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Atomic spectra calculations from Newton's equation

Abstract: Atomic spectra are not a Proof of the existence of an atomic shellstructure but a measure of Earth's spin. Atomic spectra wavelength is an

inversion of Earth's radius. Standard refractive index measurements are

taken at yellow Sodium D line, with a wave length of 598 nanometers

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The proof:

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This line or the atomic spectrum line used as a proof of the existence of an

atomic shell structure with a nucleus and electrons surrounding is the

thought of modern and Nobel not apes but dumb apes.

Sodium flame and this line as its base

Newton's law is:

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G = 6.6729 x 10-11

[1] Measurement techniques volume 39, No.10, 1996

Measurement of the gravitational constant using torsion balance

O.V. Karagioz; V.P.Izmailov; UDC 528.27; G = (6.6729 +/- 0.0005) x 10-11 

G has a dimensional unit [G] = [1/ρ, 1/T2]

G = 1/ (8π /15) ρ ε a Te2 = 6.3729882 x 10-11

Or G = 1/ (8π /15) ρ n2 a Te2

And ε a = 1.00058986 and n a2 = 1.0002932 = 1.000586

The all time experimental error in this equation is (8π /15)

The error is a multiplication of two errors: (4π /3) (2/5)

1- Spherical volume: (4π /3) r3

2 - Spherical moment of inertia: (2/5) m r2

Using inverse square Newton's law and not inverse cube allows a distance

and when distance error is made it is Earth's radius because we are not

measuring from Earth's center but we are measuring from Earth's surface.

(8 π / 15) r is the error in measurement location: The wavelength of the

atomic spectrum is λ = 2 π k; k = wave length number or in practice it is the

error k = (8 π / 15) r; λ = 2 π/ (8 π / 15) r = 1/ (4/ 15) r = 15/4 r

And r = Earth's radius = 6371000 meters

And λ= 15/4x 6371000 = 0.0000005886 meters

The error is 0.0000000001and that is excellent because two readings by two

different persons can differ by 0.0000000001

= 588.6 x 10-9 meters = 588.6 nanometers; the second one is due to

different location of the slit. Light spectra from space are measured wrong.

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First Proof : Real time astronomy solution of Earth's spin - effect of axial tilt illusion

In July 21, 1969 at age 11 and after graduating from 5th grade and on the same day a manlanded a foot on the moon I watched Apollo 11 take off and disappear into the skies on its way tothe moon and then saw Neil Armstrong land human's first step on the moon. I wondered how someone sees and measures distances in space and how someone sees and measures sizes of objects in space. Apollo 11 rocket looked like it is shrinking in size while moving up into the skies;not as if the rocket shrunk in size but as if the rocket visual changed in size indication a differentlocation and not a different size. Apollo 11 looked similar to a moving car moving away andshrinking in size. I realized that objects location and size has to do with how we see things (eye asan instrument). I imagined two snap shot of Apollo 11 at different distances A and B. At snap shotdistance A Apollo 11 looked like it has a size C, and at snaps shot distance B Apollo 11 looked like

it has a shrunk size D as shown below:

C

D

Eye ------------------------------ A ------------------------------ B

I asked myself the question: how A, B, C, and D are related?

1 = 1 is self evident; 2 = 2 is self evident; A = A is self evident

If A = A; add and subtract B; then A = B + (A - B)

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Divide by B; then (A/B) = 1 + (A - B)/B

Multiply by C; then (A/B) C = C + [(A - B)/B] C Equation - 1

 Also D = D; add and subtract C; then D = C + (D - C) Equation - 2

Comparing Equation - 1 and Equation - 2

(A/B) C = D; or, AC = BD

C = C

D - C = [(A - B)/B]

The answer is AC = BD = is how distances A and B related to sizes C and D

 And D - C = [(A - B)/B] C = visual contraction of C when moved from A to B

 AC = BD = actual distance x actual size = visual distance x visual size = constant

The initial condition solution to AC = BD is: A = B and C = D

Nature's measurement law is: (A/B) C = C + [(A - B)/B] C Equation - 1

In motion = actual + visual effects

Progressive = static + visual effects

(3/2) mc2 = (1/2) mc2 + visual illusion {[(3/2) - (1/2)] mc2 = mc2}

The Scientific method said:

If theory matches experiment, then science is correct.

If theory = measured

In motion = actual + measured data

In motion = actual + science theory 

Science theory ≠ actual

D - C = Modern and Nobel science = relativistic

= visual Illusion called modern and Nobel science!

Conclusion: If science theory matches science experiment then science is all wrong

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 AC = BD = constant = actual distance x actual size

 AC = BD = k; Taking BD = k 

Differentiating with respect to time

Then d B/ d t + d D/ d t = d k/ d t = 0

 And d B/ d t = - d D/ d t = λ + í ω; method of separation of variables

B = B0 e (λ + i ω t) = A e (λ + i ω t)

D = D0 e - (λ + i ω t) = C e (λ + i ω t)

 A real number C has a visual complex number D = C e (λ + i ω t)

Or Natural Transform: e (λ + i ω t) = e  λ e i ω t = translation + rotation

Translation = e  λ ; rotation = e i ω t

 All there is in the Universe is objects of mass m at a location r. The state of any objects in the Universe S = m r.

Take S = m r and make the first measurement S0. A measurement is freezing or a snapshot and S0 is a number and equal to a constant k. S0 = m r = k 

 And m r = k = constant, solve the equation using separation of variables method

Take the derivative

 And d (m r)/d t = d k/d t = 0;

Or, d (m r)/d t = d k/d t = 0

 And (d m/ d t) r + m (d r/ d t) = 0

Or (d m/ d t) r = - m (d r/ d t)

Separating r from m is like dividing with m r

 And (1/m) (d m/ d t) = - (1/r) (d r/ d t) = (+/-) (λ + í ω)

Then, m = m0 e (λ + í ω) t and r = r0 e - (λ + í ω) t ------------------------------------- I

Or, m = m0 e - (λ + í ω) t and r = r0 e (λ + í ω) t ------------------------------------- II

Note: All data in this article is from NASA or CODATA 

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Distance forward equation in real time is r = r0 e - (λ + í ω) t

Distance backward equation in real time is r = r0 e (λ + í ω) t

S = m r = m0 e (λ + í ω) t r0 e - (λ + í ω) t = m0 r0 = m1 r1= m3 r4 = m4 r4 = Etc

(Cosine ω t) + í sine ω t) e λ = e (λ + í ω) t

(Cosine ω t) - í sine ω t) e - λ = e - (λ + í ω) t

(Cosine ω t) + í sine ω t) e λ (Cosine ω t) - í sine ω t) e - λ = e (λ + í ω) t e - (λ + í ω) t = 1

Then, m = m0 e (λ + í ω) t and r = r0 e - (λ + í ω) t ------------------------------------- I

Or, m = m0 e - (λ + í ω) t and r = r0 e (λ + í ω) t ------------------------------------- II

Equation I mean:

Earth's spin e í ω t will produce equal and opposite visual tilt e - í ω t 

Equation II mean:

Earth's spin e - í ω t will produce equal and opposite visual tilt e í ω t 

1 - Earth's spin deviate data by 2 π; modern and Nobel uses an inverse square law and the data deviates by 1/√ (2 π) and vertical tilt will produce a celestial sphere tiltangle sine -1 [1/√ (2 π)]; using Lab clock of 24 hours = Ts = 48600 second instead of 

Earth's clock Te = 86164 seconds [NASA's Earth Data] will change the value of the angleto tilt angle (Ts/ Te) sine -1 [1/√ (2 π)]. Measuring in air and not in vacuum will changethe value of the tilt angle (1/n a); n a = 1.000293 = air index of refraction

Tilt angle ω t = (1/n a) (Ts/ Te) sine -1 [1/√ (2 π)] = 23.44° arc degree

= Exact value given by NASA of 23.44° arc degree

If planet Mercury angular velocity θ '0 given in arc seconds per century is to be measured:

 Where θ'0 = (2π/T m) (180/π) (100 x 1 year/T m) x (3600) = 70.75 arc sec per century 

T m = 88 days = 88 x 86400 second; 1 year = 365.26 days; 100 years = 36526 days

Then an error of (1 - sine 23.44°) θ'0 = 43 arc second per century 

Celestial sphere error 1: (1/n a) (Ts/ Te) sine -1 [1/√ (2 π)] = 23.440 

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Proof two: Real time astronomy spherical measurements

Modern and Nobel spherical errors are (8π/15) = (4 π/ 3) (2/5)

The (4 π/ 3) is from spherical volume

 And (2/5) is from spherical moment of inertia

Light visual incident and reflected of a sphere is not the same as lightincident on and reflected of a plane and the light incident of a sphere should be tracing a circle. This fact had been known since 10th century by Abbasid era Araband Persian physicists and astronomers in their description of fire sphere where balland not lens were used to make fires like fire is/ can be started with a lens.

 A circle has a circumference of 2 π and the moment of inertia of a sphere is (2/5) mr2.The error in visualizing light from a celestial object is 1 / (2 π). The spring method used where an error of 2/5 is multiplied, then the total error 1/ (2 π) (2/5) = 5/4 π. The vertical projection angle of light coming from a celestial object like the Sun would bemeasured in modern and Nobel terms as an error of axial tilt and not an actual tilt.

 Also, sine -1 (5/4 π) = 23.44617328. The index of refraction of air is n a = 1.000293

 And sine -1 (5/4 π) /n a = 23.4393°; Time was not used

Celestial sphere error 2: (1/ n a) sine -1 (5/4 π) = 23.4393°

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Proof three: Real time atmospheric gravity solution

Pressure P is defined as force/area = F/A; P = F/A; the mass of the atmosphere is m = 5.1352±0.0003×1018 kg and the force of the atmosphere F = m g; g = atmospheric acceleration orstandard acceleration = 9.80665 meters/second. The atmospheric pressure around Earth is

101.3kP = 101300 Pascal; The area of a sphere like Earth is 2 x π x r2

=2 x π x 6371, 0002

meters2; r = radius of Earth = 6371,000 meters.

Force F = mg =5.1352 ×1018 x 9.80665

Pressure P =101300

 Area A = 2 x π x 6371, 0002

P = F/A = 5.1352 ×1018 x 9.80665/ 2 x π x 6371, 0002 

= 197461.74 but P = 101300

197461.74/101300 =1.949276407

Measurements are made with distance and not distance squared and that means theerror is square root of 1.949276407 where measurement shift is done wrong. The squareroot of 1.949276407 is 1.396164892 and the shift is by the amount of 1.396164892 - 1 =0.396164892 and sine-1 (0.396164892) = 23.33865407 by Earth's recording time.

In human recording time period Ts = 24 hours = 86400 seconds and divide by Earth'srecording time period Te = 86164 seconds squared.

(23.33865407)(Ts/Te) 2.

Newton's law: F = G m M/r2

Where F = force between two objects with masses m and M and r is the

distance between their centers and G the proportionality constant with state of 

the art measurement of 6.6729 x 10-11 with experimental error of +/-0.0005

[1] Measurement techniques volume 39, No.10, 1996

Measurement of the gravitational constant using torsion balance

O.V. Karagioz; V.P.Izmailov; UDC 528.27; G = (6.6729 +/- 0.0005) x 10-11 

G =1/ (8 x π /15) Te2 na

2 d a; or, G =1/ (8 x π /15) Te2 εa

2 d a

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Where Cavendish made a spherical error of (4 x π /3) and moment of inertia

error of (2/5) and (4 x π /3) (2/5) and an index of refraction squared error na2

= 1.0002932. G has a unit [1/d a, Te2]; d a = air density = 1.2041; ε a is airpermittivity = 1.00058986

The index of refraction squared error na2 = 1.0002932 is the error that would

be attached to (23.33865407) (Ts /Te)2 na

2 

= (23.33865407) (86400/86164) 2 1.0002932 = 23.439°

P = F/A = me (Earth's mass) g e/A e = area (Earth's area)

P = 5.1352 ×1018 x 9.80665/ 2 x π x 6371, 0002 

P = 197461.74 kilo Pascal and P0 = 101300 kilo Pascal = normal atmospheric pressure

Or (Ts /Te)2 na

2 sine-1 (P/P0) = 1.0002932 (86400/86164) 2 sine-1 (P/P0)

Celestial sphere error 3: (Ts /Te)2 na

2 sine-1(me g e/A e)

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