The “Evil Baby Orphanage” is an idea this guy and his brother have about going back in time, taking babies that grow up to do a lot of bad things to society to some facility, and raising them there instead of going back and actually killing them/their parents to stop whatever it is they end up doing as adults. The relative time travel stuff starts at about 1:15.

What the bloody hell are you going on about? Quack.

im not saying that there are but if there are time travelers here, what do you think would happen if they said they came from the year 3000? they would be locked away in an insane asylum. also, if they said anything about them being time travelers, it could very likely cause a major paradox.

also in bttf part 2, wouldnt that be a paradox? the alternate 1985 where doc is in an insane asylum. he wouldnt have built the time machine.

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