Object at r —————- Light sensing of moving objects ——————- (seen as) S
r ———————– Cosine (wt) + i sine (wt) ——– S = r [cosine (wt) + i sine (wt)]
Particle ————————- Light —————————— Wave
Newton —————– Kepler’s Time dependent ———- Newton’s Time dependent

1- Advance of Perihelion of mercury

What is the visual effect for angular velocity along the line of sight? At Perihelion It is called the Advance of perihelion. Let us derive that

Areal velocity is constant: r² θ’ =h Kepler’s Law

h = 2π a b/T; b=a√ (1-ε²); a = mean distance value; ε = eccentricity
S = r Exp (ỉ wt); r² θ’= h = S² w’

h = S²w’= [r² Exp (2iwt)] w’=r²θ’; w’ = (θ’) exp [-2(ỉ wt)]
And w’= (h/r²) [cosine 2(wt) - ỉ sine 2(wt)] = (h/r²) [1- 2sine² (wt) - ỉ sin 2(wt)]

With w’ = w’ (x) + ỉ w’(y); w’(x) = (h/r²) [1- 2sine² (wt)]
Δ w’= w’(x) – (h/r²) = – 2(h/r²) sine² (wt) = – 2(h/r²) (v/c) ² v/c=sine wt

Angular velocity (h/ r²) (Perihelion/Periastron) = [2πa.a√ (1-ε²)]/Ta² (1-ε) ²= [2π√ (1-ε²)]/T (1-ε) ²

Δ w’ = [w'(x) – h/r²] = -4π {[√ (1-ε²)]/T (1-ε) ²} (v/c) ² radian per second
[180/π; degrees][100years=36526days; century] x [3600; seconds in degree]

Δ w” = (-720×36526x3600/T) {[√ (1-ε²]/(1-ε)²} (v/c)² seconds of arc per century

The circumference of an ellipse: 2πa (1 – ε²/4 + 3/16(ε²)²- –.) ≈ 2πa (1-ε²/4); R =a (1-ε²/4)
v=√ [G m M / (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; m<<M; Solar system

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:
v =48.14km/sec; [√ (1- ε²)] (1-ε) ² = 1.552

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

2- DI Herculis Apsidal motion solution: derived from S= r exp ỉ [ω (r) + ω (m)] t

(See other articles by Joe Nahhas)

W° (ob) = (-720×36526/T) x {[√ (1-ε²)]/ (1-ε) ²} [(v*/c) + (v°/c)] ² degrees/ century

Where v* = v (center of mass) = 106.38km/sec; v° (spin difference) = 0
T = orbital period; ε = eccentricity; c =light speed

Application 3: Gravitational red shift: Pound Rebka Experiment
S = r Exp [î ω t]
1/S = 1/r Exp [-ỉ ω t]
And λ (S) = λ (r) Exp [-ỉ ω t]; λ = wavelength

Then υ(s) = υ(r) Exp [ỉ ω t]; υ = frequency
And υ(S) = υ (r, t) = υ(r, 0) υ (0, t) = υ(r) υ (0, t)
With sin ω(r) t = v/c; cosine ω(r) t = √ [1-(v/c) ²]

Then υ (r, t) = υ(r, 0) {√ [1-(v/c) ²] + ỉ (v/c)} = Real {υ(r, t)} + Imaginary {υ(r, t)}
Real {υ (r, t)} = υ (r, 0) √ [1-(v/c) ²] ≈ υ (r, 0) [1 - 1/2(v/c) ²]

Δ υ (r, t) = real {υ (r, t)} – υ (0, t)
Δ υ (r, t) = -υ (r, 0)/2 [(v/c) ²]
Δ υ(r, t)/υ(r, 0) = -1/2(v/c)²[up]-{1/2(v/c)²[down]} = – (v/c) ²
v² = 2gh; g = 9.81km/s² gravitational acceleration; h = height
Δ υ/υ [Total] =-[2gh/c²]

4- Light bending: Lord Edenton experiment

S = r Exp [ỉ ω t]; From Kepler’s Equation: r² θ’ = h = 2A/T
h = S²(r, t) θ’(r, t) = r² (θ, t) θ’ (θ, t) = r² (θ, 0) Exp [2ỉ ω t] θ’ (θ, t) = 2A/t
And θ’ (θ, t) = θ’ (θ, 0) θ’(0, t) = [h/ r² (θ, 0)] Exp [-2ỉ ω(r) t]
Then θ ‘(θ, t) = [2A/t r² (θ, 0)] {1 – 2sin²ω(r) t – 2ỉ sin ω(r) t cosine ω(r) t}
Now [t θ'(θ, t)] = [2A/r² (θ' 0)] [1 - 2sin²ω(r) t] -2ỉ [2A/r² (θ, 0)] [sin ω(r) t cosine ω(r) t]
= Δ x + i Δ y
Δ θ = Δ x – [A/r² (θ, 0)] = – [A/r² (θ, 0)][4sin²ω(r)t]; sin ω(r)t = v/c
Δ θ = – [A/r² (θ, 0)](v/c) ²
(v/c) ² ≈ 1.75″; v² = GM/R; G = Gravitational constant; M = Sun mass; R = sun radius
Δ θ = [A/r² (θ, 0)] [1.75"]; A = area

The values depend on near by stars and the measured values fit this equation.

Russians in 1936; Δ θ = 2.74
[A/r² (θ, 0)] = π/2
Δ θ = π/2(1.75″) = 2.74″

Application 5: Shapiro time delay (Vikings 6, 7; 1977)

Mars ————————— Middle—- Sun ————- Earth

The center of mass is the sun. The sun produces a velocity field given by
v = √ [GM/a (1- ε²/4)]

From above t =2 arc length/c=2d Δ w/c = (8π r/c) (v/c) ²; Δ w=4π (v/c) ²; r = 2a=d
t = 16πGM/c³ (1-ε²/4); ε = [a (1) -a (2)]/ [a (1) + a (2)] = .2075

t = (8πd/c) (v/c) ²= 8π (377,536,987.5/299792.458) (26.6575872/299792.458)²=250μs
If d = 2a (1-ε²/4), then t = 247.597μs value theorized actual measured value is 250μs

All this is not due to space-time but due to light aberration caused by moving planets.

θ’(0,0) = h(0,0)/r²(0,0) = 2π/T
θ’ (0,t) = θ’(0,0)Exp(-2ỉwt)={2π/T} Exp (-2iwt)
θ’(0,t) = θ’(0,0) [cosine 2(wt) - ỉ sine 2(wt)] = θ’(0,0) [1- 2sine² (wt) - ỉ sin 2(wt)]
θ’(0,t) = θ’(0,t)(x) + θ’(0,t)(y); θ’(0,t)(x) = θ’(0,0)[ 1- 2sine² (wt)]
θ’(0,t)(x) – θ’(0,0) = – 2θ’(0,0)sine²(wt) = – 2θ’(0,0)(v/c)² v/c=sine wt; c=light speed
T [θ'(0, t) - θ'(0, 0)] = -4π (v/c) ²

Δ θ = -4π (v/c) ² Earth-Mars

Sun-Photon:

The circumference of an ellipse: 2πa (1 – ε²/4 + 3/16(ε²)²—) ≈ 2πa (1-ε²/4); R =a (1-ε²/4)
v=√ [Gm M/ (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; m<<M; Solar system

ΔΓ = 2 arc length/c = 2[Δ θ] 2d/c = 2[- 4π (v/c) ²] 2d/c; ΔΓ = -8πd/c (v/c) ²;

ΔΓ = 8πd/c³ [GM/a (1-ε²/4)] =16πGM/c³ (1-ε²/4) = Γ0 (1 – ε²/4)

ε = [a (planet 1) - a (planet 2)]/ [a (planet 1) + a (planet 2)] =0.2075 Mars-Earth
Γ0 = 16 πGM/c³= 247.5974607μs=universal constant; ΔΓ = 250μs Mars-Earth.

Joe nahhas1958@yahoo.com All right reserved

The motion puzzle that Einstein MIT Harvard Cal-tech NASA and all others could not solve.

Introduction: For 350 years Physicists Astronomers and Mathematicians missed Kepler’s time dependent equation that changed Newton’s equation into a time dependent Newton’s equation and together these two equations combine classical mechanics and quantum mechanics into one mechanics explains “relativistic” effects as the difference between time dependent measurements and time independent measurements of moving objects and solve all motion in all of Mechanics posted on Smithsonian NASA website SAO/NASA that Einstein and all 100,000 space-time “physicists” could not solve by space-time physics or any published physics.

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/dt) + (dm/dt) 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/dt)(d r/d t) + (d²m/dt²)r
= mγ + 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’ + rθ”)θ(1)
Proof:
r = r [cosθ î + sinθĴ] = r r (1); r (1) = cosθ î + sinθ Ĵ
v = d r/d t = r’ r (1) + r d[r (1)]/d t = r’ r (1) + r θ’[- sinθ î + cos θĴ] = r’ r (1) + r θ’ θ (1)

θ (1) = -sinθ î +cosθ Ĵ; r(1) = cosθî + sinθĴ

d [θ (1)]/d t= θ’ [- cosθî - sinθĴ= - θ' r (1)
d [r (1)]/d t = θ’ [ -sinθ'î + cosθ]Ĵ = θ’ θ(1)

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)

F = m[(r"-rθ'²)r(1) + (2r'θ' + rθ")θ(1)] + 2m’[r'r(1) + rθ'θ(1)] + (m”r) r(1)

= [d²(mr)/dt² - (mr)θ'²]r(1) + (1/mr)[d(m²r²θ')/dt]θ(1) = [-GmM/r²]r(1)

d²(mr)/dt² – (mr)θ’² = -GmM/r² Newton’s Gravitational Equation (1)
d(m²r²θ’)/dt = 0 Central force law (2)

(2) : d(m²r²θ’)/d t = 0 m²r²θ’ = [m²(θ,0)φ²(0,t)][ r²(θ,0)ψ²(0,t)][θ'(θ, t)]
= [m²(θ,t)][r²(θ,t)][θ'(θ,t)]
= [m²(θ,0)][r²(θ,0)][θ'(θ,0)]
= [m²(θ,0)]h(θ,0);h(θ,0)=[r²(θ,0)][θ'(θ,0)]
= H (0, 0) = m² (0, 0) h (0, 0)
= m² (0, 0) r² (0, 0) θ’(0, 0)
m = m (θ, 0) φ (0, t) = m (θ, 0) Exp [λ (m) + ì ω (m)] t; Exp = Exponential
φ (0, t) = Exp [ λ (m) + ỉ ω (m)]t

r = r(θ,0) ψ(0, t) = r(θ,0) Exp [λ(r) + ì ω(r)]t
ψ(0, t) = Exp [λ(r) + ỉ ω (r)]t

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
Kepler’s time dependent equation that Physicists Astrophysicists and Mathematicians missed for 350 years that is going to demolish Einstein’s space-jail of time

θ’(0,t) = θ’(0,0) Exp{-2{[λ(m) + λ(r)]t + ỉ[ω(m) + ω(r)]t}}

(1): d² (m r)/dt² – (m r) θ’² = -GmM/r² = -Gm³M/m²r²

d² (m r)/dt² – (m r) θ’² = -Gm³ (θ, 0) φ³ (0, t) M/ (m²r²)

Let m r =1/u

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
d²(m r)/dt² = -Hθ’d²u/dθ² = – Hu²[d²u/dθ²]

-Hu² [d²u/dθ²] -(1/u)(Hu²)² = -Gm³(θ,0)φ³(0,t)Mu²
[d²u/ dθ²] + u = Gm³(θ,0)φ³(0,t)M/H²

t = 0; φ³ (0, 0) = 1
u = Gm³(θ,0)M/H² + Acosθ =Gm(θ,0)M(θ,0)/h²(θ,0)

mr = 1/u = 1/[Gm(θ,0)M(θ,0)/h(θ,0) + Acosθ]
= [h²/Gm(θ,0)M(θ,0)]/{1 + [Ah²/Gm(θ,0)M(θ,0)][cosθ]}

= [h²/Gm(θ,0)M(θ,0)]/(1 + εcosθ)
mr = [a(1-ε²)/(1+εcosθ)]m(θ,0)

r(θ,0) = [a(1-ε²)/(1+εcosθ)] m r = m(θ, t) r(θ, t)
= m(θ,0)φ(0,t)r(θ,0)ψ(0,t)

r(θ,t) = [a(1-ε²)/(1+εcosθ)]{Exp[λ(r)+ω(r)]t} Newton’s time dependent Equation ——–II

If λ (m) ≈ 0 fixed mass and λ(r) ≈ 0 fixed orbit; then

θ’(0,t) = θ’(0,0) Exp{-2ì[ω(m) + ω(r)]t}

r(θ, t) = r(θ,0) r(0,t) = [a(1-ε²)/(1+εcosθ)] Exp[i ω (r)t]

m = m(θ,0) Exp[i ω(m)t] = m(0,0) Exp [ỉ ω(m) t] ; m(0,0)

θ’(0,t) = θ’(0, 0) Exp {-2ì[ω(m) + ω(r)]t}

θ’(0,0)=h(0,0)/r²(0,0)=2πab/Ta²(1-ε)²

= 2πa² [√ (1-ε²)]/T a² (1-ε) ²; θ’(0, 0) = 2π [√ (1-ε²)]/T (1-ε) ²

θ’(0,t) = {2π[√(1-ε²)]/T(1-ε)²}Exp{-2[ω(m) + ω(r)]t

θ’(0,t) = {2π[√(1-ε²)]/(1-ε)²}{cos 2[ω(m) + ω(r)]t – ỉ sin 2[ω(m) + ω(r)]t}

θ’(0,t) = θ’(0,0) {1- 2sin² [ω(m) + ω(r)]t – ỉ 2isin [ω(m) + ω(r)]t cos [ω(m) + ω(r)]t}

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

– 2ỉ θ’(0, 0) sin [ω (m) + ω(r)] t cos [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[sin ω (m) t cos ω(r) t + cos ω (m)t sin ω(r)t]²}

W(ob) = Real Δ θ (0, t) – θ’(0, 0) = – 2 θ’(0, 0){(v°/c)√ [1-(v*/c) ²] + (v*/c)√ [1- (v°/c) ²]}²

v ° = spin velocity; v* = orbital velocity; v°/c = sin ω (m)t; v*/c = cos ω (r) t

v°/c << 1; (v°/c)² ≈ 0; v*/c << 1; (v*/c)² ≈ 0

W (ob) = – 2[2π √ (1-ε²)/T (1-ε) ²] [(v° + v*)/c] ²

W (ob) = (- 4π /T) {[√ (1-ε²)]/ (1-ε) ²} [(v° + v*)/c] ² radians
W (ob) = (-720/T) {[√ (1-ε²)]/ (1-ε) ²} [(v° + v*)/c] ² degrees; Multiplication by 180/π

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

W” (ob) = (-720×26526x3600/T) {[√ (1-ε²)]/ (1-ε) ²} [(v° + v*)/c] ² seconds /100 years

The circumference of an ellipse: 2πa (1 – ε²/4 + 3/16(ε²)²- –.) ≈ 2πa (1-ε²/4); R =a (1-ε²/4)
v (m) = √ [GM²/ (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; m<<M; Solar system
v (M) = √ [Gm² / (m + M)a(1-ε²/4)] ≈ 0; m<<M

Application 1: 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; 1-ε²/4 = 0.989391
ρ (m) = 0.696×10^9m; ρ(m)=2.44×10^6m; T(sun) = 25days
v° (M) = 2km/sec ; v° = 2meters/sec
v *= v(m) = √ [GM/a (1-ε²/4)]; v(M) = √[Gm²/(m + M)a(1-ε²)] ≈ 0
v°(m) = 2m/sec (Mercury) v°(M)= 2km/sec(sun)
Calculations yields: v = v* + v° =48.14km/sec (mercury); [√ (1- ε²)] (1-ε) ² = 1.552
W” (ob) = (-720×36526x3600/T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ²
W” (ob) = (-720×36526x3600/88) x (1.552) (48.14/299792)² = 43.0”/century

V1143Cgyni Apsidal Motion Solution

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

v° = -v°(m) + v°(M)
v* = 2v(cm) + σ
v°(m) = spin velocity of primary
v°(M) = spin velocity of secondary
v(cm) = [m v(m) + M v(M)]/(m + M) center of mass velocity
σ = √ {{[v(m) - v(cm)]² + [v(M) - v(cm)]²}/2} = standard deviation
W° = 3.36°/century as reported in many articles

Visual Effects and the confusions of “Modern” physics

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

Line of Sight: r cosine wt

r ——————- r cosine (wt) line of sight light aberrations

A moving object with velocity v will be visualized by
light sensing through an angle (wt);w = constant and t= time
Also, sine wt = v/c; cosine wt = √ [1-sine² (wt) = √ [1-(v/c) ²]

A visual object moving with velocity v will be seen as S

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

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 the visual effects caused by light aberrations of a
moving object along the line of sight.

In a right angled velocity triangle A B C: Angle A = wt; angle B = 90°; Angle C = 90° -wt
AB = hypotenuse = c; BC = opposite = v; CA= adjacent = c √ [1-(v/c) ²]

Ending Einstein’s space jail of time in 2009 that led to fraud Symbol E=mc²

Areal velocity is constant: r² θ’ =h Kepler’s Law

h = 2π a b/T; b=a√ (1-ε²); a = mean distance value; ε = eccentricity
r² θ’= h = S² w’

S = r exp (ỉ wt); h = [r² Exp (2iwt)] w’=r²θ’
w’ = (θ’) exp [-2(i wt)]

w’= (h/r²) [cosine 2(wt) - ỉ sine 2(wt)] = (h/r²) [1- 2sine² (wt) - ỉ sin 2(wt)]
w’ = w’(x) + ỉ w’(y) ; w’(x) = (h/r²) [ 1- 2sine² (wt)]

Δ w’= w’(x) – (h/r²) = – 2(h/r²) sine² (wt) = – 2(h/r²) (v/c) ² v/c=sine wt
(h/ r²)(Perihelion/Periastron)= [2πa.a√ (1-ε²)]/Ta² (1-ε) ²= [2π√ (1-ε²)]/T (1-ε) ²
Δ w’ = [w'(x) – h/r²] = -4π {[√ (1-ε²)]/T (1-ε) ²} (v/c) ² radian per second

{x [180/π;degrees]x[100years=36526days;century]x[3600;seconds in degree]
Δ w” = (-720×36526x3600/T) {[√ (1-ε²]/(1-ε)²} (v/c)² seconds of arc per century

This Kepler’s Equation solves all the problems Einstein and all physicists could not solve
DI Her Binary starts systems

The circumference of an ellipse: 2πa (1 – ε²/4 + 3/16(ε²)²- –.) ≈ 2πa (1-ε²/4); R =a (1-ε²/4) v=√ [G m M / (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; m<<M; Solar system

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:
v =48.14km/sec; [√ (1- ε²)] (1-ε) ² = 1.552
Δ w”= (-720×36526x3600/88) x (1.552) (48.14/299792)²=43.0”/century

Conclusions: The 43″ seconds of arc of advance of perihelion of Planet Mercury (General relativity) is given by Kepler’s equation better than all of Published papers of Einstein. Kepler’s Equation can solve Einstein’s nemesis DI Her Binary stars motion and all the other dozens of stars motions posted for past 40 years on NASA website SAO/NASA as unsolved by any physics

Anyone dare to prove me wrong?

Lets look forward to a day when a very well written article ends with respect for mankind and muslims in general. May Allah Honor Alhle-Baith, Asharathu Mubashara, Ansar, Muhajireen, Khulafa-ur-Rashideen, and the rest of Sahabas and muslims who helped our prophet throughout his life and after.
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An Ummathee from the Ummath of the Rahmathun-lil-Aalameen