Plot Orbits and Tell Time

1. Plot Dots

Plot orbital positions with Sol based Cartesian Coordinates.
Determine as follows:

X☉ =Rν cos(ν)	Y☉ = Rν sin(ν)

Examples for ν values of 160° and 180°:

X☉ = R160 cos(160°) = -2.3 AU	Y☉= R160 sin(160°) = 0.84 AU
X☉ = R180 cos(180°) = -2.7 AU	Y☉ = R180 sin(180°)  = 0.00 AU

To determine length of radius vector, R, from Sol to satellite, use following polar equation:

Rν	=	l 1 + e × cos(ν)

TRUE ANOMALY: ν, is angle of the R vector. Ref. Angle (ν= 0°) is the ray connecting Sol to perihelion.
(In table, ν independently varies from 0° to 180° in 20° increments).
SEMI-LATIS RECTUM: l, is key to radius formula; distance from Sol to cycler at ν = 90°.

l = 1.0 AU =149,597,870.7 km (arbitrarily chosen to be same distance as from Sol to Terra).

SEMI-MAJOR AXIS: a, is longest distance from center to orbit; cycler at ν = 0°.
let a = 1.66 AU (arbitrarily chosen to give orbit a syndodic period, 2.135 years).
SEMI-MINOR AXIS: b, shortest distance from center to orbit. √(a×l)
b=√(1.587AU×1.0AU)=1.288 AU

FOCUS: c, distance from orbit center to Sol; defined: √(a²-b²)
c= √(a²-b²) = 1.045 AU
ECCENTRICITY: e, measure of orbit's "flatness".
e = c/a = .623

ν Rν X☉  Y☉ Deg AU AU AU 0° 0.614 0.61 0.00 20° 0.628 0.59 0.21 40° 0.674 0.52 0.43 60° 0.760 0.38 0.66 80° 0.901 0.16 0.89 100° 1.123 -0.19 1.11 120° 1.460 -0.73 1.26 140° 1.933 -1.48 1.24 160° 2.451 -2.30 0.84 180° 2.703 -2.70 0.00

2. Connect Dots

Determine straight line distance increments with Pythagorean formula.

Δd	=	√((ΔX)2+(ΔY)2)	=√(	(X2-X1)2	+	(Y2-Y1)2	)

μ_Sol, Solar standard gravitational parameter, = 132,712,440,018 km³s⁻²

Use following formulas to approximate incremental times.

Vν =√( 2μSol Rν + μSol a )

Δt = Δd Vν × day 86,400 sec

ν	Rν	X☉	Y☉	Δd		V	Δt
Deg	AU	AU	AU	AU	km	km/sec	secs	days
160°	2.451	-2.30	0.84
180°	2.703	-2.70	0.00	0.9286	138,914,962	11.021	12,604,896	145.89
Given	See above	See above	See above	See above	ΔdAU * 149,597,871	See above.	See above.	ΔtSee/86,400

To approximate half of orbital period, summ increments from ν = 0° to 180°. ν Rν __X☉__ Y☉ Distance (Δd) V Δt Σt deg AU AU AU AU km km/sec days days 0° 0.614 0.61 0.00 n/a 48.549 n/a 20° 0.628 0.59 0.21 0.2161 32,326,752 47.849 7.82 7.82 40° 0.674 0.52 0.43 0.2308 34,521,127 45.776 8.73 16.55 60° 0.760 0.38 0.66 0.2632 39,369,196 42.404 10.75 27.29 80° 0.901 0.16 0.89 0.3202 47,903,956 37.859 14.64 41.94 100° 1.123 -0.19 1.11 0.4137 61,882,247 32.328 22.15 64.09 120° 1.460 -0.73 1.26 0.5579 83,464,320 26.083 37.04 101.13 140° 1.933 -1.48 1.24 0.7510 112,341,873 19.569 66.44 167.57 160° 2.451 -2.30 0.84 0.9165 137,100,294 13.743 115.46 283.04 180° 2.703 -2.70 0.00 0.9286 138,914,962 11.021 145.89 428.93 Validate this approximation with Kepler's 3rd Law a3 = k * T2

3. Use Kepler's Anamolies to compute precise times for any angle.
Use previous table's approximate orbit times to validate this model.
Recall Kepler's Second Law: For undisturbed Solar orbits, the line joining the object to the Sun sweeps out equal areas in equal intervals of time.
Recall simple geometry gives the total area inside an orbit (A_Ellipse = π a b) and Kepler's Third Law gives us the total orbit( T= √(a³/μ).  Thus, finding times for parts of orbits should be a simple proportionality problem.
PROBLEM: Finding the "True Anomaly Area" is not a straightforward task. The vector, "r", from the Sun to the orbiting object does not sweep easy circular sectors; instead, it sweeps focus centered elliptical sectors.
Four hundred years ago, Kepler solved this problem by deriving his famous "Kepler's Equation". Since then, his equation has been derived by many different methods. Some pertinent concepts are briefly discussed below:
Aux. circle
Translate X-Y coordinates
---from Sun based where Sol is origin. (X_☉,Y_☉) = (0,0)
---to Auxilliary Circle (AC) based, where origin is AC center.   (X_AC,Y_AC) = (0,0)
X_AC = X_☉ xxxx
Y_AC = Y_☉ xxxxxx
The AC based coordinates enable us to determine Ecentric Anomaly (E).
Thru Kepler's Equation, E enables us to determine transition duration for any ν.

True Anom. Radius Sun Based Coordinates Aux. Circle Coordinates Eccen. Anom. Total Time Incr. Time ν Rν X☉ Y☉ XAC YAC E tν Δt 0° 0.614 0.61 0.00 1.66 0.00 0.0° 0.00 n/a 20° 0.628 0.59 0.21 1.63 0.28 9.6° 7.76 7.76 40° 0.674 0.52 0.43 1.56 0.56 19.7° 16.30 8.54 60° 0.760 0.38 0.66 1.42 0.85 30.8° 26.64 10.35 80° 0.901 0.16 0.89 1.20 1.14 43.6° 40.50 13.86 100° 1.123 -0.19 1.11 0.85 1.42 59.2° 61.04 20.54 120° 1.460 -0.73 1.26 0.31 1.63 79.1° 94.49 33.45 140° 1.933 -1.48 1.24 -0.44 1.60 105.2° 152.54 58.05 160° 2.451 -2.30 0.84 -1.26 1.08 139.4° 251.03 98.48 180° 2.703 -2.70 0.00 -1.66 0.00 180.0° 389.91 138.89 200° 2.451 -2.30 -0.84 -1.26 -1.08 220.6° 528.80 138.89 220° 1.933 -1.48 -1.24 -0.44 -1.60 254.8° 627.28 98.48 240° 1.460 -0.73 -1.26 0.31 -1.63 280.9° 685.33 58.05 260° 1.123 -0.19 -1.11 0.85 -1.42 300.8° 718.78 33.45 280° 0.901 0.16 -0.89 1.20 -1.14 316.4° 739.32 20.54 300° 0.760 0.38 -0.66 1.42 -0.85 329.2° 753.18 13.86 320° 0.674 0.52 -0.43 1.56 -0.56 340.3° 763.53 10.35 340° 0.628 0.59 -0.21 1.63 -0.28 350.4° 772.06 8.54 360° 0.614 0.61 0.00 1.66 0.00 360.0° 779.82 7.76 See above "Plot Dots" See below tν-prev. XAC = X☉ -  c YAC = Y☉× a ÷ b E = tan-1(YAC / XAC) t = (E-e×Sin(E)) × a × √(a/μ)

4. Use numerical methods to plot orbit times on this orbit.

Numerical Methods

Given True Anom. Radius Sun Based Coordinates Aux. Circle Coordinates Eccen. Anom. Total Time Incr. Time t ν R? X? Y? XAC YAC E t Δt Days Deg AU AU AU AU AU Deg Days Days 0 0° 0.614 0.61 0.00 1.66 0.00 0.0° 0.0 30.0 30 65.49° 0.793 0.33 0.72 1.37 0.93 34.1° 30.0 30.0 60 99.18° 1.112 -0.18 1.10 0.87 1.41 58.5° 60.0 30.0 90 117.86° 1.417 -0.66 1.25 0.38 1.61 76.7° 90.0 30.0 120 130.19° 1.685 -1.09 1.29 -0.04 1.66 91.5° 120.0 30.0 150 139.32° 1.915 -1.45 1.25 -0.41 1.61 104.2° 150.0 30.0 180 146.62° 2.110 -1.76 1.16 -0.72 1.49 115.6° 180.0 30.0 210 152.77° 2.274 -2.02 1.04 -0.98 1.34 126.1° 210.0 30.0 240 158.16° 2.408 -2.24 0.90 -1.19 1.15 135.9° 240.0 30.0 270 163.03° 2.516 -2.41 0.73 -1.36 0.95 145.2° 270.0 30.0 300 167.54° 2.598 -2.54 0.56 -1.49 0.72 154.2° 300.0 30.0 330 171.82° 2.657 -2.63 0.38 -1.59 0.49 162.9° 330.0 30.0 360 175.95° 2.691 -2.68 0.19 -1.64 0.24 171.5° 360.0 30.0 390 180.01° 2.703 -2.70 0.00 -1.66 0.00 180.0° 390.0 30.0 420 184.07° 2.691 -2.68 -0.19 -1.64 -0.25 188.5° 420.0 30.0 450 188.20° 2.656 -2.63 -0.38 -1.58 -0.49 197.1° 450.0 30.0 480 192.48° 2.598 -2.54 -0.56 -1.49 -0.72 205.9° 480.0 30.0 510 197.00° 2.516 -2.41 -0.74 -1.36 -0.95 214.8° 510.0 30.0 540 201.87° 2.408 -2.23 -0.90 -1.19 -1.15 224.1° 540.0 30.0 570 207.27° 2.273 -2.02 -1.04 -0.98 -1.34 234.0° 570.0 30.0 600 213.42° 2.109 -1.76 -1.16 -0.72 -1.50 244.4° 600. 30.0 630 220.73° 1.914 -1.45 -1.25 -0.41 -1.61 255.8° 630.0 30.0 660 229.87° 1.684 -1.09 -1.29 -0.04 -1.66 268.6° 660.0 30.0 690 242.23° 1.416 -0.66 -1.25 0.39 -1.61 283.4° 690.0 30.0 720 260.96° 1.110 -0.17 -1.10 0.87 -1.41 301.7° 720.0 30.0 750 294.79° 0.791 0.33 -0.72 1.38 -0.92 326.1° 750.0 30.0 779.8 360.00° 0.614 0.61 0.00 1.66 0.00 360.0° 779.8 29.8 Numerical Methods See R? below. R?cos(ν) R?sin(ν) XO +c (a/b)*YO tan-1 YAC XAC See tEarth below. ti-ti-01 Methods to Improve Precision Increase Granularity: Instead of computing a row for every 30 days, compute one for every day.  "Brute Force" method. Increase Elegance: Instead of linear interpretation of durations between degrees, try weighted interpretation.  Weigh values according to relevant velocities which is ever changine on an eccentric orbit. Previous work enables us to readily compute positions and durations for any orbiting body given the body's angular position. Thus, one can create a table with values for all integer degree values from 0° to 360°. One can also create a much smaller table with angular positions of 65° and 66°. True Anom. Radius Sun Based Coordinates Aux. Circle Coordinates Eccen. Anom. Comp. Time Incr. Time ν R☉ X☉ Y☉ XAC YAC E t Δt Deg AU AU AU AU AU Deg days days 65 0° 0.790 0.33 0.72 1.38 0.92 33.77° 29.69 n/a 66 0° 0.796 0.32 0.73 1.37 0.94 34.39° 30.33  0.64 One can see respective durations of 29.69 days and 30.33 days. However, one might want to conduct reverse calculations to determine what values would result at a duration of exactly 30.0 days duration. Unfortunately, considerable work by many scholars on Kepler's Equation still show that a straight forward solution is not obvious; thus, we must use numerical methods to approximate angles given an arbitrary duration. Therefore, following tables use linear interpolation to approximate true anomaly values for given durations. Value Between Two Successive Degrees Given Time True Anom. Radius Sun Based Coordinates Aux. Circle Coordinates Eccen. Anom. Comp. Time Incr. Time t ν R☉ X☉ Y☉ XAC YAC E t Δt Days Deg AU AU AU AU AU Deg days days 65.0° 0.790 0.33 0.72 1.38 0.92 33.77° 29.69 n/a 30 65.49° 0.793 0.33 0.72 1.37 0.93 34.1° 30.0 0.31 66 0° 0.796 0.32 0.73 1.37 0.94 34.39° 30.33 0.64 ν . . . tν n/a tGiven νt . . . tComp Δtν ν+1 . . . tν+1 Δtν+1 Approximate νt  Arbitrarily choose tGiven between tν and tν+1. Example:  29.69 < 30.00 < 30.33 Note that True Anonmaly, ν= 65°, results in duration of 29.69 days, and ν+1 (66°) results in 30.33 days. Best guess νt such that resulting tComp = tGiven  Δtν    = tGiven - tν Δtν+1 = tν+1      - tν Understood but not shown is Δν = 1° = 66° - 65°.  νt = ν + Δtν Δtν+1 (Δν)