Einstein's Relativity Stupidity on U Tube

8/14/2019 Einstein's Relativity Stupidity on U Tube

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Nahhass 21original solutions of Mercurys apparent advance of perihelion

Arabs real time physicsI/O = 1 + (I O)/OBy the greatest Physicist of all time Joe Nahhas July 4th 1973

[email protected]

I am the greatest physicist of all time and nothing anyone can do about it

Page 1

Chapter One
mailto:[email protected]:[email protected]


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Arabs real time physics: We can not see or measure something that hadnot happened. We can only see and measure something that had happened.What we see and measure in not what happened. We measure in present timean event that happened in past time. That is we measure past events inpresent timePresent time = present timePresent time = past time + [present time past time]

Present time = past time + time differenceReal time = event time + time delayReal time physics = Event time physics + time delays physics

Time 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 scale

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

Real time distance scaleMeasured 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)/r0

Page 2Real time Velocity Scale v


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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 r is the radius.In 1969 I knew about circular speed v0 = 2 r0/T where T is the period ofrotation. Also, in 1969 I knew about angular speed 0 = 2 /T = (v0/ r0).

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

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

.Page 3


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If 1= 12 = 23 = 3

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

Actual = actual

- Actual = - actual----------------------- AddVisual = Actual + (Visual actual)

Divide by actual

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

Or

Visual/actual = 1 + (Visual actual)/actual

Image/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 planetMercury from 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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The Visual Illusions and modern Physicists confusion of Planet Mercuryis:[(I O)/O] 0 = [(r m r e)/ r e] 0

Planet

r

Sun


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

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

Johannes Kepler Tyco Brahe

Kepler took Tyco Brahes observations and came out with 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 v e)/ v e](v m/r m) [(180/) (3600) (26526/T m)]

[(I O)/O] 0 = [(Tm/ Te) 2/3 1] (v m/r m) [(180/) (3600) (26526/T m)]

Then Urbain Jean Joseph Le Verrier came

The 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

Page 5If it is measured for planet Mercury from the sun thenThen it is m' = v m /r m


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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)]

Page 6[(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 v e)/ v e](v m/r m) [(180/) (3600) (26526/T m)]

SM

E


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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/r

Physicists calculated: ' c m = v c m/r

And Astronomers observed: ' s = v s/r

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 = (v s/v c m) 1

And ('cm

- 's

)/ 's

= (Tc m

/Ts

) 1

And 2 [(' cm - ' s) /' s] 2 (m/2 M) = m/MAnd 2 [(' cm - ' s) /' s] 2 (r M/ 2 r m) = r M/ r mAnd 2 [(' cm - ' s) / ' s]= 2 [(v s/v c m) 1]And 2 (' cm - ' s)/ ' s]= 2 [(T c m/T s) 1]

Multiplying by [(180/) (3600) (26526/T m)][(I O)/O] 0 = ( m/M) [(180/) (3600) (26526/T m)][(I O)/O] 0 = ( r c M/ r c m) [(180/) (3600) (26526/T m)]

[(I O)/O] 0 = 2 [(v s/v c m) 1] [(180/) (3600) (26526/T m)][(I O)/O] 0 = 2 [(T c m/T s) 1] [(180/) (3600) (26526/T m)]

Page 7Nicoklaus CopernicusDistance Illusion


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[(I O)/O] 0 = [(r m r e)/ r e](v m/r m) [(180/) (3600) (26526/Tm)]Or GalileosVelocity Illusion[(I O)/O] 0 = [(v m v e)/ v e](v m/r m) [(180/) (3600) (26526/Tm)]

Or Tyco Brahes / 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)]

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

Fig. 1. Newtons gravitational law

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

Newton is a Mathematical idiot

Newton's Gravitational Equation is: F = -GmM/r

With 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

Planet

r

Sun


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Page 8Proof:

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

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


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Page 9If m is constant then d (mr)/d t = 0; and d (r)/d t = 0Or, r = h = 2 a b/T; a = mean distance from sun and is calledsemi major axis and b is the semi minor axis.

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

r

Sun

Mercury

M

m


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Page 10Tyco Brahe Logged observational Data of Planets P motion around the Sun Sand then Kepler stated the areal velocity law: If Planet p observed from the sunthen the trajectory of planet p will cut equal areas in equal times.

Page 11

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]}DegreesAnd multiplication by 1 century = 36526 days and using T in days


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Page 12W (ob) = (-720x36526/Tdays) {[ (1-)]/ (1-) } x

sine {Inverse tan [(v* + v)/c]} degrees/100 years

ApproximationsWith v


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The circumference of an ellipse: 2a (1 - /4 + 3/16()- --.) 2a (1-/4); R =a (1-/4)Page 13

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

m


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Double throwing one stone

Inverse Cube equations F = m = - k/r r (1), then in polarcoordinatesWith m [d r/dt - 'r] = - k /r Inverse Cube Gravitational law (1)And d (r')/d t = 0 Kepler's Areal Velocity Equation

(2)

Page 14These 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) rAnd d r/ d - r [1 - (k/mh)] = 0

And 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/h)]} + t]

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

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


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If k = Gm M ; h = 2 a b/TThen ' = - 2 ' (0, 0)] sine [{ [1 - (GMT/4ab)]} + t]

Taking Kepler's: T/4a = 1/GMAnd GM = 4a/ T; GM T = 4a

And (GMT/4ab) = a /bThen ' = - 2 ' (0, 0) sine [{ (1 - a /b)} + t]

Page 15If = mb/aM

Then ' = - 2 ' (0, 0)] sine [{ [1 - (m/ M)]} + t]If = 0Then ' = - 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 t

With T = arc tan v/c


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Then = m/ M radiansAnd = m/ (M) [180/] [36526/T] [3600] arc second/100 yearsOr = (m/ M) (180) [36526/T] [3600] arc second/100 years

= 43.0344"/100 yearsOr ' = - 720 [36526/T] (3600) (1 - )/ (1 - ) ] (v/c) arc

second/100 years= 43.0"/100 years

Page 16

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 ArealVelocity (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] e k/r Nuclear Gravity(1)

With ek/r 1+ k/r; k/r


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And u = [GM/ h]/ {(1 GMk/h) + A cosine [ (1 - GMk/h)] }

And r = 1/u = 1/ {[GM/ h]/ {[1 GMk/h] + A cosine { [1 GMk/h]}}

= [1 GMk/h]/ (M/ h)]/ {1 + cosine { [1 - GMk/h]}

Where [1 GMk/h]/ (GM/h) = a (1 )And [1 GMk/h] = (GM/h) a (1-)And h - GM k = G M a (1- )

Page 17Then h/GM a (1- ) = kAnd [1 GMk/h] = {[1 (GM/h) [(h/GM) a (1 - )]}

= {[1 1 + (GM/h) a (1 - )]}= [(GM/h) a (1 - )]= {[GM/4a4 (1 - )] a (1 - )]}= (GM/4a)= (1/2a) (GM/a) = v/2a = 2/2T= 1/T = f = frequency

What is the accumulated value of 2 f per century for planet mercuryseen from Earth?The angular frequency is = 2 fHow I would see of planet mercury turning around the sun from earth?The answer is there will be a frequency change of

W = 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 days

If 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 3600

W = [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]


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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 is

With = 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]

= red shiftPage 18

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 nucleargravity force is the crudely approximated Newtonian gravity force (-GmM/r) explains planetary motion around the sun as a rotatingellipse with a rotation rate = [ m/(m + M)](180/)(36526/T)(3600)= 43.03 seconds of an arc per century for the most talked aboutplanet of mercury; m = 3.2 x 10^24 kg; M = 2x10^30 kg; T = 88days


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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/rLet m r =1/uThen d r/d t = -u'/u = - (1/u) (') d u/d = (- '/u) d u/d = - h du/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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Multiplication by 180/, then = 180m/ (m + M) degrees/second

Multiplication by (36526/T), then = [m/ (m + M)] (180)(36526/T)degrees/century

Multiplication by 3600, then = [m/ (m + M)] (180) (36526/T)(3600)seconds/century

With Planet Mercury: m = 3.2x1024kg; M = 2x1030kg

Then = [3.2x1024/ 2x 1030] (180) (36526/T)(3600) seconds of an arcper century

And = 43.03426909''/century

= 2 m/2(m + M) = m/ (m + M)And [(I O)/O] 0 = [ m/ (m + M)] (180/ ) (36526/T) (3600)]

Page 20

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} ' 2


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This is the angular time delay and will be seen as angular visualIllusionThe 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 secper century. 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 beseen 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" = {[ (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 visualIllusionThe 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", then


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And W" = (v/r) (180/) (3600) (26526/T) = angular velocity in arc secper century. 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 beseen 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] [-70.29] = 43.0" arc per century[(I O)/O] 0= [(v e/ v m) - 1] (v m /r m)] [(180/) (3600) (26526/T)

Page 22

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 tEquation (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

Planet

r

Sun


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= - 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 + )]

Or, 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)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)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)Page 23

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

Or Tyco Brahes / 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)]Or Newtons mathematically stupid[(I O)/O] 0= (-720x36526x3600/T) {[ (1-]/ (1-) } (v*m + v m/c) seconds of arcper centuryOr Yukawas

[(I O)/O] 0 =15 x (36526/T (days)] [(v* m - v*e)/v* e]Or, Nuclear gravity[(I O)/O] 0 = [ m/ (m + M)] (180/ ) (36526/ T m) (3600)]Or, Boundary value law[(I O)/O] 0= {[GM/r]} 1/2 [(v* m - v*e)/v* e] [(180/ ) (3600) (36526/ T m)Or, Constant Fore law[(I O)/O] 0 =


25/26

{{[r e/ r m]} - 1} (v m /r m)] [(180/) (3600) (26526/T)Or, constant areal velocity law[(I O)/O] 0= [(v e/ v m) - 1] (v m /r m)] [(180/) (3600) (26526/T)Or, 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)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)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)WithNewtons Inverse Square force mass Illusion three more can be deduced

[(I O)/O] 0 = ( m/M) [(180/) (3600) (26526/T m)]Or, nuclear gravity three more can be deduced[(I O)/O] 0 = [ m/ (m + M)] (180/ ) (36526/ T m) (3600)]

There is a total of at least 2125

Read my lips

I am not saying Einstein andAlfred Nobel physics andphysicists are wrong and stupid tosay the least but what I amsaying in writing is that the entire

western civilization 500 years ofphysics and 500,000 physicistsare wrong and stupid to be exact.


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Arabs real time physics is the pastpresent and future of physics with

Arabs best regards and my twomiddle fingers to 500 years ofModern Physics and 500,000Modern Physicists. The name is

the one and only Joe AlexanderNahhasJuly 4th 1973

[email protected] all rights reserved
mailto:[email protected]:[email protected]

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