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	<title>Comments on: Newton &#8211; &#8220;The greatest alchemist of them all&#8221;</title>
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	<link>http://blog.oup.com/2005/12/newton___the_gr/</link>
	<description>Academic insights for the thinking world.</description>
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		<title>By: Alexander Nahhas</title>
		<link>http://blog.oup.com/2005/12/newton___the_gr/comment-page-1/#comment-149244</link>
		<dc:creator>Alexander Nahhas</dc:creator>
		<pubDate>Wed, 11 Feb 2009 10:09:34 +0000</pubDate>
		<guid isPermaLink="false">http://216.110.190.15/2005/12/newton_-_the_greatest_alchemist_of_them_all/#comment-149244</guid>
		<description>V1143Cgyni Binary Stars Apsidal motion Puzzle solution

 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&#039;s time dependent equation that changed Newton&#039;s equation into a time dependent Newton&#039;s equation and together these two equations combine classical mechanics and quantum mechanics into one mechanics explains &quot;relativistic&quot; 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 &quot;physicists&quot; 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&#039; r; v = velocity = d r/d t; m&#039; = 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&#039;v +m&quot;r; γ = acceleration; m&#039;&#039; = mass acceleration rate

In polar coordinates system

r = r r(1) ;v = r&#039; r(1)  + r θ&#039; θ(1) ; γ = (r&quot; - rθ&#039;²)r(1) + (2r&#039;θ&#039; + rθ&quot;)θ(1)
Proof:
r = r [cosθ î + sinθĴ] = r r (1); r (1) = cosθ î + sinθ Ĵ 
v = d r/d t = r&#039; r (1) + r d[r (1)]/d t = r&#039; r (1) + r θ&#039;[- sinθ î + cos θĴ] = r&#039; r (1) + r θ&#039; θ (1)

θ (1) = -sinθ î +cosθ Ĵ; r(1) = cosθî + sinθĴ

d [θ (1)]/d t=  θ&#039; [- cosθî - sinθĴ= - θ&#039; r (1)  
d [r (1)]/d t = θ&#039; [ -sinθ&#039;î + cosθ]Ĵ = θ&#039; θ(1) 

γ = d [r&#039;r(1) + r θ&#039; θ (1)] /d t = r&quot; r(1) + r&#039; d[r(1)]/d t + r&#039; θ&#039; r(1) + r θ&quot; r(1) +r θ&#039; d[θ(1)]/d t

γ = (r&quot; - rθ&#039;²) r(1) +  (2r&#039;θ&#039; + r θ&quot;) θ(1)

F = m[(r&quot;-rθ&#039;²)r(1) + (2r&#039;θ&#039; + rθ&quot;)θ(1)] + 2m&#039;[r&#039;r(1) + rθ&#039;θ(1)] + (m&quot;r) r(1)
  
 = [d²(mr)/dt² - (mr)θ&#039;²]r(1) + (1/mr)[d(m²r²θ&#039;)/dt]θ(1) = [-GmM/r²]r(1)

d²(mr)/dt² - (mr)θ&#039;² = -GmM/r²   Newton&#039;s Gravitational Equation     (1)
d(m²r²θ&#039;)/dt = 0                           Central force law                              (2)

(2) : d(m²r²θ&#039;)/d t = 0   m²r²θ&#039; = [m²(θ,0)φ²(0,t)][ r²(θ,0)ψ²(0,t)][θ&#039;(θ, t)] 
                                                         = [m²(θ,t)][r²(θ,t)][θ&#039;(θ,t)] 
                                                         = [m²(θ,0)][r²(θ,0)][θ&#039;(θ,0)]
                                                         = [m²(θ,0)]h(θ,0);h(θ,0)=[r²(θ,0)][θ&#039;(θ,0)]
                                                         = H (0, 0) = m² (0, 0) h (0, 0) 
                                                         = m² (0, 0) r² (0, 0) θ&#039;(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&#039;(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&#039;s time dependent equation that Physicists Astrophysicists and Mathematicians missed for 350 years that is going to demolish Einstein&#039;s space-jail of time  

θ&#039;(0,t) = θ&#039;(0,0) Exp{-2{[λ(m) + λ(r)]t + ỉ[ω(m) + ω(r)]t}}

(1): d² (m r)/dt² - (m r) θ&#039;² = -GmM/r² = -Gm³M/m²r²

      d² (m r)/dt² - (m r) θ&#039;² = -Gm³ (θ, 0) φ³ (0, t) M/ (m²r²)     

Let m r =1/u

d (m r)/d t = -u&#039;/u² = -(1/u²)(θ&#039;)d u/d θ = (- θ&#039;/u²)d u/d θ = -H d u/d θ
d²(m r)/dt² = -Hθ&#039;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&#039;s time dependent Equation --------II

If λ (m) ≈ 0 fixed mass and λ(r) ≈ 0 fixed orbit; then 

θ&#039;(0,t) = θ&#039;(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&#039;(0,t) = θ&#039;(0, 0) Exp {-2ì[ω(m) + ω(r)]t}

θ&#039;(0,0)=h(0,0)/r²(0,0)=2πab/Ta²(1-ε)²
         
         = 2πa² [√ (1-ε²)]/T a² (1-ε) ²; θ&#039;(0, 0) = 2π [√ (1-ε²)]/T (1-ε) ²   

θ&#039;(0,t)  = {2π[√(1-ε²)]/T(1-ε)²}Exp{-2[ω(m) + ω(r)]t 

θ&#039;(0,t)  = {2π[√(1-ε²)]/(1-ε)²}{cos 2[ω(m) + ω(r)]t - ỉ sin 2[ω(m) + ω(r)]t}

θ&#039;(0,t) = θ&#039;(0,0) {1- 2sin² [ω(m) + ω(r)]t - ỉ 2isin [ω(m) + ω(r)]t cos [ω(m) + ω(r)]t}  

θ&#039;(0,t) =  θ&#039;(0,0){1 - 2[sin ω(m)t cos ω(r)t + cos ω(m) sin ω(r) t]²}

         - 2ỉ θ&#039;(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&#039;(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&#039;(0, 0) = - 2 θ&#039;(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 &lt;&lt; 1; (v°/c)² ≈ 0; v*/c &lt;&lt; 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) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²} [(v°+ v*)/c] ² degrees/100 years  

W” (ob) = (-720x26526x3600/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&lt;&lt;M; Solar system    
v (M) = √ [Gm² / (m + M)a(1-ε²/4)] ≈  0; m&lt;&lt;M 


Application 1: Advance of Perihelion of mercury.

G=6.673x10^-11; M=2x10^30kg; m=.32x10^24kg; ε = 0.206; T=88days
c = 299792.458 km/sec; a = 58.2km/sec; 1-ε²/4 = 0.989391
ρ (m) = 0.696x10^9m; ρ(m)=2.44x10^6m; 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&quot; (ob) = (-720x36526x3600/T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ²
W&quot; (ob) = (-720x36526x3600/88) x (1.552) (48.14/299792)² = 43.0”/century

V1143Cgyni Apsidal Motion Solution

W° (ob) = (-720x36526/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</description>
		<content:encoded><![CDATA[<p>V1143Cgyni Binary Stars Apsidal motion Puzzle solution</p>
<p> The motion puzzle that Einstein MIT Harvard Cal-tech  NASA and all others could not solve.  </p>
<p>Introduction: For 350 years Physicists Astronomers and Mathematicians missed Kepler&#8217;s time dependent equation that changed Newton&#8217;s equation into a time dependent Newton&#8217;s equation and together these two equations combine classical mechanics and quantum mechanics into one mechanics explains &#8220;relativistic&#8221; 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 &#8220;physicists&#8221; could not solve by space-time physics or any published physics.                  </p>
<p>All there is in the Universe is objects of mass m moving in space (x, y, z) at a location<br />
r = r (x, y, z). The state of any object in the Universe can be expressed as the product </p>
<p>S = m r; State = mass x location: </p>
<p>P = d S/d t = m (d r/dt) + (dm/dt) r = Total moment<br />
   = change of location + change of mass<br />
  = m v + m&#8217; r; v = velocity = d r/d t; m&#8217; = mass change rate</p>
<p>F = d P/d t = d²S/dt² = Total force<br />
   = m(d²r/dt²) +2(dm/dt)(d r/d t) + (d²m/dt²)r<br />
   = mγ + 2m&#8217;v +m&#8221;r; γ = acceleration; m&#8221; = mass acceleration rate</p>
<p>In polar coordinates system</p>
<p>r = r r(1) ;v = r&#8217; r(1)  + r θ&#8217; θ(1) ; γ = (r&#8221; &#8211; rθ&#8217;²)r(1) + (2r&#8217;θ&#8217; + rθ&#8221;)θ(1)<br />
Proof:<br />
r = r [cosθ î + sinθĴ] = r r (1); r (1) = cosθ î + sinθ Ĵ<br />
v = d r/d t = r&#8217; r (1) + r d[r (1)]/d t = r&#8217; r (1) + r θ&#8217;[- sinθ î + cos θĴ] = r&#8217; r (1) + r θ&#8217; θ (1)</p>
<p>θ (1) = -sinθ î +cosθ Ĵ; r(1) = cosθî + sinθĴ</p>
<p>d [θ (1)]/d t=  θ&#8217; [- cosθî - sinθĴ= - θ' r (1)<br />
d [r (1)]/d t = θ&#8217; [ -sinθ'î + cosθ]Ĵ = θ&#8217; θ(1) </p>
<p>γ = d [r'r(1) + r θ' θ (1)] /d t = r&#8221; r(1) + r&#8217; d[r(1)]/d t + r&#8217; θ&#8217; r(1) + r θ&#8221; r(1) +r θ&#8217; d[θ(1)]/d t</p>
<p>γ = (r&#8221; &#8211; rθ&#8217;²) r(1) +  (2r&#8217;θ&#8217; + r θ&#8221;) θ(1)</p>
<p>F = m[(r"-rθ'²)r(1) + (2r'θ' + rθ")θ(1)] + 2m&#8217;[r'r(1) + rθ'θ(1)] + (m&#8221;r) r(1)</p>
<p> = [d²(mr)/dt² - (mr)θ'²]r(1) + (1/mr)[d(m²r²θ')/dt]θ(1) = [-GmM/r²]r(1)</p>
<p>d²(mr)/dt² &#8211; (mr)θ&#8217;² = -GmM/r²   Newton&#8217;s Gravitational Equation     (1)<br />
d(m²r²θ&#8217;)/dt = 0                           Central force law                              (2)</p>
<p>(2) : d(m²r²θ&#8217;)/d t = 0   m²r²θ&#8217; = [m²(θ,0)φ²(0,t)][ r²(θ,0)ψ²(0,t)][θ'(θ, t)]<br />
                                                         = [m²(θ,t)][r²(θ,t)][θ'(θ,t)]<br />
                                                         = [m²(θ,0)][r²(θ,0)][θ'(θ,0)]<br />
                                                         = [m²(θ,0)]h(θ,0);h(θ,0)=[r²(θ,0)][θ'(θ,0)]<br />
                                                         = H (0, 0) = m² (0, 0) h (0, 0)<br />
                                                         = m² (0, 0) r² (0, 0) θ&#8217;(0, 0)<br />
m = m (θ, 0) φ (0, t) = m (θ, 0) Exp [λ (m) + ì ω (m)] t; Exp = Exponential<br />
φ (0, t) = Exp [ λ (m) + ỉ ω (m)]t  </p>
<p>r = r(θ,0) ψ(0, t) = r(θ,0) Exp [λ(r) + ì ω(r)]t<br />
ψ(0, t) = Exp [λ(r) + ỉ ω (r)]t</p>
<p>θ&#8217;(θ, t) = {H(0, 0)/[m²(θ,0) r(θ,0)]}Exp{-2{[λ(m) + λ(r)]t + ì [ω(m) + ω(r)]t}} &#8212;&#8212;I<br />
Kepler&#8217;s time dependent equation that Physicists Astrophysicists and Mathematicians missed for 350 years that is going to demolish Einstein&#8217;s space-jail of time  </p>
<p>θ&#8217;(0,t) = θ&#8217;(0,0) Exp{-2{[λ(m) + λ(r)]t + ỉ[ω(m) + ω(r)]t}}</p>
<p>(1): d² (m r)/dt² &#8211; (m r) θ&#8217;² = -GmM/r² = -Gm³M/m²r²</p>
<p>      d² (m r)/dt² &#8211; (m r) θ&#8217;² = -Gm³ (θ, 0) φ³ (0, t) M/ (m²r²)     </p>
<p>Let m r =1/u</p>
<p>d (m r)/d t = -u&#8217;/u² = -(1/u²)(θ&#8217;)d u/d θ = (- θ&#8217;/u²)d u/d θ = -H d u/d θ<br />
d²(m r)/dt² = -Hθ&#8217;d²u/dθ² = &#8211; Hu²[d²u/dθ²]</p>
<p>-Hu² [d²u/dθ²] -(1/u)(Hu²)² = -Gm³(θ,0)φ³(0,t)Mu²<br />
[d²u/ dθ²] + u = Gm³(θ,0)φ³(0,t)M/H²</p>
<p>t = 0; φ³ (0, 0) = 1<br />
u = Gm³(θ,0)M/H² + Acosθ =Gm(θ,0)M(θ,0)/h²(θ,0)</p>
<p>mr = 1/u = 1/[Gm(θ,0)M(θ,0)/h(θ,0) + Acosθ]<br />
     = [h²/Gm(θ,0)M(θ,0)]/{1  + [Ah²/Gm(θ,0)M(θ,0)][cosθ]}</p>
<p>     = [h²/Gm(θ,0)M(θ,0)]/(1 + εcosθ)<br />
mr = [a(1-ε²)/(1+εcosθ)]m(θ,0)</p>
<p>r(θ,0)  = [a(1-ε²)/(1+εcosθ)] m r = m(θ, t) r(θ, t)<br />
      = m(θ,0)φ(0,t)r(θ,0)ψ(0,t) </p>
<p>r(θ,t) = [a(1-ε²)/(1+εcosθ)]{Exp[λ(r)+ω(r)]t} Newton&#8217;s time dependent Equation &#8212;&#8212;&#8211;II</p>
<p>If λ (m) ≈ 0 fixed mass and λ(r) ≈ 0 fixed orbit; then </p>
<p>θ&#8217;(0,t) = θ&#8217;(0,0) Exp{-2ì[ω(m) + ω(r)]t}</p>
<p>r(θ, t) = r(θ,0) r(0,t) = [a(1-ε²)/(1+εcosθ)] Exp[i ω (r)t]</p>
<p>m = m(θ,0) Exp[i ω(m)t] = m(0,0) Exp [ỉ ω(m) t] ; m(0,0)   </p>
<p>θ&#8217;(0,t) = θ&#8217;(0, 0) Exp {-2ì[ω(m) + ω(r)]t}</p>
<p>θ&#8217;(0,0)=h(0,0)/r²(0,0)=2πab/Ta²(1-ε)²</p>
<p>         = 2πa² [√ (1-ε²)]/T a² (1-ε) ²; θ&#8217;(0, 0) = 2π [√ (1-ε²)]/T (1-ε) ²   </p>
<p>θ&#8217;(0,t)  = {2π[√(1-ε²)]/T(1-ε)²}Exp{-2[ω(m) + ω(r)]t </p>
<p>θ&#8217;(0,t)  = {2π[√(1-ε²)]/(1-ε)²}{cos 2[ω(m) + ω(r)]t &#8211; ỉ sin 2[ω(m) + ω(r)]t}</p>
<p>θ&#8217;(0,t) = θ&#8217;(0,0) {1- 2sin² [ω(m) + ω(r)]t &#8211; ỉ 2isin [ω(m) + ω(r)]t cos [ω(m) + ω(r)]t}  </p>
<p>θ&#8217;(0,t) =  θ&#8217;(0,0){1 &#8211; 2[sin ω(m)t cos ω(r)t + cos ω(m) sin ω(r) t]²}</p>
<p>         &#8211; 2ỉ θ&#8217;(0, 0) sin [ω (m) + ω(r)] t cos [ω (m) + ω(r)] t</p>
<p>Δ θ (0, t)        = Real Δ θ (0, t) + Imaginary Δ θ (0.t)</p>
<p> Real Δ θ (0, t) = θ&#8217;(0, 0) {1 &#8211; 2[sin ω (m) t cos ω(r) t + cos ω (m)t sin ω(r)t]²} </p>
<p>W(ob) = Real Δ θ (0, t) &#8211; θ&#8217;(0, 0) = &#8211; 2 θ&#8217;(0, 0){(v°/c)√ [1-(v*/c) ²] + (v*/c)√ [1- (v°/c) ²]}²</p>
<p>v ° = spin velocity; v* = orbital velocity; v°/c = sin ω (m)t; v*/c = cos ω (r) t </p>
<p>v°/c &lt;&lt; 1; (v°/c)² ≈ 0; v*/c &lt;&lt; 1; (v*/c)² ≈ 0</p>
<p>W (ob) = &#8211; 2[2π √ (1-ε²)/T (1-ε) ²] [(v° + v*)/c] ²</p>
<p>W (ob) = (- 4π /T) {[√ (1-ε²)]/ (1-ε) ²} [(v° + v*)/c] ² radians<br />
W (ob) = (-720/T) {[√ (1-ε²)]/ (1-ε) ²} [(v° + v*)/c] ² degrees; Multiplication by 180/π </p>
<p>W° (ob) = (-720&#215;36526/T) {[√ (1-ε²)]/ (1-ε) ²} [(v°+ v*)/c] ² degrees/100 years  </p>
<p>W” (ob) = (-720&#215;26526x3600/T) {[√ (1-ε²)]/ (1-ε) ²} [(v° + v*)/c] ² seconds /100 years </p>
<p>The circumference of an ellipse: 2πa (1 &#8211; ε²/4 + 3/16(ε²)²- &#8211;.) ≈ 2πa (1-ε²/4); R =a (1-ε²/4)<br />
v (m) = √ [GM²/ (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; m&lt;&lt;M; Solar system<br />
v (M) = √ [Gm² / (m + M)a(1-ε²/4)] ≈  0; m&lt;&lt;M </p>
<p>Application 1: Advance of Perihelion of mercury.</p>
<p>G=6.673&#215;10^-11; M=2&#215;10^30kg; m=.32&#215;10^24kg; ε = 0.206; T=88days<br />
c = 299792.458 km/sec; a = 58.2km/sec; 1-ε²/4 = 0.989391<br />
ρ (m) = 0.696&#215;10^9m; ρ(m)=2.44&#215;10^6m; T(sun) = 25days<br />
v° (M) = 2km/sec ; v° = 2meters/sec<br />
v *= v(m) = √ [GM/a (1-ε²/4)]; v(M) = √[Gm²/(m + M)a(1-ε²)] ≈ 0<br />
v°(m) = 2m/sec (Mercury) v°(M)= 2km/sec(sun)<br />
Calculations yields: v = v* + v° =48.14km/sec (mercury); [√ (1- ε²)] (1-ε) ² = 1.552<br />
W&#8221; (ob) = (-720&#215;36526x3600/T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ²<br />
W&#8221; (ob) = (-720&#215;36526x3600/88) x (1.552) (48.14/299792)² = 43.0”/century</p>
<p>V1143Cgyni Apsidal Motion Solution</p>
<p>W° (ob) = (-720&#215;36526/T) {[√ (1-ε²)]/ (1-ε) ²} [(v°+ v*)/c] ² degrees/100 years  </p>
<p>v° = -v°(m) + v°(M)<br />
v* =  2v(cm) + σ<br />
v°(m) = spin velocity of primary<br />
v°(M) = spin velocity of secondary<br />
v(cm) = [m v(m) + M v(M)]/(m + M) center of mass velocity<br />
σ = √ {{[v(m) - v(cm)]² + [v(M) - v(cm)]²}/2} = standard deviation<br />
W° = 3.36°/century as reported in many articles</p>
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	<item>
		<title>By: Alexander Nahhas</title>
		<link>http://blog.oup.com/2005/12/newton___the_gr/comment-page-1/#comment-149212</link>
		<dc:creator>Alexander Nahhas</dc:creator>
		<pubDate>Sun, 08 Feb 2009 05:58:25 +0000</pubDate>
		<guid isPermaLink="false">http://216.110.190.15/2005/12/newton_-_the_greatest_alchemist_of_them_all/#comment-149212</guid>
		<description>Einstein&#039;s Physics Dollar Store on Campus 
MIT Harvard Cal-Tech 
Sponsored by NASA
Why Relativity theory is not Physics and why Einstein&#039;s &quot;thought&quot; = 0 
Walking the walk and talking the talk taking on all space-time confusion of physics by 
MIT Harvard and Cal-Tech and all other Physics dollar stores departments
And why LHC burned itself

Visual Effects and the confusions of &quot;Modern&quot; 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&#039;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 = √ [1-(v/c) ²]</description>
		<content:encoded><![CDATA[<p>Einstein&#8217;s Physics Dollar Store on Campus<br />
MIT Harvard Cal-Tech<br />
Sponsored by NASA<br />
Why Relativity theory is not Physics and why Einstein&#8217;s &#8220;thought&#8221; = 0<br />
Walking the walk and talking the talk taking on all space-time confusion of physics by<br />
MIT Harvard and Cal-Tech and all other Physics dollar stores departments<br />
And why LHC burned itself</p>
<p>Visual Effects and the confusions of &#8220;Modern&#8221; physics</p>
<p>  r &#8212;&#8212;&#8212; Light sensing of moving objects &#8212;&#8212;- S<br />
Actual object&#8212;&#8211; Light &#8212;&#8212;&#8212; Visual object<br />
r &#8211; &#8212;&#8212;-cosine (wt) + i sine (wt) &#8211; S = r [cosine (wt) + i sine (wt)]<br />
Newton&#8211; Kepler&#8217;s time visual effects &#8212; Time dependent Newton<br />
Particle &#8212;&#8212;&#8212;&#8212;&#8211; Visual effects &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211; Wave</p>
<p>Line of Sight:  r cosine wt </p>
<p>r &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- r cosine (wt) line of sight light aberrations</p>
<p>A moving object with velocity v will be visualized by<br />
light sensing through an angle (wt);w = constant and t= time<br />
Also, sine wt = v/c; cosine wt = √ [1-sine² (wt) = √ [1-(v/c) ²]                                                     </p>
<p>A visual object moving with velocity v will be seen as S                                                                   </p>
<p>S = r [cosine (wt) + i sine (wt)] = r Exp [i wt]; Exp = Exponential   </p>
<p>S = r [√ [1-(v/c) ²] + ỉ (v/c)] = S x + i S y</p>
<p>S x = Visual along the line of sight = r [√ [1-(v/c) ²]</p>
<p>This Equation is special relativity length contraction formula<br />
And it is just the visual effects caused by light aberrations of a<br />
moving object along the line of sight.</p>
<p>In a right angled velocity triangle A B C: Angle A = wt; angle B = 90°; Angle C = 90° -wt<br />
AB = hypotenuse = c; BC = opposite = v; CA= adjacent = √ [1-(v/c) ²]</p>
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	<item>
		<title>By: Joe nahhas</title>
		<link>http://blog.oup.com/2005/12/newton___the_gr/comment-page-1/#comment-149088</link>
		<dc:creator>Joe nahhas</dc:creator>
		<pubDate>Mon, 26 Jan 2009 16:54:57 +0000</pubDate>
		<guid isPermaLink="false">http://216.110.190.15/2005/12/newton_-_the_greatest_alchemist_of_them_all/#comment-149088</guid>
		<description>Say NO to Einstein
Kepler (demolish) Vs Einstein&#039;s 
Areal velocity is constant: r² θ&#039; =h         Kepler&#039;s Law
 h = 2π a b/T; b=a√ (1-ε²); a = mean distance value; ε = eccentricity
r² θ&#039;= h = S² w&#039;
Replace r with S = r exp (ỉ wt); h = [r² Exp (2iwt)] w&#039;
w&#039; = (h/r²) exp [-2(i wt)] 
w&#039;= (h/r²) [cosine 2(wt) - ỉ sine 2(wt)] = (h/r²) [1- 2sine² (wt) - ỉ sin 2(wt)] 
w&#039; =  w&#039;(x) + ỉ w&#039;(y) ;  w&#039;(x) = (h/r²) [ 1- 2sine² (wt)]  
 w&#039;(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&#039; = [w&#039;(x) – h/r²] = -4π {[√ (1-ε²)]/T (1-ε) ²} (v/c) ² radian per second
Δ w&#039; = (- 4π /T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ² radians 
Δ w&#039; = (-720/T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ² degrees; Multiplication by 180/π 
Δ w&#039; = (-720x36526/T) {[√ (1-ε²)]/(1-ε)²} (v/c)² degrees/100 years  
Δ w” = (-720x3600/T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ² seconds of arc by 3600
Δ w&quot; = (-720x36526x3600/T) {[√ (1-ε²]/(1-ε)²} (v/c)² seconds of arc per century
This Kepler&#039;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&lt;&lt;M; Solar system    
 Advance of Perihelion of mercury.
G=6.673x10^-11; M=2x10^30kg; m=.32x10^24kg
 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&quot; seconds of arc of advance of perihelion of Planet Mercury (General relativity) is given by Kepler&#039;s equation better than all of Published papers of Einstein. Kepler&#039;s Equation can solve Einstein&#039;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?</description>
		<content:encoded><![CDATA[<p>Say NO to Einstein<br />
Kepler (demolish) Vs Einstein&#8217;s<br />
Areal velocity is constant: r² θ&#8217; =h         Kepler&#8217;s Law<br />
 h = 2π a b/T; b=a√ (1-ε²); a = mean distance value; ε = eccentricity<br />
r² θ&#8217;= h = S² w&#8217;<br />
Replace r with S = r exp (ỉ wt); h = [r² Exp (2iwt)] w&#8217;<br />
w&#8217; = (h/r²) exp [-2(i wt)]<br />
w&#8217;= (h/r²) [cosine 2(wt) - ỉ sine 2(wt)] = (h/r²) [1- 2sine² (wt) - ỉ sin 2(wt)]<br />
w&#8217; =  w&#8217;(x) + ỉ w&#8217;(y) ;  w&#8217;(x) = (h/r²) [ 1- 2sine² (wt)]<br />
 w&#8217;(x) – (h/r²) = &#8211; 2(h/r²)sine²(wt) = &#8211; 2(h/r²)(v/c)²  v/c=sine wt<br />
(h/ r²)(Perihelion/Periastron)= [2πa.a√ (1-ε²)]/Ta² (1-ε) ²= [2π√ (1-ε²)]/T (1-ε) ²<br />
 Δ w&#8217; = [w'(x) – h/r²] = -4π {[√ (1-ε²)]/T (1-ε) ²} (v/c) ² radian per second<br />
Δ w&#8217; = (- 4π /T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ² radians<br />
Δ w&#8217; = (-720/T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ² degrees; Multiplication by 180/π<br />
Δ w&#8217; = (-720&#215;36526/T) {[√ (1-ε²)]/(1-ε)²} (v/c)² degrees/100 years<br />
Δ w” = (-720&#215;3600/T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ² seconds of arc by 3600<br />
Δ w&#8221; = (-720&#215;36526x3600/T) {[√ (1-ε²]/(1-ε)²} (v/c)² seconds of arc per century<br />
This Kepler&#8217;s Equation solves all the problems Einstein and all physicists could not solve<br />
The circumference of an ellipse: 2πa (1 &#8211; ε²/4 + 3/16(ε²)²- &#8211;.) ≈ 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&lt;&lt;M; Solar system<br />
 Advance of Perihelion of mercury.<br />
G=6.673&#215;10^-11; M=2&#215;10^30kg; m=.32&#215;10^24kg<br />
 ε = 0.206; T=88days; c = 299792.458 km/sec; a = 58.2km/sec<br />
Calculations yields:<br />
 v =48.14km/sec; [√ (1- ε²)] (1-ε) ² = 1.552<br />
Δ w”= (-720&#215;36526x3600/88) x (1.552) (48.14/299792)²=43.0”/century<br />
Conclusions: The 43&#8243; seconds of arc of advance of perihelion of Planet Mercury (General relativity) is given by Kepler&#8217;s equation better than all of Published papers of Einstein. Kepler&#8217;s Equation can solve Einstein&#8217;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?</p>
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