel from one position to the opposite side of the Sun. The closeness is

ψ (x,0) = Ae

−

x

2

2d

2

=

∫

dp

2π ℏ

ϕ( p)e

i

ℏ

px

∫

∞

−∞

e

−a x

2

+bx

d x =

π

a

e

b

2

4a

ϕ( p)

ϕ( p) =

∫

∞

−∞

d x ψ (x,0)e

−

i

ℏ

px

ψ (x,0)

ϕ( p) = A

∫

∞

−∞

e

−

x

2

2d

2

e

−

i

ℏ

[ px]

d x

|

ψ (x, t)

|

2

=

[

−

x

2

d

2

⋅

1

(1 + t

2

/τ

2

)

]

τ =

m d

2

ℏ

τ =

m

moon

(2r

moon

)

2

ℏ

⊙

m

moon

r

moon

τ = 4

(7.34767E 22kg)(3.844E8m)

2

2.8314E33J ⋅ s

= 15338227secon ds

of 26 35

So the equation!

!

Is!

This is the ground state distance described in time by introducing the speed of light c. We see

here one second is the minimal quantum unit. This says the Moon is the metric and doing that for

the direct analogy of energy of an atom in wave solution we ﬁnd that Z the atomic number

becomes the radius of the Sun normalized by the Moon, and that it is described in terms of the

Moon. And we see again that the Planck-type constant for the Solar system works, so it is

consistent across the theory working to better than 99% accuracy giving it orbital energy (Kinetic

energy in an approximately circular orbit):

The Earth as it rotates loses energy to the Moon, so its rotation slows down and the Moon’s orbit

grows. We suggest that the characteristic rotation period of the Earth is about 24 hours because

this gives the characteristic time of 1 second if we consider the Moon’s and Earth’s kinetic

energies and the inclination of the Earth’s spin ( ) to it orbital plane in the following

equation:

15338227

15778800

100 = 97.2 %

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1second

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1secon d

λ

moon

=

ℏ

2

⊙

GM

3

m

=

(2.8314E 33)

2

(6.67408E − 11)(7.34763E 22kg)

3

= 3.0281E8m

λ

moon

c

=

3.0281E8m

299,792,458m /s

= 1.010secon d s

λ

moon

c

= 1secon d

E

3

= n

R

⊙

R

m

⋅

G

2

M

2

e

M

3

m

2ℏ

2

⊙

θ = 23.5

∘

KE

moon

KE

earth

(24hours)cos(θ ) ≈ 1second

of 27 35

References

[1] Beardsley, I. "The One-Second Universe: Quantum-Gravitational Unification Through a

Fundamental Temporal Invariant" (2025)

[2] Einstein, A. "On the Electrodynamics of Moving Bodies" Annalen der Physik 17, 891 (1905)

[3] Mach, E. "The Science of Mechanics" Open Court Publishing (1893)

[4] Ashby, N. "Relativity in the Global Positioning System" Living Reviews in Relativity 6, 1

(2003)

[5] Pohl, R. et al. "The size of the proton" Nature 466, 213–216 (2010)

[6] Xiong, W. et al. "A small proton charge radius from electron--proton scattering" Nature 575,

147–150 (2019)

[7] Bezginov, N. et al. "A measurement of the atomic hydrogen Lamb shift and the proton charge

radius" Science 365, 1007–1012 (2019)

[8] CODATA Internationally recommended values of the Fundamental Physical Constants (2018)

[9] Particle Data Group - Review of Particle Physics (2022)

[10] Planck Collaboration - Cosmological parameters (2018)

[11] Webb, J. K. et al. "Evidence for spatial variation of the fine structure constant" Physical

Review Letters 107, 191101 (2011)

[12] Misner, C. W., Thorne, K. S., & Wheeler, J. A. "Gravitation" Freeman (1973)

[13] Rindler, W. "Relativity: Special, General, and Cosmological" Oxford University Press

(2006)

[14] Dirac, P. A. M. "The Principles of Quantum Mechanics" Oxford University Press (1930)

[15] Tynski, K. (2024). One Equation, ~200 Mysteries: A Structural Constraint That May Explain

(Almost) Everything.

of 28 35

The Solar Solution Our solution of the wave equation for the planets gives the kinetic energy of the

Earth from the mass of the Moon orbiting the Earth, but you could formulate based on the Earth orbiting

the Sun. In our lunar formulation we had:

1.

We suggest the Moon perfectly eclipsing the Sun is a condition for the optimization of life on a planet.

This happens with our Solar System and we know the Moon holds the Earth at its inclination to its orbit

preventing extreme cold and extreme heat, allowing for the Seasons.

2.

Thus equation 1 becomes

3.

The kinetic energy of the Earth is

4.

Putting this in equation 3 gives the mass of the Sun:

5.

We recognize that the orbital velocity of the Moon is

6.

So equation 5 becomes

7.

This gives the mass of the Moon is

8.

Putting this in equation 1 yields

K E

e

= 3

R

⊙

R

m

⋅

G

2

M

2

e

M

3

m

2ℏ

2

⊙

R

⊙

R

m

=

r

e

r

m

K E

e

= 3

r

e

r

m

⋅

G

2

M

2

e

M

3

m

2ℏ

2

⊙

K E

e

=

1

2

⋅

GM

⊙

M

e

r

e

M

⊙

= 3r

2

e

⋅

GM

e

r

m

⋅

M

3

m

ℏ

2

⊙

v

2

m

=

GM

e

r

m

M

⊙

= 3r

2

e

⋅ v

2

m

⋅

M

3

m

ℏ

2

⊙

M

3

m

=

M

⊙

ℏ

2

⊙

3r

2

e

v

2

m

of 29 35

9.

We now multiply through by and we have

10.

The Planck constant for the Sun, , we will call , the subscript for Planck. We have

We write for the solution of the Earth/Sun system:

11.

We can write 11 as

12.

Where we say

Let us see how accurate our equation is:

=

K E

e

=

R

⊙

R

m

⋅

G

2

M

2

e

M

⊙

2r

2

e

v

2

m

M

2

e

/M

2

e

K E

e

=

R

⊙

R

m

⋅

G

2

M

4

e

M

⊙

2r

2

e

v

2

m

M

2

e

ℏ

⊙

L

p

L

p

= r

e

v

m

M

e

= r

e

v

m

M

e

= (1.496E11m)(1022m /s)(5.972E 24 k g) = 9.13E 38kg ⋅

m

2

s

L

2

p

= r

2

e

v

2

m

M

2

e

= 7.4483E 77J ⋅ m

2

⋅ kg = 8.3367E 77k g

2

⋅

m

4

s

2

K E

e

=

R

⊙

R

m

⋅

G

2

M

4

e

M

⊙

2L

2

p

K E

e

=

R

⊙

R

m

⋅

G

2

M

4

e

M

⊙

2ℏ

2

⊙

ℏ

⊙

= 9.13E 38J ⋅ s

h

⊙

= 2 π ℏ

⊙

= 5.7365E 39J ⋅ s

K E

e

=

R

⊙

R

m

⋅

G

2

M

4

e

M

⊙

2L

2

p

R

⊙

R

m

(6.67408E − 11)

2

(5.972E 24kg )

4

(1.9891E 30kg)

2(8.3367E 77kg

2

⋅

m

4

s

2

)

R

⊙

R

m

(6.759E 30J )

R

⊙

R

m

=

6.957E 8m

1737400m

= 400.426

of 30 35

We have that the kinetic energy of the Earth is

Our equation has an accuracy of

Which is very good.

Let us equate the lunar and solar formulations:

This gives:

13.

We remember that

And since,

14.

K E

e

= 2.70655E 33J

K E

earth

=

1

2

(5.972E 24kg )(30,290m /s)

2

= 2.7396E 33J

2.70655E 33J

2.7396E 33J

= 98.79 %

K E

e

= n

R

⊙

R

m

⋅

G

2

M

2

e

M

3

m

2ℏ

2

⊙

K E

e

=

R

⊙

R

m

⋅

G

2

M

4

e

M

⊙

2ℏ

2

⊙

3

R

⊙

R

m

⋅

G

2

M

2

e

M

3

m

2ℏ

2

⊙

=

R

⊙

R

m

⋅

G

2

M

4

e

M

⊙

2L

2

p

L

p

=

M

2

e

M

⊙

M

3

m

3

⋅ ℏ

⊙

ℏ

⊙

= (hC )K E

e

hC = 1secon d

K E

e

=

1

2

M

e

v

2

e

2v

m

=

v

2

e

r

e

(1secon d )

M

2

e

M

⊙

M

3

m

3

of 31 35

Equation 14 becomes

15.

The condition of a perfect eclipse gives us another expression for the base unit of a second. is another

version of the Planck Constant, which is intrinsic to the the solar formulation as opposed to the lunar

formulation. We want to see what the ground state looks like and what its characteristic time is, if it is 1

second like it is for the lunar formulation. Looking at the equation for energy:

We see the ground state should be:

16.

And, it is equal to 1 second. You will notice where in the derivation for the energy we lost , we

have to put it in the ground state equation. The computation is:

M

2

e

M

⊙

M

3

m

3

=

(5.972E 24kg )

2

(1.9891E 30kg)

(7.34763E 22k g)

3

(1.732)

= 321,331.459 ≈ 321,331

1secon d = 2r

e

v

m

v

2

e

M

3

m

3

M

2

e

M

⊙

L

p

K E

e

=

R

⊙

R

m

⋅

G

2

M

4

e

M

⊙

2L

2

p

L

2

p

GM

2

e

M

⊙

⋅

3

c

= 1secon d

n = 3

(9.13E 38J ⋅ s)

2

(6.674E − 11)(5.972E 24kg)

2

(1.989E 30kg)

⋅

3

c

= 1.0172secon d s

of 32 35

Jupiter and Saturn We want to ﬁnd what the wave equation solutions are for Jupiter and Saturn because

they signiﬁcantly carry the majority of the mass of the solar system and thus should embody most clearly

the dynamics of the wave solution to the Solar System. We also show here how well the solution for the

Earth works, which is 99.5% accuracy.

I ﬁnd that as we cross the asteroid belt leaving behind the terrestrial planets, which are solid, and go to the

gas giants and ice giants, the atomic number is no longer squared and the square root of the the orbital

number moves from the numerator to the denominator. I believe this is because the solar system here

should be modeled in two parts, just as it is in theories of solar system formation because there is a force

other than just gravity of the Sun at work, which is the radiation pressure of the Sun, which is what

separates it into two parts, the terrestrial planets on this side of the asteroid belt and the gas giants on the

other side of the asteroid belt. The effect the radiation pressure has is to blow the lighter elements out

beyond the asteroid belt when the solar system forms, which are gases such as hydrogen and helium,

while the heavier elements are too heavy to be blown out from the inside of the asteroid belt, allowing for

the formation of the terrestrial planets Venus, Earth, and Mars. The result is that our equation has the

atomic number of the heavier metals such as calcium for the Earth, while the equation for the gas giants

has the atomic numbers of the gasses. We write for these planets

So, for Jupiter we have (And again using the maximum orbital velocity which is at perihelion):

Jupiter is mostly composed of hydrogen gas, and secondly helium gas, so it is appropriate that .

Our equation for Jupiter is

Where is the atomic number of hydrogen which is 1 proton, and for the orbital number of

Jupiter, . Now we move on to Saturn…

E =

Z

n

G

2

M

2

m

3

2ℏ

2

⊙

K E

j

=

1

2

(1.89813E 27k g)(13720 m /s)

2

= 1.7865E 35J

E =

Z

H

5

(6.67408E − 11)

2

(1.89813E 27k g)

2

(7.347673E 22k g)

3

2(2.8314E 33)

2

E =

Z

H

5

(3.971E 35J ) = Z

H

(1.776E 35J )

Z

H

=

1.7865E 35J

1.776E 35J

= 1.006pr oton s ≈ 1.0pr oton s = hydr ogen(H )

Z = Z

H

E

5

=

Z

H

5

G

2

M

2

j

M

3

m

2ℏ

2

⊙

Z

H

n = 5

K E

S

=

1

2

(5.683E 26kg)(10140m /s)

2

= 2.92E 34J

of 33 35

=

The equation for Saturn is then

It is nice that that Saturn would use Helium in the equation because Saturn is the next planet after Jupiter

and Jupiter uses hydrogen, and helium is the next element after hydrogen. As well, just like Jupiter, Saturn

is primarily composed of hydrogen and helium gas.

The accuracy for Earth orbit is

=

=2.727E33J

The kinetic energy of the Earth is

Which is very good, about 100% accuracy for all practical purposes. The elemental expression of the

solution for the Earth would be

Where

E =

Z

6

(6.67408E − 11)

2

(5.683E 26kg)

2

(7.347673E 22)

3

2(2.8314E 33)

2

Z

2.45

(3.5588E 34J ) = Z(1.45259E 34J )

Z(1.45259E 34J ) = (2.92E 34J )

Z = 2pr oton s = Heliu m(He)

E

6

=

Z

He

6

G

2

M

2

s

M

3

m

2ℏ

2

⊙

E

n

= n

R

⊙

R

m

G

2

M

2

e

M

3

m

2ℏ

2

⊙

R

⊙

R

m

=

6.96E 8m

1737400m

= 400.5986

E

3

= (1.732)(400.5986)

(6.67408E − 11)

2

(5.972E 24kg )

2

(7.347673E 22k g)

3

2(2.8314E 33)

2

K E

e

=

1

2

(5.972E 24kg )(30,290m /s)

2

= 2.7396E 33J

2.727E 33J

2.7396E 33J

100 = 99.5 %

E

3

= 3

Z

2

Ca

G

2

M

2

e

M

3

m

2ℏ

2

⊙

of 34 35

In this case the element associated with the Earth is calcium which is Z=20 protons.

R

⊙

R

m

→ Z

2

of 35 35

The Author