Tuesday, March 04, 2014

RETHINKING KEPLER'S LAWS

      


Resonance Reflects
To transform an asteroid into an useful habitat,
alter the asteroid's orbit to resonate
with Earth's orbit.



If habitat returns to same orbital spot every two years,
it will conveniently rendezvous with the Earth
for a lucrative cargo exchange



Resonant orbits
have different centers;
however, all solar orbits

share a common focus: Sol.
(Solar orbits are ellipses and the Sun is at a focus.)



As the dominant body, Sol anchors orbits of all Solar objects.
Thus, common sense compels us to
use Sol as the origin (0,0) 
for a shared coordinate system
with AUs as the units.
(NOTE: Principles of this chapter apply to any orbiting object: asteroid or habitat.)
"A line joining an object and the Sun sweeps out equal areas
during equal intervals of time."



Reflector Principle
For every angle, ν,
an angle of equal size and shape reflects to other semi-orbit.

Orbit's major axis is line from aphelion through Sol to perihelion.
Divide orbit into symmetric views below and above major axis.
Top semi-orbit angles reflect to bottom view.
Corresponding angles sweep equal areas and equal times.

"Square of period is proportional to cube of semimajor axis, a."



Choose the semimajor axis to be 1.59 AU
for orbital period to be 2.0 years.
T2 = a3
22 = 4 = 1.593

q + Q = 2 × a
1.0 AU + 2.18 AU  = 3.18 AU = 2 × 1.59 AU
2 Year Orbit Shapes Can Differ

Per Kepler's 3rd Law, a 2 year orbit must have a semimajor axis (a) of 1.59 AU.
However, semiminor axis (b) can range between 0 and 1.59 AU.
Thus, 2 year orbit's shape can vary between highly oblique to near circular.
Also, different shaped orbits will have different centers (see figure).

However, Sol remains the focus of all the orbits.
Semimajor
Axis
Eccen-
tricity
FocusPeri-
helion
Ap-
helion
Semiminor
Axis 
Semilatus
Rectum 
 a ecqQ
1.59 AU1/30.53 AU1.06 AU2.75 AU1.50 AU1.41 AU
1.59 AU1/20.80 AU0.79 AU2.38 AU1.38 AU1.19 AU
1.59 AU2/31.06 AU0.53 AU2.12 AU1.19 AU0.88 AU
Constant (2 yr)2/3Givena × ea - c a + c(a2-c2)b2 ÷ a
Semiorbit Travel Time

Kepler's Second Law: equal areas give us equal times.
Symmetry about major axis [line from q (nearest to Sol) to Q (farthest)]
compels us to conclude that traveling on orbit from q to Q
will take half of the orbital period (2 years).
Thus, travel time for top half of 2 year orbit is always one year.
Same rule must also apply to bottom half.
Remainder Principle

Every True Anomaly, ν, has a supplement (180°-ν)
which is the remainder angle (∠Sup) in same semi-orbit.
 

EXAMPLE:  We might observe object orbits 45° in 45.27 days.
COMMON SENSE: Remaining time to Q is 320 days:
Δt  tSemi-Orb  -  tν
319. 98 days = (365.25 - 45.27) days
PERHAPS THERE EXISTS A MORE ELEGANT METHOD to determine Partial Orbit Times.
Thus, following steps.
1. Variable Vectors
Positional vector, r, ranges from minimum, q, to max at aphelion, Q.
Variable range increases difficulty of computing segment areas.
r(ν)=

1+e×Cos(ν)
=1.41 AU

1+.333×Cos(45°)
=r(45°) = 1.14 AU
Semilatus Rectum: ℓ = 1.41 AU || eccentricity: e = .333
Object's position can be in Polar Coordinates: (ν,r), angle and radial distance.

In contrast, Earth's circular orbit is centered on Sol;
thus, it's much easier to compute partial orbit times. 
If habitat orbit has period of 2 years and a perihelion (q) of 1.06 AU;
then, the habitat could rendezvous with Earth every 2 years near q.
The .06 AU past Earth's orbit adds a respectable safety margin to avoid collisions,
but still close enough for easy periodic payload exchanges.

  Polar Coordinates: (ν,r) can transform to Cartesian Coordinates:
X = r×Cos(ν) and Y = r×Sin(ν)
EXAMPLE: X=1.14 AU×Cos(45°)=.808 AU and Y=1.14AU×Sin(45°) = .808 AU

2. True Anomaly, ν
For every position on the orbit, there is an unique ν.
(Example in figure, ν = 45°.)

To approximate True Anomaly area, AT; divide sector into A and AΔ.
T
riangle on left, A, area easily determined.
A= .5 × Y × q =.5 × (R×Sin(ν)) × (a-c)
A= .5 × Y × q = .5 × .808 AU × 1.06 AU = .4284 AU2
Sadly, area of edge segment on right, AΔ, is not so easy (due to variable range).
3. Auxiliary Circle (AC)
Auxiliary circle circumscribes an elliptical orbit.
Thus, AC shares a common center (C) with the elliptic orbit
 as well as the two endpoints of the orbit's major axis, -a and +a.
At a constant distance from C, AC radius remains constant length, "a".

For any "x" between C and +a on semimajor axis, there is
---a corresponding "y" on the elliptic orbit directly above x. 
---corresponding YAC value of point directly above x on AC.
Previous work demonstrates following relation:
Elliptic-Y 

Aux. Circle-Y
= b

a

= y

YAC
 
 4. Eccentric Anomaly, E
Every True Anomaly, ν,has a corresponding Eccentric Anomaly, E,
determined by object's superposition on Aux. Circle (YAC).
a

b
=1.59 AU

1.499 AU
= 1.0601 AU=YAC

y
=.8566 AU

.808 AU 
E's apex at center, X position and Y position (YAC) form a right angle.
Sin(E) =OPP

HYP
=YAC

a
=0.8566 AU

1.59 AU 
= .5387
E =  Sin-1(.5387) = 32.6°
Recall:YAC=a

b
y Thus:Sin(E)=OPP

HYP
=YAC

a
 = ya

b
1

 a
=y

b
=.808 AU

1.499 AU
=.539
Therefore: E = Sin-1(y

b
)
5. Determine E's Edge Area, A
1) Area of Eccentric Anomaly (E) sector is easily determined:
AE = π a2 × E/360°

AE = 3.14 (1.59AU)2 × 32.6°/360° = 0.719 AU2
2) Like ecliptic section (for True Anomaly), E's circular sector can be
divided into a triangle and an edge segment.

AE = A + A
3) Like the ecliptic section, triangle can also be easily calculated:
A = 0.5 a × YAC = 0.5 a × a×Sin(E)
A = 0.5×(1.59 AU)2×Sin(32.6°) = 0.681 AU2
4) Gladly, edge segment can be easily calculated (unlike TA's ecliptic section): 
A= AE - A=πa2×E

360°
-a2Sin(E)

2
= a2(π×E

360°
-Sin(E)

2
)
A= AE - A= 0.719 - 0.681 = 0.038 AU2
6. Summarize Edge Areas
Sector areas can be divided into triangles and edges.
Sadly,
we can't easily calculate TA area,
because it is an off center, ecliptic portion not a circular sector.

However, we can approximate TA area with an inscribed triangle.
Triangle, q-S-Ye, contains most TA area except for a small portion of eclipse as shown.
ATriangle = 1/2 × q × Ye = .5×1.06×.808 = .428 AU²
Gladly,
circular sector can be easily calculated:
AEA =   π × a× E/360° = 3.14×1.59AU²×32.6°/360^deg; = .719 AU²
Triangle, q-C-Yc, contains most EA area: 1/2×a×Yc=.5×1.59AU×.857AU= .681 AU²
EA Edge Area  = EASector -EATriangle =.719 AU²-.681 AU²=-.038 AU²
CONCLUSION:
Given E's edge area (A), (EASY: Subtract triangle area from sector area.)
Then, easily compute TA edge area (AΔ).
RECALL circle area (a×a×π) to ellipse area (a×b×π) ratio
circle : ellipse = a : b

Thus, use a : b ratio to estimate TA edge area (AΔ).
Eccentric Anomaly Edge Area

True Anomaly Edge Area
= a

b
= A

AΔ
Thus, Thought Experiment assumes:
AΔ= b

a
×A= 1.499 AU

1.59 AU
×0.038 AU2= 0.0358 AU2
Per Kepler's 2nd Law,area within TA segment
 indicates travel time to current position (ν = 45°).
TA Time
= Orb. Per. ×TA Area

Orb. Area

= 2 years ×0.464 AU2

7.4877 AU2
=45.27 days
7. Summarize TA Area and Time
True Anomaly travel time can be determined:
I. True Anomaly (ν) is an angle from q, orbit's closest point to Sol. For every ν, there is a corresponding angle, E, Eccentric Anomaly.
II. Every E has an area readily computed as a sector of the orbit's Auxiliary Circle. The E sector can be readily separated into a triangle and an edge area; both are easily determined.
III. E's area helps determine TA area.  TA has a different apex and a different border than E; however, TA sector can also separate into a triangle and an edge.  E's edge is assumed to have following relation with TA edge:
Eccentric Anomaly (E) Edge Area

True Anomaly (TA) Edge Area
= semimajor axis (a)

semiminor axis (b)
IV. TA area has same proportion of total orbit area, as object's travel time (from  0° to ν) has of total orbit time. Thus, we assume:
TA Time
= Orbital Period ×TA Area

Orb. Area
Let ν (Greek "nu") be an orbiting object's True Anomaly, the angle starting from the ray from Sol to q, orbit's perihelion.  ν defines a sector with Sol at the apex with the object's orbit as the outside border. Travel time is along this outside border from q to yν.

Let θ (Greek "theta") define a similar sector.  θ can be another angle which ends with the ray from Sol to Q, orbit's aphelion, with travel time from yθ to Q. For every sector defined by ν, there can be another sector of equal area defined by θ. Recall Kepler's 2nd Law states that if sector areas are equal; then, the travel times for the associated orbit portions are also equal. However, the sector shape for θ will differ from ν.
EXAMPLE:  For a particular orbit, previous work shows that ν could equal 45° with area of 0.464AU² as well as a travel time of 45.27 days. Thus, we know the θ sector also has area of 0.464AU² and travel time of 45.27 days; however, we don't yet know θ's angular value.
Each sector has two distinct areas: a triangle and an edge area as shown in the figure. ν and θ sector share a common apex at Sol which is also a shared vertex of the ν and θ triangles.
  • ν-triangle's 3 vertices: Sol, q, and yν.
  • θ-triangle's 3 vertices: Sol, Q, and yθ.

For each sector, the triangle takes up most of the area; however, there is a tiny edge area outside the triangle.  Figure shows this tiny sliver between each triangle's short leg and the orbit. By inspection, one can assume following relationship:
 θEdge <  νEdge <  νTriangle <  θTriangle   
Example defines the angle, ν, as known and the angle, θ, as unknown; however, we know some things about θ. By inspection, one readily determines that angle θ is much less than angle ν; thus, θ sector arc is much shorter and flatter than ν sector's arc.  Thus, θ sector's edge area is much smaller than the corresponding edge of ν sector; itself, a very small portion of the entire sector.
COMPUTE TRIANGLE AREA
Triangle Area is one half Base × Height and approximates sector area.
θTriangle  =  0.5 × Bθ 
 ×  Hθ  =  0.5 × Q  × yθ  θSectorνSector

For any given orbit, θTriangle Base (aphelion, Q), is constant and easily determined.
For any given value of θ, 

Height (yθ) can be determined from the triangle's area.
yθ  = 2 × θTriangle  ÷ Q 
To more closely approximate triangle area,

determine θEdge area and subtract from sector.
ESTIMATE VALUE OF θEdge AREA
For initial guess of yθ , assume θEdge area to be about 1% of θSect.
Electronic spreadsheet can quickly generate 100 rows of values from 1.01% to 2.00%.

Following table shows a few selected rows.
θASect=0.464 AU2= νASect
AE%AEdgeATriQyθ
1.06%0.00492 AU20.4591 AU22.12AU0.4331 AU
1.07%0.00496 AU20.4590 AU22.12AU0.4331 AU
1.08%0.00501 AU20.4590 AU22.12AU0.4330 AU
GivenASect×AE%ASect-AEdgea+c2×ATri

Q

yθ=2×ATri

Q
=2×(ASect-AEdge)

Q
=ASect×(1-AE%)

Q
As edge area (AE%) grows,
triangle area (ATri) shrinks,
triangle height (yθ) decreases.
Validate Best Edge Value
True Anomaly Eccentric Anomaly
TAE%TAEdgeyθEθEAEdgeTAEdge
1.05%0.004870.4331016.7950.00530.00498
1.06%0.004920.4330516.7930.00530.00498
1.07%0.004960.4330116.7920.00530.00498
1.08%0.005010.4329616.7900.00530.00498
1.09%0.005060.4329216.7880.00530.00498
1.10%0.005100.4324816.7870.00530.00497
GivenTASect × TAE%(TASect-TATri)

Q
sin-1(yθ

b
)
a2×(πEθ

360⁰
-,5yθ

b
)
b×EAEdge

a
Look for intercepting value between TA TAEdge and EA TAEdge column.
Casual inspection determines 0.00498 AU to be the most suitable;
given some basic interpolation between indicated percentages.
Reconsider ECCENTRIC ANOMALY Let Eθ be angular distance from yAC to Q as shown in figure.

xxx
1.073%0.00498 AU20.459 AU22.12AU0.4330AU
% of ASectASect×AE%ASect - AEdgea + c2×ATri

Q
xx
True Anomaly
 Eccentric Anomaly
TAE%
TAEdge
yθ
Eθ
EAEdge
TAEdge
1.073%
0.00498 AU²
0.4330 AU
16.8°
0.0053 AU²
0.00498 AU²
GivenTASect × TAE%(TASect-TATri)

Q
sin-1(yθ

b
)
a2×(πEθ

360⁰
-.5yθ

b
)
b×EAEdge

a
xx
True Anomaly
 Eccentric Anomaly
TAE%
TAEdge
yθ
Eθ
EAEdge
TAEdge
1.073%
0.00498 AU²
0.4330 AU
16.8°
0.0053 AU²
0.00498 AU²
GivenTASect × TAE%(TASect-TATri)

Q
sin-1(yθ

b
)
a2×(πEθ

360⁰
-.5yθ

b
)
b×EAEdge

a
xxx
For every theta between 0 and 90 degrees there is an unique R (distance from Sol to the object) as well as an unique y value.
Inverse is also true, for every orbital y value from 0 to b, semiminor axis, there is an unique R and theta. 
Can be easily discerned through simple enumeration.

Next discussion, for every nu value between 0 to 90 degrees, there are several other associated values, remainder, reflected, theta of same area, remainder and reflected.

Final discussion, 1/4 period value.  There is a nu value from 90 degs to less than 180 degs which is half the area/time value of the semiorbit time/area from q to Q. Can be chosen as another reference to generate more remaining values.
x

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