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.
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.
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 semiorbit. Orbit's major axis is line from aphelion through Sol to perihelion. Divide orbit into symmetric views below and above major axis. Top semiorbit 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.
T^{2} = a^{3}
2^{2} = 4 = 1.59^{3} 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.
 
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 semiorbit. EXAMPLE: We might observe object orbits 45° in 45.27 days. COMMON SENSE: Remaining time to Q is 320 days: Δt_{ }= t_{SemiOrb}  t_{ν} 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.
Object's position can be in Polar Coordinates: (ν,r), angle and radial distance.
 
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, A_{T}; divide sector into A_{∠} and A_{Δ}. Triangle on left, A_{∠}, area easily determined. A_{∠ }= .5 × Y × q =.5 × (R×Sin(ν)) × (ac) A_{∠ }= .5 × Y × q = .5 × .808 AU × 1.06 AU = .4284 AU^{2} Sadly, area of edge segment on right, A_{Δ}, 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 Y_{AC} value of point directly above x on AC. Previous work demonstrates following relation:
 
4. Eccentric Anomaly, E
Every True Anomaly, ν,has a corresponding Eccentric Anomaly, E, determined by object's superposition on Aux. Circle (Y_{AC}).
 
5. Determine E's Edge Area, A_{∇} 1) Area of Eccentric Anomaly (E) sector is easily determined: A_{E }= π a^{2} × E/360° A_{E }= 3.14 (1.59AU)^{2} × 32.6°/360° = 0.719 AU^{2} 2) Like ecliptic section (for True Anomaly), E's circular sector can be divided into a triangle and an edge segment. A_{E }= A_{∠} + A_{∇} 3) Like the ecliptic section, triangle can also be easily calculated: A_{∠} = 0.5 a × Y_{AC} = 0.5 a × a×Sin(E) A_{∠} = 0.5×(1.59 AU)^{2}×Sin(32.6°) = 0.681 AU^{2} 4) Gladly, edge segment can be easily calculated (unlike TA's ecliptic section):
 
6. Summarize Edge Areas Sector areas can be divided into triangles and edges.  
  

 
CONCLUSION: Given E's edge area (A_{∇}), (EASY: Subtract triangle area from sector area.) Then, easily compute TA edge area (A_{Δ}). 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 (A_{Δ}).
 
Per Kepler's 2nd Law,area within TA segment indicates travel time to current position (ν = 45°).
 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:
 
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 y_{ν}.
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 y_{θ} 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.
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:
θ_{Edge} < ν_{Edge} < ν_{Triangle} < θ_{Triangle}
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. θ_{Triangle} = 0.5 × B_{θ } × H_{θ} = 0.5 × Q_{ } × y_{θ} ≅ θ_{Sector}= ν_{Sector}
For any given orbit, θ_{Triangle} Base (aphelion, Q), is constant and easily determined.
For any given value of θ, Height (y_{θ}) can be determined from the triangle's area. y_{θ} = 2 × θ_{Triangle} ÷ Q_{ } To more closely approximate triangle area, determine θ_{Edge} area and subtract from sector. 
ESTIMATE VALUE OF θ_{Edge} AREA
For initial guess of y_{θ , }assume θ_{Edge} area to be about 1% of θ_{Sect}.
Electronic spreadsheet can quickly generate 100 rows of values from 1.01% to 2.00%. Following table shows a few selected rows.
triangle area (A_{Tri}) shrinks, triangle height (y_{θ}) decreases.  
Validate Best Edge Value
Casual inspection determines 0.00498 AU to be the most suitable; given some basic interpolation between indicated percentages. 
xxx
1.073%  0.00498 AU^{2}  0.459 AU^{2}  2.12AU  0.4330AU 
% of A_{Sect}  A_{Sect}×A_{E%}  A_{Sect}  A_{Edge}  a + c  2×A_{Tri} Q 

True Anomaly

Eccentric Anomaly
 
TA_{E%}

TA_{Edge}

y_{θ}

E_{θ}

EA_{Edge}

TA_{Edge}
 
1.073%

0.00498 AU²

0.4330 AU

16.8°

0.0053 AU²

0.00498 AU²
 
Given  TA_{Sect }× TA_{E%}  (TA_{Sect}TA_{Tri}) Q 

 b×EA_{Edge} a 

True Anomaly

Eccentric Anomaly
 
TA_{E%}

TA_{Edge}

y_{θ}

E_{θ}

EA_{Edge}

TA_{Edge}
 
1.073%

0.00498 AU²

0.4330 AU

16.8°

0.0053 AU²

0.00498 AU²
 
Given  TA_{Sect }× TA_{E%}  (TA_{Sect}TA_{Tri}) Q 

 b×EA_{Edge} a 

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.
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.
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