THE HELIO-CENTRIC CIRCLES OF GALILEO

USING PARAMETRIC GEOMETRY TO EXPLORE POTENTIAL AND MOTIVE MECHANICAL ENERGY CURVES OF THE GRAVITY FIELD.

part one
A RETURN TO AN ORIGINAL PHILOSOPHICAL INQUIRY 
OF HUMAN KIND.

Mathematica Technology Conference
October 2013

CENTRAL FORCE FIELD CONCEPTS 
OF PLANETARY MOTION 
ARE FOUND, PROVEN, AND ACCEPTED.

JOHANN KEPLER: (Dec. 27, 1571 - Nov. 15, 1630)

Heliocentric advocate, finding an accurate geometric curve to fit orbit motion; leaving epicycle explanatories in antiquity.

ANOTHER TYPE OF MOTION WAS 
PROFFERED AND FORGOTTEN.

GALILEO GALILIE: (FEB. 15, 1564 - JAN. 8, 1642)

Heliocentric advocate, died believing in circular motion, rejected elliptical curves.

Basic North Pole Solar View of our Planet Group. Yellow Vector represents the collective tangential orbit direction of our planet group about our sun.

AEC ENERGY SYMBOL

Planet Level Accretion Zone and SPR (solar plane of rotation).

WHAT IS A CURVED SPACE DIVISION ASSEMBLY?

+

+  =

FINDING THE UNIT PARABOLA (UNITS OF r).

Parametric  diagram  of differential  geometry  definition  for  
RoC for  loci point (1, 3/4) on parabola  curve.

{(1/4) (5 (Sqrt[5])) Cos[t]+1,(1/4) (5 (Sqrt[5])) Sin[t]+(3/4)}

2×(focalradius)×(Sqrt[focal radius])=(((((1/2)/(((1+((((1/(-2))))^2)))^(3/2)))))^(-1))= ((5 (Sqrt[5]))/4)

A CSDA comparative is built using curves. Focal radii u1 and u2 connect variable radii (perihelion and aphelion) with field center.

MASS VOLUME CONTENT AND CENTRAL FORCE F

TRADITIONALORBIT METER AND ENERGY OF ORBIT METER CREATE 
GRAVITY FIELD SPACE AND TIME SQUARE

CSDA unity ratio

ParametricPlot[{{1Cos[t],1Sin[t]},{t,((t^2)/(-4(1)))+1}},{t,-3,+3},PlotRange->{{-3,3},{((-3)/2),(3/2)}}]

CSDA Comparative:

central property inverse square space,
focal radii (r, 𝒻(r)) following curved space.

PART 2 (unit circle content)

CSDA GRAVITY FIELD ENERGY DISTRIBUTION 
AND CENTRAL FORCE (F)

TWO TYPES OF G-FIELD ENERGY ARE EVIDENT
1. FALLING, exploring physical properties of constant acceleration.
2. ORBIT, exploring physical properties of changing acceleration.

SPACE AND TIME SQUARE FOR FALLING (CONSTANT ACCELERATION) FINDING POINT MASS ACCELERATION CURVATURE OF FIELD POTENTIAL.

Let vector u be 1 second free fall to surface acceleration curve of F in any gravity field. (u^(-1))will equal acceleration curvature of point mass F. Tangent TD is separation tangent; and tangent DS is reclaim tangent

METHODS TO USE CURVES FOR BASIC HIGH SCHOOL PHYSICS SOLUTIONS OF FALLING.

GIVEN: acceleration curvature of planet earth point mass is ((1/16)).

PROBLEM:	Find impact velocity and time to impact for an object, released via free fall, 400 feet above the earth’ s surface.

The displaced position curve with respect to surface curve of gravity field potential is presented as 1/400.

The acceleration curvature of earths point mass F is 1/16

Establish a direct proportion using both curves, (g-field acceleration curvature of point mass (F) = displacement curvature × (t^2), and solve for (t):

TIME OF FREE FALL
Solve[(1/16)==(1/400)*(t^2),t]
{{t->-5},{t->5}}
We see time of free fall is five seconds. To find impact velocity, double the radius of position displacement curve  and divide by unit time : 
TERMINAL VELOCITY
((400*2)/5)=160ft/sec

Principal Equations used to meter field potential of central force F.

free fall equation template for unit time.

(1/2)(((32ft)/(Sec^2))(((t))^2))/. t->1Sec = 16 ft

set t = 1 Sec and we see 1 Sec free fall displacement is 16 feet, acceleration curvature central force of Earth = ((1/16)).

terminal velocity per unit time

32((ft/(Sec^2)))(t)/.t->1Sec ⟶ ((32 ft)/Sec)

High school falling problem for 400 ft. above surface earth:

(1/2)(((32ft)/(Sec^2))(((t))^2))/. t->5Sec ⟶ 400 ft

set t as 5 Sec in terminal velocity problem and find v = 160 ft/Sec.

32((ft/(Sec^2)))(t)/.t->5Sec ⟶ ((160 ft)/Sec)

PART3

METHODS TO CONSTRUCT AND
READ CSDA  ENERGY CURVES
 OF GALILEO

A COMPUTER DOCUMENT FORMAT DEMONSTRATING CHANGING ORBIT ENERGY (f (r)) FOLLOWING INVERSE SQUARE RADIUS (r).

Step1: Construct a basic CSDA

{t,2},{t,1}}

Step2: Collect preliminary data for parametric description of orbit Mars

({{MARS, \[Placeholder], KM PARAMETERS, \[Placeholder], CSDA PARAMETERS, CSDA EVALUATION}, {perihelion, \[Placeholder], 206620000, \[Placeholder], 1.8131, (206620000/227925000)}, {aphelion, \[Placeholder], 249230000, \[Placeholder], 2.1870, (249230000/227925000)}, {average, \[Placeholder], 227925000, \[Placeholder], 2, (227925000/227925000)*2}, {ASI, \[Placeholder], 113962500, \[Placeholder], 1, (227925000/227925000)*1}, {\[Placeholder], \[Placeholder], \[Placeholder], \[Placeholder], \[Placeholder], \[Placeholder]}, {f(β), \[Placeholder], 20309300, \[Placeholder], 0.1782, \[Placeholder]}, {f(α), \[Placeholder], -22300700, \[Placeholder], -0.1957, \[Placeholder]}, {average v, \[Placeholder], 24.13, \[Placeholder], 24.13, \[Placeholder]}, {focal radius (β), \[Placeholder], 207615700, \[Placeholder], 1.8218, \[Placeholder]}, {focal radius (α), \[Placeholder], 250225700, \[Placeholder], 2.1957, \[Placeholder]}})

perihelion = (206620000/227925000)= 1.8131  ;  aphelion = (249230000/227925000) = 2.1870

FIELD POTENTIAL (ASI):  (227925000/227925000) × 1

ROTATING AVERAGE CURVE:  (227925000/227925000) × 2

Step 3: construct space and time square; central property position and energy levels of position. High energy perihelion is red; low energy aphelion is blue.

ParametricPlot[{{Cos[t],Sin[t]},{t,((t^2)/(-4))+1},{1.8131,t},{2.1870,t},{t,0.1782},{t,(-0.1957)}},{t,-π,2π},PlotRange->{{-2,3},{-1,1}}]

Establish Mathematica template for energy level:

((t^2)/(-4))+1/.t->1.8131 -> 0.1782

((t^2)/(-4))+1/.t->2.1870 ->(-0.1957)

STEP4: CONSTRUCT FOCAL PROPERTIES TO ORBIT LIMITS: m =((𝒻(r))/r)

{t,0.1782},{t,(-0.1957)},{t,0.1782 +0.0983(-1.8131 +t)}{t,-0.1957-0.0895 (-2.187+t)}

Solve[y-0.1782==(0.1782/((1.8131)))(x-(1.8131)),y] → {y->0.1782 +0.0983(-1.8131+t)}

Solve[y-(-0.1957)==(((-0.1957))/((2.1870)))(x-(2.1870)),y]  → {y->-0.1957-0.0895 (-2.187+t)}

Find size of focal radius to orbit parameter limits (2p - f (r) = magnitude) :

2 - 0.1782 = 1.8218  ⟶           focal property r = 1.8218;             central property inverse square r = 1.8131
2 - (-0.1957) = 2.1957 ⟶        focal property r = 2.1957;              central property inverse square r = 2.1870

unity curves; curvature and radius of curvature = 1.

PROPOSAL: LET THERE BE TWO CURVES COMPOSING ORBIT ENERGY EXCHANGE BETWEEN POTENTIAL AND MOTION:

1ST CURVE IS INVERSE SQUARE POTENTIAL (FIXED).

2ND CURVE IS INVERSE SQUARE MOTIVE PROPERTIES OF POTENTIAL, 
ORBIT MOMENTUM.

Theorem (On the Potential and Motive Circles of Galileo)
1).  Conserved sum of available energy for system motion is stored on CSDA Latus Rectum Diameter. When central force potential curvature =1, and focal radius motive curvature =1; then CSDA Inverse Square Radius 2 will balance, center to center, 2 unity curves (curvature and RoC =1). First curve is about center F of potential and second curve is at center of motive event at (slope m = -1) of energy tangent happening (where?) on CSDA period time curve (when?) at dependent curve latus rectum rotating diameter.
2). Motive curve + energy level (f (r)) = potential curve
3). Potential curve - (Motive curve + energy level (f (r))) = zero

Step 5: Prove shape of average motive curve = shape of potential curve.
1. Construct range of potential as a tangent limit through orbit space.
           2. Construct shape of potential curve (curvature = 1) about center F; given. 
      3. Compute and construct shape of motive curve at event (m = -1), using
a) focal property differential; 
	                           b) Sir Isaac Newton inverse square law.

a)  radius of motive curve = (focal radius mag - potential) ⟶  2 - 1 = 1

b) radius of motive curve using inverse square-> [(((((((1/2)))^2)×4(1)))^(-1))⟶1]

QED; (curvature of potential curve = 1); and (curvature of motive curve + 𝒻 (r) = 1).

Step 6: CONSTRUCT HIGH ENERGY MOTIVE CURVE OF FIELD

All CSDA motive curve radius = (focal property magnitude - field potential)

1.8218 - 1 ⟶ radius of motive curve = (0.8218);          focal radius magnitude = (2p - 𝒻 (r))⟶(2 - 0.1782 = 1.8218)

{.8218Cos[t]+1.8131,.8218Sin[t]+0.1782}},{t,-π,2π},PlotRange->{{-2,3},{-1,(3/2)}}]

Central Property Cartesian Parameter for high energy: ({1.8131,t};      f (r) ={t,0.1782})

(((((((1/1.8131)))^2)×(4*1)))^(-1)) ⟶ inverse square motive radius = (.8218)

Step 7: CONSTRUCT LOW ENERGY MOTIVE CURVE OF FIELD

{2.1870,t},{t,(-0.1957)},{t,1},{t,-0.1957-0.0895 (-2.187+t)},{1.1957Cos[t]+2.1870,.1957Sin[t]+(-0.1957)}

Central Property Cartesian Parameter for low energy: ({2.1870,t}; mv ={t,(-0.1957)})

(((((((1/2.1870)))^2)×4(1)))^(-1)) → 1.1957

Step 8: Construct High Energy Tangent Following Orbit Motion for Planet Mars on CSDA time curve.

Energy Tangent. Tangent slope of central property position: Slope term = ((-(t))/(2p)) first derivative of dependent curve.
Solve[y-0.1782==((-1.8131)/2)(x-1.8131),y] → {{y->0.1782 -0.9065 (-1.8131+t)}}

{t,0.1782 +0.0983(-1.8131 +t)},{t,1},{.8218Cos[t]+1.8131,.8218Sin[t]+0.1782}},

Step 9: Construct High Energy Insulator Tangent

compute abscissa of insulator tangent

Solve[(1/x)==(1.8218/1.8131),x] → {{x->0.9952}}

compute ordinate of insulator tangent

Solve[(1/x)==(1.8218/0.1782),x] →  {{x->0.0978}}

compute insulator tangent:

Solve[y-.0978==(((((-0.1782)/1.8131)))^(-1))(x- .9952),y] → {y->0.0978 -10.1745 (-0.9952+t)}

CONVERGENCE POINT OF MOTION ENERGY: All energy tangent phenomena (motive and insulator) connect at curved space directrix at (central property radius/2) point B.

set energy tangent equal to insulator tangent to find energy distribution on curved space directrix.  AB = BC

Solve[0.0978 - 10.1745 (-0.9952 + t) == 0.1782 - 0.9065 (-1.8131 + t), t] → {t = 0.906526}

central property radius: r = 1.8131; →   (1.8131/2) → 0.90655`

CONVERGENCE POINT B OF MOTION ENERGY EXCHANGE ON FIELD POTENTIAL TANGENT IMPLIES EQUAL ENERGY DISTRIBUTION : Motion of M2 is sourced from fixed field potential, and is not equal. Motive energy curves change shape to accommodate conserved angular momentum.

Step 10: Construct Low Energy Orbit Tangents (Mars).

{1.1957Cos[t]+2.1870,1.1957Sin[t]+(-0.1957)},{t,-0.1957-0.0895(-2.187+t)},
{t,-0.1957-1.0935 (-2.187+t)},t,-0.0891+11.1753 (-0.996+t)}

motive radius (inverse sq law):      (((((((1/2.1870)))^2)*4))^(-1)) ⟶ (1.1957)

insulating tangent  ordinate:        Solve[(1/x)==(2.1957/(-0.1957)),x] ⟶  x->-0.0891

insulating tangent abscissa ⟶   Solve[(1/x)==(2.1957/2.1870),x] ⟶  {x->0.9960}

linear eq of insulating tangent ⟶ Solve[y-(-0.0891)==((((0.1957/2.1870)))^(-1))(x-0.9960),y] →{y->-0.0891+11.1753 (-0.996+t)}

intersection of energy tangent and insulator tangent = B

Solve[-0.1957 - 1.0935 (-2.187 + t) == -0.0891 + 11.1753 (-0.996 + t), t]  ⟶  {t->1.0935}

central property radius midpoint:   (2.1870/2)    ⟶   1.0935 = B

finding max period momentum using space and time square   ({{Mean orbital velocity, □, 24.13}, {Max.orbital velocity, □,  26.50}})

(∂_t)(((t^2)/(-4(1)))+1) ⟶ -(t/2)

{1.813053,t},{2.186947,t},{2,t},{t,0.178211},{t,-0.195685},{t,0.1782 -0.9065 (-1.8131+t)}

Solve[1/24.13==(((1.8218/2))^(-1))/v,v]⟶ {v->26.4904}

finding min period momentum using space and time square   ({{Mean orbital velocity, ⟶, 24.13}, {Min.orbital velocity, ⟶, 21.97}})

(∂_t)(((t^2)/(-4(1)))+1) ⟶ -(t/2)

{1.813053,t},{2.186947,t},{2,t},{t,0.178211},{t,-0.195685},{t,-0.1957-1.0935 (-2.1869+t)}}

Solve[1/24.13==(((2.1957/2))^(-1))/v,v] ⟶ {v->21.9795}