peri-er (near expensive water?) ]]>

The problem that the 100,000 PHD Einsteinists could not solve

This is the solution to the “Quarter of a century” Smithsonian-NASA Posted motion puzzle that Einstein and the 100,000 space-time physicists including 109 years of Nobel prize winner physics and physicists and 400 years of astronomy and Astrophysicists could not solve and solved here and dedicated to Drs Edward Guinan and Frank Maloney

Of Villanova University Pennsylvania who posted this motion puzzle and started the search collections of stars with motion that can not be explained by any published physics

For 350 years Physicists Astrophysicists and Mathematicians and all others including Newton and Kepler themselves missed the time-dependent Newton’s equation and time dependent Kepler’s equation that accounts for Quantum – relativistic effects and it explains these effects as visual effects. Here it is

Universal- Mechanics

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 = 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’ + rθ”)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)

F = [d²(m r)/dt² – (m r)θ’²]r(1) + (1/mr)[d(m²r²θ’)/d t]θ(1) = [-GmM/r²]r(1)

d² (m r)/dt² – (m r) θ’² = -GmM/r²; d (m²r²θ’)/d t = 0

Let m =constant: M=constant

d²r/dt² – r θ’²=-GM/r² —— I

d(r²θ’)/d t = 0 —————–II

r²θ’=h = constant ————– II

r = 1/u; r’ = -u’/u² = – r²u’ = – r²θ'(d u/d θ) = -h (d u/d θ)

d (r²θ’)/d t = 2rr’θ’ + r²θ” = 0 r” = – h d/d t (du/d θ) = – h θ'(d²u/d θ²) = – (h²/r²)(d²u/dθ²)

[- (h²/r²) (d²u/dθ²)] – r [(h/r²)²] = -GM/r²

2(r’/r) = – (θ”/θ’) = 2[λ + ỉ ω (t)] – h²u² (d²u/dθ²) – h²u³ = -GMu²

d²u/dθ² + u = GM/h²

r(θ, t) = r (θ, 0) Exp [λ + ỉ ω (t)] u(θ,0) = GM/h² + Acosθ; r (θ, 0) = 1/(GM/h² + Acosθ)

r ( θ, 0) = h²/GM/[1 + (Ah²/Gm)cosθ]

r(θ,0) = a(1-ε²)/(1+εcosθ) ; h²/GM = a(1-ε²); ε = Ah²/GM

r(0,t)= Exp[λ(r) + ỉ ω (r)]t; Exp = Exponential

r = r(θ , t)=r(θ,0)r(0,t)=[a(1-ε²)/(1+εcosθ)]{Exp[λ(r) + ì ω(r)]t} Nahhas’ Solution

If λ(r) ≈ 0; then:

r (θ, t) = [(1-ε²)/(1+εcosθ)]{Exp[ỉ ω(r)t]

θ'(r, t) = θ'[r(θ,0), 0] Exp{-2ỉ[ω(r)t]}

h = 2π a b/T; b=a√ (1-ε²); a = mean distance value; ε = eccentricity

h = 2πa²√ (1-ε²); r (0, 0) = a (1-ε)

θ’ (0,0) = h/r²(0,0) = 2π[√(1-ε²)]/T(1-ε)²

θ’ (0,t) = θ'(0,0)Exp(-2ỉwt)={2π[√(1-ε²)]/T(1-ε)²} 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

Δ θ’ = [θ'(0, t) – θ'(0, 0)] = -4π {[√ (1-ε) ²]/T (1-ε) ²} (v/c) ²} radians/second

{(180/π=degrees) x (36526=century)

Δ θ’ = [-720×36526/ T (days)] {[√ (1-ε) ²]/ (1-ε) ²}(v/c) = 1.04°/century

This is the T-Rex equation that is going to demolished Einstein’s space-jail of time

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 = v (center of mass); v is the sum of orbital/rotational velocities = v(cm) for DI Her

Let m = mass of primary; M = mass of secondary

v (m) = primary speed; v(M) = secondary speed = √[Gm²/(m+M)a(1-ε²/4)]

v (cm) = [m v(m) + M v(M)]/(m + M) All rights reserved. joenahhas1958@yahoo.com

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’

Replace r with S = r exp (ỉ wt); h = [r² Exp (2iwt)] w’

w’ = (h/r²) 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)]

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’ = (d w/d t – h/r²] = -4π {[√ (1-ε²)]/T (1-ε) ²} (v/c) ² radian per second

Δ w’ = (- 4π /T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ² radians

Δ w’ = (-720/T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ² degrees; Multiplication by 180/π

Δ w’ = (-720×36526/T) {[√ (1-ε²)]/(1-ε)²} (v/c)² degrees/100 years

Δ w” = (-720×3600/T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ² seconds of arc by 3600

Δ w” = (-720x36526x3600/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

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”= (-720x36526x3600/88) 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?

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