I am a real person in real space in real time in real fight with modern and Nobel

Modern and Nobel physicists and astronomers measure in present time (real time = present time) and calculate in past time (event time = past) and puzzled by the difference between present time and past time (shift time = present time - past

time). Shift time is the time difference between when an event happens and when an event is measured. Shift time in labs is a measurement error or visual effects that make up make - believe modern and Nobel physics and astronomy. I am not

produced all wrong modern physics and astronomy that can be deleted in its entirety with real time physics and astronomy since July 21 1969.

I am Joe Nahhas founder of real time physics astronomy and mathematics

Page 1

Introduction

In July 21, 1969 at age 11 and after graduating from 5th grade and on the same day a man landed a foot on the moon I watched Apollo 11 take off and disappear into the skies on its way to the moon and then saw Neil Armstrong land human's first step on the moon. I wondered how someone sees and measures distances in space and how someone sees and measures sizes of objects in space. Apollo 11 rocket looked like it is shrinking in size while moving up into the skies; not as if the rocket shrunk in size but as if the rocket visual changed in size indication a different location and not a different size. Apollo 11 looked similar to a moving car moving away and shrinking in size. I realized that objects location and size has to do with how we see things (eye as an instrument). I imagined two snap shot of Apollo 11 at different distances A and B. At snap shot distance A Apollo 11 looked like it has a size C, and at snaps shot distance B Apollo 11 looked like it has a shrunk size D as shown below:

C

D

Eye ------------------------------ A ------------------------------ B

I asked myself the question: how A, B, C, and D are related1 = 1 is self evident; 2 = 2 is self evidentA = A is self evident

If A = A; add and subtract B; then A = B + (A - B)Divide by B; then (A/B) = 1 + (A - B)/B

Multiply by C; then (A/B) C = C + [(A - B)/B] C Equation - 1

Also D = D; add and subtract C; then D = C + (D - C) Equation - 2

Comparing Equation - 1 and Equation - 2(A/B) C = D; or, AC = BDC = C D - C = [(A - B)/B]

The answer is AC = BD = is how distances A and B related to sizes C and D And D - C = [(A - B)/B] C = visual contraction of C when moved from A to BAC = BD = actual distance x actual size = visual distance x visual size = constant

Page 2The initial condition solution to AC = BD is A = B and C = D

The general solution to AC = BD = constant = actual distance x actual size

AC = BD = constant = actual distance x actual size

(Cosine ω t + í sine ω t) = e i ω t; (Cosine ω t - í sine ω t) = e - i ω t (Cosine ω t + í sine ω t) (Cosine ω t - í sine ω t) = e i ω t e - i ω t = 1

AC = BD = kTaking AC = kDifferentiating with respect to timeThen d A/ d t + d C/ d t = d k/ d t = 0 And d A/ d t = - d C/ d t = λ + í ω; method of separation of variablesA = e (λ + i ω t) = e λ e i ω t = C e i ω t

C = e - (λ + i ω t) = e - λ e - i ω t = D e - i ω t

A = B e i ω t

C = D e - i ω t

AC = B e i ω t D e - i ω t = B D e i ω t e - i ω t = BD x 1 = BD

A real number C has a visual complex number D = C e i ω t

D along the line of sight = D x = C cosine ω t = C - 2 C sine2 (ω t/2)

D x - C = - 2 C sine2 (ω t/2) D x = C √ (1 - sine2 ω t); with ω t = tan -1 (v/c) = aberration angle D x = C √ [1 - sine2 tan -1 (v/c)]; (v/c) << 1

D x = C √ [1 - (v/c) 2]

D x = C cosine ω tD x / C = cosine ω t And ω t= cosine-1 (D x / C) = 1 - 2 sine 2 cosine-1 (D x / C)

D x / C = cosine ω t = 1 - 2 sine 2 cosine-1 (D x / 2 C)

[(D x / C) -1] = - 2 sine 2 cosine (D x / 2 C)

D - C = [(A - B)/B] C D - C = - 2 C sine 2 [cosine-1 (A / 2 B)]

Page 3

In practice: physicists and astronomers measure orbits of planets around the Sun not from the Sun (distance A) but from Earth (distance B)

A = B e i ω t

C = D e - i ω t

D - C = [(A - B)/B] C D - C = - 2 C sine 2 [cosine-1 (A / 2 B)]D - C = Einstein's numbers without Einstein = Illusions

Astronomers measure planet Mercury orbit around the Sun (distance A) not from the Sun but from Earth (distance B) and that means the orbit has visually shrunk and not actually shrunk by the quantity [(A - B)/B] C

A = Sun - Mercury distance = 5.82 x 109 meters; B = Sun Earth distance = 149.6 x 109 metersSun - Mercury Period in seconds = 88 days x 24 hours x 60 minutes x 60 seconds

Planet Mercury angular velocity around the Sun Is θ0'= 2 x π/88 x 24 x 60 x 60 radians per period Planet Mercury angular velocity around the Sun in arc second per century δθ0'= (2 x π /88 x 24 x 60 x 60) (180/ π) (36526/88) (3600) = 70.75 arc sec per century. If C = δθ0' = 70.75 arc sec per century measured from the Sun, then how much it is diminished if measured from Earth?A = 5.82 x 109; B = 149.6 x 109; C = 70.75And the answer is [(A - B)/B] C = [(5.82 x 109 - 149.6 x 109)/ 149.6 x 109] 70.75 = 43 arc sec per/100 years same numbers as Einstein's numbers Defining distance r = r x + í r y = r0 e í ω t And r = r0 [cosine ω t + sine í ω t] and r x = r0 [cosine ω tAnd r x - r0 = r0 [cosine ω t - 1] = - 2 r0 sine2 ω t/2; ω t = cosine-1 (r x/r0)And [(r x - r0)/r0] = - 2 sine2 {[cosine-1 (r x/r0)]/2} And [(r x - r0)/r0] δθ0'= - 2 sine2 {[cosine-1 (58.2/149.6)]/2} 70.75 = 43 1 - A real object (classical) of size C has a visual (quantum) of size D = C e í ω t

2 - D - C = visual illusions = (Einstein's relativity theory)

The fight is about this

Everything modern and Nobel said in physics and astronomy is all wrong! Modern and Nobel deletion is in progress and this book is 1st replacement

Page 4

Chapter One: light signal processing

1.1 - Light signal wave length shift Naming: r = Visual distance And r 0 = actual distance Then, r0 = r0; add and subtract visual distance rAnd, r0= r + (r0 - r); divide by r

Then (r0/ r) = 1 + [(r0 - r)/ r] multiply by δθ0'And (r0 / r) δθ0' = δθ0' + [(r0 - r)/ r] δθ0'Modern and Nobel error # 1 is: [(r0 - r)/ r] δθ0' = 70.75 [(r0 - r)/ r]= 70.75 [(58,200,000,000 - 149,600,000)/ 149,600,000] = 43 arc sec/ 100y

1.2 - Light signal period shift: T0 = r0/c = 58,200,000,000/300,000 (light velocity km/sec) = 194 secondsT = r/c = 149,600,000/300,000 = 498.67 secondsNaming: T = visual time And T0 = actual time Then, T0 = T0; add and subtract real time TT0= T + (T0 - T); divide by T (T0/ T) = 1 + [(T0 - T)/ T]; multiply by δθ'0(T0 / T) θ'0 = θ'0 + [(T0 - T)/ T] θ'0(T0 / T) δθ'0 = δθ'0 + [(T0 - T)/ T] δθ'0Modern and Nobel error # 2 is: [(T0 - T)/ T] δθ'0 = 70.75 [(T0 - T)/ T] = 70.75 [(194 - 498.67)/ 498.67] = 43 arc second per century

1.3 - Light signal argument shift; tan θ0 = v0/c; tan θ = v/c Naming: v = visual velocity And v 0 = actual velocityThen, v0 = v0; add and subtract vAnd, v0= v + (v0 - v); divide by vThen (v0/ v) = 1 + [(v0 - v)/ v] multiply by δθ'0And (v0 / v) δθ'0 = δθ'0 + [(v0 - v)/ v] δθ'0Modern and Nobel error # 3 is: Is: 70.75 [(v0/c) - (v/c)]/ (v/c) = (tan θ0 - tan θ)/ tan θ = [(v0 - v)/ v] δθ'0 = 70.75 [(v0 - v)/ v] = 70.75 [(48.1 - 29.8)/ 29.8] = 43 arc sec/100y

1.4 - Light signal surface acceleration frequency shift; f0 = γ 0/c; f = γ /c Naming: γ = g = Earth gravitational acceleration = 9.8And γ 0 = g 0 = Mercury gravitational acceleration = 3.8 Then: g 0 = g 0

And g 0 = g + (g 0 - g)Page 5

And (g 0 / g) = 1 + [(g 0 - g)/ g] multiply by δθ'0And (g 0 / g) δθ'0 = δθ'0 + [(g 0 - g)/ g] δθ'0And [(g 0 /c)/ (g/c) δθ'0 = δθ'0 + {(g 0/c) - (g/c)]/ (g/c)} δθ'0And [(f 0 - f)]/ f] δθ'0Modern and Nobel error # 4 is: [(f 0 - f)]/ f] δθ'0 [(g 0 - g)/ g] δθ'0 = 70.75 [(3.8 - 9.8)/ 9.8] = 43 arc sec/100y

1.5 Light signal momentum P shift

With λ p = h; differentiating d λ/λ = - d p/p = [(r0 - r)/ r]Alfred Nobel Prize winner's and Einstein's error # 5 is: [(f 0 - f)]/ f] δθ'0= (70.75) x 0.61 = 43 arc sec/100y

1.6 Light signal Energy E shift E0 = h f0; E = h f[(E0 - E)/E] δθ'0 = [(f0 - f)/f] δθ'0Modern and Nobel error # 6 is: [(E0 - E)/E] δθ'0= [(f0 - f)/f] δθ'0 = 70.75 (0.61) = 43 arc sec/100y

Planet Distance/km Signal period

Orbital period

Spin velocity

Mass Eccentricity

Mercury 58,200,000 194 sec 88days .003km/sec 0.33 x 1024 0.206Earth 149,600,000 498.67 sec 375.26 0.4651km/sec 5.97X 1024 0.00167Sun radius 0.696x106 km Sun mass 2X 1030

Page 6

Chapter two: Orbit processing 2.1 - Distance shiftr = Visual distance; r 0 = actual distance Then, r0 = r0; add and subtract visual distance rAnd, r0= r + (r0 - r); divide by rThen (r0/ r) = 1 + [(r0 - r)/ r] multiply by δθ0'And (r0 / r) δθ0' = δθ0' + [(r0 - r)/ r] δθ0'Modern and Nobel error # 7 is: [(r0 - r)/ r] δθ0' = 70.75 [(r0 - r)/ r]= 70.75 [(58,200,000,000 - 149,600,000)/ 149,600,000] = 43 arc sec/ 100y

2.2 - Velocity shiftLight signal argument shift; tan θ0 = v0/c; tan θ = v/c Naming: v = visual velocity; v 0 = actual velocityThen, v0 = v0; add and subtract vAnd, v0= v + (v0 - v); divide by vThen (v0/ v) = 1 + [(v0 - v)/ v] multiply by δθ'0And (v0 / v) δθ'0 = δθ'0 + [(v0 - v)/ v] δθ'0Modern and Nobel error # 8 is: [(v0 - v)/ v] δθ'0 = 70.75 [(v0 - v)/ v] = 70.75 [(48.1 - 29.8)/ 29.8] = 43 arc sec/100y

2.3 - Acceleration shiftNaming: γ = g = Earth gravitational acceleration = 9.8And γ 0 = g 0 = Mercury gravitational acceleration = 3.8 Then: g 0 = g 0

And g 0 = g + (g 0 - g)And (g 0 / g) = 1 + [(g 0 - g)/ g] multiply by δθ'0And (g 0 / g) δθ'0 = δθ'0 + [(g 0 - g)/ g] δθ'0Modern and Nobel error # 9 is: [(g 0 - g)/ g] δθ'0 = 70.75 [(3.8 - 9.8)/ 9.8] = 43 arc sec/100y

2.4 - Linear distance shiftNaming: d = Real time linear distance And d 0 = event time linear distance Then, d 0 = d 0; add and subtract real time distance dAnd, d 0= d + (d 0 - d); divide by dThen (d 0/ d) = 1 + [(d 0 - d)/ d]; multiply by δθ'0And (d 0 / d) δθ'0 = δθ'0 + [(d 0 - d)/ d] δθ'0Modern and Nobel error # 10 is: [(d0 - d)/ d] δθ'0 = 70.75 [(d 0 - d)/ d]With d 0 = v 0 τ 0= 47.9 x 88, and d = v τ = 29.8 x 365.26 = 70.75 [(v0 τ 0 - v τ /)/ v τ]= 70.75 [(47.9 x 88 - 29.8 x 365.26)/ 29.8 x 365.26] = 43 arc sec/100y

Page 7Chapter 3: light aberrations processing

3.1 - Light signal period Aberration:Modern and Nobel error # 11is: = {[Tan-1 (T0/ τ 0)/tan-1 (T/ τ)] -1} δθ'0 = 70.75 {[Tan-1 (194/ 88 x 24 x 3600)/tan-1 (498.67/ 88 x 24 x 3600)] -1}= 70.75 x 0.61 = 43 arc sec/100 y

3.2 - Light signal distance Aberration:Modern and Nobel error # 12 is: = {[tan-1 (r0/ c τ 0)/tan-1 (r/ c τ)] -1} δθ'0= 70.75 {[Tan-1 (58,200,000/ 300,000x88 x 24 x 3600)/tan-1 (149,600,000/ 88 x 24 x 3600)] -1} = = 70.75 x 0.61 = 43 arc sec/100 y

3.3 - Light signal velocity Aberration:Modern and Nobel error # 13 is: = {[tan-1 (v0/ c)/tan-1 (v/ c)] -1} δθ'0= 70.75 {[Tan-1 (48.1/ 300,000)/tan-1 (29.8/300,000)] -1} = 70.75 x 0.61 = 43 arc sec/100 y

3.4 - Light signal frequency Aberration:Modern and Nobel error # 14 is: = [(tan-1 f0/tan-1 f) -1] δθ'0 = {[tan-1 (γ0/ c)/tan-1 (γ0/ c)] -1} δθ'0= 70.75 {[Tan-1 (3.8/3x108)/tan-1 (9.8/3x108)] -1} = 70.75 x 0.61 = 43 arc sec/100 y

3.5 Image processingOrbit and planet sizes are governed by AC = BD = constant(A/B) -1 = (D/C) - 1

Radius of Earth is 6731 km and radius of Mercury is 2440 km The fact we measure from Earth then velocity of earth is 29.8 km/secThe fact that we measure Earth orbit and confused with Mercury orbit the velocity is 29.8 - 0.465 = 29.335 km/secThe mistake astronomers are doing is taking Mercury's orbit as Earth orbit but with less orbital speed of 0.465 when they use Sun as reference point and measure from Earth. They are looking at Earth orbit and not mercury's orbit

Modern and Nobel error # 15 is: [(D /C) (29.8/29.335) - 1] δθ'0[(D x 29.8/C x 29.335) - 1] δθ'0 = [(2440 x 29.8/6371x 29.335) - 1] δθ'0 = 0.61 x 70.75 = 43 arc sec/ century

8Chapter four: Mistakes of Kepler, Newton

4.1 - Kepler's Period historical mistakeKepler’s law: a³/T² = k = constantOr, a1³/ T1² = a2³/ T2² = k Or, a1/ a2 = (T1/ T2)2/3

And (a1 - a2)/ a2 = (τ0/ τ) 2/3 – 1Or (a 0 – a)/ a = (τ 0/ τ) 2/3 – 1And [(am – a e)/ a e] δ θ’0 = [(a0 – a)/ a] (v 0 /r 0) δθ'0= [(a0 – a)/ a] δθ'0Modern and Nobel error # 16 is: [(τ 0/ τ) 2/3 – 1] δθ'0= 70.75 [(88/365.26) 2/3 -1] = 43 arc sec/100y

4.2 - Newton's distance historical mistakeNewton force is F = k/r2 = G m M/r2 = m v2/r And velocity of a celestial object measured from the sun is = (GM/r) 1/2

With r3 0 = (GM/v2 0); r0 = (GM/v2 0)And r3

= (GM/v2); r = (GM/v2)Modern and Nobel error # 17 [(r 0 - r)/r] δθ'0 = {[(GM/v2 0)/ (GM/v2)] - 1} δθ'0= [(v/v0)2 - 1] δθ'0 = [(29.8/47.9)2 - 1] x 70.75 = 43 arc sec/100y

4.3 - Newton's velocity historical mistakeNewton force is F = k/r2 = G m M/r2 = m v2/r And v 0 = (GM/r 0) ½; v = (GM/r) 1/2

Modern and Nobel error # 18 [(v 0 - v)/v] δθ'0 = [(GM/r 0) 1/2/ (GM/r) ½ -1] δθ'0= [(r/r 0) 1/2

- 1] δθ'0 = [(149.6/58.2) 1/2 - 1] x 70.75 = 43 arc sec/100y

4.4 - Newton's surface gravity historical mistakeNewton force is F = k/r2 = G m M/R2 = m g And g 0 = Gm0/R2 0; g = Gm/R2

Modern and Nobel error # 19 = [(g 0 - g)/ g] δθ'0 = {[(Gm0/R0

2) - (Gm/R2)]/ (Gm/R2)} δθ'0 = {[(m0/ R02) - (m/R2)]/ (m/R2)} δθ'0

= [(g 0 - g)/ g] δθ'0 = [(3.8/9.8) -1] 70.75 = 43 arc sec/100 years Note: [(m0/ R0

2)/ (m/R2) (29.8/29.335)2 -1] δθ'0 =43 arc sec/100y

4.5 - Newton's signal time historical mistakeNewton force is F = k/r2 = G m M/r2 = m v2/r; v 0 = (GM/r 0) ½; v = (GM/r) 1/2

Modern and Nobel error # 20 [(r/r 0) 1/2 - 1] δθ'0 = {[(r/c)/(r 0/c) ½] - 1]} δθ'0

= [(T/T 0) 1/2 - 1] δθ'0 = 70.75 [(498/194) 1/2

- 1] = 43 arc sec/100yPage 9

Chapter five: mistakes of Le Verrier

5.1- Le Verrier velocity historical mistakeThe angular velocity of Mercury around the Sun is: θ 0' = v 0 /r 0

If it is measured for planet Mercury from the sun, then θ 0' = v 0 /r 0

If planet Mercury around the sun measured from earthThen θ m' (Earth) = (v 0 + v)/r 0

And θ m' (Earth) = v 0 /r 0 + v /r 0; not v 0 /r 0

Le Verrier mistake is: v /r 0

The angular speed shift: v /r 0; taking into account Earth rotation v º Le Verrier mistake: = [(v +/- vº) /r 0] In arc second per century multiplying by (180/ π) (3600) (100 τ / τ 0)= [(v +/- v º) /r 0] [(180/ π) (3600) (100 τ / τ 0)] = [(v +/- v º) /v 0) (v 0 /r 0)] (180/ π) (3600) (100 τ / τ 0)Modern and Nobel error # 21 is: [(v +/- v º) /v 0)] δθ'0

[(29.8 - 0.465) /47.9)] 70.75 = 43 arc sec/100y5.2- Le Verrier distance mistake And v= [G M2 / (m + M) r] 1/2 = (G M / r) 1/2; m <<M; Solar systemAnd v 0 = [G M2 / (m0 + M) r] 1/2 = (G M / r0) 1/2; m0 <<M; Solar systemAnd v = (G M / r) ½; v 0 = (G M / r0) 1/2

And (v/ v 0) = (G M / r) ½/ (G M / r0) ½ = (r0 / r) 1/2 With [(v - v º) /v 0)] δθ'0 = (r0 / r) 1/2 [1- (v º/v)] δθ'0Modern and Nobel error # 22 is: (r 0/r) 1/2

[1- (v º/v)] δθ'0= (58.2/149.6) 1/2

x 0.98 x 70.75 = 0.61 x 70.75 = 43 arc sec/100y 5.3 - Le Verrier signal time mistake With (r/r 0) 1/2

[1- (v º/v)] δθ'0 = [(r 0/c) /(r /c)] 1/2 [(1- (v º/v)] δθ'0 Modern and Nobel error # 23 is: (T 0 /T) 1/2 [(1- (v º/v)] δθ'0= (194 /498.67) 1/2 x 0.98 x 70.75 = 43 arc sec/100y5.4 - Le Verrier orbital period mistakeKepler’s law: a³/T² = k = constantOr, r 0³/ τ0² = r³/ τ ² = k; r 0/ r = (τ 0 / τ) 2/3

With (r 0 /r) 1/2 [(1- (v º/v)] δθ'0

Modern and Nobel error # 24 is: (τ 0 / τ) 1/3 [(1- (v º/v)] δθ'0= (88/365.26) 1/3 x 0.98 x 70.75 = 43 arc sec/100y5.5 - Le Verrier gravity mistakeAnd γ 0 = g 0 = Gm0/r2

0; γ = g = Gm/r2 And (γ / γ 0) = (g/g0) = (Gm/r2)/ (Gm0/r2

0) = (m r20/ m0 r2)

And (r0/r) = (m0g/g0m) ½; (r0/r) 1/2 = (m0g/g0m) 1/4

With (r/r 0) 1/2 [1- (v º/v)] δθ'0

Modern and Nobel error # 25 is: (m0g/g0m) 1/4 [1- (v º/v)] δθ'0 = (0.33 x 9.8/3.8 x5.97) 1/4 0.98 x 70.75 = 43 arc sec/100y

Page 10

Chapter six: Real time Vision

6.1 Distance: AC = BD = kDifferentiation with respect to time of BD = kGives D (dB/d t) + B (d D/d t) = 0And D (dB/d t) = - B (d D/d t) And [(dB/d t)/B] = - [(d D/d t)/D] = - í ω

B = A e - í ω t D = C e í ω t

Or, r = r0 e í ω t; r0 = r e - í ω t And r0 = r x + í r y = r (cosine ω t + í sine ω t); ω t = cosine-1 (r x/r)And r x = r cosine ω t = r [1 - 2 sine2 (ω t/2)]And (r x - r)/r = - 2 sine2 (ω t/2)And (r x - r)/r = - 2 sine2 {[cosine-1 (r x/ r)]/2}= [(r x - r)/r] δθ'0 = - 2 sine2 {[cosine-1 (r x/ r)]/2} δθ'0

Modern and Nobel error # 26 is: - 2 sine2 {[cosine-1 (r x/ r)]/2} δθ'0= - 2 sine2 {[cosine-1 (58.2/ 149.6)]/2} x 70.75 = 43

6.2 Time r = r0 e í ω t and r = c T and r0 = c T0 Then T = T0 e í ω t; T0 = TAnd T = T x + í T y = T0 (cosine ω t + í sine ω t); ω t = cosine-1 (T/ T0)And T x = T0 cosine ω t = T0 [1 - 2 sine2 (ω t/2)]And (T x - T0)/T0 = - 2 sine2 (ω t/2)And (T x - T0)/T0 = - 2 sine2 {[cosine-1 (T/ T0)]/2}= [(T x - T0)/T0] [(v 0 /r 0)] [(180/π) (3600) (100 τ/ τ0)] Modern and Nobel error # 27 is: - 2 sine2 {[cosine-1 (T x/ T0)]/2} δθ'0= - 2 sine2 {[cosine-1 (194/ 498.67)]/2} x 70.75 = 43

6.3 Velocity v = 2 π r / τ; and v 0 = 2 π r0/ τ0 Then r = (v τ /2 π) and r 0 = (v 0 τ 0/2 π) And (r /r 0) = (v τ /2 π)/ (v 0 τ 0/2 π) = (v τ)/ (v 0 τ 0) = [(r - r0)/r0] [(v 0 /r 0)] [(180/π) (3600) (100 τ/ τ0)] = - 2 sine2 {[cosine-1 (r/ r0)]/2} [(v 0 /r 0)] [(180/π) (3600) (100 τ/ τ0)]

Modern and Nobel error # 28 is - 2 sine2 {[cosine-1 (v τ)/ (v 0 τ 0)]/2} δθ'0= - 2 sine2 {[cosine-1 (48.1 x 88/ (29.8 - 0.465) x 365.26]/2} x 70.75 = 43

Page 11

6.4 Areal velocity r v = h = r0 v0

The r/r0 = v0/v; taking Earth spin into accountThen r/r0 = (v0 - v0)/vModern and Nobel error # 29 is: - 2 sine2 {[cosine-1 (v0 - v0)/v]/2} δθ'0=- 2 sine2 {[cosine-1 (29.8 - 0.451)/48.14]/2} δθ'0

6.5 Acceleration g = GM/r2; g0 = GM/r0

The r/r0 = (g0/g) 1/2; taking Earth spin into accountThen r/r0 = (g0/g) ½ (1 - v0/ v)Modern and Nobel error # 30 : - 2 sine2 {[cosine-1 (g0/g) ½ (1 - v0/ v)/2} δθ'0=- 2 sine2 {cosine-1 {(3.8/9.8)[1 - (0.451/29.8]}/2} 70.75 = 43 arc sec/100y

6.6 Argument r/r0 = (v0/c)/ (v/c) same as Le Verrier The r/r0 = (v0/c)/ (v/c); taking Earth spin into accountThen r/r0 = [(v0 - v0)/c]/ (v/c)Modern and Nobel error # 31: - 2 sine2 {[cosine-1 {[(v0 - v0)/c]/ (v/c)}/2} δθ'0=- 2 sine2 {[cosine-1 {{[(29.8 - 0.451)]/ 3x108}/ (48.14/3x108)]}/2} 70.75

= 43 arc sec/100y

6.7 Frequency [(g0/c)/ (g/c)] ½ (1 - v0/ v)The r/r0 = v0/v; taking Earth spin into accountThen r/r0 = (v0 - v0)/v

Modern and Nobel error # 32: - 2 sine2 {{cosine-1 [(g0/c)/ (g/c)] ½ (1 - v0/ v)}/2} δθ'0=- 2 sine2 {{cosine-1 {[(3.8/3x108)/ (9.8/3x108)]1/2

x0.98}/2}} 70.75= 43 arc sec/100y

6.8 - Kepler's Period historical mistake IIKepler’s law: a³/T² = k = constantOr, a1³/ T1² = a2³/ T2² = k Or, a1/ a2 = (T1/ T2)2/3

And (a1 - a2)/ a2 = (τ0/ τ) 2/3 – 1Or (a 0 – a)/ a = (τ 0/ τ) 2/3 – 1And [(am – a e)/ a e] δ θ’0 = [(a0 – a)/ a] (v 0 /r 0) δθ'0= [(a0 – a)/ a] δθ'0Modern and Nobel error # 16 is: [(τ 0/ τ) 2/3 – 1] δθ'0= 70.75 [(88/365.26) 2/3 -1] = 43 arc sec/100yModern and Nobel error # 33 is: - 2 sine2 {[cosine-1 (τ 0/ τ) 2/3]/2} δθ'0= - 2 sine2 {[cosine-1 (88/365.26)2/3]/2} 70.75 = 43 arc sec/100y

Page 12

6.9 - Newton's distance historical mistake IINewton force is F = k/r2 = G m M/r2 = m v2/r And velocity of a celestial object measured from the sun is = (GM/r) 1/2

With r3 0 = (GM/v2 0); r0 = (GM/v2 0)And r3

= (GM/v2); r = (GM/v2)Modern and Nobel error # 17 = [(r 0 - r)/r] δθ'0 = {[(GM/v2 0)/ (GM/v2)] - 1} δθ'0= [(v/v0)2 - 1] δθ'0 = [(29.8/47.9)2 - 1] x 70.75 = 43 arc sec/100y

Modern and Nobel error # 34: = - 2 sine2 {[cosine-1 (v/v0)2]/2} δθ'0= - 2 sine2 {[cosine-1 (v/v0)2]/2} δθ'0

6.10 - Newton's velocity historical mistake IINewton force is F = k/r2 = G m M/r2 = m v2/r And v 0 = (GM/r 0) ½; v = (GM/r) 1/2

Modern and Nobel error # 18 = [(v 0 - v)/v] δθ'0 = [(GM/r 0) 1/2/ (GM/r) ½ -1] δθ'0= [(r/r 0) 1/2

- 1] δθ'0 = [(149.6/58.2) 1/2 - 1] x 70.75 = 43 arc sec/100y

Modern and Nobel error # 35: = - 2 sine2 {[cosine-1 (r/r 0) 1/2]/2} δθ'0= - 2 sine2 {[cosine-1 (149.6/58.2) 1/2]/2} 70.75 = 43 arc sec/100y

6.11 - Newton's surface gravity historical mistake IINewton force is F = k/r2 = G m M/R2 = m g And g 0 = Gm0/R2 0; g = Gm/R2

Modern and Nobel error # 19 = [(g 0 - g)/ g] δθ'0 = {[(Gm0/R0

2) - (Gm/R2)]/ (Gm/R2)} δθ'0 = {[(m0/ R02) - (m/R2)]/ (m/R2)} δθ'0

= [(g 0 - g)/ g] δθ'0 = [(3.8/9.8) -1] 70.75 = 43 arc sec/100 years Note: [(m0/ R0

2)/ (m/R2) (29.8/29.335)2 -1] δθ'0 = 43 arc sec/100y

Modern and Nobel error # 36: = - 2 sine2 {{cosine-1 [(m0/ R0

2)/ (m/R2)]}/2} δθ'0= - 2 sine2 {{cosine-1 [(0.33/ 24402)/ (5.97/63712)]}/2} 70.75 = 43 arc sec/100y

6.12 - Le Verrier mistake II Modern and Nobel error # 21 is: [(v +/- v º) /v 0)] δθ'0 = (v /v 0) [1- (v º/v)] δθ'0Modern and Nobel error # 37: - 2 sine2 {{cosine-1 (v /v 0) [1- (v º/v)]}/2} δθ'0= - 2 sine2 {{cosine-1 (29.8 /48.14) [1- (0.4651/29.8)]}/2} 70.75= 43 arc sec/100y

Page 136.12 - Le Verrier distance mistake II And v = [G M2 / (m + M) r] 1/2 = (G M / r) 1/2; m <<M; Solar systemAnd v 0 = [G M2 / (m0 + M) r] 1/2 = (G M / r0) 1/2; m0 <<M; Solar systemAnd v = (G M / r) ½; v 0 = (G M / r0) 1/2

And (v/ v 0) = (G M / r) ½/ (G M / r0) ½ = (r0 / r) 1/2 With [(v - v º) /v 0)] δθ'0 = (r0 / r) 1/2 [1- (v º/v)] δθ'0Modern and Nobel error # 22 is: (r 0/r) 1/2

[1- (v º/v)] δθ'0= (58.2/149.6) 1/2

x 0.98 x 70.75 = 0.61 x 70.75 = 43 arc sec/100y = 43 arc sec/100y Modern and Nobel error # 38:- 2 sine2 {{cosine-1 (r 0/r) 1/2

[1- (v º/v)]}/2} δθ'0= - 2 sine2 {{cosine-1 (58.2/149.6) 1/2

[1- (0.4651/29.8)]}/2} 70.75= 43 arc sec/100y

6.13 - Le Verrier signal time mistake IIWith (r/r 0) 1/2

[1- (v º/v)] δθ'0 = [(r 0/c) /(r /c)] 1/2 [(1- (v º/v)] δθ'0 Modern and Nobel error # 23 is: (T 0 /T) 1/2 [(1- (v º/v)] δθ'0= (194 /498.67) 1/2 x 0.98 x 70.75 = 43 arc sec/100yModern and Nobel error # 39:- 2 sine2 {{cosine-1 (v /v 0) [1- (v º/v)]}/2} δθ'0= - 2 sine2 {{cosine-1 (194 /498.67) 1/2 x 0.98}/2} x 70.75

13.14 - Le Verrier orbital period mistakeKepler’s law: a³/T² = k = constant

Or, r 0³/ τ0² = r³/ τ ² = k; r 0/ r = (τ 0 / τ) 2/3

With (r 0 /r) 1/2 [(1- (v º/v)] δθ'0

Modern and Nobel error # 24 is: (τ 0 / τ) 1/3 [(1- (v º/v)] δθ'0= (88/365.26) 1/3 x 0.98 x 70.75 = 43 arc sec/100yModern and Nobel error # 40:= - 2 sine2 {{cosine-1 (τ 0 / τ) 1/3 [(1- (v º/v)}}/2} 70.75= - 2 sine2 {{cosine-1 (88 / 149.6) 1/3 [(1- (0.4561/29.8)}}/2} 70.75= 43 arc sec/100y13.15 - Le Verrier gravity mistakeAnd γ 0 = g 0 = Gm0/r2

0; γ = g = Gm/r2 And (γ / γ 0) = (g/g0) = (Gm/r2)/ (Gm0/r2

0) = (m r20/ m0 r2)

And (r0/r) = (m0g/g0m) ½; (r0/r) 1/2 = (m0g/g0m) 1/4

With (r/r 0) 1/2 [1- (v º/v)] δθ'0

Modern and Nobel error # 24 is: (m0g/g0m) 1/4 [1- (v º/v)] δθ'0 = (0.33 x 9.8/3.8 x5.97) 1/4 0.98 x 70.75 = 43 arc sec/100y Modern and Nobel error # 41:=- 2 sine2 {{cosine-1 (m0g/g0m) 1/4 [1- (v º/v)]/2} δθ'0= - 2 sine2 {{cosine-1 (0.33 x 9.8/ /3.8 x5.97) 1/4 [(1- (0.4561/29.8)}}/2} 70.75= 43 arc sec/100y

Page 14

Chapter 7: Fourier Light motion visual analysisr = r0 e í ω t

Distance r in real time is r = r0 e i (θ + ω t)

With r2θ' = h = r0 2θ'0; θ' = θ'0 e -2 i (θ + ω t)

At θ = 0; Perihelionr = r0 e í ω t

Taking r = c t and r0 = c t0

Then (t/t 0) = e i ω t

And the Fourier transform is:

Γ = (1/2 π) {-∞∫∞ e i ω t d t}

And Γ = (1/2 π) 0∫ τ0 e i ω t d t

Γ = (1/2 π) [e i ω τ0 -1]/ i ω

= (1/2 π) [cosine ω τ 0 + i sine ω τ 0 - 1]/ i ω

Along the line of sightΓx = (1/2 π) [sine ω τ 0]/ ωΓ x [100

τ / τ 0] = (100

τ /2 π) [sine ω τ 0 / ω τ 0] per 100 τ

With ω τ 0 = arc tan (v0/c); 100 τ = 1 century = 36526 days

Γ x [100 τ / τ 0] per century

= (100 τ /360) {sine [arc tan (v0/c)] / [arc tan (v0/c)]}

Γ x [100 τ /

τ 0] per century = (36526days /360degrees) {sine [arc tan (v0/c)] / [arc tan (v0/c)]} In arc seconds per century:Γ x [100

τ / τ 0] per century

= (36526x 24 x 3600 /360 x 3600) {sine [arc tan (v0/c)] / [arc tan (v0/c)]} Γ x [100

τ / τ 0] per century

= (36526x 24 x 3600/360 x 3600) {sine [arc tan (v0/c)] / [arc tan (v0/c)]}Where v0 = 47.9 km/second; c = 300,000 km/secondModern and Nobel error # 42:Γ x [100

τ / τ 0] = (36526 /15) {sine [arc tan (47.9/300,000)] / [arc tan (47.9/300,000)]}

= 42.5 arc seconds per centuryModern and Nobel error # 43:Γ x [100

τ/ τ 0] = (36526 /15) {sine [arc tan (29.8/300,000)] / [arc tan (29.8/300,000)]}

= 42.5 arc seconds per century

Page 15Angular velocity θ' in real time is θ' = θ'0 e - 2i (θ + ω t)

Taking θ'= 2 π f and θ'0 = 2 π f 0

Then (f/f 0) = e - 2i (θ + ω t)

At θ = 0; PerihelionAnd (f/f 0) = e - 2i ω t

And the Fourier transform is:

Γ = (1/2 π) {-∞∫∞ e - 2i ω t d t}

And Γ = (1/2 π) 0∫ τ0 e - 2i ω t d t

Γ = (1/2 π) [e - 2i ω t -1]/ -2 í ω = (1/2 π) [cosine 2 ω τ0 + í sine 2 ω τ0 - 1]/ -2 í ω

Along the line of sightΓx = - (1/2 π) [sine 2ω τ0]/ 2ωΓ x [100

τ / τ0] = (100

τ /2 π) [sine 2 ω τ0 / 2 ω τ0] per 100 τ

With ω τ0 = arc tan (v0/c); 100 τ = 1 century = 36526 days

Γ x [100 τ / τ0] per century

= (100 τ /360) {sine 2 [arc tan (v0/c)] / [arc tan (v0/c)]}

Γ x [100 τ /

τ0] per century = (36526days /360degrees) {sine2 [arc tan (v0/c)] / 2 [arc tan (v0/c)]}

In arc seconds per century:Γ x [100

τ / τ0] per century

= (36526x 24 x 3600 /360 x 3600) {sine 2[arc tan (v0/c)] / 2[arc tan (v0/c)]} Where v0 = 47.9 km/second; c = 300,000 km/secondModern and Nobel error # 44:Γ x [100

τ /τ 0] = (36526 /15) {sine 2[arc tan (47.9/300,000)] / 2[arc tan (47.9/300,000)]} = 42.5 arc seconds per centuryModern and Nobel error # 45:Γ x [100

τ / τ 0] = (36526 /15) {sine 2[arc tan (29.8/300,000)] / 2[arc tan (29.8/300,000)]} = 42.5 arc seconds per century

Page 16

Chapter 8: Axial tilt processing

In Chapter 7 Fourier analysis said that planet Mercury's Perihelion has nothing to do with planet Mercury but has something to do with Earth.

Astronomers measure with an axial tilt of 23.440 because they measure from Earth and use the Sun as reference point

Parallaxes are shifted by an angle of 23.440

Meaning measurements are shifted by a factor of: (1 - sine 23.440) Modern and Nobel error # 46: (1 - sine 23.440) δθ'0= (1 - sine 23.440) 70.75 = 43 arc sec/ 100y

Page 17

Chapter ten: Descartes and LaGrange failures 10.1 Γ = Γ x + Γ y = Γ 0 e í ω t Along the line of sightΓ x = Γ 0 cosine ω tThen ∆ Γ x (seconds) = Γ x - t = - 2 t sine² {[arc tan (V r m - V r e)/ (√ 2c)]/ 2} = 43And in arc ∆ Γ x (arc seconds) = Γ x - t

= - 30 t sine² {[arc tan (V r m - V r e)/ (√ 2c)]/ 2} = 43Where r’ = V r; and r θ’ = V θ; and V r ²= 2 γ r r; and V θ ² = r γ θ

The confusion is and was γ r = γ θ; V r ²= 2 V θ ²; and taking V r = (√ 2) V θ And taking V r = V r (Mercury) - V r (Earth) = V r m - V r e

And V θ = V r / √ 2

Modern and Nobel error # 47: ∆ Γ x (Arc second) = Γ x - Γ 0 = - 30 t sine² {[arc tan (V r m - V r e)/ (√ 2c)]/ 2} = 43 arc second /century

10.2 La Grange historical mistake

Angular velocity with respect to center of mass θ' c m = v c m/rAngular velocity with respect to Sun θ's = v s/rAnd Astronomers observed: θ's = v s/r Angular velocity with to the sunAnd v c m = √ [GM²/ (m + M) a]; m = 0.32 x1024 kg; and M = 2.0 x1030 kg And v s = √ (GM/a) And (θ' cm - θ’s) /θ’s = [(2 π/T c m) - (2 π /T s)]/ (2 π /T s)

= (T c m/T s) – 1 = (v s/v c m) – 1 = [√ (GM/a)] / {√ [(m + M) a/ GM²]} - 1 = √ [(m + M)/M] - 1 = √ [1 + (m/M)] - 1 ≈ 1 + (m/2 M) – 1 ≈ m/2 M

And (θ' cm - θ’s) /θ’s ≈ (m/2 M); and (θ' cm - θ’s) = (2 π /T s) (m/2 M) And (θ' cm - θ’s) T s = (2 π + δ θ - 2 π) = π m/ M; δ θ = π m/ MMultiplying by [(180/π) (3600) (26526/ τ0)]

Modern and error # 48 = π (m/ M) [(180/π) (3600) (26526/ τ0)] = 43= π (0.32 x 1024/2x 1030) [(180/π) (3600) (26526/ τ0)] = 43

Page 18Chapter 9: Newton's equation was/is solved wrong for 350 years

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/d t) + (dm/d t) r = Total moment = change of location + change of mass = m v + m' r; v = velocity = d r/d t; m' = mass change rateF = d P/d t = d²S/dt² = Total 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 rateIn polar coordinates system

First r = r r1= r [cosine θ î + sine θ Ĵ]

Define r1= cosine θ î + sine θ Ĵ

Define v = d r/d t = (d r/d t) r' r1+ r (d r1/d t) = r' r1+ r (d r1/d t) = r' r1+ r θ'[- sine θ î + cosine θ Ĵ] = r' r1+ r θ' θ1

Define θ1= -sine θ î +cosine θ Ĵ;And with r1= cosine θ î + sine θ ĴThen d θ1/d t= θ' [- cosine θ î - sine θ Ĵ= - θ' r1 And d r1/d t = θ' (-sine θ î + cosine θ Ĵ) = θ' θ1

Define γ = d (r' r1+ r θ' θ1) /d t = r" r1+ r' (d r1/d t) + r' θ' θ1+ r θ" θ1+ r θ' (d θ1/d t) = (r" - rθ'²) r1+ (2r'θ' + r θ") θ (1)

Then r = r r1; v = r' r1 + r θ' θ1; γ = (r" - rθ'²) r1+ (2r'θ' + r θ") θ1

r = location; v = velocity; γ = acceleration

F = m γ + 2m'v +m" r

F = m [(r"-rθ'²) r1+ (2r'θ' + r θ") θ1 + + 2m'(r' r1+ r θ' θ1) + (m" r) r1

= [d² (m r)/dt² - (m r) θ'²] r1 + (1/m r) [d (m²r²θ')/d t] θ1

Page 19

With d² (m r)/dt² - (m r) θ'² = F (r) And d (m²r²θ')/d t = 0 If light mass m = constant, then With m [d² r/dt² - r θ'²] = F (r) Eq-1 And d (r²θ')/d t = 0 Eq-2

A - Newton's gravitational classical or past time solution

Newton took m = constantThen F = m γ With d² (m r)/dt² - (m r) θ'² = - G m M/r² Newton's Gravitational Equation (1)And d (m²r²θ')/d t = 0 Kepler's law (2)Then m (d² r/dt² - r θ'²) = - G m M/r² Eq-1 And d (r²θ')/d t = 0 Eq-2 From Eq-2: d (r²θ')/d t = 0 Then r²θ' = h = constant With d² r/dt² - r θ'² = 0Let u = 1/r; r = 1/u; r²θ' = h = θ’ /u² And d r/d t = (d r/ d u) (d u /d θ) (d θ/ d t) = (- 1/u ²) (θ’) (d u/ d θ) = - h (d u/ d θ)And d² r/ d t² = - h (θ’) (d² u/ d θ ²)

= (- h²/r²) (d² u/ d θ ²) = - h² u² (d² u/ d θ ²)With d² r/dt² - r θ'² = - G M/r2 E q – 1 And - h² u² (d² u/ d θ ²) – (1/u) (h u²) ² = - G M u2

Then (d² u/ d θ ²) + u = G M/h2

And u = G M/h2 + A cosine θThe r = 1/u = 1/ (G M/h2 + A cosine θ); divide by G M/h2 And r = (h2/G M)/ [1 + (A h2/G M) cosine θ]With; h2/G M = a (1 - ε2); (A h2/G M) = εOr, r = a (1 - ε2)/ (1 + ε cosine θ); definition of an ellipseThis is Newton's equation classical solutionMeasuring planetary orbit in real time using Newton's equation classical solution does not match Newton's equation classical solution but solving Newton's equation in real time or solving Newton's equation in present time will match measurements of planetary orbits in real time. Solving an equation in real or present time is solving it in complex numbers system. Solving Newton's equation in complex numbers produces quantum mechanics solution. The difference between real numbers classical solution and real time or complex numbers solution will produce relativistic effects as visual effects. In short: Real (Complex numbers solution) = real numbers solution + relativistic effects

Page 19

B - Real time solution or complex numbers solution of Newton's equation is:

F = m [(r"-rθ'²) r1+ (2r'θ' + r θ") θ1 + + 2m'(r' r1+ r θ' θ1) + (m" r) r1

= [d² (m r)/dt² - (m r) θ'²] r1 + (1/m r) [d (m²r²θ')/d t] θ1

With d² (m r)/dt² - (m r) θ'² = F (r) = - G m M/r2 E q - 1 And d (m²r²θ')/d t = 0 E q - 2

From E q - 2; d (m²r²θ')/d t = 0 Then m²r²θ' = constant = H = m² h; h = r² θ' With m²r²θ' = constant Differentiate with respect to timeThen 2mm'r²θ' + 2m²rr'θ' + m²r²θ" = 0Divide by m²r²θ' Then 2 (m'/m) + 2(r'/r) + θ"/θ' = 0This equation will have a solution 2 (m'/m) = 2(λ m + ì ω m)And 2(r'/r) = 2(λ r + ì ω m)And θ"/θ' = -2[λ m + λ r + ỉ [ω m + ω r)]

Then (m'/m) = (λ m + ì ω m)

Or d m/m d t = (λ m + ì ω m)And dm/m = (λ m + ì ω m) d tThen m = m 0 ℮ (λ m + ì ω m) t And m = m (θ, 0) m (0, t); m 0 = m (θ, 0) And m = m (θ, 0) ℮ (λ m + ì ω m) t

And m (0, t) = ℮ (λ m + ì ω m) t

Finally, m = m 0 ℮ (λ m + ì ω m) t

Similarly we can get (r'/r) = (λ r + ì ω r)Or d r/r d t = (λ r + ì ω r)And d r/r = (λ r + ì ω r) d tThen r = r 0 ℮ (λ r + ì ω r) t And r = r (θ, 0) r (0, t); r 0 = r (θ, 0) And r = r (θ, 0) ℮ (λ r + ì ω r) t

And r (0, t) = ℮ (λ r + ì ω r) t

Finally, r = r 0 ℮ (λ r + ì ω r) t

Page 20

And θ'(θ, t) = θ' (θ, 0)] ℮ -2 [(λ m + ỉ ω m) + (λ r + ỉ ω r)] t

And, θ'(θ, t) = θ' (θ, 0) θ' (0, t)

And θ' (0, t) = ℮ -2 [(λ m + λ r) + ỉ (ω m + ω r)] t

Also θ' = H / m² r²

From (1): d² (m r)/dt² - (m r) θ'² = - G m M/r² = -G M m³/m²r² = - G M m3 u2

Let m r =1/u

Then 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

And d² (m r)/dt² = -H θ'd²u/dθ² = - Hu² [d²u/dθ²]

- Hu² [d²u/dθ²] - (1/u) (Hu²)² = -G M m³ u²And (d²u/ dθ²) + u = G M m³/ H²

And [d²u (θ, 0)/ dθ²] + u (θ, 0) = G M (θ, 0) m³ (θ, 0)/ H² (θ, 0)Then u (θ, 0) = G M m³ (θ, 0)/ H² (θ, 0) + A cosine θ = G m 0 M 0/ h² + A cosine θ

And m0 r = 1/u (θ, 0) = 1/ [G m 0 M 0/ h² + A cosine θ]Or, r = 1/ [G M 0/ h² + A cosine θ] And r = h²/ G M0/ [1 + (Ah²/ G M0) cosine θ] Then r (θ, 0) = a (1-ε²)/ (1+ ε cosine θ)

This is Newton's gravitational law classical solution of two body problem which is the equation of an ellipse of semi-major axis of length a and semi minor axis b = a √ (1 - ε²) and focus length c = ε aThen, r (θ, t) = [a (1-ε²)/ (1+ε cosine θ)] ℮ (λ r + ì ω r) t ------------- IThis is real time solution or present solution of Newton's equation It is the math formula that matches astronomical measurementsIf time is frozen that is t = 0 Then r (θ, 0) = a (1-ε²)/ (1+ε cosine θ) or classical or event time solution -- II Relativistic is the difference between Real I and II And it is the visual difference motion and motion measurement

Page 20The difference between and event and its measurement in real time

Real time solution = Event time solution + time shift solution Real of a complex orbit solution = real numbers orbit solution + shift solution

We Have θ' (0, 0) = h (0, 0)/r² (0, 0) = 2πab/ τ0 a² (1- ε) ² = 2πa² [√ (1- ε²)]/ τ0a² (1- ε) ²] = 2π [√ (1- ε²)]/ τ0 (1- ε) ²]

Then θ'(0, t) = 2 π √ [(1- ε²)/ τ0 (1- ε) ²] ℮ -2 [(λ m + λ r) + ỉ (ω m + ω r)] t

Assuming that λ m + λ r = 0; or λ m = λ r = 0

Then θ'(0, t) = 2 π √ [(1-ε²)/ τ0 (1-ε) ²] ℮ -2 ỉ (ω m + ω r) t

= 2 π √ [(1-ε²)/ τ0 (1- ε) ²] [cosine 2 (ω m+ ω r) t - ỉ sine 2 (ω m+ ω r) t]

Real θ'(0, t) = 2 π √ [(1- ε²)/ τ0 (1-ε) ²] cosine 2 (ω m+ ω r) t

Real θ'(0, t) = 2 π √ [(1-ε²)/ τ0 (1-ε) ²] [1 - 2sine² (ω m+ ω r) t] Naming θ' = θ'(0, t); θ'0 = 2 π √ [(1-ε²)/ τ0 (1-ε) ²]

Then θ' = 2 π √ [(1- ε²)/ τ0 (1- ε) ²] [1 - 2 sine² (ω m+ ω r) t] And θ' = θ'0 [1 - 2sine² (ω m+ ω r) t]

And θ' - θ'0 = - 2 θ'0 sine² (ω m+ ω r) t] = -2{2 π √ [(1-ε²)/ τ0 (1-ε) ²]} sine² (ω m+ ω r) t]

And θ' - θ'0 = -4 π √ [(1-ε²)/ τ0 (1-ε) ²] sine² (ω m+ ω r) t]

If this apsidal motion is to be found as visual effects, then With, v ° = spin velocity; v0 = orbital velocity; τ0 = orbital period And ω m τ0 = tan-1 (v°/c); ω r τ0 = tan-1 (v0/c)Δ θ' = θ' - θ'0 = - 4 π √ [(1-ε²)]/ τ0 (1-ε) ²] sine² [tan-1 (v°/c) + tan-1 (v0/c)] radians per τ0 In degrees per period is multiplication by 180/ π Δ θ' = (-720) √ [(1-ε²)/ τ0 (1-ε) ²] sine² {tan-1 [(v°+ v0)/c]/ [1 - v° v0/c²]} The angle difference in degrees per period is: Δ θ = (Δ θ') τ0

Δ θ = (-720) √ [(1-ε²)/ (1-ε) ²] sine² {tan-1 [(v°+ v0)/c]/ [1 - v° v0/c²]} calculated in degrees per century is multiplication = 100 τ; τ = Earth orbital period = 100 x 365.26 = 36526 days and dividing by using τ0 in days

Page 21Δ θ (100 τ / τ0) = Δ θ in degrees per century = (-72000 τ / τ0) √ [(1-ε²)/ (1-ε) ²] sine² {tan-1 (v°+ v0)/c]/ [1 - v° v0/c²]} In arc second per century is multiplying by 3600 Δ θ = - 3600 x 720 (100 τ / τ0) √ [(1-ε²)/ (1-ε) ²] x Sine² {tan-1 [(v°+ v0)/c]/ [1 - v° v0/c²]} Approximations I With v° << c and v0 << c, then v° v0 <<< c² and [1 - v° v*/c²] ≈ 1Δ θ ≈ - 3600 x 720 (100 τ / τ0) [√ (1-ε²)]/ (1-ε) ²] Sine² tan-1 [(v°+ v0)/c] Arc second per centuryApproximations II With v° << c and v* << cThen Sine² tan-1 [(v°+ v0)/c] ≈ (v° + v0)/cΔ θ (Calculated in arc second per century) = (-720x36526x3600/ τ0 days) √ [(1-ε²)/ (1-ε) ²] [(v° + v0)/c] ²Approximations III The circumference of an ellipseIs: 2 π a (1 - ε²/4 + 3/16(ε²)²- --.) ≈ 2 π a (1- ε²/4); r0 = a (1- ε²/4)From Newton's laws for a circular orbit: F = [M/m F = - Gm M/r0

2 = m v0²/ r0 Then v0² = GM/ r0 For planet MercuryAnd v0 = √ [GM/ r] = √ [GM/a (1-ε²/4)]G = 6.673 x 10 -11; M = 2 x1030 kg; a = 58.2 x 109 meters; ε = 0.206 Then v0 = √ [6.673 x 10 -11 x 2 x1030 /58.2 x 109 (1- 0.206 ²/4)]And v0 = 48.14 km/sec [Mercury]; c = 300,000 Δ θ (Calculated in arc second per century) = (-720x36526x3600/ τ0 days) √ [(1-ε²)/ (1-ε) ²] [(v° + v0)/c] ²

With ε = 0.206; √ [(1-ε²)/ (1-ε) ²] = 1.552; v° = 3 meters per second Δ θ = (-720x36526x3600/88) 1.552 (48.143/300,000) ² Δ θ = 43 arc second per centuryModern Nobel error # 49: = (-720x36526x3600/ τ0 days) √ [(1-ε²)/ (1-ε) ²] [(v° + v0)/c] ²= (-720x36526x3600/88) 1.552 (48.143/300,000) ² = 43 arc second per centuryWith θ' = θ'0 e - 2i ω t Then τ = τ0 e 2i ω t; Δ τ = - 2 τ0 sine2 arc tan (v/c) Modern and Nobel error # 50 is Δ θ (per century) = - 2 x 15 τ0 sine2 arc tan (v/c); τ0 = 100 years = - 30 (100 x 36526 x 24 x 3600) sine2 tan-1 (48.14/300,000) = 43 arc sec/100y

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