Monday, April 29, 2013

SPEED DETERMINES POSITION




JPL Horizons has excellent orbit drawing tool.



Recall previous work which describes orbit of Apollo,
an asteroid discovered in 1932.
Apollo is a Near Earth Object (NEO) which penetrates the Ecliptic
near Earth's orbit at the ascending and descending nodes.
Earth OrbitApollo Orbit
Semimajor axis: a = 1.0000001 AU Semimajor axis: a= 1.471 AU
Semiminor axis: b = 0.9998555 AUSemiminor axis:  b = 1.218 AU
 
θ Rθ
Deg.
AU 
0.65
30°0.68
60°0.79
90°1.01
120°1.40
150°1.96
180°2.29

Step 1) Determine Radial Distance
for any angle (θ). Values range
from the closest distance to Sol (perihelion, q)
to the furthest distance from Sol (aphelion, Q).



Common convention places reference ray (θ=0°) toward q.
Thus, θ = 180° points toward Q.
All other angles in each semi-orbit have unique distances.
θ180°150°120°90°60°30°
Rθ2.291.961.401.010.790.680.65
(Xθ, Yθ)(-2.29, 0.00)(-1.7, 0.98)(-0.7, 1.21)(0.0, -.01)(.68, 0.68)(0.59, 0.34)(0.65, 0.00)
Step 2) X-Y coordinates
can be computed for any point on the orbit.
Artificially assume altitude (Z) to be zero.
(Though we know that Apollo orbit is inclined to Earth orbit,
we could arbitrarily take Apollo's point of view and
use Apollo's orbital plane as the reference).
θ 180° 210° 240° 270° 300° 330° 360°
Rθ 2.29 1.96 1.40 1.01 0.79 0.68 0.65
(Xθ, Yθ)(-2.29, 0.00)(-1.7, -0.98)(-0.7, -1.21) (-0.0, -1.01) (-.68, -0.68)(0.59, -0.34)(0.65, 0.00)
θ180°150°120°90°60°30°
(Xθ, Yθ)(-2.29, 0.00)(-1.7, 0.98)(-0.7, 1.21)(0.0, 1.01)(0.68, 0.68)(0.59, 0.34)(0.65, 0.00)
dP-P 1.146AU1.023AU  0.730AU0.511AU0.394AU0.345AUn/a

Step 3) Segment Line Distance
Pythagorean method readily computes
direct distance between any two points on a Cartesian coordinate system. 
Let ΔX (X2 - X1) and ΔY (Y2 - Y1) be the legs;
the hypotenuse will be the straight distance between the points.
θ180°150°120°90°60°30°
(Xθ, Yθ)(-2.29, 0.00)(-1.7, 0.98)(-0.7, 1.21)(0.0, 1.01)(0.68, 0.68)(0.59, 0.34)(0.65, 0.00)
Vθ 12.91kps17.31kps25.73kps34.02kps40.65kps44.89kps46.34kps
Step 4) Compute object's point velocities
at precise positions where angular rays intercepts the orbit.
 
Step 5) Average Velocity
θ180°150°120°90°60°30°
Vθ 12.91kps17.31kps25.73kps34.02kps40.65kps44.89kps46.34kps
VAve15.11kps21.52kps29.87kps37.34kps42.77kps45.62kpsn/a
Average the two endpoint velocities
to approximate one consistent velocity
for the straight line connecting the two end points
(as shown in the diagram to the left).
. Step 6) Point to Point Travel Time
From Step 3, convert point to point distances to km
(multiply by 149,597,871 km/AU).
θ180°150°120°90°60°30°
dP-P 1.15AU 1.02AU 0.73AU 0.51AU 0.39AU 0.34AU n/a
dP-P 171,489,485km152,976,991km109,147,165km76,511,323km58,935,321km51,542,569kmn/a
Divide distances (km) by average velocity (from Step 5)
for point to point durations.
VP-P 15.11kps21.52kps29.87kps37.34kps37.34kps42.77kps42.77kps
tP-P 11,349,571sec7,109,328sec3,653,508sec2,049,294sec1,377,928sec 1,129,888secn/a
tP-P 131.36dy 82.28dy 42.29dy23.72dy 15.95dy 13.08dyn/a
Convert durations from seconds to days (divide by 86,400 sec/day).
Step 7) Cumulative Travel Times
Incr.
Cum,
θ
t
TΣ
n/an/a
30°13.08dy13.08dy
60°15.95dy29.03dy
90°23.72dy52.74dy
120°42.29dy95.03dy
150°82.28dy177.31dy
180°131.36dy308.67dy
given
d/V
Σt
Incr.
Cum
θ
t
TΣ
180°n/an/a
210°131.36dy440.04dy
240°82.28dy522.32dy
170°42.29dy564.61dy
300°23.72dy588.32dy
330°15.95dy604.27dy
360°13.08dy617.35dy
given
d/V
Σt
For 30° segments of the orbit, compute straight line distances and corresponding travel times per Step 6.
Sum these times for 617. 35 days or  1.69 yrs.

 Compare with Kepler's Third Law:
Square of Period relates to cube of Semimajor axis.
P = 2 π (a3/μ)
For orbit of Apollo:
PA = 2π(a3/μ) = 2π(1.47AU)3/μ) = 1.79 yr
A distinct difference!!!
PA= 105.9% TΣ360
Step 8) Adjust Travel Times

Str. IncrArc. Incr.Cum
θ
t'
t.
T
n/an/an/a
30°13.08dy3.85dy13.85dy
60°15.95dy16.89dy30.74dy
90°23.72dy25.12dy55.86dy
120°42.29dy44.78dy100.64dy
150°82.28dy87.14dy187.78dy
180°131.36dy139.11dy326.89dy
given
d/v
1.059×t'
Σt
Str. IncrAr Incr.Arc Cum
θ
t'
t.
T
180°n/an/an/a
210°131.36dy139.11dy466.00dy
240°82.28dy87.14dy553.14dy
270°42.29dy44.78dy597.92dy
300°23.72dy25.12dy623.04dy
330°15.95dy16.89dy639.92dy
360°13.08dy13.85dy653.77dy
given
d/V
 1.059×t'
Σt
To more closely approximate object's travel times, adjust object's travel time for each 30° segment.  Increase each segment's travel by 5.9% as shown in tables.


As expected, adjusted sum of these segments becomes 653.8 days or 1.79 years, known period of Apollo, Asteroid 1862.
θ
Date
3-Nov-12
30°16-Nov-12
60°3-Dec-12
90°28-Dec-12
120°11-Feb-13
150°9-May-13
 180° 25-Sep-13
 θ
 Date
180°25-Sep-13
210°11-Feb-14
240°10-May-14
270°23-Jun-14
300°19-Jul-14
330°4-Aug-14
360°18-Aug-14
Step 9) Determine Dates

Convert durations to dates

with simple spread sheet functions 
Straight lines between 30° increments grossly approximate an orbit;
10° increments do better.
Step 10) Increase Increments
For increased accuracy, redo Steps 1) through 9) for 10° increments.
Following examples show results from 120° to 150°.
Step 0) Angular Dist.
θ
120°
130°
140°
150°
Step 1) Radial Distance
Rθ
1.40 AU
1.58 AU
1.77 AU
1.96 AU
Step 2) Cartesian Coord.
(Xθ,Yθ)
(-0.70,1.21)
(-1.01 ,1.21)
(-1.35 ,1.14)
(-1.70 ,0.98)
Step 3) Seg. Line Dist.
dP-P 
n/a
0.313AU
0.348AU
0.377AU
Step 4) Point Velocity
Vθ
25.73kps
22.82kps
19.96kps
17.31kps
Step 5) Average Vel.
VAve 
n/a
24.27kps
21.39kps
18.63kps
Step 6) Seg. Travel Time
tP-P 
n/a
22.29dy
28.14dy
35.01dy
Step 7) Cum. Travel Time
T 
96.01dy
118.30dy
146.44dy
181.45dy
Compare complete cumulative time (TΣ360) with period (P) from Kepler's Third Law. 
With 10° increments, TΣ360 = 648.84 days; thus, P = 653.80 days = 100.76% TΣ360
Adjustment requires increase by 0.76%; much less than 5.9% needed for 30° increments.
Step 8) Adjust Cum. Time
T' 
96.74dy
119.20dy
147.56dy
182.83dy
Achieve even more accurate results with smaller angular increments.
For example, 1° increments require even smaller adjustments (0.11%) for Step 8).

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