How the Islamic world increased scientific knowledge between the 8th and 14th centuries.

Episode 1 parts 1-6

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```# The 'first true scientist'

By Professor Jim Al-Khalili

University of Surrey

An artist's impression of al-Hassan Ibn al-Haytham

**Isaac Newton is, as most will agree, the greatest physicist of all time.**

At the very least, he is the undisputed father of modern optics, or so we are told at school where our textbooks abound with his famous experiments with lenses and prisms, his study of the nature of light and its reflection, and the refraction and decomposition of light into the colours of the rainbow.

Yet, the truth is rather greyer; and I feel it important to point out that, certainly in the field of optics, Newton himself stood on the shoulders of a giant who lived 700 years earlier.

For, without doubt, another great physicist, who is worthy of ranking up alongside Newton, is a scientist born in AD 965 in what is now Iraq who went by the name of al-Hassan Ibn al-Haytham.

Most people in the West will never have even heard of him.

As a physicist myself, I am quite in awe of this man's contribution to my field, but I was fortunate enough to have recently been given the opportunity to dig a little into his life and work through my recent filming of a three-part BBC Four series on medieval Islamic scientists.

**Modern methods**

Popular accounts of the history of science typically suggest that no major scientific advances took place in between the ancient Greeks and the European Renaissance.

But just because Western Europe languished in the Dark Ages, does not mean there was stagnation elsewhere. Indeed, the period between the 9th and 13th Centuries marked the Golden Age of Arabic science.

Great advances were made in mathematics, astronomy, medicine, physics, chemistry and philosophy. Among the many geniuses of that period Ibn al-Haytham stands taller than all the others.

Ibn-al Haytham conducted early investigations into light

Ibn al-Haytham is regarded as the father of the modern scientific method.

As commonly defined, this is the approach to investigating phenomena, acquiring new knowledge, or correcting and integrating previous knowledge, based on the gathering of data through observation and measurement, followed by the formulation and testing of hypotheses to explain the data.

This is how we do science today and is why I put my trust in the advances that have been made in science.

But it is often still claimed that the modern scientific method was not established until the early 17th Century by Francis Bacon and Rene Descartes.

There is no doubt in my mind, however, that Ibn al-Haytham arrived there first.

In fact, with his emphasis on experimental data and reproducibility of results, he is often referred to as the "world's first true scientist".

**Understanding light**

He was the first scientist to give a correct account of how we see objects.

**It is incredible that we are only now uncovering the debt that today's physicists owe to an Arab who lived 1,000 years ago**
Prof Jim Al-Khalili

He proved experimentally, for instance, that the so-called emission theory (which stated that light from our eyes shines upon the objects we see), which was believed by great thinkers such as Plato, Euclid and Ptolemy, was wrong and established the modern idea that we see because light enters our eyes.

What he also did that no other scientist had tried before was to use mathematics to describe and prove this process.

So he can be regarded as the very first theoretical physicist, too.

He is perhaps best known for his invention of the pinhole camera and should be credited with the discovery of the laws of refraction.

He also carried out the first experiments on the dispersion of light into its constituent colours and studied shadows, rainbows and eclipses; and by observing the way sunlight diffracted through the atmosphere, he was able to work out a rather good estimate for the height of the atmosphere, which he found to be around 100km.

**Enforced study**

In common with many modern scholars, Ibn-al Haytham badly needed the time and isolation to focus on writing his many treatises, including his great work on optics.

He was given an unwelcome opportunity, however, when he was imprisoned in Egypt between 1011 and 1021, having failed a task set him by a caliph in Cairo to help solve the problem of regulating the flooding of the Nile.

While still in Basra, Ibn al-Haytham had claimed that the Nile's autumn flood waters could be held by a system of dykes and canals, thereby preserved as reservoirs until the summer's droughts.

But on arrival in Cairo, he soon realised that his scheme was utterly impractical from an engineering perspective.

Yet rather than admit his mistake to the dangerous and murderous caliph, Ibn-al Haytham instead decided to feign madness as a way to escape punishment.

This promptly led to him being placed under house arrest, thereby granting him 10 years of seclusion in which to work.

**Planetary motion**

He was only released after the caliph's death. He returned to Iraq where he composed a further 100 works on a range of subjects in physics and mathematics.

While travelling through the Middle East during my filming, I interviewed an expert in Alexandria who showed me recently discovered work by Ibn al-Haytham on astronomy.

It seems he had developed what is called celestial mechanics, explaining the orbits of the planets, which was to lead to the eventual work of Europeans like Copernicus, Galileo, Kepler and Newton.

It is incredible that we are only now uncovering the debt that today's physicists owe to an Arab who lived 1,000 years ago.

*Professor Jim Al-Khalili presents Science and Islam on BBC Four at 2100GMT on Monday 5, 12 & 19 January*

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Joe Nahhas

said:Ending Einstein in 2009

There is One and Only One Mechanics: Universal Mechanics:

Introduction to 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 = 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)

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}}

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

(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)] —————————————————————– II

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} Nahhas’ Equation ——————-III

If λ (m) ≈ 0 and λ(r) ≈ 0; then

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

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

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

S = r(θ) = Exp [ì ω(r)t]; m = m(0) Exp[i ω(m)t]

θ'(0,0)=h(0,0)/r(0,0)=2πab/Ta²(1-ε)²

= 2πa²[√(1-ε²)]/Ta²(1-ε)²

θ'(0,0)= 2π[√(1-ε²)]/(1-ε)²

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

1- E = 1/2(mc²)

S = r Exp [i ω(r) t] = r [cosω(r) t + ìsinω(r) t]

P =d S/d t = v Exp [ì ω(r) t] + ì r w Exp (ì w t); v =d r/d t; v = ω(r)r

= v (1+ ì) {Exp [ì w(r) t)] = visual velocity

E (definition) = mv²/2; E = mc²/2 If v = c

E (visual) = mp²/2 = mv²/2(1+ì) ² Exp 2[ì w(r) t]

E (visual) = mv²/2(1 + 2ì -1) [cos2w(r) t + ì sin2w(r)t]

E (visual) = ì (mv²) {1-2sin²w(r) t +2i[sinw(r)t][cos(r)t]}

E (visual) = ±ì (mc²); -ì = negative complex unit number; w(r) t = (n+1) π/2

If w(r)t = π/4

E (visual) = ỉ(mc²)[1-1+ỉ]= -mc²

If w(r)t = 3/4π

E (visual) = ỉ(mc²)[1-1-ỉ]= mc²

If w(r)t = 5π/4

E (visual) = ỉ(mc²)[1-1+ỉ]= mc²

If w(r)t = 7π/4

E (visual) = ỉ(mc²)[1-1+ỉ] = -mc²

E= mc² is illusion of E = 1/2 mc²

1-1 E=1/2mc²

E=1/2m [(m v + m’ r) ²]; v = 0; m’ r = m c

E=1/2m (mc) ² = 1/2mc²

E=/12mc²

1-2 E=1/2mv²=1/2mc² (1+i) ²; v = c (1+i) Michelson-Morley

E = 1/2mc² (2i) = imc²

1-4 If v = (1+ ỉπ/4) c; v² = (√2/2 + ỉ√2/2)² c²

E = 1/2mv² = 1/2m (1/2 -1/2 + ỉ) c² = ỉ/2mc²

2- Advance of perihelion of Planets ω (m) = 0

Central force law Areal velocity is constant: r² (d θ/d t) =h

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

r² (d θ/d t) = h = S² (d w/d t)

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

(d w/d t) = (h/r²) exp [-2(i wt)]

d w/d t= (h/r²) [cosine 2(wt) – ỉ sine 2(wt)] = (h/r²) [1- 2sine² (wt) – ỉ sin 2(wt)]

d w/d t = d w(x)/d t + d w(y)/d t; d w(x)/d t = (h/r²) [ 1- 2sine² (wt)]

d w(x)/d t – (h/r²) = – 2(h/r²)sine²(wt) = – 2(h/r²)(v/c)² v/c=sine wt

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

Δ w/d t = (d w/d t – h/r²] = -4π {[√ (1-ε²)]/T (1-ε) ²} (v/c) ² radian per second

Δ w/d t = (- 4π /T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ² radians

Δ w°/d t = (-720/T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ² degrees; Multiplication by 180/π

Δ w°/d t = (-720×36526/T) {[√ (1-ε²)]/(1-ε)²} (v/c)² degrees/100 years

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

Δ w”/d t = (-720x36526x3600/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=√ [GM²/ (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; m<<M; Solar system

Relativity explained without relativity

Application 3: 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

v = √ [GM/a (1-ε²/4)]

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

Δ w"= (-720x36526x3600/T) {[√ (1-ε²)]/(1-ε)²}(v/c)²

Δ w"= (-720x36526x3600/88) x (1.552) (48.14/299792)²=43.0”/century

Space-time continuum is as good as garbage

Application 4: Gravitational red shift: Pound Rebka Experiment

S=r exp (î wt); 1/S= (1/r) {Exp [-(î wt)]}

υ = υ (0) Exp (-ì wt) will yield; Δ υ/υ(0)=1/2(v/c)²[up]; Δ υ/υ[down]=-1/2(v/c)²

Δυ/υ(0)[up]-Δ υ/υ(0)[down]=(v/c)²; v² = 2gh; Δ υ/υ(0) [total] = (v/c) ²

v² = 2gh=2(9.806) (22.6m)

Δ υ/υ [Total]=[2×9.806×22.6/299792.458]=4.93169×10^-15

Observed value: 5.1±5×10^-15

5- Light bending: Lord Eddington experiment

Δ w/d t = – 2(h/r²) (v/c)² =-2 (2A/r²T)(v/c)²

T [Δ w/d t] = (A/r²) [-4(v/c) ²]; v² = GM/R

T [Δ w/d t] = (A/r²) [1.75"]

The values depends on near by stars measurements.

If you collect all values measured they will fit this equation.

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

Δ w/d t = -2(h/r²) (v/c)²=-2(2A/r²T)(v/c)²

T(Δ w/d t) = -4π(v/c)²; Δ w = -4 π (v/c)²

a (earth) = 149,597,887.5km; a(mars)= 227,939,100 km

a = [a (earth) + a (mars)]/2 = 377,536, 987.5/2=188,768,493.8km

ε = [a (earth) – a (mars)]/ [a (earth) + a (mars)] = .2075

1- ε²/4 = 0.989235814; a (1-ε²/4) =18676554.6km

v = √ [GM/a (1-ε²/4)]=26.6575872km/sec

d= distance between mars and earth

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 = [

(1) -a(2)]/[a(1) + a(2)]

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.

Problems that Einstein and All 100,000 space-time physicists can not solve by any Physics and Solved by Universal Mechanics:

7- Binary stars apsidal motion

r = r (θ,0) Exp [λ(r) + ì ω(r)] t; m = m (θ,0) Exp [λ (m) + ì ω (m)] t

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

λ(m) ≈ 0; λ(r) ≈ 0 then

r(θ,t) = r(θ,0) Exp [ì ω (r)t]; m = m(θ,0) Exp [ ì ω (m)t]

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

θ'(θ,t) = θ'(θ,0) {cos2[ω(r) + ω(m)]t – ì sin2[ω(r) + ω(m)]t}

θ'(θ,t) = h/r²(0,0){1 – 2sin²[ω(r) + ω(m)]t -2 ì sin {[ω(r) + ω(m)]t}cos{[ω(r) + ω(m)]t}}

θ'(θ,t) -h/r²(0,0) = – 2(h/r²){sin²[ω(r) + ω(m)]t – ì sin {[ω(r) + ω(m)]t}cos{ω(r) + ω(m)]t}

W° = -2(h/r²)sin²[ω(r) + ω(m)]t; sin[ω(r)t] = v/c= orbital speed/light speed

sin[ω(m)t] = v°/c = spin speed/light speed

W° = -2(h/r²) [sinω(r)tcosω(m)t + cosω(r)tsinω(m)t]²

W° = -2(h/r²) {(v/c) {√[1-(v°/c)²]} + {√[1-(v/c)²]}(v°/c)}²

W° = -2(h/r²) [(v/c) + (v°/c)]²; h/r² = 2πab/Ta²(1- ε)²; b=√(1-ε²)

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

DI Her Solution

Data: T=10.55days r(m) = 0.0621 m=5.15M(0) R(m)=2.68R(0) [v°(m),v°(M)]=[45,45]

ε = 0.4882; r(M) = 0.0574 M=4.52M(0) R(M) =2.48 m + M=9.67M(0)

1-ε = 0.5118; (1-ε²/4) = 0.94; [√ (1-ε²)] / (1-ε) ² = 3.33181

G=6.673×10^-11; M(0) = 1.98892×19^30kg; R(0) = 0.696×10^9m

Calculations

V (m) = √ [GM²/ (m + M) a (1-ε²/4)] = 99.88 km/sec

V(M) = √ [Gm²/ (m + M) a (1-ε²/4)] = 113.9km/sec

Apsidal motion is given by this formula:

W°(observed) = (-720×36526/T) {[√ (1-ε²)]/ (1-ε)²]}[(v/c) + (v°/c)]²

v° (spin) = (45km-45km)= 0

v(k) (Orbit)=v(cm)/c

v(k) = v(cm)=∑ m v/∑m=[(5.15×99.88)+ (4.52×113.8)]/9.67 = 106.38km/s

W° (observed) = (-720×36526/T) {[√ (1-ε²)]/ (1-ε) ²} {[v (k) + v°]/c] ²

= (-720×36526/10.55) (3.33181) (106.38/300,000)²

W° (observed) = 1.04°/century same as 1994 article

AS Cam Solution T=3.431; r(m) =0.1499 m=3.3M(0) R(m) =2.57R(0) [v°(m),v°(M)]=[40,30]

ε = 0.1695; 1-ε = 00.8305; r(M) =0.1111 M=2.5M(0) R(M) = 2.5R(0) ;m + M=5.8M(0)

G=6.673×10^-11; M (0) = 1.98892×10^30kg; R (0) = 0.696×10^9m

1- ε²/4 = 0.9928; [√(1-ε²)]/(1-ε)² = 1.43

a = [R(m)/r(m)] = (2.57/0.1499)(0.696×10^9)m

a(1-ε²/4)=(2.57/0.1499)(0.696×10^9)(0.9988)=11.8470×10^9m

v(m) = √[GM²/(m + M)a(1-ε²/4)] = 110.1786325km/sec

v(M) = √[Gm²/(m + M)a(1-ε²/4)] = 145.435795km/sec

v(cm) = ∑m v/∑m = 125.3756853km/sec

σ =√{∑[v-v(cm)]²/2}=√{[(110.1786325-125.3756853)² +(145.435795-125.7356853)²]/2}

σ = 25.1659669

Spin: v° = 40 + 30=70km/sec

v(m) = 110.1786325km/sec; v(M) = 145.435795km/sec; v(cm)= 125.3756853km/sec

σ = 25.1659669km/sec

1- v (k) = v (m) + v(M) +σ=350,7803944km/sec; [√ (1-ε²)]/(1-ε)² = 1.43; T = 3.431days

W°= (-720×36526/T) x {√ [(1-ε²)] (1-ε)²}{[v(k) + v°]/c}²= 15.0°/century

Dr Guinan : W°= 15°/century 1989

2- v (k) = 2v(cm) + σ = 275.9176729km/sec;

v(k) + v° = 275.9176729 + 70 = 345.9176729km/sec

W° = (-720×36526/T)x{[√(1-ε²)]/(1-ε)²}{[ v(k) + v°]/c}²= 14.6°/100 years

Khailullin: 1983 v (p) =110.4; v(s) = 145.8; σ=25.2685; 2∑ m v/∑m + σ + 70=346.0185

W°= 14.6 °/century same as reported

V1143 Solution

T= 7.641days r(m)=0.059 m=1.391M(0) R(1)=1.346R(0) [v°(m) ,v°(M) ]=[18,28]

ε = 0.54 r(M)=0.058 M=1.347M(0) R(2)=1.323R(0) [v°(m),v°(M)]=[21,28]

1-ε=0.46 1-ε²/4=0.9721 v (m)=77.5126;v(M)=80.00448; v(cm)=78.7385

σ =1.246; T =7.641; [√ (1-ε²)]/(1-ε)²= 3.977722951

∑m v/∑m=78.7385km/sec; [√(1-ε²)]/(1-ε)²=3.977622971

2 ∑m v/∑m + σ = 158.723km/sec

Spin: v° = 28-18=10km/sec

v (total) = 158.723km/sec-10=148.723km/sec

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

W°/century= (-720×36526/7.641) (3.977622951) (148.723/300,000)²=3.36°/century

W (observed) = 3.36°/100 years

v(m) + v(M) + σ = 77.5126 + 80.00448 + 1.24=158.75708km/sec

158.75708 – 10 = 148.75708km/sec

V541Solution: Apsidal motion catalogue

T = 15.338days r(m)=0.0441 m=2.69M(0) R(m)=2.0R(0) [v°(m),v°(M)]=[20±5,20±5]

ε = 0.474 r(M) =0.0425 M=2.60M(0) R(M)=1.92R(0) [v°(m),v°(M)]=[24±2,24±2]

1- ε = 0.526 1-ε²/4=0.943831 [√(1-ε²)]/(1-ε)²=3.1825

a = [R(m)/r(m)]R(0)= (2/0.0441)0.696×10^9m= 31.565×10^9m

a(1-ε²/4)=29.79167238×10^9m

v(m)= √[GM²/(m + M)a(1-ε²/4)]= 75.66km/sec; v(m)°=20

v(M)= √[Gm²/a(m + M)(1-ε²/4)]=78.28km/sec; v(M)°=20

v(cm)= ∑mv/∑m= 76.95km/sec; σ = √[∑(v-v(cm))²/2]=1.31

v(m) + v(M) + σ = 75.66 + 78.28 + 1.31= 155.19km/sec

2∑mv/∑m + σ = 2×76.95 + 1.31 = 155.21km/sec

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

v°=20 + 20 = 40km/sec

v(k) + v°= 155.19- 40=115.19km/sec

W°=0.80°/century

v(k) + v°= v(m) + v°(M) =75.66 + 24=99.66km/sec—-W°=0.60°/century Lacy

v + v°= [2v(cm) + σ(m)] -[2v° – σ°(=5)]

=155.19 – (40-5)=119.19km/sec ……………………..W°=0.86°Khalullin

Different answers are due to different methods of studying the motion

General rules:

Spin Addition/subtraction

primary(m) /secondary(M) + v°(m) – v°(m)

+v°(M) + v°(m) + v°(M) [As Cam] v(m) – v(M) [v1143]

-v°(M) v(m) – v(M)[DI her] -[v°(m) +v°(M)] [v541]

Orbit:

primary/secondary ∑m v/∑m + σ/2 ∑m v/∑m -σ/2

∑m v/∑m + σ/2 2(∑m v/∑m) + σ [As Cam], [v1143] 1/2= ∑m v/∑m

[DI Her]

∑m v/∑m -σ/2 -1/2= -∑m v/∑m DI Her 2(∑ m v/∑m) – σ[v541]

Spin – Orbit: Add or Subtract: Spin and orbit relative orientations can be told from apsidal motion

What is missing in all of Physics is self-referenced motion when added to relative motion it can produce the blue prints of motion. Energy is tiny particles in motion at high speed.

Finding and controlling these tiny particles is energy research. Four dimensional Universes of space-imaginary time claims have/had crippled energy research because it does not allow the visualization of such a motion and the basis of such Universes is Naive enough to not keep digging the wrong hole to find energy. Energy spent is based on distance traveled and not relative distance and distance traveled is self referenced distance.

Definition: self referenced motion: Applied to Earth

Can planet Earth measure the distance traveled by planet Earth after the passage of a specific time span τ from where it was at the beginning of the same specific time span τ?

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