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March 27th, 2017 | #1 |
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A funny thing happened on the Physics Forums
I was censored on the Physics Forums for a reason so persnickety that it seems suspicious. Read the contents of the thread that I posted (to follow), only to have it removed just as I was finishing up.
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March 27th, 2017 | #2 | ||
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The observations in this exposition of the method of Gauss (for finding preliminary Keplerian elements of a planet's orbit) are fictional. The data were taken from JPL's Horizons interface program at
https://ssd.jpl.nasa.gov/horizons.cgi Quote:
Quote:
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March 27th, 2017 | #3 |
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First Observation
At 0h UT on 20 July 2018, an observer on Earth sees Mars at 20h 39m 21.03s right ascension and −24° 50' 00.0" declination. The time, t₁, converted to Julian Date: t₁ = JD 2458319.5 The obliquity of the ecliptic at time t₁ ε₁ = 0.40905071279426947 radians Earth's heliocentric position in ecliptic coordinates at time t₁ x⊕₁ = +0.4618841708199115 AU y⊕₁ = −0.8984126365037669 AU z⊕₁ = −0.00005165456609240359 AU The sun's position in geocentric celestial coordinates at time t₁ is calculated by a rotation and a translation of the coordinate system. X๏₁ = −0.4618841708199115 AU Y๏₁ = +0.8242719747171781 AU Z๏₁ = +0.3573807210716430 AU A unit vector in the apparent direction of Mars in geocentric celestial coordinates at time t₁ a₁ = +0.5813786066519955 b₁ = −0.6968612501535729 c₁ = −0.4199801349609095 |
March 27th, 2017 | #4 |
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Second Observation
At 0h UT on 27 July 2018, Mars can be seen at 20h 31m 47.37s right ascension and −25° 32' 35.7" declination. The time, t₂, converted to Julian Date: t₂ = JD 2458326.5 The obliquity of the ecliptic at time t₂ ε₂ = 0.4090506693018622 radians Earth's heliocentric position in ecliptic coordinates at time t₂ x⊕₂ = +0.5637478350603771 AU y⊕₂ = −0.8380340129569841 AU z⊕₂ = −0.00005336797521227511 AU The sun's position in geocentric celestial coordinates at time t₂, therefore, is X๏₂ = −0.5637478350603771 AU Y๏₂ = +0.7688739928199521 AU Z๏₂ = +0.33336735425957914 AU A unit vector in the apparent direction of Mars in geocentric celestial coordinates at time t₂ a₂ = +0.5548334913212998 b₂ = −0.7115005281989506 c₂ = −0.43119229501561324 |
March 27th, 2017 | #5 |
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Third Observation
At 0h UT on 03 August 2018, Mars can be seen at 20h 23m 48.06s right ascension and −26° 06' 15.8" declination. The time, t₃, converted to Julian Date: t₃ = JD 2458333.5 The obliquity of the ecliptic at time t₃ ε₃ = 0.4090506258094553 radians Earth's heliocentric position in ecliptic coordinates at time t₃ x⊕₃ = +0.6578217701347128 AU y⊕₃ = −0.7659716360160346 AU z⊕₃ = −0.00005554489987022230 AU The sun's position in geocentric celestial coordinates at time t₃, therefore, is X๏₃ = −0.6578217701347128 AU Y๏₃ = +0.702755994303274 AU Z๏₃ = +0.3047073394868152 AU A unit vector in the apparent direction of Mars in geocentric celestial coordinates at time t₃ a₃ = +0.5271965424026115 b₃ = −0.7269503439654071 c₃ = −0.44000795798179343 |
March 27th, 2017 | #6 |
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Useful constants and conversions
GM = 1.32712440018e+20 m³ sec⁻² k = 0.01720209895 A = 1/c = 0.00577551833109 days/AU 1 AU = 1.49597870691e+11 meters 1 AU/day = 1731456.8368056 m/s |
March 27th, 2017 | #7 |
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First Approximation for the heliocentric and geocentric distances (equations)
τ₁ = k(t₃−t₂) τ₂ = k(t₃−t₁) τ₃ = k(t₂−t₁) n₁° = τ₁/τ₂ n₃° = τ₃/τ₂ ν₁ = τ₁τ₃ (1+n₁°) / 6 ν₃ = τ₁τ₃ (1+n₃°) / 6 D = a₂ (b₃c₁−b₁c₃) + b₂ (c₃a₁−c₁a₃) + c₂ (a₃b₁−a₁b₃) For i = 1 to 3 dᵢ = X๏ᵢ (c₁b₃−c₃b₁) + Y๏ᵢ (a₁c₃−a₃c₁) + Z๏ᵢ (b₁a₃−b₃a₁) R๏ᵢ = √(X๏ᵢ² + Y๏ᵢ² + Z๏ᵢ²) qᵢ = −2(aᵢX๏ᵢ + bᵢY๏ᵢ + cᵢZ๏ᵢ) K₀ = (d₂ − d₁n₁° − d₃n₃°)/D L₀ = (d₁ν₁ + d₃ν₃)/D r₂ = the distance from the sun to the planet at time 2. ρ₂ = the distance from Earth to the planet at time 2. We must solve these two equations simultaneously for r₂ and ρ₂. ρ₂ = K₀ − L₀/r₂³ r₂ = √(R๏₂² + q₂ρ₂ + ρ₂²) This leads to an 8th degree polynomial in r₂. F{r₂} = r₂⁸ − (R๏₂² + q₂K₀ + K₀²) r₂⁶ + L₀ (q₂ + 2K₀)r₂³ − L₀² F{r₂} = 0 We will use Newton's method to get the answer. dF/dr₂ = 8 r₂⁷ − 6 (R๏₂² + q₂K₀ + K₀²) r₂⁵ + 3 L₀ (q₂ + 2K₀) r₂² As an initial guess at r₂, r₂(j=0) = 5.0 Then... Repeat over j r₂(j+1) = r₂(j) − F{r₂(j)} / dF/dr₂(j) Until | r₂(j+1) − r₂(j) | < 1.0e−15 Assign to r₂ the converged value of r₂ from the loop. r₂ = r₂(j+1) ρ₂ = K₀ − L₀/r₂³ n₁ = n₁° + ν₁/r₂³ n₃ = n₃° + ν₃/r₂³ Q₁ = n₁X๏₁ − X๏₂ + n₃X๏₃ + a₂ρ₂ Q₂ = n₁Y๏₁ − Y๏₂ + n₃Y๏₃ + b₂ρ₂ Q₃ = n₁Z๏₁ − Z๏₂ + n₃Z๏₃ + c₂ρ₂ Q₄ = a₁ − a₃c₁/c₃ Q₅ = b₁ − b₃a₁/a₃ Q₆ = b₃ − b₁a₃/a₁ Q₇ = a₃ − a₁c₃/c₁ ρ₁ = ½ { (Q₁−a₃Q₃/c₃) / (n₁Q₄) + (Q₂−b₃Q₁/a₃) / (n₁Q₅) } ρ₃ = ½ { (Q₂−b₁Q₁/a₁) / (n₃Q₆) + (Q₁−a₁Q₃/c₁) / (n₃Q₇) } r₁ = √(R๏₁² + q₁ρ₁ + ρ₁²) r₃ = √(R๏₃² + q₃ρ₃ + ρ₃²) |
March 27th, 2017 | #8 |
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First Approximation for the heliocentric and geocentric distances (example)
τ₁ = 0.12041469265 τ₂ = 0.24082938530 τ₃ = 0.12041469265 n₁° = 0.5 n₃° = 0.5 ν₁ = 0.003624924551498492 ν₃ = 0.003624924551498492 D = −0.00008086527122733167 d₁ = −0.007999178173536615 d₂ = −0.007285847860243794 d₃ = −0.006470768826225836 R๏₁ = 1.0101892175982878 R๏₂ = 1.0100062530777665 R๏₃ = 1.0096742204936617 q₁ = 1.9860511558039105 q₂ = 2.0071717322553524 q₃ = 1.983487457488723 K₀ = 0.6291249579754575 L₀ = 0.648640205396262 F{r₂} = r₂⁸ − 2.6786726757084853 r₂⁶ + 2.1180837685979137 r₂³ − 0.42073411605650496 r₂(j=0) = 5.0 r₂(j=1) = 4.3929125172136540 r₂(j=2) = 3.8645627337430860 r₂(j=3) = 3.4056413590103425 r₂(j=4) = 3.0081300591803277 r₂(j=5) = 2.6651781699814600 r₂(j=6) = 2.3710083783224194 r₂(j=7) = 2.1208545400998022 r₂(j=8) = 1.9109403021959235 r₂(j=9) = 1.7385176091531582 r₂(j=10) = 1.6019958403129406 r₂(j=11) = 1.5011460045069969 r₂(j=12) = 1.4368768296650747 r₂(j=13) = 1.4079301074280257 r₂(j=14) = 1.4021251234310480 r₂(j=15) = 1.4019087711519191 r₂(j=16) = 1.4019084781470565 r₂(j=17) = 1.4019084781465196 r₂(j=18) = 1.4019084781465194 r₂(j=19) = 1.4019084781465194 r₂ = 1.40190847814652 AU ρ₂ = 0.39370413259633 AU n₁ = 0.5013156488339009 n₃ = 0.5013156488339009 Q₁ = +0.22086196320358292 Q₂ = −0.28347167403935020 Q₃ = −0.17121443708688030 Q₄ = +0.07817844626231640 Q₅ = +0.10480064268380263 Q₆ = −0.09503365936122998 Q₇ = −0.08190658470375745 ρ₁ = 0.40112680805014 AU ρ₃ = 0.39332223978467 AU r₁ = 1.40642928447905 AU r₃ = 1.39796070946325 AU |
March 27th, 2017 | #9 |
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Correction for Planetary Aberration (equations)
Light doesn't travel instantaneously. We must correct the initial times for the amount of time it took light to travel from the planet to our eyes. This is done only once, after the first approximation, but before the 2nd approximation. It will not be done between any other successive approximations. For i = 1 to 3 tᵢ° = tᵢ − A ρᵢ Where A is the time (in days) required for light to travel 1 astronomical unit. τ₁ = k(t₃°−t₂°) τ₂ = k(t₃°−t₁°) τ₃ = k(t₂°−t₁°) Where k is the mean motion of Earth in radians per day. n₁° = τ₁/τ₂ n₃° = τ₃/τ₂ |
March 27th, 2017 | #10 |
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Correction for Planetary Aberration (example)
t₁° = JD 2458319.4976832850 t₂° = JD 2458326.4977261545 t₃° = JD 2458333.4977283604 τ₁ = 0.12041473059503525 τ₂ = 0.24083016069401889 τ₃ = 0.12041543009898362 n₁° = 0.4999985477235360 n₃° = 0.5000014522764639 |
March 27th, 2017 | #11 |
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Recursive Procedure for Successive Approximations (equations)
For i = 1 to 3 xᵢ = aᵢρᵢ − X๏ᵢ yᵢ = bᵢρᵢ − Y๏ᵢ zᵢ = cᵢρᵢ − Z๏ᵢ K₁ = √{ 2 (r₂r₃ + x₂x₃ + y₂y₃ + z₂z₃) } K₂ = √{ 2 (r₁r₃ + x₁x₃ + y₁y₃ + z₁z₃) } K₃ = √{ 2 (r₁r₂ + x₁x₂ + y₁y₂ + z₁z₂) } h₁ = τ₁² / { K₁² [K₁/3 + (r₂+r₃)/2] } h₂ = τ₂² / { K₂² [K₂/3 + (r₁+r₃)/2] } h₃ = τ₃² / { K₃² [K₃/3 + (r₁+r₂)/2] } For i = 1 to 3 Begin ζ₁ = (11/9) hᵢ ζ₂ = ζ₁ Repeat ζ₃ = ζ₂ ζ₂ = ζ₁ / ( 1 + ζ₂ ) Until |ζ₂ − ζ₃| / ζ₃ < 1e−15 Ψᵢ = 1 + (10/11) ζ₂ End n₁ = n₁° Ψ₂/Ψ₁ n₃ = n₃° Ψ₂/Ψ₃ ν₁ = n₁° r₂³ (Ψ₂/Ψ₁−1) ν₃ = n₃° r₂³ (Ψ₂/Ψ₃−1) K₀ = (d₂ − d₁n₁° − d₃n₃°) / D L₀ = (d₁ν₁ + d₃ν₃) / D ρ₂ = K₀ − L₀/r₂³ r₂ = √(R๏₂² + q₂ρ₂ + ρ₂²) F{r₂} = r₂⁸ − (R๏₂² + q₂K₀ + K₀²) r₂⁶ + L₀ (q₂ + 2K₀) r₂³ − L₀² F{r₂} = 0 dF/dr₂ = 8r₂⁷ − 6 (R๏₂² + q₂K₀ + K₀²) r₂⁵ + 3 L₀ (q₂ + 2K₀) r₂² An initial guess at r₂ is made. r₂(j=0) = 5.0 Repeat over j r₂(j+1) = r₂(j) − F{r₂(j)} / dF/dr₂(j) Until | r₂(j+1) − r₂(j) | < 1.0e−15 Assign to r₂ the converged value from the loop. r₂ = r₂(j+1) ρ₂ = K₀ − L₀/r₂³ Q₁ = n₁X๏₁ − X๏₂ + n₃X๏₃ + a₂p₂ Q₂ = n₁Y๏₁ − Y๏₂ + n₃Y๏₃ + b₂p₂ Q₃ = n₁Z๏₁ − Z๏₂ + n₃Z๏₃ + c₂p₂ Q₄ = a₁ − a₃c₁/c₃ Q₅ = b₁ − b₃a₁/a₃ Q₆ = b₃ − b₁a₃/a₁ Q₇ = a₃ − a₁c₃/c₁ ρ₁ = (1/2) { (Q₁−a₃Q₃/c₃) / (n₁Q₄) + (Q₂−b₃Q₁/a₃) / (n₁Q₅) } ρ₃ = (1/2) { (Q₂−b₁Q₁/a₁) / (n₃Q₆) + (Q₁−a₁Q₃/c₁) / (n₃Q₇) } r₁ = √(R๏₁² + q₁ρ₁ + ρ₁²) r₃ = √(R๏₃² + q₃ρ₃ + ρ₃²) Repeat this whole section (Recursive Procedure for Successive Approximations) until your values for r converge. |
March 27th, 2017 | #12 |
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2nd Approximation
x₁ = +0.6950907155748627 AU y₁ = −1.1038017036451089 AU z₁ = −0.5258460120529784 AU x₂ = +0.7821880734964239 AU y₂ = −1.0489946911163521 AU z₂ = −0.5031295427509228 AU x₃ = +0.8651798949992420 AU y₃ = −0.9886817318039844 AU z₃ = −0.4777722550432934 AU K₁ = 2.7978740465198397 K₂ = 2.7964458183802980 K₃ = 2.8063597760501007 h₁ = 0.0007940909165695694 h₂ = 0.0031771970467915564 h₃ = 0.0007869223244419950 Ψ₁ = 1.0008814685551377 Ψ₂ = 1.0035166156963282 Ψ₃ = 1.0008735187996662 n₁ = 0.5013149570937037 n₃ = 0.5013218511700457 ν₁ = 0.0036270200092790930 ν₃ = 0.0036380120924249812 K₀ = 0.6291524070013451 L₀ = 0.6498947413724790 [Run the loop to converge r₂ with these improved values for K₀ and L₀.] r₂ = 1.40104915810870 AU ρ₂ = 0.39284197048693 AU Q₁ = +0.22037984626244440 Q₂ = −0.28285445669627170 Q₃ = −0.17084103674551712 Q₄ = +0.07817844626231640 Q₅ = +0.10480064268380263 Q₆ = −0.09503365936122998 Q₇ = −0.08190658470375745 ρ₁ = 0.40023158610550 AU ρ₃ = 0.39248103668010 AU r₁ = 1.40554188199514 AU r₃ = 1.39712727023708 AU |
March 27th, 2017 | #13 |
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3rd Approximation
x₁ = +0.6945702526880453 AU y₁ = −1.1031778581616043 AU z₁ = −0.5254700366198498 AU x₂ = +0.7817097170831797 AU y₂ = −1.0483812623201196 AU z₂ = −0.5027577850922944 AU x₃ = +0.8647364156310531 AU y₃ = −0.9880702189177708 AU z₃ = −0.4774021189830026 AU K₁ = 2.7961791000789040 K₂ = 2.7947160071365147 K₃ = 2.8046107513968490 h₁ = 0.0007955352634619711 h₂ = 0.0031830908801306914 h₃ = 0.0007883948794899752 Ψ₁ = 1.0008830702760896 Ψ₂ = 1.0035231140784986 Ψ₃ = 1.0008751518307448 n₁ = 0.5013174011504072 n₃ = 0.5013242795711852 ν₁ = 0.0036270759743293342 ν₃ = 0.0036380048010743680 K₀ = 0.6291524070013451 L₀ = 0.6498996939776956 r₂ = 1.40104553690590 AU ρ₂ = 0.39283833730013 AU [Skipping the Q values to save space.] ρ₁ = 0.40021552948029 AU ρ₃ = 0.39247674784955 AU r₁ = 1.40552596570965 AU r₃ = 1.39712302101410 AU |
March 27th, 2017 | #14 |
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4th Approximation
[Hereafter showing only the resulting values for r and ρ] r₁ = 1.40551810179968 AU r₂ = 1.40103775126222 AU r₃ = 1.39711525882522 AU ρ₁ = 0.40020759623016 AU ρ₂ = 0.39283052589046 AU ρ₃ = 0.39246891330763 AU 5th Approximation r₁ = 1.40551793902222 AU r₂ = 1.40103770073617 AU r₃ = 1.39711520326079 AU ρ₁ = 0.40020743201740 AU ρ₂ = 0.39283047519720 AU ρ₃ = 0.39246885722527 AU 6th Approximation r₁ = 1.40551786734425 AU r₂ = 1.40103763002107 AU r₃ = 1.39711513274520 AU ρ₁ = 0.40020735970740 AU ρ₂ = 0.39283040424807 AU ρ₃ = 0.39246878605239 AU 7th Approximation r₁ = 1.40551786570544 AU r₂ = 1.40103762940236 AU r₃ = 1.39711513208127 AU ρ₁ = 0.40020735805414 AU ρ₂ = 0.39283040362731 AU ρ₃ = 0.39246878538226 AU 8th Approximation r₁ = 1.40551786505173 AU r₂ = 1.40103762875969 AU r₃ = 1.39711513144030 AU ρ₁ = 0.40020735739466 AU ρ₂ = 0.39283040298252 AU ρ₃ = 0.39246878473532 AU 9th Approximation r₁ = 1.40551786503536 AU r₂ = 1.40103762875261 AU r₃ = 1.39711513143281 AU ρ₁ = 0.40020735737816 AU ρ₂ = 0.39283040297541 AU ρ₃ = 0.39246878472776 AU 10th Approximation r₁ = 1.40551786502943 AU r₂ = 1.40103762874680 AU r₃ = 1.39711513142702 AU ρ₁ = 0.40020735737217 AU ρ₂ = 0.39283040296958 AU ρ₃ = 0.39246878472191 AU 11th Approximation r₁ = 1.40551786502925 AU r₂ = 1.40103762874670 AU r₃ = 1.39711513142691 AU ρ₁ = 0.40020735737199 AU ρ₂ = 0.39283040296948 AU ρ₃ = 0.39246878472181 AU 12th Approximation r₁ = 1.40551786502921 AU r₂ = 1.40103762874666 AU r₃ = 1.39711513142687 AU ρ₁ = 0.40020735737195 AU ρ₂ = 0.39283040296944 AU ρ₃ = 0.39246878472177 AU 13th Approximation r₁ = 1.40551786502920 AU r₂ = 1.40103762874665 AU r₃ = 1.39711513142687 AU ρ₁ = 0.40020735737195 AU ρ₂ = 0.39283040296944 AU ρ₃ = 0.39246878472177 AU The geocentric and heliocentric distances of Mars have converged. |
March 27th, 2017 | #15 |
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The positions of the object in heliocentric celestial coordinates, and the velocity at time t₂
For i=1 to 3 xᵢ = aᵢρᵢ − X๏ᵢ yᵢ = bᵢρᵢ − Y๏ᵢ zᵢ = cᵢρᵢ − Z๏ᵢ x₁ = +0.6945561666206916 AU y₁ = −1.1031609740960513 AU z₁ = −0.5254598610330625 AU x₂ = +0.7817032990370643 AU y₂ = −1.0483730320253148 AU z₂ = −0.5027527972678799 AU x₃ = +0.8647299564409844 AU y₃ = −0.9880613123524494 AU z₃ = −0.47739672802383704 AU The obliquity of the ecliptic at the times corrected for planetary aberration ţᵢ = (tᵢ°−2451545)/3652500 ţ₁ = 0.001854756381460590 ţ₂ = 0.001856672888748669 ţ₃ = 0.001858589384903587 εᵢ" = 84381.448" − 4680.93" ţ − 1.55" ţ² + 1999.25" ţ³ − 51.38" ţ⁴ − 249.67" ţ⁵ − 39.05" ţ⁶ + 7.12" ţ⁷ + 27.87" ţ⁸ + 5.79" ţ⁹ + 2.45" ţ¹⁰ [Source: J. Laskar via Wikipedia.] εᵢ = 4.84813681109536e-6 ε" ε₁ = 0.4090507128082722 radians ε₂ = 0.4090506693155986 radians ε₃ = 0.4090506258231779 radians The positions of the object in heliocentric ecliptic coordinates Xᵢ = xᵢ Yᵢ = zᵢ sin qᵢ + yᵢ cos qᵢ Zᵢ = zᵢ cos qᵢ − yᵢ sin qᵢ X₁ = +0.6945561666206916 AU Y₁ = −1.2211445112611437 AU Z₁ = −0.043339161761816236 AU X₂ = +0.7817032990370643 AU Y₂ = −1.1618451636648004 AU Z₂ = −0.04429678439054979 AU X₃ = +0.8647299564409844 AU Y₃ = −1.0964241450956385 AU Z₃ = −0.04502096119374227 AU Estimating the velocity at time 2 using Lagrange interpolation w₁ = (t₂°−t₃°) / { (t₁°−t₂°) (t₁°−t₃°) } w₂ = (2t₂°−t₁°−t₃°) / { (t₂°−t₁°) (t₂°−t₃°) } w₃ = (t₂°−t₁°) / { (t₃°−t₁°) (t₃°−t₂°) } w₁ = −0.07142792651844013 w₂ = −0.0000008298695931895751 w₃ = +0.07142875638803331 Here's a conversion factor that converts a velocity from AU/day to meters per second. λ = 1731456.8368056 Vx₂ = λ {w₁X₁ + w₂X₂ + w₃X₃} Vy₂ = λ {w₁Y₁ + w₂Y₂ + w₃Y₃} Vz₂ = λ {w₁Z₁ + w₂Z₂ + w₃Z₃} Vx₂ = +21046.2558357560 m/s Vy₂ = +15424.8069263065 m/s Vz₂ = −207.9965286294 m/s |
March 27th, 2017 | #16 |
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From the state vector to the orbital elements (equations)
We remove the subscripts and wingdings from the position and velocity of the object at time t₂. The distance of the object from the sun r = √( x² + y² + z² ) The speed of the object, relative to the sun V = √[ (Vx)² + (Vy)² + (Vz)² ] The semimajor axis of the object's orbit around the sun a = 1 / { 2/r − V²/(GM) } The angular momentum, h, per unit mass (m²/sec), in the orbit hx = y Vz − z Vy hy = z Vx − x Vz hz = x Vy − y Vx h = √[ (hx)² + (hy)² + (hz)² ] The eccentricity of the object's orbit around the sun e = √[ 1 − h² / (GMa) ] The inclination of the object's orbit, relative to the ecliptic i = arccos( hz / h ) The longitude of the ascending node, Ω, of the object's orbit Ω = Arctan ( hx , −hy ) The true anomaly, θ, of the object in the orbit at time t₂ cos θ = h²/(rGM) − 1 sin θ = h (x Vx + y Vy + z Vz) / (rGM) θ = Arctan ( sin θ , cos θ ) (Arctan is the two-dimensional arctangent function.) Argument of the perihelion, ω, of the orbit cos ω'' = (x cos Ω + y sin Ω) / r If sin i = 0 then sin ω'' = (y cos Ω − x sin Ω) / r If sin i ≠ 0 then sin ω'' = z / (r sin i) ω'' = Arctan( sin ω'' , cos ω'' ) ω' = ω'' − θ If ω' > 0 then ω = ω' If ω' < 0 then ω = ω' + 2π The eccentric anomaly, u, of the object in the orbit at time t₂ sin u = (r/a) sin θ / √(1−e²) cos u = (r/a) cos θ + e u = Arctan( sin u , cos u ) The mean anomaly, m, of the object in the orbit at time t₂ m = u − e sin u (Note: u must be entered in radians, and m will return in radians.) The period of the orbit, P, days P = (π / 43200) √[ a³/(GM) ] The mean motion, μ, in the orbit, radians per day μ = 2π / P The time of perihelion passage, T, of the object in the orbit, Julian Date T = t − m/μ |
March 27th, 2017 | #17 |
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From the state vector to the orbital elements (example)
t = JD 2458326.4977261545 x = +1.16941149055e+11 meters y = −1.73809562567e+11 meters z = −6.626704624e+9 meters Vx = +21046.2558357560 m/s Vy = +15424.8069263065 m/s Vz = −207.9965286294 m/s r = 2.09592246031e+11 meters = 1.4010376287467 AU V = 26094.3061983615 m/s The estimated semimajor axis of Mars' orbit a = 1.51523 AU The estimated angular momentum per unit mass for Mars in its orbit hx = +1.3836742502626161e+14 m²/sec hy = −1.1514396779369656e+14 m²/sec hz = +5.4618351660801300e+15 m²/sec The estimated eccentricity of Mars' orbit e = 0.085238 The estimated inclination of Mars' orbit to the ecliptic i = 1.88766° The estimated longitude of the ascending node of Mars' orbit Ω = 50.23409° The estimated true anomaly of Mars in its orbit (at time t₂) θ = 329.77123° The estimated argument of the perihelion for Mars' orbit ω = 283.93619° The estimated eccentric anomaly of Mars in its orbit (at time t₂) u = 332.14649° The estimated mean anomaly of Mars in its orbit (at time t₂) m = 334.42826° The estimated time of perihelion passage for Mars T = JD 2458374.890 |
March 27th, 2017 | #18 |
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The official NASA values of the Keplerian elements for Mars' orbit are
a = 1.52365 AU e = 0.093327 i = 1.84815° Ω = 49.50637° ω = 286.65831° T = JD 2458378.005 Using the Keplerian elements that I've calculated (shown in the previous post), an observer on Earth would find Mars, at 0h UT on Christmas 2018, at Right ascension: 23h 40m 54s Declination −2° 38' 2" The position for this same object at the same time, from JPL's Horizons interface is Right ascension: 23h 42m 19.67s Declination: −2° 25' 10.2" Last edited by Jerry Abbott; March 27th, 2017 at 10:28 AM. |
March 27th, 2017 | #19 |
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The notation used in this demonstration of the method of Gauss closely follows that used by A.D. Dubyago in chapter five of his book The Determination of Orbits, until the post headed "From the state vector to the orbital elements (equations)." From that point forward, the procedure is the one I learned as an undergraduate.
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March 27th, 2017 | #20 |
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Physics Forums moderator fresh_42 wrote:
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