Mercury’s “apparent” advance of perihelion of 43″ seconds of an arc per century

By Professor Joe Nahhas

Abstract: This is the solution to the 150 years old “apparent” Advance of Perihelion motion puzzle solution using Newtonian mechanics and not space-time confusions of physics. Planetary advance of perihelion or rate of orbital axial rotation is visual effects along the line of sight of moving objects applied to the angular velocity at perihelion. This rate of “apparent” axial rotation is given by this new equation
W° (ob) = (-720×36526x3600/T) {[√ (1-ε²)]/ (1-ε) ²]} [(v° + v*)/c] ² seconds/100 years
T = period; ε = eccentricity; v° = spin velocity effect; v*= orbital velocity effect
For Mercury: v* = 48.14km/s
And v° = 0.002km/s = spin velocity of Planet

Universal Mechanics Solution:

For 350 years Physicists Astronomers and Mathematicians and philosophers missed Kepler’s time dependent Areal velocity wave equation solution that changed Newton’s classical planetary motion equation to a Newton’s time dependent wave orbital equation solution and these two equations put together combines particle mechanics of Newton’s with wave mechanics of Kepler’s into one time dependent universal mechanics equation that explain “relativistic” as the difference between time dependent measurements and time independent measurements of moving objects and in practice it amounts to light aberrations along the line of sight of moving objects

All there is in the Universe is objects of mass m moving in space (x, y, z) at a location
r = r (x, y, z). The state of any object in the Universe can be expressed as the product

S = m r; State = mass x location:

P = d S/d t = m (d r/d t) + (dm/d t) r = Total moment
= change of location + change of mass
= m v + m’ r; v = velocity = d r/d t; m’ = mass change rate

F = d P/d t = d²S/dt² = Total force
= m (d²r/dt²) +2(dm/d t) (d r/d t) + (d²m/dt²) r
= m γ + 2m’v +m” r; γ = acceleration; m” = mass acceleration rate

In polar coordinates system

r = r r (1) ;v = r’ r(1) + r θ’ θ(1) ; γ = (r” – rθ’²)r(1) + (2r’θ’ + r θ”)θ(1)
r = location; v = velocity; γ = acceleration
F = m γ + 2m’v +m” r
F = m [(r"-rθ'²) r (1) + (2r'θ' + r θ") θ (1)] + 2m’[r' r (1) + r θ' θ (1)] + (m” r) r (1)
= [d² (m r)/dt² - (m r) θ'²] r (1) + (1/mr) [d (m²r²θ')/d t] θ (1)
= [-GmM/r²] r (1) ——————————- Newton’s Gravitational Law
Proof:
First r = r [cosine θ î + sine θ Ĵ] = r r (1)
Define r (1) = cosine θ î + sine θ Ĵ
Define v = d r/d t = r’ r (1) + r d[r (1)]/d t
= r’ r (1) + r θ’[- sine θ î + cosine θĴ]
= r’ r (1) + r θ’ θ (1)

Define θ (1) = -sine θ î +cosine θ Ĵ;
And with r (1) = cosine θ î + sine θ Ĵ

Then d [θ (1)]/d t= θ’ [- cosine θ î - sine θ Ĵ= - θ' r (1)
And d [r (1)]/d t = θ’ [-sine θ î + cosine θ Ĵ] = θ’ θ (1)

Define γ = d [r' r (1) + r θ' θ (1)] /d t
= r” r (1) + r’d [r (1)]/d t + r’ θ’ r (1) + r θ” r (1) +r θ’d [θ (1)]/d t
γ = (r” – rθ’²) r (1) + (2r’θ’ + r θ”) θ (1)

With d² (m r)/dt² – (m r) θ’² = -GmM/r² Newton’s Gravitational Equation (1)
And d (m²r²θ’)/d t = 0 Central force law (2)

(2): d (m²r²θ’)/d t = 0
Then m²r²θ’ = constant
= H (0, 0)
= m² (0, 0) h (0, 0); h (0, 0) = r² (0, 0) θ’(0, 0)
= m² (0, 0) r² (0, 0) θ’(0, 0); h (θ, 0) = [r² (θ, 0)] [θ'(θ, 0)]
= [m² (θ, 0)] h (θ, 0); h (θ, 0) = [r² (θ, 0)] [θ'(θ, 0)]
= [m² (θ, 0)] [r² (θ, 0)] [θ'(θ, 0)]
= [m² (θ, t)] [r² (θ, t)] [θ' (θ, t)]
= [m²(θ, 0) m²(0,t)][ r²(θ,0)r²(0,t)][θ'(θ, t)]
= [m²(θ, 0) m²(0,t)][ r²(θ,0)r²(0,t)][θ'(θ, 0) θ' (0, t)]
With m²r²θ’ = constant
Differentiate with respect to time
Then 2mm’r²θ’ + 2m²rr’θ’ + m²r²θ” = 0
Divide by m²r²θ’
Then 2 (m’/m) + 2(r’/r) + θ”/θ’ = 0
This equation will have a solution 2 (m’/m) = 2[λ (m) + ì ω (m)]
And 2(r’/r) = 2[λ (r) + ì ω (r)]
And θ”/θ’ = -2{λ (m) + λ (r) + ỉ [ω (m) + ω (r)]}

Then (m’/m) = [λ (m) + ì ω (m)]
Or d m/m d t = [λ (m) + ì ω (m)]
And dm/m = [λ (m) + ì ω (m)] d t
Then m = m (0) Exp [λ (m) + ì ω (m)] t
m = m (0) m (0, t); m (0, t) Exp [λ (m) + ì ω (m)] t
With initial spatial condition that can be taken at t = 0 anywhere then m (0) = m (θ, 0)
And m = m (θ, 0) m (0, t) = m (θ, 0) Exp [λ (m) + ì ω (m)] t; Exp = Exponential
And m (0, t) = Exp [λ (m) + ỉ ω (m)] t
Similarly we can get
Also, r = r (θ, 0) r (0, t) = r (θ, 0) Exp [λ (r) + ì ω (r)] t
With r (0, t) = Exp [λ (r) + ỉ ω (r)] t

Then θ’(θ, t) = {H(0, 0)/[m²(θ,0) r(θ,0)]}Exp{-2{[λ(m) + λ(r)]t + ì [ω(m) + ω(r)]t}} —–I
And θ’(θ, t) = θ’ (θ, 0)]} Exp {-2{[λ (m) + λ (r)] t + ì [ω (m) + ω (r)] t}} ——————–I
And, θ’(θ, t) = θ’ (θ, 0) θ’ (0, t)
And θ’ (0, t) = Exp {-2{[λ (m) + λ(r)] t + ì [ω (m) + ω(r)] t}
Also θ’(θ, 0) = H (0, 0)/ m² (θ, 0) r² (θ, 0)
And θ’(0, 0) = {H (0, 0)/ [m² (0, 0) r (0, 0)]}

With (1): d² (m r)/dt² – (m r) θ’² = -GmM/r² = -Gm³M/m²r²
And d² (m r)/dt² – (m r) θ’² = -Gm³ (θ, 0) m³ (0, t) M/ (m²r²)
Let m r =1/u
Then d (m r)/d t = -u’/u² = – (1/u²) (θ’) d u/d θ = (- θ’/u²) d u/d θ = -H d u/d θ
And d² (m r)/dt² = -Hθ’d²u/dθ² = – Hu² [d²u/dθ²]

-Hu² [d²u/dθ²] – (1/u) (Hu²)² = -Gm³ (θ, 0) m³ (0, t) Mu²
[d²u/ dθ²] + u = Gm³ (θ, 0) m³ (0, t) M/ H²

t = 0; m³ (0, 0) = 1
u = Gm³ (θ, 0) M/ H² + A cosine θ =Gm (θ, 0) M (θ, 0)/ h² (θ, 0)

And m r = 1/u = 1/ [Gm (θ, 0) M (θ, 0)/ h (θ, 0) + A cosine θ]
= [h²/ Gm (θ, 0) M (θ, 0)]/ {1 + [Ah²/ Gm (θ, 0) M (θ, 0)] [cosine θ]}
= [h²/Gm (θ, 0) M (θ, 0)]/ (1 + ε cosine θ)

Then m (θ, 0) r (θ, 0) = [a (1-ε²)/ (1+εcosθ)] m (θ, 0)
Dividing by m (θ, 0)
Then r (θ, 0) = a (1-ε²)/ (1+εcosθ)
This is Newton’s Classical Equation solution of two body problem which is the equation of an ellipse of semi-major axis of length a and semi minor axis b = a √ (1 – ε²) and focus length c = ε a
And m r = m (θ, t) r (θ, t) = m (θ, 0) m (0, t) r (θ, 0) r (0, t)
Then, r (θ, t) = [a (1-ε²)/ (1+εcosθ)] {Exp [λ(r) + ỉ ω (r)] t} ———————————- II
This is Newton’s time dependent equation that is missed for 350 years
If λ (m) ≈ 0 fixed mass and λ(r) ≈ 0 fixed orbit; then
Then r (θ, t) = r (θ, 0) r (0, t) = [a (1-ε²)/ (1+ε cosine θ)] Exp i ω (r) t
And m = m (θ, 0) Exp [i ω (m) t] = m (θ, 0) Exp ỉ ω (m) t

We Have θ’(0, 0) = h (0, 0)/r² (0, 0) = 2πab/ Ta² (1-ε) ²
= 2πa² [√ (1-ε²)]/T a² (1-ε) ²
= 2π [√ (1-ε²)]/T (1-ε) ²

Then θ’(0, t) = {2π [√ (1-ε²)]/ T (1-ε) ²} Exp {-2[ω (m) + ω (r)] t
= {2π [√ (1-ε²)]/ (1-ε) ²} {cosine 2[ω (m) + ω (r)] t – ỉ sin 2[ω (m) + ω (r)] t}
And θ’(0, t) = θ’(0, 0) {1- 2sin² [ω (m) + ω (r)] t}
– ỉ 2i θ’(0, 0) sin [ω (m) + ω (r)] t cosine [ω (m) + ω (r)] t

Then θ’(0, t) = θ’(0, 0) {1 – 2sine² [ω (m) t + ω (r) t]}
– 2ỉ θ’(0, 0) sin [ω (m) + ω(r)] t cosine [ω (m) + ω(r)] t

Δ θ’ (0, t) = Real Δ θ’ (0, t) + Imaginary Δ θ (0, t)
Real Δ θ (0, t) = θ’(0, 0) {1 – 2 sine² [ω (m) t ω(r) t]}

Let W (ob) = Δ θ’ (0, t) (observed) = Real Δ θ (0, t) – θ’(0, 0)
= -2θ’(0, 0) sine² [ω (m) t + ω(r) t]
= -2[2π [√ (1-ε²)]/T (1-ε) ²] sine² [ω (m) t + ω(r) t]

If this apsidal motion is to be found as visual effects, then
With, v ° = spin velocity; v* = orbital velocity; v°/c = tan ω (m) T°; v*/c = tan ω (r) T*
Where T° = spin period; T* = orbital period

And ω (m) T° = Inverse tan v°/c; ω (r) T*= Inverse tan v*/c
W (ob) = -4 π [√ (1-ε²)]/T (1-ε) ²] sine² [Inverse tan v°/c + Inverse tan v*/c] radians
Multiplication by 180/π

W (ob) = (-720/T) {[√ (1-ε²)]/ (1-ε) ²} sine² {Inverse tan [v°/c + v*/c]/ [1 - v° v*/c²]} degrees and multiplication by 1 century = 36526 days and using T in days

W° (ob) = (-720×36526/Tdays) {[√ (1-ε²)]/ (1-ε) ²} x
sine² {Inverse tan [v°/c + v*/c]/ [1 - v° v*/c²]} degrees/100 years

Approximations I

With v° << c and v* << c, then v° v* <<< c² and [1 - v° v*/c²] ≈ 1
Then W° (ob) ≈ (-720×36526/Tdays) {[√ (1-ε²)]/ (1-ε) ²} x sine² Inverse tan [v°/c + v*/c] degrees/100 years

Approximations II

With v° << c and v* << c, then sine Inverse tan [v°/c + v*/c] ≈ (v° + v*)/c

W° (ob) = (-720×36526/Tdays) {[√ (1-ε²)]/ (1-ε) ²} x [(v° + v*)/c] ² degrees/100 years
This is the equation that gives the advance of perihelion rates ———————–III

The circumference of an ellipse: 2πa (1 – ε²/4 + 3/16(ε²)²- –.) ≈ 2πa (1-ε²/4); r =a (1-ε²/4)

Where v (m) = √ [GM²/ (m + M) a (1-ε²/4)]
And v (M) = √ [Gm² / (m + M) a (1-ε²/4)]
From Newton’s laws for a circular orbit: m v²/ r (cm) = GmM/r²; r (cm) = [M/m + M] r
Then v² = [GM r (cm)/ r²] = GM²/ (m + M) r
And v = √ [GM²/ (m + M) r = a (1-ε²/4)]
And v* = v (m) = √ [GM²/ (m + M) a (1-ε²/4)] = 48.14 km

Advance of Perihelion of mercury.

G=6.673×10^-11; M=2×10^30kg; m=.32×10^24kg
ε = 0.206; T=88days; c = 299792.458 km/sec; a = 58.2km/sec

Calculations yields:
With v* =48.14km/sec; v° = spin = 2 meters per second
Then v* + v° = 48.14 km/sec
And [√ (1- ε²)] (1-ε) ² = 1.552

W” (ob) = (-720×36526x3600/Tdays){[√(1-ε²)]/ (1-ε) ²} [(v° + v*)/c] ²seconds /100 years

= (-720×36526x3600/88) x (1.552) [(48.14/299792)]²=43.0”/century

Joenahhas1958@yahoo.com all right reserved

Visual Effects and the confusions of “Modern” physics

r ————————- Light sensing of moving objects ——————————- S
Actual ———————————– Light ———————————————- Visual
Object at r ———- Light aberration = cosine (wt) + i sine (wt) —– S = r [cosine (wt) + i sine(wt)]
Newton space particle ——— Kepler’s time dependent visual effects —- Time dependent Newton
Particle ——————————– Visual effects —————————– Wave

Line of Sight: rcosine wt

r ————————————————————————- r cosine (wt) due light aberrations

B
A moving object with velocity v [C-B] will ^
meet light at B hypotenuse . ^ Object
sine wt = v/c light . ^ Velocity
cosine wt = √ [1-(v/c)²] Velocity . ^ v
c . Angle ^
.A= wt ^
. . . . . . .
A c √ [1-(v/c) ²] C

S = r [cosine (wt) + i sine (wt)] = r Exp [i wt]; Exp = Exponential

Also, sine ω(r) t= v/c; cosine ω t = √ [1-sin²ω(r) t] = √ [1-(v/c) ²]

S = r [√ [1-(v/c) ²] + ỉ (v/c)] = S x + i S y

S x = Visual along the line of sight = r [√ [1-(v/c) ²]
This Equation is special relativity length contraction formula and it is just visual effects along the line of sight.

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Though thinking this through, it would be cool to have iTunes like sync for laptops connected to desktop machines where you select the folders and items you want synced each time you connect it.