## New Problem: The Kepler Exhibit

At the Exploratorium in San Francisco, you can play with this exhibit:

What should the height profile of the bowl be so that balls that roll without slipping (or, so that blocks sliding without friction) would reproduce 2-D Kepler orbits when viewed in projection from above?

### 10 Responses to “New Problem: The Kepler Exhibit”

1. Nik Says:

Would you calculate this in terms of theta where theta is the average inclination of the bowl?

2. meichenl Says:

What I was looking for was some function h = h(r), where “h” is the height of the bowl above the floor of the room, and “r” is the distance from the center of the bowl.

3. Answer: The Kepler Problem « Arcsecond Says:

[…] The Kepler Problem Here is the question I’m […]

4. Joe Nahhas Says:

Kepler (demolish) Vs Einstein’s

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

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?

5. meichenl Says:

Joe, I cannot understand your derivation without some further explanation. Why does $\Delta \omega '$ represent the advancement of the perihelion point? I am confused beginning from your first substitution, which shows that $\omega$ is some complicated, time-dependent, complex expression with no physical interpretation that is obvious to me.

6. Alexander Says:

give me an e-mail. I will send you the whole perihelion derivation

7. Alexander Says:

Here it is. Apsidal motion is for two stras where the center of mass moves but it is the same for advance of perihelion of mercury
There is one and only one Mechanics
Universal Mechanics: Apsidal motion of two body system
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 posted puzzles in all of Mechanics 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) = (-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<<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) = (-720x36526x3600/T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ²
W” (ob) = (-720x36526x3600/88) x (1.552) (48.14/299792)² = 43.0”/century

8. Alexander Says:

Relativity theory experimental proofs are just light aberrarions along the line of sight. Example of length contraction derivation as light aberration along the line of sight

Visual Mechanics (Demolish) Vs Einstein’s Space-Time

Visual Effects and the Confusions of “Modern” Physics

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

Line of Sight: r cosine wt: light aberrations

A moving object with velocity v will when 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 location along the line of sight = r [√ [1-(v/c) ²]

This Equation is special relativity distance contraction formula and it is just the visual effects and 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]

And what is the visual effect for angular velocity of Perihelion of Mercury along the line of sight?

9. Alexander Says:

Relativity theory experimental proofs claims are visual effects. Yes I am saying that Harvard MIT Cal-tech NASA and all others are wrong including 400 years od astronomy and 109 years of noble prize winners physics and physicists because they can not read a telescope.
It is like this:
1×12=12
2×6=12
3×4= 12
4×3=12
6×2=12
12×1=12
everytime physicist look at the same thing they get different answers for spacial and time but it is all visual effects

S = r [cosinewt +isinwt]
in absolute value they are the same thing but when r rotating it could give differet value for each wt value.
At the end of 19 century astronomers and physicists not familiar with measurements of moving objects reported these visual effects as real measurements without accounting for light aberrations and classrooms ended with fiction called relativity

10. The Crank Continuum « Arcsecond Says:

[…] By Mark Eichenlaub I’ve had one true crank on this blog. He jumped into the comments on this post with mathematical gibberish he claimed disproved relativity. Another time I saw a crank letter […]