Astrodynamics 103 Part 1: Gauss/Laplace+ Algorithm unknown orbit Observer 1b Observer 1a r 3 object at t 3 x = r v object at t r 1 Angles-only Problem Given: 1a. Ground Observer coordinates: ( latitude, longitude, altitude ) at t i or 1b. Space Observer position vectors: ( x, y, z ) with respect to Sun/Earth at t. Angles data of unknown object: ( Right ascension, Declination ) at t object i at t 1 i = 1,, 3 i Sphere/ Spheroid AMOS September 14, 01 Find: x is the range at t giving r, v, and ( TLE if desired ) of the unknown orbit Gim Der DerAstrodynamics July 1, 013
Analytic Astrodynamics Overview Astrodynamics 101: Kepler+ Algorithm Part1: Analytic Prediction Algorithms Part: Verifications Astrodynamics 10: Lambert+ Algorithm Part1: Analytic Multi-revolution Targeting Algorithms (Orbit Determination for Radar Data) Part: Verifications Astrodynamics 103: Gauss/Laplace+ Algorithm Part1: Analytic Angles-only Algorithms (Orbit Determination for Optical Sensor Data) Part: Verifications
Astrodynamics 103: Gauss/Laplace+ Algorithm Part 1. Analytic Angles-only Algorithms 1. What, Why, How. Physics 3. New Angles-only Algorithms 4. Details and Hints 5. Applications for SSA Part. Verifications 6. Numerical Examples
How Close Will It Get? Feb 15, 013 Asteroid Earth DA14 GEO Moon Planet Asteroid and Space Debris Collisions Happened and Will Happen Again
Range Guessing Current Angles-only Algorithms can process 10% or less of current Space Catalog ~,500 GEO and deep space objects GEO and Deep-Space Objects, if guessed correctly
Using optical sensors for 10% of cataloged space objects is not good enough! Range Solving 90% of the space objects is Near-Earth Gauss/Laplace+ can process optical sensor data for 100% of cataloged and unknown space objects in any Orbit Regime
Bigger Means Better? The European/World s Biggest Optical Telescope 39.3-metre dish, to be built in Chile by 0 Size Matters But Powerful Angles-only Algorithm Matters More
1. What, Why, How What? Why? How? Cataloging of 100,000+ space debris and satellites is a challenging Space Situation Awareness (SSA) problem Need new Angles-only algorithm to process and catalog efficiently millions optical sensor measurements of known and unknown objects Understand the Physics and Mathematics of Astrodynamics for orbit determination using angles-only data Building New Space Catalog of 100,000+ Objects Requires Analytic Gauss/Laplace+ Algorithm
Range Guessing (out of thin air) x Gauss or Laplace or Double-r -Body "known" orbit Observer 1b Observer 1a Improve with f and g r 3 Angles-only and Gauss/Laplace+ Algorithms x Angles-only Problem Range Solving Gauss/Laplace Given: 1a. Ground Observer coordinates: th 8 deg. Polynomial Equations ( latitude, longitude, altitude ) at t i or 1b. Space Observer position vectors: x ( x, y, z ) with respect to Sun/Earth at t Find: object at t 3 = r v. Angles data of unknown object: ( Right ascension, Declination ) at t x is the range at t, giving r, v and ( TLE if desired ) of the unknown orbit object at t r 1 object at t 1 unknown orbit Observer 1b Observer 1a i Gauss and Laplace and Double-r add perturbations analytically ( J, J, J, object at t 3 4 3 J, J 31, Sun, Drag,.. ) i, i = 1,, 3 r 3 x = r v object at t r 1 object at t 1 Sphere Angles-only (-Body) solution (inaccurate but fast) Spheroid Gauss/Laplace+ solution for SSA (accurate and fast)
. Physics Physics and Primary Objective
Ranging with Old Optical Telescopes 18 th Century Optical Telescopes: Poor angles measurement accuracy th no use for the Gauss/Laplace 8 degree polynomial range equations FOV Resolution: Poor ~ 1.0 degree
1.0 degree Why Optical Sensors and Telescopes? Field-of-View Comparison (typical) Radar / Laser (very small FOV) 1.5 degree Optical Sensor (bigger FOV) Optical Sensor provides much larger FOV and angles data but no range. Until now Angles-only algs cannot consistently find range Optical Sensor is at least 10x cheaper to build and operate than comparable Radar or Laser systems
Angles Data: Right ascension and Declination North Celestial Pole Input angles ( 3 sets ): a, d i, i = 1,, 3 @ t i i Celestial Equator Object Declination d d Vernal equinox a Right ascension a The problem: Angles-only, no range! Optical Detection (D obs data)
The Classical Angles-only Problems Finding range of any space object by optical sensor A Non-Moving Object A Moving Object (a star) (a satellite or asteroid) B p x A star unknown orbit Observer r 3 object at t 3 v x = object at t r object at t 1 Earth (Jan) A 1 AU Sun B Earth (July) Earth r 1 Parallax: x = 1 AU p 00-year Angles-only riddle: Pick the correct range, x
Ranging with Modern Optical Sensors GEODSS Space Surveillance Telescope SST Accurate FOV Resolution: angles arc-seconds measurements sub- arc-second Very accurate angles measurements Gauss/Laplace+ Algorithm (Range Solving IOD) Cataloging to Near-Real-Time SSA
Primary Objective of Angles-only Algorithms Range, Range, Range 1. Given optical detection (D obs angles data). Innovative Range Solving Gauss/Laplace 8 th degree polynomial equations for any known or unknown object in any orbit regime 3. New analytic Angles-only Algorithm for fast and accurate IOD, TLE, Cataloging,.., SSA
3. New Angles-only Algorithms Angles-only Algorithms for Initial Orbit Determination
Incomplete Textbook Solutions Moulton, 1904 since Gauss 1801,... and in spite of hundreds of papers on orbit determination, very little is new... Prussing and Conway, 1993 The most significant of which is probably the solution of the eight-degree polynomial in the unknown range ( x )... 8 6 3 f(x) = x a x b x c = 0 Vallado, 001 -- 01 multiple roots may exist... Selecting the correct root can be very difficult and many others Textbook Solutions: Multiple Roots Correct Range not found
Classical Gauss/Laplace Angles-only Problem unknown orbit Observer r 3 object at t 3 v x = object at t r r 1 object at t 1 Gauss / Laplace Angles-only Problem Input: 1. Observer coordinates: ( latitude, longitude, altitude ) at t. Angles data of unknown object: ( Right ascension, declination ) at t i = 1,, 3 Accurate angles are required to give accurate coefficients: a, b, c i i Sphere/ Spheroid th Gauss/Laplace 8 degree polynomial equations: 8 6 3 f(x) = x a x b x c = 0 Output: x is the range at t, which gives ( r, v ) and TLE set of the unknown orbit
Guidelines for Angle/Time Spacing Normal object time spacing within an arc of the orbit Unknown Orbit object at t 3 object at t r 3 x = r object at t 1 Unknown Orbit 1 revolution object at t 3 Improper Obs data spacing: 1. Closely spaced data (Time span too close) Observer r 3 r r 1 object too close at t and t 1 Observer r 1 Given: * 3 sets of angle t a i i d i i = 1,, 3,.. * 1, and/or 3 sets of Observer coordinates Unknown Orbit revolutions r Observer object in first rev at t and t 1 r 1 Proper Obs data spacing: (1) Geocentric: minutes apart (max: 100 minutes) () Heliocentric: days or months apart (max: to be determined) object in second rev at t 3 r 3. Multi-revolutions data (Time span too far apart)
Guidelines for Total Time Span unknown orbit Observer r 3 object at t 3 x = r v t r 1 total time span object at t 1 t 3 t (Reasonable total angle spacing) 1 Sphere/ Spheroid Total Time Spacing for any Geocentric object in any orbit regime: 0 to 40 minutes
Step 1 of New Algorithm: Range Solving unknown orbit Observer r 3 object at t 3 x = r v t r 1 object at t 1. Proper spacing total time span t 3 t 1 1. Advanced optical sensors accurate angles Accurate coefficients: a, b, c Gauss/Laplace 8 th degree polynomial eqns: Sphere/ Spheroid Picking the correct root (range), x Unknown -body Keplerian orbit in terms of ( r, ) v
Step of New Algorithm: Adding Perturbations Kepler + solution (Vinti and other perturbations) (3) Vinti spheroidal solution only Numerical accurate solution (desired) V (t ) () Kepler solution with (other perturbations) a d General Equations of motion: d R d t = R 3 R a d, a d = 0 R (t) (1) Kepler solution only General method of solution => Kepler+ solution (1) Kepler solution only () Kepler + a d perturbed solution (3) Vinti spheroidal solution only Central Body R (t ) 0 V (t ) 0 object at t 0
New Angles-only IOD Algorithm unknown orbit Observer r 3 object at t 3 v x = r Sphere/ Spheroid object at t r 1 Gauss/Laplace 8 8 6 3 f(x) = x a x b x c = 0 object at t 1 th Range solving ( The 00-year Riddle to find one correct root ) deg poly eqn: Range guessing ( Multiple or no solution ) 1. The Correct Root (-body or Keplerian) New angles-only algorithms find the correct root consistently without guesswork in a few micro-seconds on a computer. Fast and Accurate ( Kepler + ) Add analytically perturbations ( J, J, J, Sun/Moon, Drag,.. ) 3 4 resulting in accurate solution for any object in any orbit regime
Perturbations Compliant Gauss/Laplace+ Classical Kepler (-Body) Astrodynamics 101 Vinti (J, J3, J4 included) Kepler+ (J, J3, J4 and other perturbations)) Classical Lambert (-Body) Astrodynamics 10 Targeting by Kepler+ Lambert+ (J, J3, J4 and other perturbations)) Classical Gauss/Laplace (-Body) Astrodynamics 103 + Targeting by Kepler+ and Lambert+ Gauss/Laplace+ (J, J3, J4 and other perturbations))
4. Details and Hints Insight To New Angles-only Algorithm
Myths of Angles-only Algorithms 1. The straight edge of the Ocean is difficult to see (illusion vs knowledge) 1. Myth. The range of a very faraway Star is difficult to observe (perception vs knowledge) 3. The correct roots of Gauss and Laplace equations are difficult to select (guessing vs knowledge) 4. Analytic Kepler+ solutions with perturbations are difficult to believe (ignorance vs knowledge). Myth 3. Myth 4. Myth
The Correct Root of the 8 th Degree Poly. Eqn. Gauss/Laplace New Angles-only: It is all about picking the range, x 8 6 3 f(x) = x a x b x c = 0, x is the range at t Simple facts: (1) For real trajectory, x must be a positive real root of f(x) () Only three terms can change signs, Descartes' Rule of Signs gives at most three real roots [1, ] (3) c is always negative, only values of a and b can change Unknown Orbit Observer r 3 object at t 3 x = r t a i i d i i = 1,, 3,.. r 1 object at t object at t 1 Reducing 8 roots to at most 3 positive real roots References: 1. Charlier, C. V. L., On Multiple Solutions in the Determ... From three obs., RAS, 1900&1911. Plummer, H. C., An Introductory Treatise on Dynamical Astronomy, Dover, 1918
Descartes Rule of Signs Down to either 1 or 3 positive real root(s) The second theorem of Descartes Rule of Signs implies that positive real roots cannot exist If a root solver is used, then you will not know how to pick the correct root when there are 3 roots
Determining the positive real root(s) Angles-only 8th degree polynomial (range) equation of Gauss/Laplace 8 6 3 f(x) = x a x b x c = 0, x is the range at t Unknown Orbit object at t 3 f(x) f(x) not in closed-form, iterative method required x = i r [ 1 + ( i 1 ) /10 ] Earth i = 1,,... object at t object at t 1 x 1 one solution: f(x) = 0 x Observer three solutions: f(x) = 0 Need to pick! The 00-year angles-only IOD problem is solved: The correct root of f(x) is found without guesswork
Gauss Geometric Method 8 6 3 Transform: f(x) = x a x b x c = 0 r = R + object at t 4 to: f( ) = sin M sin ( m = 0 M and m are computed from a, b and c Gauss: Determine the correct root r > R > r Earth and < (180 ) < 90 o o Observer R L Earth r Unknown Orbit r sin r = ECI position vector to object R = ECI position vector to observer = range vector from observer L = Line of sight vector from observer = = sin ( + R sin Hint: Laplace agreed? The correct positive real root can be identified
Double-r Algorithms (Der vs Gooding ) Der range solving th Gauss 8 - degree polynomial equation Unknown Orbit Observer r 3 object at t 3 x = r r 1 object at t object at t 1 Gooding range guessing 1. r and r estimates from range solving instead of range guessing 1 3. r, r and r are optimized instead of assumed with no errors 1 3. 3. r by Gibbs gives direct or retrograde instead of direct and retrograde 4. Analytic Kepler + perturbed solution instead of -Body solutions Double _ r (Der) Lambert + and Kepler + _ Double r (Gooding) Lambert and -Body 5. One (near SP) solution versus Multiple (-body) solutions
Angles-only Algorithms (Der vs Others ) 1. Start: Range solving (No initial guess needed and no Newton iterative) 1. Start: Range guessing (Laplace/other methods to initiate and Newton iterative). Separated root resolution o 0 < < 90 (Ground, up) o o 90 < < 180 (Space, down) 3. Input data accuracy & req. (5+ significant digits for error-free, sparse data only) 4. Output Real orbit & motion rd (include: J, J, J, 3 body,.. ) 3 4 5. Tests and comparison (00+ objects, all orbit regimes, 1000+ test cases, simulated & real data, Accuracy: ASTAT-, 100%, ASTAT-1, 90% + ) 6. One Accurate solution. Confused roots o 0 < < 180 (up or down?) 3. Input data (No accuracy definition, sparse and/or dense data) 4. Output -body orbit & motion (Perturbations: none) 5. Tests and comparison ( lucky to find one, what accuracy? ) 6. Multiple guessed solutions
Unsuitable for real-time automatic processing Orbit Determination and Prediction/Propagation Osculating Orbital Elements at t o [ r (t ), v (t )] o Radars / Optical sensors o Future look angles ( pointing prediction ) Others...... Object at t r Ephemerides Raw Observation data v Orbit Determination (Estimated future/past) position and velocity vectors Rise/Set Site visibility Astrodynamics 10: Lambert / 103: Gauss/Laplace Initial Orbit Determination Processed Observation data Differential Correction Batch / KF TLE conversion difficulties Singularity difficulties OscMean SGP4 Close approach (miss distance) Analytic algorithms in OD and P/P need to process 100,000+ Objects in less than 1 hours with accuracy of 10 km to centimeters for SSA (Estimated initial) positin and velocity vectors Orbital element set r and v Prediction / Propagation 1 SP 3 Numerical Integration Osculating Orbital Elements at t [ r, v ] Astro 101: Kepler Kepler+ 1 3 SGP4 needs TLE conversion (not efficient for SSA) SP is accurate, but slow for SSA Kepler+ is accurate and fast for SSA
SSA Applications 5. Applications
Radar Data Multi-sensor Multi-object UCT Cataloging SF SF I Correlating 90+ % of objects to catalog J K Fence or Radar UCT processing Solve by New angles-only algorithm: Example: I = J = K = 100 Correlation combination = I J K = 1,000,000 Takes a few seconds for a million combinations 3 computed? ranges = 3 detected ranges
Optical Sensor Data Multi-sensor Multi-object UCT Cataloging SF SF Correlating 90+ % of objects to catalog UCT processing Example: I = J = K = 100 Correlation combination = I J K = 1,000,000 Takes a few seconds for a million combinations A computed orbit = Solve by New angles-only algorithm:? Orbit type,,,...
Gauss/Laplace+ for SSA New Angles-only IOD Algorithms provide Future Sensors new capabilities to Track, Search and UCT cataloging for any unknown object in any orbit regime New Analytic and Semi-analytic Astrodynamics Algorithms are Key Tools for SSA with Collision Prediction for All Objects in the Space Catalog 10% 90% or more are Near-Earth Optical Sensor Track and Search UCT Cataloging: 50,000+ objects Near Real-Time SSA 100,000+ objects
Next Astrodynamics 103 Part : Verifications for SSA (Also please download iorbit: http://derastrodynamics.com/index.php?main_page=index&cpath=1_7 and run gau for Astrodynamics 103 Verifications)
6. Numerical Examples w and w/o Input Errors Geocentric Objects More examples in iorbit application software Der, G. J., New Angles-only Algorithms for IOD, AMOS, 01 Readers should solve the example problems without errors first More examples can be provided upon request Examples of multiple sensors can be provided upon request
Notes on the Examples ANSWER in Numerical examples = Reference Data Simulated data from scenario generators GEODSS observation data (Geocentric objects) JPL DE405 and IAU data (Heliocentric objects) Input data Angles data > 5 significant digits => accurate Total time span between t1 and t3 reasonable Range Solving Objective Compute position ( r ) and velocity ( v ) at t Compare ( r, v ) of Answer against ( r, v ) of Gauss, Laplace, Double-r and Gauss/Laplace+
Example 1: LEO Object Input: (no errors in Right asc & Declination) One site, # 1: (lat, lon, alt) = ( 3.5781 deg, 115.089 deg, 0.0 km) a Time (s) year mo day hr min sec Right asc, deg d Declination, deg t1 = 0 006 1 16 3 46 6.368 33.93588 0.887495 t = 150 006 1 16 3 48 36.368 4.11939 31.98153 t3 = 450 006 1 16 3 53 36.368 76.140070 61.34941 Output: DerAstro Algs r_eci (t) (km) v_eci (t) (km/s) Gauss (-Body) 401 850 46 75 71 654 37867 5 053643 5 068175 Laplace (-Body) 4665 930 4669 110 39 061 0 433840 6 085380 6 996756 Double-r (-Body) 4018 495 463 739 734 16 361143 5 0600 4 991948 Gauss/Laplace+ 4045 368 466 01 706 07 414199 5 15750 5 180705 ANSWER 4048 910 464 673 705 159 44601 5 16619 5 186867 LEO object ( x = eci range(t) = 6715. km ) Coefs of Gauss 8 th deg. Polynomial Eqn: a = 0.997879, b = 0.07330, c = 0.059373 Estimated: semi-major = 6704 km, ecc = 0.001, incl = 51.70 deg, approx. alt = 30 (km)
Example : Sun Synchronous Debris Input: (no errors in Right asc & Declination) One site, # 1: (lat, lon, alt) = ( 3.5781 deg, 115.089 deg, 0.0 km) a Time (s) year mo day hr min sec Right asc, deg d Declination, deg t1 = 0 010 1 9 0 39 13.788 38.95835 80.845661 t = 300 010 1 9 0 44 13.788 45.533066 67.961395 t3 = 600 010 1 9 0 49 13.788 40.3895 50.177345 Output: DerAstro Algs r_eci (t) (km) v_eci (t) (km/s) Gauss (-Body) 1331 165 6831 47 1810 509 0 78133 1 931007 6 981687 Laplace (-Body) 1311 584 6874 460 197 301 1 040096 1 963111 7 80136 Double-r (-Body) 1339 591 6835 03 1791 308 0 768793 1 909651 6 984801 Gauss/Laplace+ 1314 998 6866 83 1906 997 0 800165 1 970689 7 104399 ANSWER 1315 18 6866 668 1906 156 0 800168 1 970613 7 1063 MEO Sun-synchronous object (x = eci range(t) = 747. km) Coefs of Gauss 8 th deg. Polynomial Eqn: a = 1 000630, b = 0 74894, c = 1 61135 Estimated: semi-major = 746 km, ecc = 0.0198, incl = 98.66 deg, approx. alt = 900 (km)
Example 3: MEO Iridium Satellite Input: (no errors in Right asc & Declination) One site, # 1: (lat, lon, alt) = ( 0.70801448 deg, 156.576333 deg, 3.055454785 km) a Time (s) year mo day hr min sec Right asc, deg d Declination, deg t1 = 0 013 4 10 14 0 9.491 11.61550 70.141643 t = 300 013 4 10 14 5 9.491 147.515117 69.97690 t3 = 750 013 4 10 14 3 39.491 179.146654 5.4900 Output: DerAstro Algs r_eci (t) (km) v_eci (t) (km/s) Gauss (-Body) 317 461 4566 805 4357 137 153778 4 0739 5 665166 Laplace (-Body) 3134 604 4590 909 434 313 085790 3 980679 6 1744 Double-r (-Body) 3160 761 4597 369 436 538 096736 4 09331 5 714888 Gauss/Laplace+ 31 507 4535 891 4514 678 19473 4 16339 5 768801 ANSWER 30 607 4536 150 4513 34 16535 4 163684 5 77450 MEO (Iridium SID 479) object (x = eci range(t) = 7164. km) Coefs of Gauss 8 th deg. Polynomial Eqn: a = 1.005801, b = 1.367533, c =.66005 Estimated: semi-major = 715 km, ecc = 0.0017, incl = 86.41 deg, approx. alt = 900 (km)
Example 4: MEO Global Star Satellite Input: (no errors in Right asc & Declination) One site, # 1: (lat, lon, alt) = ( 0.70801448 deg, 156.576333 deg, 3.055454785 km) a Time (s) year mo day hr min sec Right asc, deg d Declination, deg t1 = 0 013 4 9 6 39 4.883 45.01667 39 001651 t = 450 013 4 9 6 47 1.883 96.560740 11.81938 t3 = 1050 013 4 9 6 57 1.883 170.716391 54.570871 Output: DerAstro Algs r_eci (t) (km) v_eci (t) (km/s) Gauss (-Body) 4990 896 5381 70 6 645 4 417670 1 718065 4 800083 Laplace (-Body) 5101 70 6341 415 84 685 3 565660 0 771036 3 64098 Double-r (-Body) 508 904 5391 441 673 756 4 301948 1 655669 4 59995 Gauss/Laplace+ 5003 653 5495 93 647 968 4 650900 1 787398 5 064394 ANSWER 5004 109 5496 614 646 833 4 65471 1 788077 5 06988 MEO (GlobalStar SID 516) object (x = eci range(t) = 7890. km ) Coefs of Gauss 8 th deg. Polynomial Eqn: a = 0. 99813, b = 0. 74736, c = 0. 85399 Estimated: semi-major = 7883 km, ecc = 0.0011, incl = 5.01 deg, approx. alt = 1,400 (km)
Example 5: GPS Satellite Input: (Data with unknown errors in Right asc & Declination) One site, # 1: (lat, lon, alt) = (38.88713 deg, 104.0159 deg, 1.941865 km) a Time (s) year mo day hr min sec Right asc, deg d Declination, deg t1 = 0 01 6 0 6 1 35.946 31.3046 5.5899 t = 656.58 01 6 0 6 3 3.56 3.9158 53.5985 t3 = 875.66 01 6 0 6 36 11.605 36.505 53.7139 Output: DerAstro Algs r_eci (t) (km) v_eci (t) (km/s) Gauss (-Body) 953 608 160 649 133 686 577690 830949 0 538971 Laplace (-Body) 9358 773 1478 057 0953 44 54748 79366 0 50415 Double-r (-Body) 953 599 160 65 133 688 577709 830948 0 538964 Gauss/Laplace+ 953 600 160 653 133 688 577710 830949 0 538964 ANSWER 956 84 1618 99 189 446 58954 84484 0 536548 GPS (SID 3575) object (x = eci range(t) = 6465. km) Coefs of Gauss 8 th deg. Polynomial Eqn: a = 4.00897, b = 594.6693, c = 4955.18 Estimated: semi-major = 669. km, ecc = 0.0081, incl = 54.48 deg, approx. alt = 1,000 (km)
Example 6: Molniya Satellite Input: (Data with unknown errors in Right asc & Declination) One site, # 1: (lat, lon, alt) = (38.88713 deg, 104.0159 deg, 1.941865 km) a Time (s) year mo day hr min sec Right asc, deg d Declination, deg t1 = 0 01 6 0 4 59 54.406 44.0775 61.003 t = 35.38 01 6 0 5 5 19.786 46.8508 59.8794 t3 = 590.35 01 6 0 5 9 44.751 49.19 58.8410 Output: DerAstro Algs r_eci (t) (km) v_eci (t) (km/s) Gauss (-Body) 6816 554 14641 890 38 649 1 89018 0 74700 3 05171 Laplace (-Body) 6819 468 14648 707 395 49 1 889139 0 758671 3 038718 Double-r (-Body) 6816 545 14641 890 38 651 1 89016 0 74705 3 05175 Gauss/Laplace+ 6816 545 14641 890 38 651 1 89016 0 74705 3 05175 ANSWER 6781 67 14571 366 3158 58 1 87707 0 746038 3 098 Molniya (SID 09880) object (x = eci range(t) = 8336. km) Coefs of Gauss 8 th deg. Polynomial Eqn: a = 8.8517, b = 381.311, c = 418784.95 Estimated: semi-major = 7139 km, ecc = 0.7033, incl = 63.7 deg, approx. alt = 8,000 (km)
Example 7: GEO Satellite Input: (Data with unknown errors in Right asc & Declination) One site, # 1: (lat, lon, alt) = (38.88713 deg, 104.0159 deg, 1.941865 km) a Time (s) year mo day hr min sec Right asc, deg d Declination, deg t1 = 0 01 6 0 3 53 41.09 33.4567 6.10440 t = 617.67 01 6 0 4 3 58.766 36.0375 6.10480 t3 = 960.39 01 6 0 4 9 41.478 37.469 6.10500 Output: DerAstro Algs r_eci (t) (km) v_eci (t) (km/s) Gauss (-Body) 4315 895 3448 988 5 000 510807 1 778381 0 001087 Laplace (-Body) 430 13 34489 398 5 87 516568 1 769798 0 000018 Double-r (-Body) 4315 895 3448 988 4 999 510806 1 778380 0 001089 Gauss/Laplace+ 4315 895 3448 988 4 999 510806 1 778380 0 001089 ANSWER 49 801 3446 711 0 875 51311 1 771484 0 00004 GEO (SID 3018) object (x = eci range(t) = 4194. km) Coefs of Gauss 8 th deg. Polynomial Eqn: a = 0.519904, b = 303.19713, c = 371464. Estimated: semi-major = 483 km, ecc = 0.0030, incl = 0.014 deg, approx. alt = 35,000 (km)
Examples with and without Angles Errors Heliocentric Objects More examples in DerAstrodynamics iorbit application software iorbit will allow you to build your own examples verifiable with JPL or IAU data Der, G. J., New Angles-only Algorithms for IOD, AMOS, 01 Visit JPL and IAU website, planets and minor planets angles data can be easily obtained for verification with the computed values of these examples. Need help? See the document, IAU Minor planet ephemeris data extraction procedures, which is part of the iorbit package. km to AU conversion, use DE405 value: 1 AU = 149,597,870.691 km
Example 8: Saturn Input: (no angular errors in Right asc & Declination) a Time (day) Year mo day hr min sec Right asc, deg d Declination, deg t1 = 0 01 1 5 0 0 0.0 07.536715011 8 6177163740531 t = 10 01 1 15 0 0 0.0 07.7067334087 8 7460613139881 t3 = 0 01 1 5 0 0 0.0 08.0504067551 8 81905633807838 Time (day) r_earth_sci (t), (km) source JPL DE405 t1 = 0 3555503. 13107107.6 5680595 7 t = 10 59975833.5 137965.0 5344454.0 t3 = 0 8851904.0 111707451.0 4846993.7 Output: sci= Sun Centered Inertial DerAstro Algs. r_sci (t) (km) v_sci (t) (km/s) Gauss (-Body) 1333576603 545451505 167863101 3 50175 8 179054 3 51836 Laplace (-Body) 13334306 545373845 16783750 3 31043 8 145966 3 503335 Gauss/Laplace+ 1333576689 545451569 16786315 3 5001 8 179185 3 51898 ANSWER 1333553680 545439469 167859117 3 51758 8 177989 3 517889 Helio object, Saturn ( x = sci range(t) = 1,450,559,81. km ) Coef of Gauss 8 th deg. Polynomial Eqn: a = 94.131, b = 176.090, c = 83.408 Estimated: semimajor = 14485411. km, ecc = 0.0559, incl =.48764 deg wrt Ecliptic
Example 9: Jupiter Input: (no angular errors in Right asc & Declination) a Time (day) Year mo day hr min sec Right asc, deg d Declination, deg t1 = 0 1990 1 13 0 0 0.0 94.13563043390 3 97933744955 t = 160 1990 6 0 0 0.0 108.93508668434 4958016358460 t3 = 360 1991 1 8 0 0 0.0 134.07413976346 18 046997655977 Time (day) r_earth_sci (t), (km) source JPL DE405 t1 = 0 567158.1 145561.9 54004949 4 t = 160 149033.7 13948439.0 60477803.1 t3 = 360 4378190.0 18851037.9 55867435. Output: sci= Sun Centered Inertial DerAstro Algs. r_sci (t) (km) v_sci (t) (km/s) Gauss (-Body) 75300870 666640663 9459456 1 381444 3 79534 1 33937 Laplace (-Body) 1416950 48845987 14433598 1 15515 8 790566 3 478197 Gauss/Laplace+ 75856958 66817786 9313335 1 406058 3 80191 1 37508 ANSWER 75779769 66803663 93070634 1 40066 3 803151 1 38044 Helio object, Jupiter (x = sci range(t) = 780,053,495. km) Coefs of Gauss 8 th deg. Polynomial Eqn: a =.869, b = 640.31, c = 4616.1931 Estimated: semi-major = 78004158. km, ecc = 0.048417, incl = 1.304659 deg wrt Ecliptic
Input: (no angular errors in Right asc & Declination) a Time (day) Year mo day hr min sec Right asc, deg d Declination, deg t1 = 0 007 9 1 11 4 0.0 53.8601490 9 3379799 t = 55.89 007 11 16 8 1 0.0 45.6454769 7 994311 t3 = 97.39 007 1 7 0 8 0.0 38.9896167 9 9789989 Time (day) r_earth_sci (t), (km) source JPL DE405 t1 = 0 15011957 6 483669 096694 6 t = 55.89 88037190 4 10909350 4 4795458 1 t3 = 97.39 14334465 0 134330178 4 5836334 3 Output: Example 10: Ceres, Asteroid sci= Sun Centered Inertial DerAstro Algs. r_sci (t) (km) v_sci (t) (km/s) Gauss (-Body) 79537585 30370533 85506536 13 653531 8 74334 6 90335 Laplace (-Body) 31363694 33678588 947111 15 70406 6 387 6 461303 Gauss/Laplace+ 79350995 30191603 85469860 13 661384 8 75040 6 90365 ANSWER 79388198 3019760 8547678 13 653910 8 755670 6 904600 Helio object, Ceres (x = sci range(t) = 40,571,834. km) Coefs of Gauss 8 th deg. Polynomial Eqn: a = 8.3765, b = 7.49807, c = 1.7151 Estimated: semi-major = 41405868. km, ecc = 0.080135, incl = 10.5959 deg wrt Ecliptic
a Time (day) Year mo day hr min sec Right asc, deg d Declination, deg t1 = 0 1990 1 13 0 0 0.0 94.14 3 9 t = 160 1990 6 0 0 0.0 108.90 50 t3 = 360 1991 1 8 0 0 0.0 134.10 18 05 Time (day) r_earth_sci (t), (km) source JPL DE405 t1 = 0 567158.1 145561.9 54004949 4 t = 160 149033.7 13948439.0 60477803.1 t3 = 360 4378190.0 18851037.9 55867435. Output: Example 11: Jupiter (with large errors) Input: (Data with approximately > 0.% errors in Right asc and Declination) sci= Sun Centered Inertial DerAstro Algs. r_sci (t) (km) v_sci (t) (km/s) Gauss (-Body) 43401035 575080050 5371618 10 5476 19859 0 67353 Laplace (-Body) 190147965 419538943 1847869 10 7170 8 101543 3 05068 Gauss/Laplace+ 44008906 57754039 53319468 10 5750 058150 0 61170 ANSWER 75779769 66803663 93070634 1 40066 3 803151 1 38044 Helio object, Jupiter (x = sci range(t) = 675,968,435. km) Coefs of Gauss 8 th deg. Polynomial Eqn: a = 14.7014, b = 467.3589, c = 3736.63 Estimated: semi-major = 48036754. km, ecc = 0.43566, incl = 1.3650 deg wrt Ecliptic
a Time (day) Year mo day hr min sec Right asc, deg d Declination, deg t1 = 0 007 9 1 11 4 0.0 53.85 9 336 t = 55.89 007 11 16 8 1 0.0 45.6 7 99 t3 = 97.39 007 1 7 0 8 0.0 38.98 9 96 Time (day) r_earth_sci (t), (km) source JPL DE405 t1 = 0 15011957 6 483669 096694 6 t = 55.89 88037190 4 10909350 4 4795458 1 t3 = 97.39 14334465 0 134330178 4 5836334 3 Output: Example 1: Ceres, Asteroid (with small errors) Input: (Data with approximately 0.1% errors in Right asc and Declination) sci= Sun Centered Inertial DerAstro Algs. r_sci (t) (km) v_sci (t) (km/s) Gauss (-Body) 7954647 30348774 85493999 13 7090 8 6938 6 8935 Laplace (-Body) 313714456 33688559 930957 15 75435 6 39106 6 45180 Gauss/Laplace+ 7943916 3046460 85475486 13 703815 8 701644 6 893003 ANSWER 79388198 3019760 8547678 13 653910 8 755670 6 904600 Helio object, Ceres (x = sci range(t) = 40,410,759. km) Coefs of Gauss 8 th deg. Polynomial Eqn: a = 8.370, b = 7.49756, c = 1.71498 Estimated: semi-major = 4143160. km, ecc = 0.08359, incl = 10.5958 deg wrt Ecliptic