arabs real time physics

8/9/2019 Arabs Real Time Physics

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Arabs real time physics(I/O) 0 = 0 + [(I O)/O] 0Quantum = classical + relativistic

[email protected]

I am the greatest physicist of all time since July 4 th 1973

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mailto:[email protected]:[email protected]


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Chapter One

Arabs Real time physics: We can not see or measure something that had nothappened. We can only see and measure something that had happened. What wesee and measure in not what happened. We measure in present time an eventthat happened in past time. That is we measure past events in present timePresent time = present timePresent time = past time + [present time past time]Present time = past time + time difference

Real time = event time + time delayReal time physics = Event time physics + time delays physicsTime dependent measurements = time independent measurements +[Time dependent measurements time independent measurements]

Relative = absolute + [relative absolute]Measured = actual + [measured actual]

In common Terms: Image = object + [Image object]I = O + (I O)I/O = O/O + (I O)/OI/O = 1 + (I O)/O

In mathematical terms:

Real time scaleReal time = event time + [real time event time] = t + ( t)/t = t/t + ( t)/t/t = 1 + ( t)/t

Real time distance scale

Measured distance = actual distance + [measured distance actual distance]Or, r = r0 + (r r0)And r/r0 = r0/r0 +(r r0)/r0And r/r0 = 1+(r r0)/r0And (r/r0) 0= 0+ [(r r0)/r0] 0

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Real time Velocity Scale v

With r = r0 + (r r0)Divide by t

Then r/t = r0/t+ (r/t r0/t)And v = v0 + (v v0)

Real time Angular velocity scale

With v = v0 + (v v0)Divide by r0

Then v/ r0 = v0/r0 + (v/r0 v0/ r0)And = 0 + ( 0)

In 1969 there was a man on the moon. In 1969 I finished 5th grade. In1969 I knew all need to be known about circles in 5 th grade arithmetic.

I knew that the circumference of a circle C = 2 r0 where r0 is the radius.In 1969 I knew about circular speed v0 = 2 r0/T where T0 is the period ofrotation. Also, in 1969 I knew about angular speed 0 = 2 /T0 = (v0/ r0).

Or C = 2 r0; v0 = 2 r0/T0; 0 = 2 /T0 = (v0/ r0)

Angular velocity is 0 = 2 /T0 = (v0/r0)In arc second per century 0 = (v0/r0) (180/ ) [36526/T0 (days)] (3600)

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If 1= 1

2 = 23 = 3

Then we can write anything that is equal to itself as equal to itself like:Visual = VisualActual = actual

- Actual = - actual----------------------- Add

Visual = Actual + (Visual actual)

Divide by actual

Visual/actual = Actual/actual + (Visual actual)/actual

Or

Visual/actual = 1 + (Visual actual)/actualImage/object = 1 + (Image object)/object

I/O = 1+ (I O)/O

We do not see object O but we see Image I. If we to look at planet Mercuryfrom Earth and not from the Sun, then we are not looking at 0 = 2 /T =(v0/r0) but we are looking:

At 0 I/O = 1 0+ [(I O)/O] 0Or 0 I/O = 0+ [(I O)/O] 0

The visual illusion and Modern Physicists confusion is: [(I O)/O] 0Where I is Mercury Sun distance = 58,200,000 kilometers = r mAnd O is Earth Sun distance = 149,600,000 kilometers = r e

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Planet

r

Sun


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The Visual Illusions and modern Physicists confusion of Planet Mercury is:[(I O)/O] 0 = [(r m r e)/ r e] 0

Multiplying by (t/t) = 1[(I O)/O] 0 = [(r m/t r e/t)/ r e/t] 0

Then [(I O)/O] 0 = [(v m v e)/ v e] 0

[(I O)/O] 0 = [(r m r e)/ r e] 0 = [(r m r e)/ r e](v0/r0)[(I O)/O] 0 = [(v m v e)/ v e] 0 = [(v m v e)/ v e](v0/r0)With v0 = v mAnd r0 = r mCopernicus Illusion[(I O)/O] 0 = [(r m r e)/ r e](v m/r m) [(180/) (3600) (26526/T m)]Or Galileo Illusion; r m = v m T m; r e = v e T e[(I O)/O] 0 = [(v m T m v e T e)/ v e T e](v m/r m) [(180/) (3600) (26526/T m

Tyco Brahe: [(r m r e)/ r e]=[(r m /t) (r e/ t)]/ (r e/t) = (v m v e)/ v e[(I O)/O] 0 = [(v m v e)/ v e](v m/r m) [(180/) (3600) (26526/T m)]

Johannes Kepler: Keplers law:

Of: a/T = k = constantOr, a1/ T1 = a2/ T2

Or, a1/ a2 = (T1/ T2)2/3

And (a1 - a2)/ a2 = (T1/ T2)2/3 1Or (am a e)/ a e = (Tm/ Te) 2/3 1[(I O)/O] 0 = [(r m r e)/ r e](v m/r m) [(180/) (3600) (26526/T m)]Or[(I O)/O] 0 = [(v m T m v e T e)/ v e T e](v m/r m) [(180/) (3600) (26526/T mOr[(I O)/O] 0 = [(v m v e)/ v e](v m/r m) [(180/) (3600) (26526/T m)]Or[(I O)/O] 0 = [(Tm/ Te) 2/3 1] (v m/r m) [(180/) (3600) (26526/T m)]

Then Urbain Jean Joseph Le Verrier cameThe angular velocity of Mercury around the Sun is: m' = v m /r mAnd 0 = (v m /r m) (180/) (3600) (26526/Tm); Tm = 88 days

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If it is measured for planet Mercury from the sun thenThen it is m' = v m /r m

If planet Mercury around the sun measured from earth then =Then m' (Earth) = (v m + v e)/r m

And m' (Earth) = v m /r m + v e /r mAnd not v m /r m

Le Verrier [1] mistake is: v e /r mThe angular speed delay is: v*e /r mOr, [(v e/v m) (v m /r m)]Taking into account Earth rotation v eLeverrier mistake: Then the angular speed delay:

Is: v e /r m = [v* e /r m +/- v e /r m]

In arc second per century multiplying by [(180/) (3600) (26526/T)][(I O)/O] 0= [(v* e +/- v e) /r m] [(180/) (3600) (26526/T)][(I O)/O] 0= [(v* e +/- v e) /v m) (v m /r m)] [(180/) (3600) (26526/T)]

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SM

E


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[(I O)/O] 0 = [(r m r e)/ r e](v m/r m) [(180/) (3600) (26526/T m)]Or[(I O)/O] 0 = [(v m T m v e T e)/ v e T e](v m/r m) [(180/) (3600) (26526/T mOr[(I O)/O] 0 = [(v m v e)/ v e](v m/r m) [(180/) (3600) (26526/T m)]Or[(I O)/O] 0 = [(Tm/ Te) 2/3 1] (v m/r m) [(180/) (3600) (26526/T m)]Or[(I O)/O] 0= [(v* e +/- v e) /v m) (v m /r m)] [(180/) (3600) (26526/Tm)]

When Newton came with F = - GmM/rPhysicists calculated: ' c m = v c m/r;Angular velocity with respect to center of massAnd Astronomers observed: ' s = v s/rAngular velocity with to the sun

And v c m = [GM/ (m + M) a]And v s = (GM/a)And (' cm - ' s) /' s = [(2 /T c m) - (2 /T s)]/ (2 /T s)

= (T c m/T s) - 1 = (v s/v c m) 1

= [ (GM/a)] x { [(m + M) a/ GM]} - 1= [(m + M)/M] - 1= [1 + (m/M)] - 1 1 + (m/2 M) 1 m/2 M

And (' cm - ' s) /' s (m/2 M)And (' cm - ' s) / ' s = (2 /T s) (m/2 M)And [(' cm - ' s) ] T s = (m/ M)Multiplying by [(180/) (3600) (26526/T m)][(I O)/O] 0 = ( m/M) [(180/) (3600) (26526/T m)]Nicoklaus CopernicusDistance Illusion[(I O)/O] 0 = [(r m r e)/ r e](v m/r m) [(180/) (3600) (26526/T m)]Or GalileosVelocity Illusion[(I O)/O] 0 = [(v m T m v e T e)/ v e T e](v m/r m) [(180/) (3600) (26526/T mOr Tyco Brahes[(I O)/O] 0 = [(v m v e)/ v e](v m/r m) [(180/) (3600) (26526/T m)]Johannes Kepler Period Illusion[(Tm/ Te) 2/3 1] (v m/r m) [(180/) (3600) (26526/T m)]Or Le Verrier Frame Illusions[(I O)/O] 0= [(v* e +/- v e) /r m) [(180/) (3600) (26526/T m)]Or Newtons Inverse Square force mass Illusion[(I O)/O] 0 = ( m/M) [(180/) (3600) (26526/T m)]

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Newton said there is gravity Force F = -GmM/r whose solution is this

Fig. 1. Newtons gravitational law

And r (, t) = [a (1-)/ (1+ cosine )]

Newton equation is a solved wrong for 350 years

Newton's Gravitational Equation is: F = -GmM/rWith d (m r)/dt - (m r) ' = -GmM/r (1)And d (mr')/d t = 0 (2)

The solution is notAnd r (, t) = [a (1-)/ (1+cos)]But this solution

Real time solution is: r (, t) = [a (1-)/ (1+ cosine )] [ (r)+ (r)] t

"Apparent advance of perihelion"

[(I O)/O] 0 = (-720x36526x3600/T) {[ (1- )]/ (1-) } x[(v+ v*)/c] arc second per century

Proof: All there is in the Universe is objects of mass m moving in space (x, y, z) at alocation r = r (x, y, z) = r [length, width, height].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 rate

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Planet

r

Sun


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F = d P/d t = dS/dt = Total force

= m (dr/dt) +2(dm/d t) (d r/d t) + (dm/dt) r

= m + 2m'v +m" r; = acceleration; m'' = mass acceleration rate

In polar coordinates system

Location = r = r r (1)

Velocity = v = r' r (1) + r ' (1)

Acceleration = = (r" - r') r (1) + (2r'' + r ") (1)

F = m + 2m'v +m" r

F = m [(r"-r') r (1) + (2r'' + r ") (1)]+2m'[r' r (1) + r ' (1)] ++ (m" r) r (1)

F = [d (m r)/dt - (m r) '] r (1)

+ (1/mr) [d (mr')/d t] (1)

F = F1 + F2

F1 = [d (m r)/dt - (m r) '] r (1)

F2 = (1/mr) [d (mr')/d t] (1)

Newtons force law is F1 = [-GmM/r] r (1)

Kepler's force law is:

F2 = (1/mr) [d (mr)/d t] (1) = 0

If m is constant then d (mr)/d t = 0; and d (r)/d t = 0

Or, r = h = 2 a b/T; a = mean distance from sun and is called semimajor axis and b is the semi minor axis.

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And the motion of a planet m around the Sun M measured in real time is a rotating ellips

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r

Sun

Mercury

M

m


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Tyco Brahe Logged observational Data of Planets P motion around the Sun S andthen Kepler stated the areal velocity law: If Planet p observed from the sun then trajectory of planet p will cut equal areas in equal times.

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A

AA

A

S

A

A

A

S

T

T

TTT

T

T

P

P


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When the areas size A are sliced equally it was found that the times spent byplanets orbiting around the Sun and making areas A each are equal also.Or, r (0) ' (0) = r (1) ' (1) = r (2) ' (2) = r (3) ' (3) = --

= location x [angular speed] = constant= Areal velocity

If r '= hThen differentiating with respect to timeThen d (r ')/ d t= d h/ d tAnd 2 r r + r = 0Or 2 (r /r) = - (/)

The r = r0 e t

And = 0 e 2 t

With r = r0 e t

In real time r (n) = r (0) e t

With r (0) ' (0) = r (1) ' (1)Then ' (1) = [r (0)/ r (1)]' (0)

And ' (1) = {[e 2 t] ' (0)And (1) = {[cosine 2 t + sine 2 t] 1} ' (0) (1) = (x) + (y)

= [cosine 2 t + sine 2 t] ' (0) (x) = [cosine 2 t] ' (0) (x) = [1 2 sine t] ' (0)

(x) - ' (0) = - 2' (0) sine t

W = (x) - ' (0) = - 2' (0) sine t= -2(2) [ (1-)]/T (1-) ] sine tAnd ' (0) = h/r (0) = 2 a b/T a = 2 (1-)]/T (1-)

W = - 4 ) [ (1-)]/T (1-) ] sine t

If this apsidal motion is to be found as visual effects, thenWith, v = spin velocity; v* = orbital velocityAnd v/c = (v* + v)/c = tan t

W = -4 [ (1-)]/T (1-) ] sine {Inverse tan [(v* + v)/c]} radians

Multiplication by 180/ to change to degrees

W = (-720/T) {[ (1-)]/ (1-) } sine {Inverse tan [(v* + v)/c]} DegreeAnd multiplication by 1 century = 36526 days and using T in days

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W (ob) = (-720x36526/Tdays) {[ (1-)]/ (1-) } xsine {Inverse tan [(v* + v)/c]} degrees/100 years

ApproximationsWith v


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Where v* (p) = [G M / (m + M) a (1-/4)] [GM/a (1-/4)];m


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These two equations give an axial rotation rate:One: = (m/ M) (180) [36526/T] [3600] arc second/100 years

= 43.0344 seconds of arc / century for MercuryTwo: ' = - 720 [36526/T] (3600) (1 - )/ T (1 - ) (v/c) arcsecond/100 years

= 43.0" seconds of arc /century for Mercury

Solution:With m = constantThen d r/dt - 'r = - k/ r (1)And d (r')/d t = 0 (2)From (2) d (r')/d t = 0; r' = hFrom (1), ' d r/ d - 'r = - k/ mrAnd ' [d r/ d - r] = - k/ mrAnd d r/ d - r = - (k/mh) r

And d r/ d - r [1 - (k/mh)] = 0And r (, 0) = r (0, 0) e { [1 - (k/mh)]}

From (2) d (r')/d t = 0; r' = hThen 2rr'' + r'' = 0Dividing by r'

We get 2 (r'/r) + (''/') = 0And 2 (r'/r) = - ''/' = 2 t

And r = r (0, 0) e { [1 - (k/mh)]}e t

And ' = ' (, 0) e- 2 tOr r = r (0, 0) e { [1 - (k/mh)]}+ t

And ' = ' (0, 0) e-2 [{ [1 - (k/mh)]} + t]

And ' = ' (0, 0) e-2 [{ [1 - (k/mh)]} + t]And ' = (' (0, 0) {cosine 2 [{ [1 - (k/mh)]} + t]

- sine 2 [{ [1 - (k/mh)]} + t]}

And ' - ' (0, 0) = - 2 ' (0, 0) sine [{ [1 - (k/mh)]} + t]And ' = - 2 ' (0, 0) sine [{ [1 - (k/mh)]} + t]If k = Gm M (1 - ); h = 2 a b/T

Then: k/mh = Gm M a (1 - )/m 4 a b/TAnd k/mh = GM T (1 - )/4 a b; multiply by (a/a)Then [a (1 - )/ b] [GM T/4 a] = [GM T/4 a]

Then ' = - 2 ' (0, 0)] sine [{ [1 - (GM T/4 a)]} + t]Taking Kepler's: GM T/4a = 1

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Then ' = - 2 ' (0, 0) sine tAnd ' (0, 0) = h/r = 2ab/Ta (1 - )

= 2a (1 - )/Ta (1 - ) = 2 (1 - )/T (1 - ) And ' = - 4 (1 - )/ T (1 - ) sine tWith T = arc tan v/c


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Chapter two

Nuclear Gravity F = (-GmM/r) ek/r

With d (m r)/dt (m r) ' = [-GmM/r] e k/ r Nuclear gravity Equation (1)And (mr')/d t = 0 Kepler's Areal Velocity (2)

(2): d (mr')/d t = 0Then mr' = constant; if m is taken as constant then r' = h

And (1): d r/dt - r ' = [-GmM/r] ek/rLet m r =1/uThen d r/d t = -u'/u

= - (1/u) (') d u/d = (- '/u) d u/d = - h d u/d

And d r/dt = -h'du/d= - hu [du/d]

With d (m r)/dt - (m r) ' = [-GmM/r] ek/r Nuclear Gravity (1)

With ek/r

1+ k/r; k/r


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Then, r = 1/u = (GM/ h) + A cosine If, T/ a = 4/ G (M +m) NewtonsThen, [1 GMk/h] = (GMT/4a) = [M/ (M + m)]= 1/ [1 + (m/M)] 1/ [1 + (m/2M)] = 1 (m/ 2M)And [1 GMk/h]/ (M/ h)]/ {1 + cosine { [1 - GMk/h]}

= a (1- )/ [1 + cosine { [1 (GMk/h)]} = a (1- )/ {[1 + cosine [(1 m/2M) ]}= a (1- )/ [1 + cosine ( )]And 2 = 2 (m/ 2M) = (m/M)If Newtons law were to be F = - Gm (M + m)/rThe 2 = 2 [m/ 2(M + m)] = [m/ (m + M)]

Frequency GamesWhat is the accumulated value of 2 f per century for planet mercury seenfrom Earth?The angular frequency is = 2 f

How I would see of planet mercury turning around the sun from earth?The answer is there will be a frequency change ofW = 2 f [v* m - v*e]/v* e; radians per secondWhere v*e = Earth orbital velocity around the Sun = 29.8 km/secAnd v e = Earth spin speed = 0.465 km/secAnd v* m = Mercury orbital velocity around the Sun = 47.9 km/secWith f = 1/T; f = frequency; T = Period = 88 daysIf W is wanted in degrees multiply by: 180/If W is wanted in degree per century multiply by (180/) x (36526 days/ T)If W is wanted in arc second per century multiply by

(180/) x (36526 days/ T) x 3600W = [2 f [(v* m - v*e)/v* e] x (180/) x [36526 days/ (T days)] x 3600With f = 1/T (seconds)W = [360 x 3600 x (36526/T (days)] [1/T (seconds)] [(v* m - v*e)/v* e]

W = [360 x 3600 x (36526/T (days)] [1/ 24 x 3600 x T (days)] [(v* m - v*e)/v* e]W = 15 x (36526/T (days)] [(v* m - v*e)/v* e][(I O)/O] 0 =15 x (36526/T (days)] [(v* m - v*e)/v* e]For MercuryW = 15 x [36526/ (88)] [(47.9 29.8)/29.8] = 43.0 arc second per century

The conclusion isWith = 2 f = 2 /T angular frequency in event timeAnd (real time) = (event time) + = (real time) - (event time)= 2 f [(v* m - v*e)/v* e]= 2 f Z; Z = [(v* m - v*e)/v* e]

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And = 2 f (1 + Z)The Advance of Planet Mercury Perihelion is2 f Z = 2 Z/ T = (2 /T) [(v* m - v*e)/v* e]In arc seconds per century: Multiply by: (180/) [36526/T (days)] (3600)And = [2 / T (seconds)][(v* m - v*e)/v* e] x

(180/) [36526/T (days)] (3600);T (seconds) = T days x 24x 3600 = [2 / T days x 24x 3600][(v* m - v*e)/v* e] x

(180/) [36526/T (days)] (3600) = [15 x 36526 / T (days)] [(v* m - v*e)/v* e]

Visual force: F = - Gm M/r - Gm Mk/r

With m [d r/dt - 'r] = - Gm M/r - Gm Mk/r Visual Gravitational law (1)And d (r')/d t = 0 Kepler's Areal Velocity Equation (2)

Gives an axial rotation rate ofW = 15 x (36526/T (days)] [(v* m - v*e)/v* e]

Nuclear Gravity (-GmM/r) e (m / M r)

Abstract: Yukawa Gravity (-GmM/r) e (m / M r) or the nuclear gravityforce is the crudely approximated Newtonian gravity force (-GmM/r)explains planetary motion around the sun as a rotating ellipse with arotation rate = [ m/(m + M)](180/)(36526/T)(3600) = 43.03 secondsof an arc per century for the most talked about planet of mercury; m = 3.2x 1024 kg; M = 2x1030 kg; T = 88days

With d (m r)/dt (m r) ' = (-GmM/r) e (m / M r) Gravity Force (1)And d (mr')/d t = 0 Kepler's Areal Velocity (2)

Then mr' = constant; if m is taken as constant then r' = hAnd (1): d r/dt - r ' = -GmM/r + Gm/r

Let m r =1/uThen d r/d t = -u'/u = - (1/u) (') d u/d = (- '/u) d u/d = - h d u/d And d r/dt = -h'du/d = - hu [du/d]

-hu [du/d] - (1/u) (hu) = -G M u [1 ( m /M) u]; m /M r


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= [1 Gm/h]/M/ h]/ {1 + cosine { [1 - Gm/h]}

Where [1 Gm/h]/GM/h = a (1 )And [1 Gm/h] = (GM/h) a (1-)

If = a (1 ), the h = G (m + M) a (1- )Then Gm/h = G ma (1 - ) /G (m + M) a (1- )

= m/ (m + M)With r (, 0) = [1 Gm/h]/M/ h]/ {1 + cosine { [1 Gm/h]} Then r (, 0) = a (1 )/ {1 + cosine { [1 Gm/h]}

And r (, 0) = a (1 )/ {1 + cosine { [1 m/ (m + M)]} And m


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Boundary value LawWith d (m r)/dt - (m r) ' = - k/ r Inverse square Gravitational (1)And d (mr')/d t = 0 Kepler's law (2)At Perihelion: d r/d t - r ' = - GM/r = - r '; d r/d t = 0Then r ' = GM/rA quick answer by Newton would be: First ' = [GM/r]And = [GM/r] [(v* m - v*e)/v* e]

In arc sec / centuryThen = {[GM/r]} 1/2 [(v* m - v*e)/v* e] [(180/) (3600) (36526/Tm)

= 43"/century[(I O)/O] 0= {[GM/r]} 1/2 [(v* m - v*e)/v* e] [(180/) (3600) (36526/Tm)

Constant force law

Or, r 1 ' 1= r 2 '2= constantOr, ' 1= {[r 2/ r 1]} ' 2And ' 1 - ' 2= {[ (r 2/ r 1)] - 1} ' 2 ' = {[ (r 2/ r 1)] - 1} ' 2This is the angular time delay and will be seen as angular visual IllusionThe angular speed is ' = v/rFor Mercury: ' = v/r = (47.9km/sec)/58,200,000 km = 0.000000843radians/secIf you want the accumulation value in arc sec /century W", then

And W" = (v/r) (180/) (3600) (26526/T) = angular velocity in arc sec percentury. If it is measured for planet Mercury thenW" = (47.9/58,200,000) (180/) (3600) (26526/88)W"= 70.29 arc second per centuryOr, W" = {[ (r 2/ r 1)] - 1} W" (2)What is the angular visual Illusion for planet Mercury that would be seenwhen measured from Earth with Earth location r (1) = Earth = 149.6 x 106

And r (2) = Mercury = 58.2 x 106

And W" (2) = - 70.29 arc sec /century

W" = {[ (r 2/ r 1)] - 1} W" (2) W" = {{ [149.6/58.2]} - 1} [-70.29] = 43.0" arc per century

[(I O)/O] 0 ={{[r e/ r m]} - 1} (v m /r m)] [(180/) (3600) (26526/T)

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Constant Areal velocity Law

Or, r 1 '1= r 2 '2 = location x speed = constant = Areal velocity

Or, r 1 v 1= r 2 v2

Or, ' 1= (r 2/ r 1) ' 2And ' 1 - ' 2= [(r 2/ r 1) - 1] ' 2

' = ' 1 - ' 2= [(r 2/ r 1) - 1] ' 2 ' = [(v 1/ v 2) - 1] ' 2

This is the angular time delay and will be seen as angular visual IllusionThe angular speed is ' = v/rFor Mercury: ' = v/r = (47.9km/sec)/58,200,000 km = 0.000000843radians/sec

If you want the accumulation value in arc sec /century W", thenAnd W" = (v/r) (180/) (3600) (26526/T) = angular velocity in arc sec percentury. If it is measured for planet Mercury thenW" = (47.9/58,200,000) (180/) (3600) (26526/88)W"= 70.29 arc second per century

Or, W" = [(v 1/ v 2) - 1] W" (2)

What is the angular visual Illusion for planet Mercury that would be seen

when measured from Earth with Earth location r (1) = Earth = 149.6 x 106

And r (2) = Mercury = 58.2 x 106

And W" (2) = - 70.29 arc sec /century

W" = [(v 1/ v 2) - 1] W" (2)

W" = [(v 1/ v 2) - 1] (v m /r m)] [(180/) (3600) (26526/T)

W" = [(29.8/ 47.9) - 1] [-70.75] = 43.0" arc per century

[(I O)/O] 0

= [(v e/ v m) - 1] (v m /r m)] [(180/) (3600) (26526/T)

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No Force solutionNewton said there is gravity Force F = -GmM/r whose solution is this

Fig. 1. Newtons gravitational law

With m [d r/dt - 'r] = 0 Zero force lawequation (1)And d (r')/d t = 0 Kepler's Areal Velocity Equation (2)

Equation (2) r = r0 e t; '= '0 e- 2 t

Equation (1) - h d / d u - h u = 0Or d / d u + u = 0

Then u = u0 e - And r0 = r0 (0) e + Or, r = r0 (0) e ( + t) [h/ '0] e ( + t)

And '= '0 e- 2 t

' = - 2 '0 sine t= - 2[2/T] x [180/ ] [38526/T] [3600] sine t= - 2 [2/88 x 24 x 3600] x [180/ ] [38526/T] [3600] sine t= - 141.5 sine t

And t = arc tan [(r m - r e)/ (r m + r e)]

= arc tan [(149.6 58.2)/ (149.6 + 58.2)] ' = - 141.5 sine arc tan [(149.6 58.2)/ (149.6 + 58.2)]

= 43 arc seconds per century

Or t = arc tan [(a - c)/ (a + c)]

=arc tan [(1 )/ (1 + )]

[(I O)/O] 0 = - 2 '0 sine t= -2 sine {arc tan [(r m - r e)/ (r m + r e)] } x(v m /r m)] [(180/) (3600) (26526/T)[(I O)/O] 0 = - 2 '0 sine t= -2 sine {arc tan arc tan [(1 )/ (1 + )]

} x(v m /r m)] [(180/) (3600) (26526/T)

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Planet

r

Sun


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Newtons mathematically wrong [(I O)/O] 0 =(-720x36526x3600/88) (1.552) (48.14/300,000) arc sec per century

= 43This particular equation structure is the basis of stars studying. Originally, Iderived it from Arabic real time physics literature. A visual Illusion alongthe line of sight is constructed as follows:Instead of seeing r0 we see r0 cosine arc tan (v/c) and perpendicular tothe line of sight we see r0 sine arc tan (v/c)Or r = r0 [cosine arc tan (v/c) + sine arc tan (v/c)]

Or r = r0 e arc tan (v/c)

Or r = r0 e tAnd areal velocity law gives the same resultsIf r '= hThen differentiating with respect to timeThen d (r ')/ d t= d h/ d tAnd 2 r r + r = 0Or 2 (r /r) = - (/)

The r = r0 e t

And = 0 e 2 t

With r = r0 e t

In real time r (n) = r (0) e tWith r (0) ' (0) = r (1) ' (1)Then ' (1) = [r (0)/ r (1)]' (0)

And ' (1) = {[e 2 t] ' (0)And (1) = {[cosine 2 t + sine 2 t] 1} ' (0)

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Planet Distance rX 106km

PlanetOrbit T

Orbitspeed vinkm/sec

LessEarthspeed

Spinspeedkm/sec

Angular velocity;v/rarc sec/ century

Mercury 58.2 88 47.9 18.1 .002 70.75Venus 108.2 224.7 35.05 5.7 6.52 10.86Earth 149.6 365.26 29.8 .46511 4.1Mars 227.936 687 24.14 0.2411Jupiter 778.412 4333 13.06 12.6Saturn 1,426.725 10760 9.65 9.87Uranus 2,870.97 30690 6.80 2.59Neptune 4,498 60180 5.43 2.68Pluto 5906.4 90730 4.74


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(1) = (x) + (y)= [cosine 2 t + sine 2 t] ' (0)

(x) = [cosine 2 t] ' (0) (x) = [1 2 sine t] ' (0) (x) - ' (0) = - 2' (0) sine t

W = (x) - ' (0) = - 2' (0) sine t= -2(2) [ (1-)]/T (1-) ] sine t

And ' (0) = h/r (0) = 2 a b/T a = 2 (1-)]/T (1-) W = - 4 ) [ (1-)]/T (1-) ] sine tIf this apsidal motion is to be found as visual effects, thenWith, v = spin velocity; v* = orbital velocityAnd v/c = (v* + v)/c = tan t W = -4 [ (1-)]/T (1-) ] sine {Inverse tan [(v* + v)/c]} radiansMultiplication by 180/ to change to degrees W = (-720/T) {[ (1-)]/ (1-) } sine {Inverse tan [(v* + v)/c]} Degree

And multiplication by 1 century = 36526 days and using T in daysW (ob) = (-720x36526/Tdays) {[ (1-)]/ (1-) } xsine {Inverse tan [(v* + v)/c]} degrees/100 years

Abstract: light measurements are done along the line of sight and we measurelight projections to our eye. Meaning, that we do not measure distance r but wemeasure r cosine arc tan (v/c) when light is used and as a consequence insteaof measuring time T we measure T cosine arc tan (v/c) and here is theexperimental proof of it.

Your eyes

A000 0

0000 00 0 0 0 00 0 0 0 00 0 0 0 00 0 (2) 0 0 00 (1) 0 0 (3) 0 00 0 0 0 00 0 0 0 00 0 0 0 00 0 0 0 00 0 0 0 0B------------------------- 0 --------------- 0 ------------- 0

1 2 3

The projections of light is what we measure and the projectionof 1 and 2 and 3 is AB

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If the angle between ray 1 and AB is (1)If the angle between ray 1 and AB is (2)If the angle between ray 1 and AB is (3)

Then C (1) cosine (1) = C (2) cosine (2) = C (3) cosine (3) = c = ABYou want experimental proof?Here is my iron clad proof of my theory: This is using the lightprojections to be equal c then the advance of perihelion andstars to solve mercury Venus and DI Herculis binary stars betterthan all isIf an object at location A sends light signals in all directions at time at t = 0where B1 is;And B1 is moving to B2 then the projection of AB2 of AB1 is measured.AB1 is theorized and AB2 is measured; or AB2 = AB1 cosine

And = arc tan (v/c)B1B2 /AB1 = sine ; AB2/AB1 = cosine ; B1B2/AB2 = tan Instead of measuring time t we theorize time t and we measure time = t

cosine

A Light V E L O C I T Y B2O O O O O O O O O O

O C OO O

O O

O V OO OO O

O OO O

OB1

= t cosine arc tan (v/c) = t {1- 2 sine {[arc tan (v/c)]/2}} - t = - 2 t sine {[arc tan (v/c)]/2}With r2 = h and I/T= / = - 2r/r = 2/tOr T = t/2 or the time measurements changes are twice period changes

Or ( t)/2 = - t sine {[arc tan (v/c)]/2} is the time delays for periodsWith t = 1 century and t in arc seconds is 15 times time seconds then( t)/2 = - 15 T sine {[arc tan (v/c)]/2}; T = One centuryWith v = 47.9 (mercury) 29.8 (Earth) =18.1 km/sec( t)/2 = - 15 (36526) sine {[arc tan (18.1/300,000)]/2] = 43

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Nicoklaus CopernicusDistance Illusion

[(I O)/O] 0 = [(r m r e)/ r e](v m/r m) [(180/) (3600) (26526/T m)] = 43

M r Copernicus you are wrong

Or GalileosVelocity Illusion

[(I O)/O] 0 = [(v m T m v e T e)/ v e T e](v m/r m) [(180/) (3600) (26526/T m)] = 43

M r Galileo you are wrong

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Or Tyco Brahes[(I O)/O] 0 = [(v m v e)/ v e](v m/r m) [(180/) (3600) (26526/T m)] = 43

M r Tyco Brahe you are ok but not good enough

Or Johannes Kepler Period Illusion[(Tm/ Te) 2/3 1] (v m/r m) [(180/) (3600) (26526/T m)] = 43

M r Kepler you are OK but not good enoughPage 28


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Or Le Verrier Frame Illusions[(I O)/O] 0= [(v* e +/- v e) /r m) [(180/) (3600) (26526/T m)] = 43

IdiotPage 29


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Or Newtons Inverse Square force mass Illusion[(I O)/O] 0 = ( m/M) [(180/) (3600) (26526/T m)] = 43

Mr Newton you are not a physicist

Or Laplaces wrong [(I O)/O] 0 =-720x36526x3600/T) {[ (1-]/ (1-) } [(v*m + v m)/c] seconds of arc per centur

43

Mr Newton you are not a good mathematician either

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Or Yukawas:[(I O)/O] 0 Mr Yukawa I am definitely not talking about you and time travel is not physics

= 15 x (36526/T (days)] [(r m r e)/ r e] = 43= 15 x (36526/T (days)] [(v m T m v e T e)/ v e T e] = 43= 15 x (36526/T (days)] [(v m v e)/ v e] = 43= 15 x (36526/T (days)] [(Tm/ Te) 2/3 1] = 43Or, Center of massgravity equation

[(I O)/O] 0 = [ m/ (m + M)] (180/ ) (36526/ T m) (3600)] = 43

M r Pierre LaplaceThe solution to F = -GmM/r is notr = [a (1-)/ (1+ cosine )] and it is time you get corrected

Real time solution is: r (, t) = [a (1-)/ (1+ cosine )] [ (r)+ (r)] t

"Apparent advance of perihelion"Page 31


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Or, Boundary value law[(I O)/O] 0 == {[GM/r]} 1/2 [(r m r e)/ r e] [(180/ ) (3600) (36526/ T m) = 43= {[GM/r]} 1/2 [(v m v e)/ v e] [(180/ ) (3600) (36526/ T m) = 43= {[GM/r]} 1/2 [(Tm/ Te) 2/3 1] [(180/) (3600) (26526/T m)] = 43

= {[GM/r]} 1/2 [(v m T m v e T e)/ v e T e][(180/) (3600) (26526/T m) = 43

Or, Constant Fore law[(I O)/O] 0 =

{{[r e/ r m]} - 1} (v m /r m)] [(180/) (3600) (26526/T) = 43

Or, constant areal velocity law[(I O)/O] 0

= [(v e/ v m) - 1] (v m /r m)] [(180/) (3600) (26526/T) = 43Or, No force law one[(I O)/O] 0 = - 2 '0 sine t

= -2 '0sine {arc tan [(r m - r e)/ (r m + r e)] } x(v m /r m)] [(180/) (3600) (26526/T) = 43

Or, No force law two[(I O)/O] 0 = - 2 '0 sine t= -2 sine {arc tan arc tan [(v m T m v e T e)/ (v m T m + v e T e)] } x

(v m /r m)] [(180/) (3600) (26526/T) = 43

Or, No force law three[(I O)/O] 0 = - 2 '0 sine t

= -2 sine {arc tan arc tan [(1 )/ (1 + )] } x(v m /r m)] [(180/) (3600) (26526/T) = 43

With Newtons mathematically wrong

[(I O)/O] 0 =(-720x36526x3600/T) {[ (1-]/ (1-) } (v*m + v m/c) Seconds of arc per centuryTime: ( t)/2 = - 15 T sine {[arc tan (v/c)]/2}For v


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Chapter three

Arabs real time physics in outer Space

Apparent apsidal motion of binary (two) stars

The wrong solutions of Newtons equation have/had led to stupidity calledmodern physics. Newtons equation was solved wrong and modified to fitexperimental data here and there but the correct solution of Newtonsequation deletes the modifications and the incorrect solution and theincorrect conclusions and explains planetary motions with unprecedentedaccuracy including Apsidal motion of binary stars that no one can solve andGPS systems and interplanetary telecommunications systems of wrongShapiros time delays.A- This is the solution to the 150 years apsidal motion puzzle that is not

solvable by space-time physics or any said or published physics including500 years of Modern physics and 500,000 modern physicists. Binary starsapsidal motion or "Apparent" rate of orbital axial rotation is a visual effectalong the line of sight of moving objects applied to the angular velocity atApses. From the thousands of close binary stars astronomers picked adozen sets of binary stars systems that would be a good test of relativitytheory and modern physics collected data for all of past century andrelativity theory failed every one of them. This rate of "apparent" axialrotation is given by this new equationW (ob) = (-720x36526/T) {[ (1-)]/ (1-) ]} (v* + v/c)

degrees/100 yearsT = period; = eccentricity; v = spin velocity effect; v*= orbital velocity effectsThe simplest problem in physics that all 500 years modern physicsand 500,000 modern physicists could not solve by any physics:1 As Camelopardalis; 2 DI Herculis; 3 V1143 Cygni; 4 V 541 Cygni5 - AI Hydra; 6 - V 731 Cephei; 7 - SW Canis Majoris; 8 - NV Canis Majoris9 - GG Orion; 10- CM Draconis; 11- VV PYX; 12 EW Oriomis13- V1147 Sagittari; 14 PSR 1913 + 16; 12 PSR 0737 3039Solution: Location r = r r (1) ; Velocity v = r' r (1) + r ' (1)Acceleration = (r" - r') r (1) + (2r'' + r ") (1)S = m r; State = mass x distance

P = d S/ d t = d (m r)/d t = m (d r/d t) + (d m/d t) rVelocity = v = (d r/d t); mass rate change = m' = (d m/d t)P = m v + m' r; Momentum = change of state = change in location orchange in massF =d P/d t = d S/d t = d [m (d r/d t) + (d m/d t)]/d t

= m d r/d t + (d m/d t) (d r/d t) + (d m/d t) (d r/d t) + (d m/d t) rPage 33


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F = m d r/d t + 2 (d m/d t) (d r/d t) + (d m/d t) rForce = Change of momentumF = m a + 2 m ' v + m" rF = - GmM/rOr, Newton's Kepler's equation: F = - GmM/r

ThenWith d (m r)/dt - (m r) ' = -GmM/r Newton's Gravitational Equation (1)And d (mr')/d t = 0 Kepler's force law (2)With m = constant, then m can be taken out from both equations (1) and (2)With d r/d t - r ' = - GM/r Newton's Gravitational Equation (1)And d (r') /d t = 0 Kepler's force law (2)From 2: With m = constant; then d (mr')/d t = 0And m d (r')/d t = 0And d (r')/d t = 0And r' = hWith (1): d (m r)/dt - (m r) ' = -GmM/rWith m = constantThen m [d r/ d t - r '] = - Gm M/ rAnd [d r/ d t - r '] = - G M/ rLet r =1/uThen d r/d t = -u'/u = - (1/u) (') d u/d = (- '/u) d u/d = - h d u/d And d r /d t = - h 'du/d = - h u [du/d]And - h u [du/d] - (1/u) (hu) = - G MuOr, [du/ d] + u = G M/ h u = G M/ h + A cosine And r = 1/u = 1/ [G M / h + A cosine ]

= (h/ G M)/ [1 + (Ah/ GM) cosine ]= (h/GM)/ (1 + cosine )

Then r (, 0) = a (1-)/ (1+ cosine )This is Newton's Classical Equation solution of two body problem. Wesolved this equation and we got the motion equation:Is: r (, t) = [a (1-)/ (1+ cosine )] which is the equation of an ellipsewith eccentricity and semi - major axis a, and a semi- minor axis whosevalue is b = a [1 - ] and two foci one equals to c = a, and the otherfoci location equals to - c = - a. Or the motion of one ball around theother ball should be an ellipse with the other ball at one of the foci +/- a,

of the ellipse with semi major axes (- a, a) on x-axis and semi minor axes(-b, b) on y- axis and foci (- c, c) on x- axis again with angle of rotation .If this law to work on two planets instead of two balls like planetary motionaround the sun, then Astronomers should see this motion of a planetmoving around the sun in an ellipse with the sun at one of its foci.

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What astronomers saw was not an ellipse but a rotating ellipse like theellipse below wth rotating angle

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M

m

M

m


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When the areas size A are sliced equally it was found that the times spent byplanets orbiting around the Sun and making areas A each are equal also.Or, r (0) ' (0) = r (1) ' (1) = r (2) ' (2) = r (3) ' (3) = --

= location x [angular speed] = constant= Areal velocity

If r '= hThen differentiating with respect to timeThen d (r ')/ d t= d h/ d tAnd 2 r r + r = 0Or 2 (r /r) = - (/)

The r = r0 e t

And = 0 e 2 t

With r = r0 e t

In real time r (n) = r (0) e t

With r (0) ' (0) = r (1) ' (1)Then ' (1) = [r (0)/ r (1)]' (0)

And ' (1) = {[e 2 t] ' (0)And (1) = {[cosine 2 t + sine 2 t] 1} ' (0) (1) = (x) + (y)

= [cosine 2 t + sine 2 t] ' (0) (x) = [cosine 2 t] ' (0) (x) = [1 2 sine t] ' (0)

(x) - ' (0) = - 2' (0) sine t

W = (x) - ' (0) = - 2' (0) sine t= -2(2) [ (1-)]/T (1-) ] sine tAnd ' (0) = h/r (0) = 2 a b/T a = 2 (1-)]/T (1-)

W = - 4 ) [ (1-)]/T (1-) ] sine t

If this apsidal motion is to be found as visual effects, thenWith, v = spin velocity; v* = orbital velocityAnd v/c = (v* + v)/c = tan t

W = -4 [ (1-)]/T (1-) ] sine {Inverse tan [(v* + v)/c]} radians

Multiplication by 180/ to change to degrees

W = (-720/T) {[ (1-)]/ (1-) } sine {Inverse tan [(v* + v)/c]} DegreeAnd multiplication by 1 century = 36526 days and using T in days

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W (ob) = (-720x36526/Tdays) {[ (1-)]/ (1-) } xsine {Inverse tan [(v* + v)/c]} degrees/100 years

Astronomy re - written

W (calculated) = (-720x36526/T) {[ (1- )]/ (1- ) ]} [(v + v*)/cdegrees/100 years

1- As Cameloppardalis: Binary stars SystemAS Cam Data T=3.431; r (m) =0.1499; m=3.3 M (0); M=2.5 M (0)R (m) =2.57 R (0); [v (m), v (M)] = [40, 30]; = 0.1695; 1- = 00.8305R (M) = 2.5 R (0); r (M) =0.1111;m + M=5.8 M (0); G=6.673x10-11

M (0) = 2 x 1030 kg; R (0) = 0.696x109m;The circumference of an ellipse: 2a (1 - /4 + 3/16()- --.) 2a (1-/4); R =a (1-/4)

Finding orbital velocitiesFrom Newton's inverse square law of an ellipse motion applied to a circularorbit gives the following: m v/ r (cm) = GmM/rPlanet --- r (cm) ----- Center of mass ------- r (CM) --------- Mother SunPlanet ------------------- r -------------------------------------- Mother SunCenter of mass law m r (cm) = M r (CM); m = planet mass; M = sun massAnd r (cm) = distance of planet to the center of mass

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Primary Secondary

v(p) v* (p) v (p) v* (p) v (p) v* (p) v (p) V* (p)

v(s) v* (s) Spin=[,][,]=orbit

[,][,] [,][,] [,][,]

Spin results v(p) + v(s) v(p) + v(s) - v(p) + v(s) -v(p) + v(s)Orbit results v*(p) + v*(s) -v*(p) + v*(s) v* (p) + v*(s) -v* (p) + v* (s)Examples AS CAMv (s) v* (s) [,][,] [,][,] [,][,] [,][,]Spin results v(p) + v(s) v(p) + v(s) -v(p) + v(s) -v(p) + v(s)Orbit results v*(p) - v*(s) -v*(p) - v*(s) v*(p) - v*(s) -v* (p) - v* (s)

Examplesv (p) v*(s) [,][,] [,][,] [,][,] [,][,]Spin results v(p) - v(s) v(p) - v(s) -v(p) - v(s) -v(p) - v(s)Orbit results v*(p) + v*(s) -v*(p) + v*(s) v*(p) + v*(s) -v* (p) + v* (s)Examples AS CAM

v (s) V*(s) [,][,] [,][,] [,][,] [,][,]Spin results v(p) - v(s) v(p) - v(s) -v(p) - v(s) -v(p) - v(s)Orbit results v*(p) - v*(s) -v*(p) - v*(s) v*(p) - v*(s) -v* (p) - v* (s)Examples AS CAM


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And r (CM) = distance of sun to center of massAnd r (cm) + r (CM) = r = distance between sun and planetSolving to get: r (cm) = [M/ (m + M)] rAnd r (CM) = [m/ (m + M)] rThen v = [GM r (cm)/ r] = GM/ (m + M) r

And v = [GM/ (m + M) r = a (1-/4)]Planet orbital velocity or primary velocity:

And v* (m) = v (m) = [GM/ (m + M) a (1-/4)]R =a (1-/4)

Calculations: (1- /4) = 0.9928[ (1-)]/ (1-) = 1.43With a = [R (m)/r (m)]= (2.57/0.1499) (0.696x109) = 11.9327x109

And v* (m) = [GM/ (m + M) a (1- /4)] = 110km/sec

= [6.673x10-11

(2.5) (2 x 1030

)

/ (5.8) 11.932x109

(0.9928)] = 110.178km/sec

And v* (M) = [GM/ (m + M) a (1- /4)]= [6.673x10-11 (3.3) (2 x 1030) / (5.8) 11.932x109 (0.9928)] = 145.435km/sec

Spin: v = 40 + 30=70km/secAnd v* (e) = v* Earth = 29.8 km/secAnd v (e) = 0.465 km/secThen v* + v = v* (m) + v* (M) + v* (e) + v (m) + v (M) + v (e)

= 110.178 + 145.435 + 70 + 29.8 0.465 = 355 m/sec

[ (1-)]/ (1-) = 1.43; T = 3.431days

W= (-720x36526/T) x { [(1-)] (1-) } {[v* + v]/c}

W= (-720x36526/3.431) x (1.43) (355/ 300,000) = 15.0/centuryDr Guinan and DR Maloney 1989: W= 15/century 1989DR Khailullin and Dr Kozyreva: 1983 W = 14.6 /centuryEinstein: 44.3/ century

2 - Binary stars System: DI Her Apsidal Motion SolutionDI Her Apsidal motion solution:Data: T=10.55days r(m) = 0.0621 m=5.15M(0) R(m)=2.68R(0)[v(m),v(M)]=[45,45]And = 0.4882; r (M) = 0.0574 M=4.52 M (0) R (M) =2.48; m +

M=9.67 M (0)L = 2000 +/- 200 Ly

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Calculations1- = 0.5118; (1-/4) = 0.94; [ (1-)] / (1-) = 3.33181; 1 + =1.4882; G=6.673x10-11; M (0) = 1.98892x19^30kg; R (0) = 0.696x109mV* (p) = [GM/ (m + M) a (1-/4)] = 99.88 km/secV* (s) = [Gm/ (m + M) a (1-/4)] = 113.8km/sec

A- Apsidal motion is given by this formula:W (cal) = (-720x36526/T) {[ (1-)]/ (1-) ]} [(v + v*)/c] degrees/100 yearsAnd v = - 45 km/s - 45 km/sec = 90km/secNow let us calculate v* (cm) = m v/m = 106.38km/secWith v* = 2 v*(cm) = 212.76 km/secAnd v = -90 km/secThen v* + v = 212.76 - 90 = 122.76 km/sec andW (cal) = (-720x36526/T) {[ (1-)]/ (1-) } sine [Inverse tan122.76/300,000]

= (-720x36526/10.55) (3.33181) (122.76/300,000)W (cal) = 1.39/centuryWith = { [v* - v* (cm)] /2}

= {[106.36 - 99.88]/2 + [106.36 - 113.8]/2} = 6.975 km/secThen W (cal) = 1.39/century +/- 0.16Observed 2004: 1.39 +/- 0.3/century; Relativity: 4.27/century

B - W (cal) = (-720x36526/T) {[ (1-)]/ (1-) ]} [(v + v*)/c] degrees/100 years

With v = - 45 km/s + 45 km/sec = 0 km/secAnd v* = v* (cm) = m v/m = 106.38km/secW (cal) = (-720x36526/10.55) (106.38.300, 000) degrees/100 yearsW (cal) = 1.044 degrees/centuryC - V* (p) = [GM/ (m + M) a (1-/4)] = 99.88 km/secWith v* x (1-/4)/ (1-) = 99.88 (.94)/ (0.5118) = 135.36And v* x (1-/4)/ (1+ ) = 99.88 (.94)/ (1.4882) = 79.38K (A) = [(135.36 + 79.38)/2] = 107.37 km/secV* (s) = [Gm/ (m + M) a (1-/4)] = 113.8 km/secWith v* x (1-/4)/ (1-) = 113.8 (.94)/ (0.5118) = 154.2254

km/secAnd v* x (1-/4)/ (1+ ) = 113.8 (.94)/ (1.4882) = 90.44312km/secK (B) = 122.3342 km/secV* = [m K (A) + M K (B)]/ (m + M) = 114.3646 km/secOr v* = [K (A) + K (B)]/2 = 114. 85 km/sec; with v* = 114.85

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W (cal) = (-720x36526/T) {[ (1-)]/ (1-) } sine [Inverse tan114.85/300,000]

= (-720x36526/10.55) (3.33181) (114.85/300,000)= 1.22 degrees/ century

Observed: 1.24 degree/century +/- 0.05

References: Go to Smithsonian/NASA website SAO/NASA and type:1- Apsidal motion of DI Her: Dr Edward Guinan and Dr Frank Maloney;1985.2- New Apsidal Motion of DI Her: Dr Edward Guinan and Dr Frank Maloney;1994.3- D. YA. Martynov; and KH. F. Khaliulullin 19804- Petrie et al.19675- Petrova - Ovlav Apsidal motion catalogue6- Riazi 20037- Maloney Guinan 2004

3 - V1143 Cgyni Apsidal Motion Solution

V1143 Cgyni dataT= 7.641days; r (m) = 0.059; m =1.391 M (0);R (1) =1.346R (0); = 0.54And [v (m), v (M)] = [18, 28]; r (M) = 0.058; M=1.347 M (0)Distance [38 +/- 2 parsec] = 123.956 +/- 6.524 Ly

CalculationsWe have 1- = 0.46 1- /4=0.9721 R (0) = 0.696x109m

With a = [R (1)/r (m)] R (0) = 15.87823729x10^9m; 1+ = 1.54With v (p -perihelion) = [GM/ (m + M) a (1- )] = 110 km/secAnd v (p- aphelion) = [GM/ (m + M) a (1+)] = 60 km/secK (A) = (110 + 60)/2 = 170/2 = 85km/secWith v (s - perihelion) = [Gm/ (m + M) a (1-)] = 113.6 km/secAnd v (s - aphelion) = [Gm/ (m + M) a (1+)] = 62 km/secK (B) = (113.6 + 62)/2 = 175.6/2 = 87.8km/sec

With v (1) = [GM/ (m + M) a] = 74.632 km/secAnd v (2) = [Gm/ (m + M) a] = 77.0699 km/secWhen spinning on opposite directions

1- With v [21, 28] = 28 - 21 = 72- With v [18, 28] = 28 - 18 = 103- Taking average 10 + 7/2 = 8.5With v (m) = [GM/ (m + M) a (1-/4)] = 77.5126 km/sAnd v (M) = [Gm/ (m + M) a (1-/4)] = 80.00448 km/sAlso, [ (1-)]/ (1-) = 3.977622971

Page 40


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Now:With 1- v + v* = 157.51648km/sec - 10 km/sec = 147.51648km/secAnd 2- v + v* = 157.51648km/sec - 8.5 km/sec = 149.01648km/secAnd 3- v + v* = 157.51648km/sec - 7 km/sec = 150.51648km/sec

W (obo) = (-720x36526/T) x {[ (1-)]/ (1-) } {[v* + v]/c}

1- W/century= (-720x36526/7.641) (3.977622951)(147.51648/300,000)=3.31/century2- W/century= (-720x36526/7.641) (3.977622951)(149.01/300,000)=3.3778/century3- W/century= (-720x36526/7.641) (3.977622951)(150.51648/300,000) = 3.44614561/century

With v* = 2 v* (cm) = 157.4770 km/secAnd v = v (p) - v (s) = 21 - 28 = -7 km/secThen v* + v (p) = 157.477 - 7 = 150.477 km/sec

W /century= (-720x36526/7.641) (3.977622951)(150.477/300,000)=3.44/centuryObserved values are: W = 3.393987698/century; W = =3.489592985Average observed: 3.44/ century

References:1-Geminez and Margrave, 1985

[0.00071/cycle] = [1 century = 36526days/7.641days] = 3.393987698/century2- Anderson and Nordstrom and Garcia and Geminez 1987: 0.00073/cycle

[0.00073/cycle] = [1 century = 36526days/7.641days] =3.489592985/century

Relativity theory: 4.254435283/ century = 0.00089/cycle

4- V541Solution: Apsidal motion catalogue

W (cal) = (-720x36526/T) {[ (1-)]/ (1-) } [(v+ v*)/c] degrees/100 yearsT = 15.3379days r(m)=0.0440 m = 2.4M(0) R(m)=1.88R(0)[v(m),v(M)]=[242,242]And = 0.479 r (M) =0.0425 M=2.4 M (0) R (M) =1.79 R (0)With 1- = 0.521 1-/4=0.94263975; [ (1-)]/ (1-) = 3.2339


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Page 41

And a = [R (m)/r (m)] R (0) = (1.88/0.0440)0.696x10^9m =29.73818182x10^9mThen a (1-/4) = 28.03x10^9mAnd v (m) = [GM/ (m + M) a (1-/4)] = 75.5883km/sec; v (m) =24And v (M) = [Gm/a (m + M) (1-/4)] = 75.883km/sec; v (M) =24With v=24 + 24= 44km/secAnd v* = 151.1766km/secWith v* + v= 151.1766 - 48 = 103.1766km/secW (ob) = (-720x36526/T) {[ (1-)]/ (1-) ]} {[v* + v]/c} W (ob) = (- 720 x 365226/15.3379) (3.2339) (103.1766/300,000)W (ob) = 0.65Notice: [v (m), v (M)] = [24 2, 24 2]If v (m) = v (M) = 24 + 2 = 26Then v* + v= 151.1766 - 52 = 99.1766km/secAnd W (ob) = (- 720 x 365226/15.3379) (3.2339) (99.1766/300,000)W (ob) = 0.60/centuryObserved is W= 0.60 +/ -0.1/century Lacy = [0.5; 0.7]

Relativity: W = 0.97/century

1- Apsidal motion of V541Cgyni Lacy 19895 - AI Hydra apsidal motion puzzle solution

W (cal) = (-720x36526/T) {[ (1-)]/ (1-) } [(v+ v*)/c] degrees/100 yearsAI Hydra Apsidal motion solution:Data: T=8.29days r(m) = 0.1418 m=2.15M(0) R(m)=3.92R(0)[v(m),v(M)]=[28,27]; and = 0.0.23; r (M) = 0.1002 M=1.98 M (0) R(M) =2.77(0); m + M=4.13 M (0); L = 575 +/- 15 Ly

Calculations1- = 0.77; (1-/4) = 0.986775; [ (1-)] / (1-) = 1.6414G=6.673x10^-11; M (0) = 1.98892x19^30kg; R (0) = 0.696x10^9mThen a = [R (m)/ r (m)] = 19.24062059 x 10 ^ 9 m

V (m) = [GM/ (m + M) a (1-/4)] = 81.11439578 km/secV (M) = [Gm/ (m + M) a (1-/4)] = 88.11km/sec


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Apsidal motion is given by this formula:

W (ob) = (-720x36526/T) {[ (1-)]/ (1-) ]} [(v + v*)/c] degrees/100 years

Page 42

With v* = v (m) + v (M) = 81.11439578km/sec + 88.11km/sec =169.2243958km/secAnd v = v (m) + v (M) = 28 + 27 = 55 km/sec

Then (v* + v) = 224.2243958 km/secW (ob) = (-720x36526/T) {[ (1-)]/ (1-) } sine [Inverse tan224.2243958/300,000]

= (-720x36526/8.29) (1.6414) (224.2243958/300,000)= 2.90/ century

W (observed) = 2.90/century as measuredObserved is 2.90/century

Space-Time Relativity theory 6.8 / centuryNow let us calculate v* (cm) = m v/m

= 2.15 x 81.11439578 + 1.98 x88.11=

84.46822 km/secWith v* = 2 v*(cm) = 168.936411 km/secAnd v = 55 km/secThen v* + v = 223.9364411 km/sec andW (observed) = (-720x36526/T) {[ (1-)]/ (1-) } sine [Inverse tan223. 936/300,000]= (-720x36526/10.55) (3.33181) (223.9364411/300,000)W (observed) = 2.90/centuryWith = { [v* - v* (cm)] /2}

= {[88.11 -84.46822]/2 + [81.11439578 - 84.46822]/2}

= {[3.64178]/2 + [3.35382422]/2} = 3.5 km/secThen W (ob) = 2.90/century; observed is W = 2.9/per centuryAnd Einstein's 100,000 space-timers 6.8 / century

References: Go to Smithsonian/NASA website SAO/NASA and type:1- Apsidal motion of AI Hydra Popper 19852- KH. F. Khaliulullin and V.S Kozyreva 19883- Petrova - Ovlav Apsidal motion catalogue 1999


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6 - V731Cehpei Apsidal Motion Solution

Next the same equation will be used to find the advance of Periastron or"apparent" apsidal motion of V731 binary stars system.

V731Cehpei Apsidal Motion SolutionW (ob) = (-720x36526/T) {[ (1-)]/ (1-) } [(v+ v*)/c] degrees/100 years

V731data [see below]T= 6.068567days; m= 2.577 M (0); M = 2.577 M (0); [v (m), v (M)] =[19+/-3, 18+/-3]; = 0.0165; a = 23.27x R (0)Calculations: M + m = 2.738; 1-=0.9835 1-/4=0.9999 R (0) = .696x10^9mWith v [21, 28] = [19 +/- 3] + [18 +/- 3] = 37 +/- 6With v* (p) = [GM/ (m + M) a (1-/4)] = 85.6111965km/secAnd v* (M) = [Gm/ (m + M) a (1-/4)] = 109.38km/secAlso, [ (1-)]/ (1-) = 1.033694356With v* (cm) = 2m v/m = 96.46688km/sec; 2 v* (cm) = 192.9337619And = { [v*-v* (cm)] /2} = {[96.46688 - 85.6] /2} +

{[109.38 - 96.46688] /2} = 11.9288422 km/secWith v* (p) = 85.6111965km/sec +/- 11.9288422 km/secAnd v* (s) = 109.38km/sec +/- 11.9288422 km/secThen v* (p) + v* (s) = [192.9337619 +/-] x 2 = 23.8567844 km/secThen v* + v = 229.9288422km/sec +/- 29.8567844 km/secNow: Taking the upper limitThen v* + v = 229.9288422km/sec + 29.8567844 km/sec =259.7856266 km/sec

Primary Secondary



[,][,] [,][,] [,][,]

Spin results v(p) + v(s) v(p) + v(s) - v(p) + v(s) -v(p) + v(s)Orbit results v*(p) + v*(s) -v*(p) + v*(s) v* (p) + v*(s) -v* (p) + v* (s)Examples V731Cepheiv (s) v* (s) [,][,] [,][,] [,][,] [,][,]Spin results v(p) + v(s) v(p) + v(s) -v(p) + v(s) -v(p) + v(s)Orbit results v*(p) - v*(s) -v*(p) - v*(s) v*(p) - v*(s) -v* (p) - v* (s)Examplesv (p) v*(s) [,][,] [,][,] [,][,] [,][,]Spin results v(p) - v(s) v(p) - v(s) -v(p) - v(s) -v(p) - v(s)

Orbit results v*(p) + v*(s) -v*(p) + v*(s) v*(p) + v*(s) -v* (p) + v* (s)Examplesv (s) V*(s) [,][,] [,][,] [,][,] [,][,]Spin results v(p) - v(s) v(p) - v(s) -v(p) - v(s) -v(p) - v(s)Orbit results v*(p) - v*(s) -v*(p) - v*(s) v*(p) - v*(s) -v* (p) - v* (s)

Examples V731Cephei


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Page 44W (obo) = (-720x36526/T) x {[ (1-)]/ (1-) } {[v* + v]/c} W/century= (-720x36526/6.068567) (1.033694356)(259.7856266 /300,000)=2.91/centuryW/century = 3.35914177/century = 0.0335914177/year

U = 360/0.0335914177 = 10717 yearsObserved values are U = 10000 +/- 2500

References: 1- Absolute dimensional and apsidal motion of V731CepV. Batkis; M.Zejda; I. Bulut; M.Wolf; S. Bilir; H. Bakis; O.Demircan:J.w.Lee: M.Slechta: B. Kucerova. 2008

8 - NV CMa apsidal motion solution:

Data: T=1.885159 days; = 0; v* (p) = 128.55 km/sec; v* (s) = 130.87km/sec[ (1-)] / (1-) = 3.33181; v (p) = 51.7 km/sec and v (s) = 52.4km/secApsidal motion is given by this formula:W (ob) = (-720x36526/T) {[ (1-)]/ (1-) ]} [(v + v*)/c] degrees/100 yearsWith v* = v* (p) + v*(s) = 259.42 km/sec and v= v (p) + v (s) = 104.1And v* + v = 363.52 km/sec

W (observed) = (-720x36526/T) {[ (1-)]/ (1-) }sine [Inverse tan 363.52/300,000]

= (-720x36526/1.885159) (1) (363.52/300,000)= 20.48333818/century = 0.2048333818/year

U = 360/0.2048333818/year;

U = 1757. 5 years

References: Go to Smithsonian/NASA website SAO/NASA and type:Absolute dimensions NV CMa; Kaluzny, J; Pych, W; Rucinski, S. M;Thompson, I.B

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Relativity theory coffin nail # 8 SW Canis Majoris

W (cal) = (-720x36526/T) {[ (1-)]/ (1-) ]} [(v + v*)/c] degrees/100 years

T = period = 10.09 days; = eccentricity = 0.3179

And v = spin velocity effect = v (p) + v(s) = 57km/sec

And v*= orbital velocity effect = v*(p) + v* (s)= 80.5 + 87.8 = 168.3 km/sec

For SW Canis Majoris: v* + v = 225.3km/sec

W (observed) = 2.99565967/century = 0.0299565967

U = 360/ 0.0299565967= 12017years

U (observed) = 12,000 yearsEinstein and space-timers 14,000 yearsSW Canis Majoris Binary stars

Primary

Secondary



[,][,] [,][,] [,][,]

Spin results v(p) + v(s) v(p) + v(s) - v(p) + v(s) -v(p) + v(s)Orbit results v*(p) + v*(s) -v*(p) + v*(s) v* (p) + v*(s) -v* (p) + v* (s)Examples Sw Canis Majorisv (s) v* (s) [,][,] [,][,] [,][,] [,][,]Spin results v(p) + v(s) v(p) + v(s) -v(p) + v(s) -v(p) + v(s)Orbit results v*(p) - v*(s) -v*(p) - v*(s) v*(p) - v*(s) -v* (p) - v* (s)Examplesv (p) v*(s) [,][,] [,][,] [,][,] [,][,]Spin results v(p) - v(s) v(p) - v(s) -v(p) - v(s) -v(p) - v(s)Orbit results v*(p) + v*(s) -v*(p) + v*(s) v*(p) + v*(s) -v* (p) + v* (s)Examplesv (s) V*(s) [,][,] [,][,] [,][,] [,][,]Spin results v(p) - v(s) v(p) - v(s) -v(p) - v(s) -v(p) - v(s)Orbit results v*(p) - v*(s) -v*(p) - v*(s) v*(p) - v*(s) -v* (p) - v* (s)Examples SW Canis majoris


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Page 46

SW CMa apsidal motion solution:Data: T=10.09 days; r (m) = 0.0942; m = 2.22 M (0); R (m) = 3.01R (0); =0.3179

And r (M)= N/A M = 2.03 M (0) R (M) =2.46 R (0); m + M = 4.25 M (0)And[v (m), v (M)] = [30+/-2, 27+/-3]K (1) = 80.5; K (2) = 87.8Calculations

1- = 0.6821; (1-/4) = 0.974734898[ (1-)] / (1-) = 2.037835646

G=6.673x10^-11; M (0) = 1.98892x19^30kg; R (0) = 0.696x10^9mThen a = [R (m)/ r (m)] = 22.23949045 x 10 ^ 9 m

V* (p) = [GM/ (m + M) a (1-/4)] = 77.26298km/sec

V* (s) = [Gm/ (m + M) a (1-/4)] = 84.4944913km/sec

And v* = v* (p) + v* (s) = 161.7574713 km/secWith v = v (p) + v (s) = 30 + 27 = 57 km/sec

And, v* + v = 218.7574713km/sec

Apsidal motion is given by this formula:


W (ob) = (-720x36526/T) {[ (1-)]/ (1-) } sine [Inverse tan218.7574713/300,000]

= (-720x36526/10.09) (2.037835646) (218.7574713/300,000)= 2.8242/ century = 0.026242/yr

U = 360/ 0.026242 = 12747 yearsTaking: v* + v = 80.5 + 87.8 +57 = 225.3 km/secW (ob) = (-720x36526/T) {[ (1-)]/ (1-) }

Sine [Inverse tan 225.3 /300,000]= (-720x36526/10.09) (2.037835646) (218.7574713/300,000)= 2.995659677/ century = 0.0299565967/yr

U = 360/ 0.0299565967 = 12017 years NahhasU (observed) = 12,000 yearsReferences: Go to Smithsonian/NASA website SAO/NASA and type:1- Lacy Apsidal motion Canis Majoris 1997


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47

9 - GG Orion Apsidal motion puzzle solution

Data: T=6.6314948; m = 2.342 M (0); M = 0.2338 M (0); R (1) =1.852 R (0); R (2) =1.830 = 0.2218; 1 - = 0.7782; r (1) = 0.0746; r (2) =.988 r (1); m + M = 4.68 M (0)And [v (p); v (s)] = [16 +/- 1; 16 +/- 1]; [v (p); v (s)] = [25 +/- 3; 24 +/-

U = 10700 +/- 4500 yearsCalculations(1-/4) = 0.9877; [ (1-)] / (1-) = 1.57G=6.673x10^-11; M (0) = 1.98892x19^30kg; R (0) = 0.696x10^9mCalculationsWith v* (p) = [GM/ (m + M) a (1-/4)] = 95.6 km/secAnd v* (s) = [Gm/ (m + M) a (1-/4)] = 95.735 km/secAnd v (p) = 16 km/sec; v (s) = 16 km/secThen v* (p) + v* (s) + v (p) + v (s) = 223.335 km/secW (ob) = (-720x36526/6.6314948) (1.57) [223.335/300,000] degrees/100 years

W (ob) = 3.45/century = 0.0345/yearU [years] = 360/[0.0345/year]U = 10,432 years NahhasU (observed) = 10700+/-4500years

References: Absolute dimensions and apsidal motion of eclipsing binary GG OrionDr Lacy; Dr Torres; Dr Claret; Dr Sabby: 2000

Primary

Secondary


v(s) v* (s) Spin=[,][,]=orbit [,][,] [,][,] [,][,]Spin results v(p) + v(s) v(p) + v(s) - v(p) + v(s) -v(p) + v(s)Orbit results v*(p) + v*(s) -v*(p) + v*(s) v* (p) + v*(s) -v* (p) + v* (s)Examples GG Orionv (s) v* (s) [,][,] [,][,] [,][,] [,][,]Spin results v(p) + v(s) v(p) + v(s) -v(p) + v(s) -v(p) + v(s)Orbit results v*(p) - v*(s) -v*(p) - v*(s) v*(p) - v*(s) -v* (p) - v* (s)Examples

v (p) v*(s) [,][,] [,][,] [,][,] [,][,]

Spin results v(p) - v(s) v(p) - v(s) -v(p) - v(s) -v(p) - v(s)Orbit results v*(p) + v*(s) -v*(p) + v*(s) v*(p) + v*(s) -v* (p) + v* (s)Examples

v (s) V*(s) [,][,] [,][,] [,][,] [,][,]Spin results v(p) - v(s) v(p) - v(s) -v(p) - v(s) -v(p) - v(s)Orbit results v*(p) - v*(s) -v*(p) - v*(s) v*(p) - v*(s) -v* (p) - v* (s)Examples GG Orion


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The time has come to send relativity theories and all four-dimensionalspace-time confusion of physics to the...

Page 48

Chapter four

Arabs real time physics in outer Space IIA- Arabs Binary stars and Binary Pulsar apsidal motion

B- Arabs GPS or Global positioning systemsC- Arabs Interplanetary Telecommunications systems

2009 total collapse of general relativity theory; CD Draconis Binary stars

10 2009 CD Draconis apsidal motion Table:Data: T=1.268389985days; m = 0.231 M (0); M = 0.2141 M (0); a =3.7634R (0); = 0.0051And [v (p); v (s)] = [9.5 +/- 1; 10.0 +/- 1]Einstein and space timers: W = 1.91x10-3/day

Calculationsm + M = 0.4451 M (0)1- = 0.9949; (1-/4) = 0.99993498; [ (1-)] / (1-) = 1.01G=6.673x10-11; M (0) = 1.98892x1030kg; R (0) = 0.696x10 9mCalculationsWith v* (p) = [GM/ (m + M) a (1-/4)] = 72.436 km/secAnd v* (s) = [Gm/ (m + M) a (1-/4)] = 78.153 km/sec

Primary Secondary



[,][,] [,][,] [,][,]

Spin results v(p) + v(s) v(p) + v(s) - v(p) + v(s) -v(p) + v(s)Orbit results v*(p) + v*(s) -v*(p) + v*(s) v* (p) + v*(s) -v* (p) + v* (s)Examples CD Draconisv (s) v* (s) [,][,] [,][,] [,][,] [,][,]Spin results v(p) + v(s) v(p) + v(s) -v(p) + v(s) -v(p) + v(s)Orbit results v*(p) - v*(s) -v*(p) - v*(s) v*(p) - v*(s) -v* (p) - v* (s)Examplesv (p) v*(s) [,][,] [,][,] [,][,] [,][,]Spin results v(p) - v(s) v(p) - v(s) -v(p) - v(s) -v(p) - v(s)Orbit results v*(p) + v*(s) -v*(p) + v*(s) v*(p) + v*(s) -v* (p) + v* (s)Examples

v (s) V*(s) [,][,] [,][,] [,][,] [,][,]Spin results v(p) - v(s) v(p) - v(s) -v(p) - v(s) -v(p) - v(s)Orbit results v*(p) - v*(s) -v*(p) - v*(s) v*(p) - v*(s) -v* (p) - v* (s)Examples CD Draconis


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And v (p) = 9.5 km/sec; v (s) = 10 km/secThen v* (p) + v* (s) + v (p) + v (s) = 170.117 km/sec

Page 49Apsidal motion is given by this formula:


W (ob) = (-720x36526/1.238389985) (1.01) [170.09/300,000] degrees/100 yearsW (ob) = 6.731598944/century = 0.06731598944/year

U [years] = 360/[0.06731598944/year]

U = 5348 years Nahhas

U (observed) = 5400+/-3200yearsEinstein's and space-timers U = 360/ [0.00191x365.26] = 516 yearsCan it get any better?It is not just about dumping relativity it is dumping relativity and AlfredNobel institution with it.

References: Absolute properties of the low-mass eclipsing binary CMDraconis; 2009By : Juan Carlos Morales; Ignasi ribas; carme jordi; Guillermo Toress; Jose

Gallardo; Edward F. Guinan; David Chardonneau; Marek wolf; Davidw.latham; Guillem Angalada Escude; David H.Bradstreet; Mark E.Everett;Francis T. O, Donavan; Georgi Mandushev; Robert D. Mathieu and other 15

11 - V 1147 Sagittari

V 1147 Sagittari: Data: U = 592.5 +/- 6 yearsWith K (p) = 117.4 km/sec; K(s) = 130.8 km/secGiving v* (p) = 117.4 + 29.8 - 0.465 = 146.735 km/secAnd v* (s) = 130.8 + 29.8 0.465 =160.135 km/sec

Orbital speed of earth = 29.8 km/sec; spin speed of earth is 0.465 km/secAnd v (p) = 80 + 29.88 0.465 km/sec = 109.335 km/secAnd v(s) =75 + 29.88 0.465 km/sec = 104.335 km/secThen v* + v = 146.735 + 160.135 + 109.335 + 104.335 = 520.54km/sec[ (1- )] (1-) = 2.62W (cal) = (-720x36526/3.28) x (2.62) (520.541/300,000)= 63/centuryU = [63/100]-1x 360 = 570 years.Observed values; 592.5


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ReferencesAbsolute dimensions of solar-type eclipsing binaries V 1147SagittariJ. Anderson, A. Gimenez 1985

Page 50

12- EW OrionisEW Orionis: Data: U =16300 +/- 3900With K (p) = 73.18km/sec; K(s) = 75.46 km/secGiving v* (p) = 73.18 + 29.8 - 0.465 = 102.515 km/secAnd v* (s) = 75.46 + 29.8 0.465 =104.795 km/secAnd v (p) = 9 + 29.8 - 0.465 = 102.515 km/secAnd v (s) = 73.18 - 29.8 + 0.465 = 102.515 km/sec

Orbital speed of earth = 29.8 km/sec; spin speed of earth is 0.465 km/secAnd v (p) = 9km/sec; v(s) =8.8km/secThen v* + v = 102.515 + 104.795 +9.0 + 8.8 =225.11km/sec

[ (1- )] (1-) = 1.167

W (cal) = (-720x36526/6.93) x (1.167) (225.11/300,000)= 2.1367/centuryU = [2.1367/100]-1x 360 = 16848 years.

13 - The Apsidal motion of VV PYX

Data: T = 4.596 days; m = 2.098 M (0); M = 2.098 M (0)With r (p) = 0.1156 and r (s) = 0.1156 and R (m) = 2.167 R (0) and R (M)= 2.167 R (0)

And = 0.0956; 1 - = .9044; U = 3200 +/- 1000And [v (p); v (s)] = [23; 23]V = 381.03095km/sec

Calculations

With m + M = 4.196 M (0)And (1-/4) = 0.99771516; [ (1-)] / (1-) = 1.216984877G=6.673x10^-11; M (0) = 1.98892x19^30kg; R (0) = 0.696x10^9mAnd a = [R (m)/r (m)] = [2.06/ 0.097] R (0) = [2.167/ 0.1156][0.696x10^9m]Then a = 13.04698962 x10^9mAnd a (1-/4) = [13.04698962 x10^9m] [0.99771516] = 13.01717934 x109mK (A) = { [GM/ (m + M) a (1- )] + GM/ (m + M) a (1+ )]}/2K (A) = { [6.673x2.098x2x10-19/ (4.196) 13.04698962x10^9x0.9044]

+ [6.673x2.098x2x10-19/ (4.196)13.04698962x10^9x1.0956]}/2

= [108.85 + 98.96511511]/2 = 103.9km/sec


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K (B) = { [Gm/ (m + M) a (1- )] + Gm/ (m + M) a (1+ )]}/2= 103.9km/sec

Page 51With v* (m) = [GM/ (m + M) a (1-/4)] = 103.70 km/sec

And v* (M) = [Gm/ (m + M) a (1-/4)] = 103.70 km/secAnd v* (cm) = [m v* (p) + M v* (s)]/ (m + M) = 103.70 km/sec

And v (p) = 23vkm/sec; v (s) = 23 km/sec

W (ob) = (-720x36526/T) {[ (1-)]/ (1-) ]} (v* + v/c) degrees/100 years

T = period; = eccentricity; v = spin velocity effect; v*= orbital velocityeffects

With v* (m) = [GM/ (m + M) a (1-/4)] = 103.70 km/secPrimary orbital speed contribution to apsidal motion

Is v* (m) = 103.7 + 29.8 0.465 = 133 km/secAnd v* (M) = [Gm/ (m + M) a (1-/4)] = 103.70 km/sec

Secondary orbital speed contribution to apsidal motionIs v* (M) = 103.7 + 29.8 0.465 = 133 km/secAnd v (m) = 23 km/sec

Primary spin speed contribution to apsidal motionIs v (m) = 23 + 29.8 0.465 = 52.335 km/secAnd v (M) = 23 km/sec

Secondary spin speed contribution to apsidal motionIs v (m) = 23 + 29.8 0.465 = 52.335 km/secThen v * + v = 370.67

With = 0.17; T = 4.596 days[ (1-)] / (1-) = 1.216984877

W (cal) = (-720x36526/4.596) (1.216984877) [370.67/300,000] U = 3386 years Nahhas'U (observed) = 3200 +/- 1000years

ReferencesAbsolute dimensions of solar-type eclipsing binaries EW OrionisJ.V. Clausen, H. Bruntt, E.H. Olsen, B.E. Helt, and A. Claret 2009


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Page 52

14 - 1993 Nobel Prize Winner Dr Joseph H. Taylor

Binary Pulsar 1913 + 16For PSR 1913 + 16Data: Joseph Taylor and Joel M. Weisberg 2004With a =2.3417725 R (0); R (0) = 0.696 x 10 9metersAnd = 0.6171338T = 0.322997448930

And ' = 4.226595 /yearWith mass m = 1.4414 M (0) and M = 1.3867 M (0); M (0) = 2 x 1030 kgV* (p) = [GM/ (m + M) a] = 235.9574664 km/sec

V* (s) = [Gm/ (m + M) a] = 245.2640841km/secW (cal) = (-720x365.26/Tdays) xsine [Inverse tan (v/c)] degrees/100 years

= (-720x365.26/0.329974489) x sine [Inverse tan (481.22/300, 00)]deg/100 years

W = 4.2/year

15- PSR J0737 3039

Data: U =21.3 years

With m = 1.34 M (0); M = 1.25 M (0); M (0) = 2 x 1030And = 0.0878; [v (p), v (s)] = [22.7 km/sec, 2km/sec]With G = 6.673 x 10 11; T =2.4 hours; a = 1.145 x 0.069 x 109

With K (p) = v* (p) = [GM/ (m + M) a (1-/4)] = 303 km/secAnd K(s) = v* (m) = [Gm/ (m + M) a (1-/4)] = 325 km/secGiving v* (p) = 303 + 29.8 - 0.465 = 332.335 km/secAnd v* (s) = 325+ 29.8 0.465 =354.335 km/secAnd v (p) = 22.7 + 29.8 - 0.465 km/secAnd v (s) = 2 - 29.8 + 0.465 km/secOrbital speed of earth = 29.8 km/sec; spin speed of earth is 0.465 km/sec

Then v* + v = 332.335 + 354.335 +22.7 + 2 =711.37km/sec[ (1- )] (1-) = 1.197

W (cal) = (-720x36526/6.93) x (1.197) (711/300,000)= 1770/centuryU = [1770/100]-1x 360 = 20.3389 years.ReferencesThe Double pulsar PSR J0737 - 3039


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Burgay, M; DAmico,N.; A.Manchester,R.N; Lyne, A.G.; Kramer, M; Mclaughlin, M.A;Lorimer, D.R; Carmilo, F.; Stairs, I.H: Freire, P.C.C; Joshi,

Page 53B- Arabs real time Global Positioning System: GPS

Abstract: The Global Positioning System or GPS 45 micro seconds per daytime delays have nothing to do with Einstein's relativity theory time travelsconfusions of physics and they are a consequence of Satellite orbital speedand Earth rotational speed given by this formula below. Even if Einstein'sformulas were correct for all practical purposes they are insignificant to theperformance to the GPS system. Earth - Satellite distance is a variable thatEngineers account for and the tiny "relativistic" effect has no significancewhatsoever because distance adjustment is far more than any relativisticadjustment. For the president of the United States President to ask forInnovations like that in an AAAS meeting is a sign of a bankrupt scientificcommunity and for those who knew the facts must have laughed to theirteeth at the president.

NASA's and AAAS laughing at the president requires a reply with laugh atAAAS and NASA and all other including the person who wrote the speechfor the president

GPS time delays of 45 micro seconds per day have nothing to do withrelativity theory or Einstein or AAAS or NASA or silly time travel physics orany theory. This time delay is due to Earth rotation and satellite orbital

speed that Scientists turn their heads away from because without Einstein'ssilly magic sock of time travel physics, Physicists will have less tricks thatallows them to say anything publish anything based on nothing for jobsmoney prestige Nobel Prizes and silly physics.

W" (ob) = (-720x3600x15) [(v +/- v*)/c] arc sec /day

T = period; = eccentricity; v = spin velocity of earth; v*= orbitalvelocity of satelliteAnd v* = 14000km/hr = 3.88888888889 km/s; = 0; T = 0.5 days and v= 0.465km/s

U = W" x (24/360) = 45.016microsecond per dayNahhas'Relativity theory silly professor of time travel accounted for 38 Microseconds and blamed the other 7 Micro seconds on weather


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1 ArabsGlobal Positioning Systems in arc second per century:W" (calculated) = [-720x36526x3600/T] {[ (1 - )]/ (1 - ) } (v/c) arc sec/ century

Page 542 ArabsGlobal Positioning Systems in arc degree /century

W" (calculated) = [-720x36528/T] {[ (1 - )]/ (1 - ) } (v/c) arc sec/ century

3 Arabs Global Positioning Systems in arc second per year:W" (calculated) = [-720x365.26x3600/T] {[ (1 - )]/ (1 - ) } (v/c) arc sec/ century

4- Arabs Global Positioning Systems in arc degree per year:W" (calculated) = [-720x365.26/T] {[ (1 - )]/ (1 - ) } (v/c) arc sec/ century

5 - Arabs Global Positioning Systems delays in arc degree per cycle:W" (calculated) = [-720/T] {[ (1 - )]/ (1 - ) } (v/c) arc sec/ century

6 Arabs Global Positioning Systems delays in arc second per cycle:W" (calculated) = [-720x3600/T] {[ (1 - )]/ (1 - ) } (v/c) arc sec/ century

7- Arabs Global Positioning Systems delays in arc second per day:W" (calculated) = [-720x3600] {[ (1 - )]/ (1 - ) } (v/c) arc sec/ century

8- Arabs Global Positioning Systems delays in second per cycle:W" (calculated) = [24/360] [-720x3600/T] {[ (1 - )]/ (1 - ) } (v/c) arc sec/ century

9 - Arabs Global Positioning Systems delays in second per day:U (seconds/day) = [-720x 3600/15] {[ (1 - )]/ (1 - ) } (v/c) sec/day

10 - Arabs Global Positioning System Circular Satellite Orbits time delays in seconds/dayU (seconds/day) = [-720x 3600/15] (v/c) sec/day

Application of Arabs Global positioning system

GPS Data: T = 0.5 days satellite orbital Period; = 0And v = 0.465km/sec Earth spin speed;And v* = 14,000 km/hr = 35/9 km/secondThen v* +/- v = 35/9 = 3.88888889km/sec - 0.465km/secondWe subtracted because satellite and motion and spin orientations are

oppositeGPS time delays are given by this formula per day in seconds of an arcW" (ob) = (-720x3600/T) {[ (1-)]/ (1-) ]} [(v +/- v*)/c] seconds/dayW" (ob) = (-720x36/0.5) (1) [3.423888889/300,000] seconds of arc /1day


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W" (ob) = 0.000675246 arc seconds per day to get this answer in secondsdivide by 24hours/360degress = 15U [seconds] = 0.000675246 x [24/360] seconds/dayU = 0.000045016 seconds/day = 45 micro seconds /day

Page 55

C - Interplanetary telecommunications around the sun

Harvard Irwin Shapiro historical mistake

The stupidity of Harvard Physics department

Universal Constant 0 =16GM/C= 247.597s

Abstract: Interplanetary time delays around the moving sun derived fromthree dimensional time-dependent Newton - Kepler's equations solutiongives a solar round trip time delay rate of:= 16GM/c [1 + (v/v)] = 0 [1 + (v/v)] 0 = 16GM/c= 247.597sG = Gravitational constant; M=Sun mass; a=mean distance from Sun. Andeccentricity; c = light speed; a = mean distance


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And v = Planet speed; v= Sum/Difference in spin between Earth andplanets. When applied to actual data it gives extremely accurate resultsbetter than Shapiro's Space-time-delay analysis and without space-timefictional forces or space-time fiction.

Page 56W (ob) = (-720x36526/Tdays) {[ (1-)]/ (1-) } x [(v + v*)/c] degrees/100 years

The circumference of an ellipse: 2a (1 - /4 + 3/16()- --.) 2a (1-/4); R =a (1-/4)

Where v (m) = [GM/ (m + M) a (1-/4)]And v (M) = [Gm / (m + M) a (1-/4)]

Then W (ob) = -4 [ (1-)]/T (1-) sine [ (m) t + (r) t] = T W (ob) = - 4 {[ (1-) ]/ (1-) } (v + v*/c) } radians; andwith = 0

= - 4 (v + v*/c) Sun-Photon; and with v = 0 = -4 (v*/c) Sun-Photon: 0 = [Sun - Photon] [Earth - Mars] = 0.2075The circumference of an ellipse: 2a (1 - /4 + 3/16()---) 2 a (1-/4); R =a (1-/4)v= [Gm M/ (m + M) a (1-/4)] [GM/a (1-/4)]; m


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Page 57

These data compared to Shapiro's time delay from NASA 1977 Vikings 6, 7Earth - Mars Telecommunications mission are more accurate because the

actual value is 250s and the value published by Doctor Irwin Shapiro ofHarvard is 247.597sAlthough this formula works the correct formula is = -4 [(v + v*)/c] Sun-Photon; and with v 0v = [GM/a] = 24.1 km/sec; v = 0.46511 0.241 = 0.224 km/sec = 2 arc length/c = 2[ ] d/c= 16GM/c [1 + (v/v)] = 247.597 x [1 + 0.224/24.1)] = 250 s For Mars; 0.4651 = Earth rotation; 0.241 = mars rotation; v= mars speed

Planet Distant Planet-Earth

Planet+Earth

Eccentricity 1-/4 0 s

Mercury 57,910 91,690 207,510 0.441858224

0.951190328

247.5974607

260.3

Venus 108,200 41,400 257,800 0.160589604

0.993552745

247.5794607

249.2

Earth 149,600 0 299,200 0 1 247.5794607

247.597

Mars 227,940 78,340 377,540 0.207501192

0.989235814

247.5794607

250.273

Jupiter 778,330 628,730 927,930 0.677561885

0.885227473

247.5794607

279.6789


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Harvard is where physics changed from science to stupidityPage 58

Physics in a capitalist country is a business and physics is notnecessarily science or scientific and if it is a western invention it

is to be at least 88.88 % fraud

After World War II and the emergence of USA as a global superpower andthe shift of power from traditional old empires to the new worldsuperpower American Universities rushed for gold and gold is reputablephysics department that can say anything publish anything based onnothing and nothing is Robert Pound and Glen Rebka confusion of anexperiment

Harvard strikes again with fraud Robert Pound and Glen Rebka (top)


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22.5 meters apart

To start with gravity is a spin effect of a planet with atmosphereThe gravitational and electric and magnetic constants all are derived fromtwo constants p and T which are air density and Earth spin

Page 59

If we took a look at the most famous physics fundamental constant G from

Newton's 350 years old Equation F = - GmM/r, or, Coulomb electric forceequation F = - Qq/4 0, one can see that, for example, Gs value has unitdimension of [G] = [1/ T]; is density and T is time rotational period. If wecan show that the gravitational constant G = G (, T) is a variable air densityand Earth rotational period dependent then we can say what Earth is sayingthat the most famous physics fundamental constants are not so universalconstants because all three fundamental constant are air density and Earth spindependent.


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arabs real time physics

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