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 Orbit | Apollo Orbit |
|---|
| Semimajor axis: a = 1.0000001 AU | Semimajor axis: a= 1.471 AU |
|---|
| Semiminor axis: b = 0.9998555 AU | Semiminor axis: b = 1.218 AU |
|---|
 |
| θ |
Rθ |
|---|
Deg.
|
AU
|
| 0° | 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.
|
|---|
| θ | 180° | 150° | 120° | 90° | 60° | 30° | 0° |
| 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, -.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) |
|
|
|---|
 |
| θ | 180° | 150° | 120° | 90° | 60° | 30° | 0° |
| (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 | 1.146AU | 1.023AU | 0.730AU | 0.511AU | 0.394AU | 0.345AU | n/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.
|
|---|
| θ | 180° | 150° | 120° | 90° | 60° | 30° | 0° |
| (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 | 12.91kps | 17.31kps | 25.73kps | 34.02kps | 40.65kps | 44.89kps | 46.34kps |
Step 4) Compute object's point velocities
at precise positions where angular rays intercepts the orbit.
|  |
|---|
 | Step 5) Average Velocity
| θ | 180° | 150° | 120° | 90° | 60° | 30° | 0° |
| Vθ | 12.91kps | 17.31kps | 25.73kps | 34.02kps | 40.65kps | 44.89kps | 46.34kps |
| VAve | 15.11kps | 21.52kps | 29.87kps | 37.34kps | 42.77kps | 45.62kps | n/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).
| θ | 180° | 150° | 120° | 90° | 60° | 30° | 0° |
| dP-P | 1.15AU | 1.02AU | 0.73AU | 0.51AU | 0.39AU | 0.34AU | n/a |
| dP-P | 171,489,485km | 152,976,991km | 109,147,165km | 76,511,323km | 58,935,321km | 51,542,569km | n/a |
Divide distances (km) by average velocity (from Step 5)
for point to point durations.
| VP-P | 15.11kps | 21.52kps | 29.87kps | 37.34kps | 37.34kps | 42.77kps | 42.77kps |
| tP-P | 11,349,571sec | 7,109,328sec | 3,653,508sec | 2,049,294sec | 1,377,928sec | 1,129,888sec | n/a |
| tP-P | 131.36dy | 82.28dy | 42.29dy | 23.72dy | 15.95dy | 13.08dy | n/a |
Convert durations from seconds to days (divide by 86,400 sec/day). | |
|---|
Step 7) Cumulative Travel Times
∠
|
Incr.
|
Cum,
|
θ
|
t
|
TΣ
|
| 0° | n/a | n/a |
| 30° | 13.08dy | 13.08dy |
| 60° | 15.95dy | 29.03dy |
| 90° | 23.72dy | 52.74dy |
| 120° | 42.29dy | 95.03dy |
| 150° | 82.28dy | 177.31dy |
| 180° | 131.36dy | 308.67dy |
given
|
d/V
|
Σt
|
∠
|
Incr.
|
Cum
|
θ
|
t
|
TΣ
|
| 180° | n/a | n/a |
| 210° | 131.36dy | 440.04dy |
| 240° | 82.28dy | 522.32dy |
| 170° | 42.29dy | 564.61dy |
| 300° | 23.72dy | 588.32dy |
| 330° | 15.95dy | 604.27dy |
| 360° | 13.08dy | 617.35dy |
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. Incr | Arc. Incr. | Cum |
θ
|
t'
|
t.
|
T
|
| 0° | n/a | n/a | n/a |
| 30° | 13.08dy | 3.85dy | 13.85dy |
| 60° | 15.95dy | 16.89dy | 30.74dy |
| 90° | 23.72dy | 25.12dy | 55.86dy |
| 120° | 42.29dy | 44.78dy | 100.64dy |
| 150° | 82.28dy | 87.14dy | 187.78dy |
| 180° | 131.36dy | 139.11dy | 326.89dy |
given
|
d/v
|
1.059×t'
|
Σt
|
| ∠ | Str. Incr | Ar Incr. | Arc Cum |
θ
|
t'
|
t.
|
T
|
| 180° | n/a | n/a | n/a |
| 210° | 131.36dy | 139.11dy | 466.00dy |
| 240° | 82.28dy | 87.14dy | 553.14dy |
| 270° | 42.29dy | 44.78dy | 597.92dy |
| 300° | 23.72dy | 25.12dy | 623.04dy |
| 330° | 15.95dy | 16.89dy | 639.92dy |
| 360° | 13.08dy | 13.85dy | 653.77dy |
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
|
| 0° | 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).
0 Comments:
Post a Comment
<< Home