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Space Flight Dynamics

by: Laila Windler

Space Flight Dynamics ASEN 5050

Laila Windler

GPA 3.96

Robert Nerem

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Robert Nerem
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This 90 page Class Notes was uploaded by Laila Windler on Friday October 30, 2015. The Class Notes belongs to ASEN 5050 at University of Colorado at Boulder taught by Robert Nerem in Fall. Since its upload, it has received 28 views. For similar materials see /class/232167/asen-5050-university-of-colorado-at-boulder in Aerospace Engineering at University of Colorado at Boulder.

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Date Created: 10/30/15
Principal Axis Rotations rX rX cos9 ry san Y Y ry rX sin 9 ry cos 9 r2 0 r2 6 ix 1 cos 9 sin 9 0 VY39 7m sin 9 cos9 0 7m Romm mquot XV 9 X 0 0 1 cos 6 0 sin6 1 0 0 ROT26 0 1 0 ROT16 0 cos6 sin6 sin6 0 cos6 0 sin6 cos6 lecmreS ppt RS Nerem 2004 1 17 and 17 From Orbital Elements Express the position and velocity in the perifocal system 2 goes through periapse 2 in the direction of Ii 7 perpendicular to 2 and 2 in the orbit plane 7 cosv Iecosv A A sinv rcosv P rsmv Q p 7 PW Q Iecosv 0 rcosv rv39 sinv Vsinv rv39E Mp1ecasvEIecosv p 7 I7 7 rsinv rv39 cosv l 0 J i W va Eesinv thus 7 V PQW 1 ecosv lecmreS ppt RS Nerem 2004 2 1quot and 17 From Orbital Elements Now we simply need to rotate into the geocentric equatorial system The order of the rotations does matter 7m zRompamawpmxaupmaw z PEWJQW vm R0T379R0T17iR0T37wTPW P127 UK sianini W 39 7 39 39 39 7 39 39 icostini sin to sini cos 9 sini 605i Algorithm 10 in book Ex 26 IzchlzeSppl RS NeanUDA Example 26 Example 25 Finding Pusilion and Valarin Vectors RANDV Test Case GIVEN p 11067790 km 1735 27 ER 2 02532 85 i 8787 n 227 80 c 5338 V 92335 FIND a 5 We have to change the rolallon llllgles if wc39ic using special orbits equatorial or circular but lliis Orbit dccsn39i have speclal Cases mm m given infumuon form he PQW posilmn and velocity vectors pCOSlr l ECUSU A 0073 867 rl QW 251va 1794 7339 ER News 1 0232 34C0592336 0 o u IzchlzeSppl RS NeanUDA Example 26 VESINM 75 9239335 075849l8 ER 5 0601 quot136 PQW FUN cow 1 1 70 832 54 cos92336 3 TU v 0 0 Ram has vectors to k 39 m using maniac le 1Rom m1011111Ror3 01FQW lm 1Ror34112071 1112013711ww 139 use the expanded matrix with a computer 0 dc Lhe many ulgonnmeln39c upcmlions which l CSull in me Iransformalinn mamx 39 O462 53S 21 0580 670 14 0669 955 57 0802 105 7 0596 023 47 0037 182 20 K 41377 736 47 0554 597 39 0741442 44 FQW hamsm 115 NeanDDA Example 26 Finally multiply each vector 0 apply the Iransformatiun39 1 W x 0377 736 47 0554 597 39 70741442 44 0073 180 7 quotW PQW PQW 70462 533 21 05110 670 14 0569 935 52 1794 7339 0802 105 71 0596 023 42 0037 182 20 0 1023 6524834 7 1076 ER 6862875 km 1011 6448296 UK A 0377 736 47 0554 597 39 41741 442 44 0758 499 8 Um PQWJUFQW 0462 533 21 0530 67014 0669 985 52 0601313 6 0802105 71 0596 023 42 0037 182 20 0 062 4901 320 070 ERTU 5533 756 kmS 025 1 975 341 hamsm 115 NeanDDA f and g Series Figure 214 Geometry for the Peril ucul Coordinate System We can break a sutelliLe s posir inn into two components x y in the orbital plane Fx yQ 7 ig170 Vx yQ vf70gv0 lectureS ppt RS Nerem znm f and g Series Start by crossing the position vector into the initial velocity vector rxvg gvgxvg fr0 x vggvg x v0 The second term is zero and the other terms are normal to the plane 3 W f xyg x y h Differentiating this last equation f xy x y h lectureS ppt RS Nerem znm f and g Series Now cross the initial position vector into the position vector ix7ix g o fr0 xr0 gr0 x v0 The first term is zero and the other terms are normal to the plane xgy xyg gh xoy xyo g h Differentiating this last equation xoy39 5 h lecmreS ppt RS Nerem 2004 9 f and g Series Look at the crossproduct E7xv g 0x gvo ma x agt fg3970 xVDgtfgltvo xa ggwo xi fgh fgh Which can only be true if 1 fg xrcosv yrsinv X Esinv yEecosv P P lecmreS ppt RS Nerem 2004 10 f and g Series Which gives f 1 10 cosAv 7 rro sinAv M f tan l COSAvli p 2 p r r g 1 1 cosm P lecmrei ppt RS Nerem 2004 f and g Series So to summarize given an initial position and velocity we can calculate a future postion and velocity given the change in the true anomaly Av FfHO a fl l COSAV r gv0 g 0 grr0sinAv FForrocosAv M eel 1 02 g 1 1 cosAv Which you can test using Example 24 in the textbook lecmrei ppt RS Nerem 2004 f and g Series State Transition Matrix We can reeXpress our f and g series representation 7 fi gig v ff gig in terms of a statevariable relationship a 7 0 a a F G a Xq X0q XCDX0 XO v v0 F G f 0 0 where F 0 f 0 etc 0 0 f and CI is the State Transition Matrix lecmrei ppt RS Nerem 2004 ASEN 5050 SPACEFLIGHT DYNAMICS Coordinate and Time Systems Dr Steve Nerem University of Colorado Boulder lecmrei ppt RS Nerem 2004 Coordinate Systems Celestial Sphere North ce estial pole North ecliptic pole Celestial poles 1ntersect bje Earth s rotation aX1s H l x Celestial equator extends F d m Lylfi Ra W n a 3 e Observer I 11quot Earth equator s Direction of objects Ce s a39e 39 measured with right WWW South eclipticpole 6 Southcelestial pole Figure 115 Geometry of the Celestial Sphere The celestial sphere is based on an observer s perceived view of objects in space A meridian or hour circle is any Circle that passes through the observer at the center I ve shown the principal direction to help us nd right ascension a and declination 6 Vallado 1997 lectureS ppt R S Nerem 2004 Coordinate Systems The Vernal Equinox de nes the reference direction The ecliptic is de ned as the mean plane of the Earth s orbit about the Sun z E 39 t i s s illIr Ecli I tic plane Earth equator The angle between the Earth s mean equator and the ecliptic is 2 called the obliquity of the o o o eel tlca 8 5 39 Figure 116 Geometry of the Verna Equinox The Earth s mean orbit around the Sun forms the ecliptic plane The Earth s equatorial plane is inclined about 235 to the ecliptic When the Sun is at the intersection of the two planes has zero declination and is at the ascending node as viewed from Earth it s the first day of spring Vallado 1997 Mgt lectureS ppt R S Nerem 2004 ASEN 5050 SPACEFLIGHT DYNAMICS Time Systems Dr Steve Nerem University of Colorado Boulder lecmre7 ppt RS Nerem 2004 Time systems Time is important Signal travel time of electromagnetic waves Altimetry GPS SLR VLBI For positioning Orbit determination One nanosecond 10 9 second is 30 cm of distance Relative motion of celestial bodies lecmre7 ppt RS Nerem 2004 Astronomical clocks 0 We commonly de ne time in terms of astronomical and geodetic periods Rotation of the Earth day Revolution of the Earth around the sun year Orbit of the moon around the Earth month Number of bodies in the solar system Visible to the naked eye or 14 of a lunar cycle week 0 These astronomical clocks are not consistent The current length ofa year is 365242190 days 1900 365242196 days 2100 365242184 day lecmre7 ppt RS Nerem 2004 Mechanical clocks When mechanical clocks were less accurate than variations in astronomical clocks astronomical clocks were used to correct mechanical clocks 7 The pendulum clock was invented in the 17th century In 1928 with the invention of the quartz clock 7 Apparent that the uncertainty in the astronomical day was 10 7 due to irregularities in Earth rotation Atomic clocks 7 The idea of using hyper ne quantum states of atoms for a clock was rst proposed by US physicist Isador Rabi in 1945 lecmre7 ppt RS Nerem 2004 Atomic clocks Certain atoms in a the magnetic field can exhibit one of two hyperfine states 7 The spin of the outermost electron of an atom either points in the same direction as the magnetic eld of the nucleus or it points opposite 7 The laws of quantum physics forbid other orientations Generally an atom remains in its hyperfine state But when prodded by electromagnetic radiation at a specific frequency it will switch to the other state undergoing the socalled hyperfine transition Essentially an electronic clock selects atoms in one hyperfine state and eXposes them to radiation which causes them to switch to the other state The frequency of the radiation causing the transition becomes the regular beat that the clock counts to register time Hydrogenmasers cesium and rubidium standards are usually used 7 Each GPS satellite has 2 Cs and 2 Ru clocks lecmre7 ppt RS Nerem 2004 International Atomic Time TAI Atomic Time with the unit of duration the Systeme International SI second defined in 1967 as the duration of 9192631770 cycles of radiation corresponding to the transition between two hyperfine levels of the ground state of cesium133 The second defined in 1967 to correspond to traditional measurement A uniform timescale of high accuracy is provided by the International Atomic Time Temps Atomique International TAI The origin of TAI was chosen to start on 1 January 1958 Oh It is estimated by a large set gt 200 of atomic clocks 7 Mostly cesium atomic clocks and a few hydrogen masers at 60 laboratories worldwide Time centers compare clocks using GPS observations as time links A weighted mean of local time center results in the TAI lecmre7 ppt RS Nerem 2004 Nonuniform due to Fundamentals of Time Sun Solar Time determined from the rotation of the Earth WRT the Sun position of the Sun WRT the Greenwich meridian Rotation Sidereal Time period of rotation of the Earth WRT the stars or the hour angle of the Vernal Equinox Greenwich Sidereal Time HOST 0cm 50 AUTl lecture7 ppt R S Nerem 2004 7 changes in Earth s Sidereal time Sidereal time is directly related to the rotation of the Earth 0 Local Apparent or true Sidereal Time LAST refers to the observer s local meridian It is equal to the hour angle of the true vernal equinox The vernal equinox is the intersection of the ecliptic and the equator Where the sun passes from the southern to the northern hemisphere The vernal equinox is affected by precession and nutation and experiences long and short period variations NOR Il l POL ECUPTM PLANE lecture7 ppt R S Nerem 2004 8 Solar time and Universal Time Solar time is used in everyday life It is related to the apparent diurnal motion of the sun about the Earth This motion is not uniform assumes a constant velocity in the motion about the Sun Mean solar time is equal to the hour angle of the mean sun plus 12 hours If referred to the Greenwich mean astronomical meridian it is termed Universal Time UT Its fundamental unit is the mean solar day the interval between two transits of the sun through the meridian Conversion of UT to GMST is defined by convention based on the orbital motion of the Earth about 360 365 days 1 mean sidereal day 1 mean solar day 3 m 5590 s 8616410 s lecmre7 ppt RS Nerem 2004 9 Fundamentals of Time Universal Time 1 1T Greenwich hour angle of a fictitious Sun uniformly orbiting in the equatorial plane augmented by 12 hours eliminates ecliptic motion of the Sun UTO is determined from motions of the stars thus function of sidereal time UT1 is UTO corrected for polar motion Closely approximates mean diurnal motion of the Sun Solar Time Enamic Time Ephemeris Time derived from planetary motions in the solar system deduce time from position of planets and equations of motion Independent variable in the equations of motion 7 Barycentric Dynamic Time BDT is based on planetary motions WRT the solar system barycenter 7 Terrestrial Dynamic Time TDT derived from satellite motions around the Earth lecmre7 ppt RS Nerem 2004 10 U T C A practical time scale as needed in navigation for instance has to provide a uniform unit of time and maintain a close relationship with UTl This led to the introduction of the Coordinated Universal Time U TC 13 time interval corresponds to atomic time TAI and is epoch differs by not more than 09 sec from UTl In order to keep the difference DUT1 UT1 rUTCl lt 09m leap seconds are introduced to UTC When necessary 7 A speci cation ofaUTC clock should alvmys differ from TAI by an int er number of seconds GPS provides easy access to UTC With an accuracy Within 100 nanosecon s 7 GPS navigation data provides the integer offset for TAI new wt k s Nenmznm 11 U T C and U T differences from TAI new wt k s Nenmznm 12 Leap seconds Since the first leap second in 1972 all leap seconds have been positive 7 Currently the Earth runs slow at roughly 2 milliseconds per day so a leap second is needed about every ve hundred days 7 There have been 32 leap seconds in the 27 years to January 1999 7 This pattern mostly re ects the general slowing trend of the Earth due to tidal braking Sometimes there is a misconception that the regular insertion of leap seconds every few years indicates that the Earth should stop rotating within a few millennia Leap seconds are not a measure of the rate at which the Earth is slowing 7 The 1 second increments are indications of the accumulated difference in time between the two systems 7 Usually you would reset a slow clock to the accurate time We can t alter the Earth s rotation so we reset the accurate clock lecmre7 ppt RS Nerem 2004 13 Fundamentals of Time Atomic Time International Atomic Time TAT is based on Vibrations of the Cesium atom TDT TAI 32184sec Coordinated Universal Time UTC equal to TAI but augmented by leap seconds to keep it close to UT TAI UTC lSOn ninteger28 in early 1994 GPS Time needs to be uniform has a constant offset of 19 secs WRT TAI and was equal to UTC at the GPS standard epoch of Jan 6 1980 lecmre7 ppt RS Nerem 2004 14 Leap Seconds Since 19 72 2450630 2451179 1972 JAN 1 2441317 1972 JUL 1 2441499 1973 JAN 1 2441683 1974 JAN 1 2442048 1975 JAN 1 2442413 1976 JAN 1 2442778 1977 JAN 1 2443144 1978 JAN 1 2443509 1979 JAN 1 2443874 198 JAN 1 2444239 1981 JUL 1 2444786 1982 JUL 1 2445151 1983 JUL 1 2445516 1985 JUL 1 2446247 1988 JAN 1 2447161 1990 JAN 1 2447892 1991 JAN 1 2448257 1992 JUL 1 2448804 1993 JUL 1 2449169 1994 JUL 1 2449534 1996 JAN 1 2450083 1 1 lecmre7 ppt RS Nerem 2004 Present time differences uuuuuuuuuuuuuuuuuuuuuuu TAI UTC TA TA I U I T UTC As of 1 January 1999 TAI is ahead of UTC by 32 seconds TAI is ahead of GPS by 19 seconds GPS is ahead of UTC by 13 seconds The Global Positioning System GPS epoch is January 6 1980 and is synchronized to UTC No positive leap second will be introduced at the end of June 2004 lecmre7 ppt RS Nerem 2004 ooooooooooooooooooooooo mmmmmmmmmmmmmmmmmmmmmmm 41317 41317 41317 41317 41317 41317 41317 41317 41317 41317 41317 41317 41317 41317 41317 41317 41317 41317 41317 41317 41317 41317 41317 ooooooooooooooooooooooo mmmmmmmmmmmmmmmmmmmmmmm Yet more definitions Terrestrial Time TT or Terrestrial Dynamical Time TDT with unit of duration 86400 SI seconds on the geoid is the independent argument of apparent geocentric ephemerides TDT TAT 32184 seconds The epoch of TDT is defined as 1977 January 1 Oh TAT Julian Day Number is a count of days elapsed since Greenwich mean noon on 1 January 4713 BC Julian calendar The Julian Date is the Julian day number followed by the fraction of the day elapsed since the preceding noon 7 The Julian Day Number for 7 February 2002 is 2452313 The Modi ed Julian Day was introduced by space scientists in the late 1950s It is defined as MJD JD 24000005 lecmre7 ppt RS Nerem 2004 Fundamentals of Time Julian Date JD defines the number of mean solar days since 4713 BC January 1 05 noon Modified Julian Date g y JD obtained by subtracting 2400000 days from JD Thus MJD commences at midnight instead of 5 noon Civilian Date Q 1980 Jan 60 24442445 GPS Standard Epoch 2000 Jan 15 24515450 J2000 Epoch Algorithm 14 in book lecmre7 ppt RS Nerem 2004 Earth rotation mm mm Earth Orientation Measurements Length OfD 7 must be measured preferably using VLBI Polar Motion 7 must be measured with space geodetic techniques Nutation 7theorymodeling can generally be used sometimes apply small corrections from VLBI Precession 7 theorymodeling can generally be used mm mm Motion of the Coordinate System For this class we Will ignore precession nutation and polar motion Luni solar precession e ect Earth s orbit about Sun Figure 132 Precession of the Earth s Equatorial Plane This figure shows the combined effects of perturbing forces on the Earth The general motion of the equinox is to the west Planetary and nutation effects are small compared to the lunisolar effects Vallado 1997 lecture7 ppt R S Nerem 2004 21 Precession and Nutation LuniSolar Precession 50 per year period of 26000 years due to the torques of the Moon and the Sun on the Earth s equatorial bulge Planetary Precession precession of 12 century and decrease the obliquity of the ecliptic of 47 century due to the planetary perturbations on the Earth s orbit causing changes in the ecliptic Nutation amplitude 9 occurs at orbital periods of the Sun and the Moon 137 days 276 days 6 months 1 year 186 years etc 186 year motion is largest 20 amplitude 05 km lecture7 ppt R S Nerem 2004 22 Motion of the Coordinate System quot JECEFWPM JECEFwoPM Figure 135 Transformation Geometry Due to Polar Motion Accounting for polar motion takes into account the actual location of the Celestial Ephemeris Pole CEP over time It moves from an ECEF system without polar motion through the CEP to an ECEF system with polar motion using the Conventional lntemational Origin ClO This correction changes the values very little but highly accurate studies should include it The inset plot shows the motion for the CIO from May 1986 to May 1996 Vallado 1997 lecture7 ppt R S Nerem 2004 23 Polar Motion Linear drift of the rotation pole of 34 milli arc secondsyear in a direction between Greenland and Hudson Bay due to post glacial rebound Long period wobble 30 years of amplitude 30 milli arc seconds cause unknown Annual Wobble amplitude of 01 are seconds 3 meters on the Earth s surface 75 caused by annual variation in the inertia tensor of the atmosphere rest by mass variations in snow ice ground water etc Chandler Wobble 430 day period 6 meters amplitude Normal mode of the Earth Caused by atmospheric and oceanic effects l are min 160 deg l are sec 160 arc min lecture7 ppt R S Nerem 2004 24 Motion of the Coordinate System Only concerned with rotation of Earth about Z 605T 60m w 39AUTl w 72921158553 x10395radssec 6 1753368560 6283319706889T UTl 67707 gtlt10396T2 45 gtlt10391 T3 UTl UTl GSTO TUT1 number of Julian centuries elapsed from epoch JZOOO lecmre7 ppt RS Nerem 2004 25 Motion of the Coordinate Systems 711K R0T36GST fECEF 17UK R0T30GST VECEF 3 0GST fECEF a X R0T30GST fECEF cos 9 sin 9 0 w sin 9 woos 9 0 ROT3 sin9 cos9 0 ROT3 wcos9 wsin9 0 0 0 1 0 0 0 lecmre7 ppt RS Nerem 2004 26 Length of day changes Decadal annual and semiannual vn39ations lecture7 ppt R s Nerem 2004 27 Variations ofLOD There is a linear increase in the LCD of 12 mseccentury 39 Increase mainly due to tidal breaking Some decrease due to decreasing moment of inertia J2 e g skater spins faster when hands are drawn in Decadal uctuations of 4 5 msec due to transfer of angular momentum between the core and the mantle Short period variability of 23 msec at period of less than 5 years mainly 2 weeks 1 month 6 months 1 year due to Earth and ocean tides the atmosphere and Winddriven ocean circulation Before modern times ancient eclipse records have been used to determine variations in LCD lecmre7 ppt RS Nerem 2004 28 AAM vs LOD changes Length of day changes are highly correlated with changes in atmospheric angular momentum AAM 08 i 04 T m Oil yl U39 r11 W Va 8 9 0 y 0 g1 X W 7 a 79 w 02 E X 04 0 8 S 39 o 39 N D lecmre7 ppt RS Nerem 2004 Reference Frame Terminology Eguato 7 the great circle on the surface of a body formed by the intersection of the surface with the plane passing through the center of the body perpendicular to the axis of rotation Celestial Eguato 7 the projection on to the celestial sphere of the Earth s equator Eclipti 7 plane of the Earth s orbit about the Sun affected by planetary precession True Equator and Equinox of Dat 7 the celestial coordinate system determined by the instantaneous positions of the celestial equator and ecliptic motion due to precession and nutation Mean Equator and Equinox of Dat 7 the celestial reference system determined by ignoring small variations of shortperiod nutation in the motions of the celestial equator motion due to only precession Mean Equator and Equinox of J20000 7 celestial reference system at 12 hours January 1 2000 lecmre7 ppt RS Nerem 2004 Reference Frame Transformation PNSM i7 A rJ 2000 ECEF P precession transformation moves state from mean equinox of epoch J2000 to mean equinox of date 0 N nutation transformation moves state from mean equinox to true equinox of date 0 S Apparent Sidereal Time Transformation referenced to true equinox ROT36GST EQ EQ GAST GMST OGST Greenwich Mean Sidereal Time equinox 37 7 u EQeqmn0X equat1on of the equ1noxes d1fference between mean and true equinoxes 1 0 xp 0 M polar motion transformation 0 1 yp xp yp I lecmre7 ppt RS Nerem 2004 31 What the Rotations Do M 7 Polar Motion rotates the terrestrial ECEF frame from the conventional pole to the celestial ephemeris pole S i Sidereal Time rotates the terrestrial frame from the Greenwich Meridian to the true equinox of date N 7 Notation rotates the celestial frame from the true equinox of date to the mean equinox of date P i Precession rotates the celestial frame from the mean equinox of date to the mean equinox of J20000 lecmre7 ppt RS Nerem 2004 32 ASEN 5050 SPACEFLIGHT DYNAMICS Lambert s Problem Dr Steve Nerem University of Colorado Boulder lecturell ppt R s Nerem2004 Lambert s Problem Short Way AV lt 180 rm 1 Long Way Av gt 180 rm 1 Figure 610 Transfer Methods tm for the Lambert problem Traveling between the two speci ed points can take the long way or the short way For the long way the change in true anomaly exceeds 180 Given two positions and the timeof ight between them determine the orbit between the two positions lecturell ppt R s Nerem2004 Lambert s Theorem The orbital transfer time depends only upon the semimaj or axis the sum of the distances of the initial and final points of the arc from the center of force and the length of the chord joining these points At fa r0r 6 Also Gaussian Battin Universal Two transfer methods short way and long way Av lt 180 tm1 and Av gt 180 tml respectively lecturell ppt R S Nerem 2004 Lambert s Problem Figure 611 Geometry for the Lambert Problem I This gure shows how we locate the sec ondary focus the intersection of the dashed circles The chord length c is the short est distance between the two position vectors The sum of the distances from a focus to any point r or re is equal to twice the semimajor axis Vallado1997 lecturell ppt R S Nerem 2004 Lambert 5 Problem Find the chord c using the cosine law and cos Av r0 Or rr 2 2 c1Hbr 2QramAv 0 Define the serniperirneter s as half the sum of the sides of the triangle created by the position vectors and the chord Qrc S 2 We know the sum of the distances from the foci to any point on the ellipse equals twice the sernirnaj or axis thus 2ar2a r lecturell ppt R s Nerem 2004 5 Lambert 5 Problem Figure 612 Geometry for the Lambert Problem II Solving Lambert s problem relies on many geometrical quantities Be sure to allow for the Earth when viewing representa tions like this I ve shown transfers between a satellite at 95672 km 15 ER and 153075 km 24 ER from the Earth s center We can use the inset figure to find the transfer semimajor axis The concentric circles are drawn for elliptical values of a When the circles of the same a touch half the sum of their radii equals a of the transfer and the intersection is the location of the second focus F 39 Vallado199 7 lecturell ppt R s Nerem 2004 6 Lambert s Problem We will look at the minimumenergy solution where the chord length equals the sum of the two radii a single secondary focus 2a r2a r0 c Thus s r0 r c a i mm 2 4 Using inset of figure 612 and the Pythagorean theorem we can write 4aimeim s I sinoc2 r s rcosa2 S rZI cos2 a r2 s rz cos a 2rs rcosoc s2 2rs r1cosoc lecturell ppt R s Nerem 2004 7 Lambert s Problem The cosine law gives r02 r2 CZ 2rc cosoc rjr2cz Thus cosa 2rc r2r2cz rrc Canshow 2ss rU rc0 using s0 2 2 2ss r Thus cosa U I rc and 4afmeim s2 s r0s r 461262 4aaez lecturell ppt R s Nerem 2004 8 Lambert s Problem But 4a2e2 4aa p 7 S Sll lce am 2 4612 62 s2 23pmm mm mm Thus s2 2spms2 lts njxs rgt C Pm ES rg XS r EU cos Av C c K 1 mm I up an S lecturell ppt R S Nerem 2004 9 Lambert s Problem If Av gt 180 then e 3e If tgt tmm then one 2n oce 2 4a 20 4a 2a 2 4 2 4a 2a t Jake sma sm Lambert s 3 Solution swan ah mam m y lecturell ppL R s Nerem 2004 10 ii N L9 Lambert s Problem For minimum energy elliptic orbit chrt g quotSIM J rm n esin e M VUMf 1 rl COSAVU rgr sin Av mm lecturell ppL R s Nerem 2004 1 1 Lambert s Problem Timeof Flight given by Lambert s Equation tmm EJ g sin32 y 32 rgr c s c sm 7 17 2 4a 2a V Mr 1 r1 comvrg 0 rgr sm Av Also mm lecturell ppL R s Nerem 2004 12 Lambert 5 Problem ytransb A Avb vrgt 39 b Target afteAr T01 Target at initial time rgt 7 rrarLra Figure 615 Geometry for Intercept Maneuvers For the targeting maneuver to take place we must account for the target s motion during the transfer Vallado l9 9 7 lecturell ppt R S Nerem 2004 Lambert 5 Problem Short way Av lt 180 Long way AV gt 180 39 Hyperbolic l I Parabolic Elliptical W Figure 616 Varying Time of Flight for Intercept As the time of ight increases for the transfer to the target short way left long way right the transfer orbit becomes less eccentric until it reaches a minimum at e O The circular orbit dashed circle is the point at which the long way and short way are equal although their change in velocity is not The eccentricity begins to increase after the circular orbit Note the position of the initial vector for each case Vallado 199 7 lecturell ppt R S Nerem 2004 Lambert s Problem Figure 617 Geometry of Orbits that Can Hit the Earth The orbit transfer on the left is valid because the transfer orbit intersects the Earth after the transfer The transfer on the right would be unacceptable Vallado 1997 lecmrell ppt R S Nerem 2004 15 Culp s Proof of Lambert s Theorem Let E0 and E be the eccentric anomalies corresponding to f0 and 17 and define 2G E E0 2g E E0 Using r a1 ecosE gives r0 r 2a 1 cosE0 cosE Using the trig identity COS cosxcosy2cosx2y xy lecmrell ppt R S Nerem 2004 16 Culp s Proof of Lambert s Theorem r0r2a1 ecochosg xacosE ybsinE CZ xxozyyoz a1 eZZsinE 612cosE cosEgZ a2I 62XsinE sinEgZ a22sinGsingZ a2I 62X2singcosGZ x x Sll lce COSJC COSy 2s ysm y 2 x x smx smy2sm yCOS y 2 2 lecturellppt RSNeremZOO4 17 Culp s Proof of Lambert s Theorem Since for elliptical motion 6 S I can define angle h as cosh e cosG 1 F coszh CZ 4a2 sin glsinz G cos2 G ez cos G 4a2 sin gsinz It So r0r can be rewritten as r0 r 2a1 e cosGcosg 2a1 cosg cosh lecturell ppt R s Nerem 2004 18 Culp s Proof of Lambert s Theorem Define hg Shig Then 8752gE7E0 and cos s 6 cosh ecosG ecosE 2E0 so r0 r c 2a1 cosg cosh 2asing sinh 2a1 cosh g 2a1 cos 8 Usmg I cos 8 2sin2 E 2 2 3 r0 rc 4asm i 2 lecturell ppt R s Nerem 2004 19 Culp s Proof of Lambert s Theorem Similarly r0 r C 2a1 cosg cosh 2a sing sz39nh 2a1 00302 g 2a1 cos 5 4 a sinz g 2 With E gt E0 and M n2a3 the time interval can be written using Kepler s Equation nt tg E Eg esz39nE sinEg 8 5 e2sinEE cosEEgj 2 2 8 3 2sz39n EE cos 86 2 2 8 5 2sin86cos 2 2 lecturell ppt R s Nerem 2004 20 Culp s Proof of Lambert s Theorem Finally nt t0 3 sins sin3 We have S and 6 as functions of r1 r2 0 and a so we have proved Lambert s Theorem Determining the Orbit I I X M X rl Get Q and i lecturellppt RSNerem2004 21 Culp s Proof of Lambert s Theorem 2r0rcosAv Compute cordcas czrjr2 2fUf 8 Also have r r2 c4asm27 2 6 rjr2 c4asm27 2 12 t tU 8 6 sins sin6 a Three equations three unknowns a e 5 Solve for a e and E using normal rV9oe equations lecturell ppt R s Nerem 2004 22 ASEN 5050 SPACEFLIGHT DYNAMICS TwoBody Motion Dr Steve Nerem University of Colorado Boulder lecmre3 ppt RS Nerem 2004 1 Kepler s First Law 1 Substituting u A 1 hZH h 2Ac0s6 w IAh c0s6 w Let v 6 cu true anomaly co argument ofperiapse p h2M semilatus rectum 0r semiparameter e Ah2M eccentricity r Thus r p a1 ez lecosv lecosv This is exactly the equation of a conic section This is sometimes called the trajectory equation and extends Kepler s First Law from ellipses to parabolas and hyperbolas lecmre3 ppt RS Nerem 2004 2 Geometry of Conic Sections Conic Section is the intersection of a plane with a cone m9 is at the primary focus of the ellipse Figure 72 u u m plant i m z mm m Wm Vauadm 1997 1mm ppt eRs Nemn 2004 Geometry of Conic Sections Parabolic Orbit V 7 in if mu ms m mnwlcrvd open Mam ma Ann39v mean i mum l39umi rs A Hmmly mum 1997 41m I2 gum l Parabolic 0min mm Tm wmwnnjnr m h mfm c lemma ppt eRs Nemn 2004 Parabolic Orbit eI 50 ri Icosv e I not necessarily a parabola need 5 0 2 s0L v 2i r oo v 2 r r Note As v gt 180 r OO A parabolic orbit is a borderline case between an open hyperbolic orbit and a closed elliptic orbit mm megs mm mm 5 Geometry of Conic Sections Hyperbolic Orbit Figure 77 Valladm 1997 mm mums Nerem 2mm 6 Hyperbolic Orbit V2 2 27 V22 7M r r as r a 00 V2 a 28 V02 27M for any other point on the hyperbola U 2 2 V or Vm 3 a 8 Excess Veloc1ty at r 00 and since a 21 V hyperbolic excess velocity at a lecmre3 ppt RS Nerem 2004 Hyperbolic Orbit One important application is planetary approach Iecosv r r gt 00 ecosvm gt I vm 6 1 But 6 21 180 6 6 90 7 90 7 2 w 2 6 6 I So cosvm smi gt smi i 2 2 6 Approach The smaller the e the greater the asymptote turning angle lecmre3 ppt RS Nerem 2004 Geometry of Conic Sections Figure 174 Ellip ml Orbih An m m ruler vi Um E Li H I nrhiv ham v0 4 wn dmmcl mm mm m puma mm r and vanapmi r m t c it Scnuumwr axis a and m wmvmmnr am 5 be pm me I Halnlm mime belwucuvlu mm M mm mm Damn mu lumlcs Lin 0 tan minim mmm from In39 way im 9 Valladu we lacth mans NeanEIUA Geometry of Conic Sections Elliptical Orbits 0 lt e lt 1 also e Sometimes attening is also used a I f 7 0 e2 2f f2 lacth mans NeanEIUA Elliptic Orbits The semilatus rectum is p5aI e25 a M Earth Sun Moon 1 A apoapsis 9 apogee 9 aphelion 9 aposelenium 9 etc 1 P periapsis 9 perigee 9 perihelion 9 periselenium 9 etc P r 7a1e A 1e P r 7a1 e P Ie lecmre3 ppt RS Nerem 2004 1 1 Properties of Conic Sections V s e a Ellipses lt 27 lt0 05elt1 agt0 r 2M Parabolas 7 0 e I a 00 r 2M Hyperbolas gt 7 gt 0 e gt I a lt 0 r lecmre3 ppt RS Nerem 2004 12 Flight Path Angle This is also a good time to define the ight path angle pfpa as the angle from the local horizontal to the velocity vector mm hori39wnlal M pmillw from perlap51s t0 apoap51s Only 8135170 i Dummy or I from apoap51s to perlapsm hqavfrpvp 0 at periapsis and apoapsis 2vwff39 39 if 4 M Always 0 for elrcular 0rb1ts qwnegam 130 ltylt360 Figure 179 GunmelryfurlheFlighlrth Angle Th nighppm angle 15 alwuyh mcnsllvrcd 39 1 v w mggmmd m mugum or mmy Vallad0 1997 lecture3 ppl R s Nerem 2004 13 Flight Path Angle Another useful relationship is hrvcos m Figure 24 Geomclry 0f the FlightPath Angle The Illglllrllalll angle 0 vs measure from hc ml hurizonlnl m Ihc velocin Vector which we cm brcaii lnln radial and Imus verse components The auxiliary angle 9 it smutlimes used m dcrivalmm I e COSV e sianallado 1997 COS sin pfpa VI2ecosveZ pfpa VI2ecosveZ lecture3 ppl R s Nerem 2004 14 Vis Viva Equation V2 M Recall the energy equation 3 3 7 Note at periapse 2 hrpr rpaIe 171 16 gt hW 51 M 2 WIN6 2 m1 e P a2Ie2 aIe So at periapse V I i M 1 6 Thus 81 16 M 2 i 2 aI e aI e aI e 2a 1 2a lecmre3 ppL RS Nerem 2004 15 Vis Viva Equation Thus rewriting the energy equation gives V2 Vis Viva Equation r a V e Foracircular orbit V E K p VaUe C r V gUe 391 1 For a parabolic orbit a 00 thus V 2i V W 8 1 r Also note we canzderive an expression for the eccentricity of the 8 orbit p a1e2 gi gt i u 2a a u 2 2 e I 128il e W M lecmre3 ppt RS Nerem 2004 16 Proving Kepler s 2nd and 3 4 Laws dA jrdvdr lrzdv g 2 lrl hr29 dt 2 dt 2 2fdA hfdt HPraves ZEL area of ellipse nab 1 r0 1030 P t t period AreaSwepl 00 byaSatellim Acc0rding 10 Kepler s second 100 50011005 sweep Z 1 0013000 000301 tquuluml 10100010 thus P2Lhab bZaZ1 ezap gt b1lap h1ma1 ez 2 3 Ex ressian a P1 al Kgler s 3E Law P 2 3 2 2 ledur pplQRS Nemm 2004 17 Proving Kepler s 2nd and 3 4 Laws If we define the mean motion n to be 271 M n then H 3 0r Shuttle 300M 90 min P 0 Earth Obs 800 km 101 min 2 3 Mean angular GPS 20 000 km N12 hrs 7 a M rate afchange 0f GEO 36 000 km N24 hrs the object in orbit v s e a 2M Elllpses lt 7 lt0 05elt1 agt0 r 2 Parabolas i 0 e I a 00 r 2M Hyperbolas gt 7 gt 0 e gt1 1 lt20 r 31m 1117 22 13 positive ledur ppl RS Nemm 2004 18 TheAnomalies p K Eccentric Anomaly 5100 True Anomaly Mean Anomaly I H4 2 ACOME KIDS ltM q p gm Gmmelry ol mum Mua on m quotmum anomaly um 0 auullary mm 3 shown llm ulmmue pm i m tleknmm In am A Wm allow us 0 mm For a coordinate system centered on write the location of the satellite in terms of E XW acosE ae 2 Eq of Ellipse XS 1 gt YSAT bsinE a lecture3 ppt R s Nerem 2004 19 The Anomalies Auxiliary Circle uSINE Figure 4 2 Geometry ofKepler s Equation The eccentric anomaly uses an auxiliary circle as hown The ullimale goal is to determine the areaY Al which allows us to calculale the time lecture3 ppt R s Nerem 2004 20 Derivation of Kepler s Equation Now VZX ATY M a 1 92 acosE aef bsinE2 a2cos2E ZecosE 62 I ezsin2E a2I ZecosEeZI sin2E a2I ZecosEezcos2E 612I ecosEZ gt r aIecosE Now we will derive Kepler s Equation lecmre3 ppt RS Nerem 2004 21 Derivation of Kepler s Equation P Remember r I e cosv he sinv r 2 7 I ecosvz p p But also r a1 ecos E f ae sin EE So aeEsinEM Note bsinErsinv P b 39 7 39 E a si Ehbs va rsm p r lecmre3 ppt RS Nerem 2004 22 Derivation of Kepler s Equation rE1laail e2aVI e2 h pa ail ezg a p WaI e2 a rE a2I e2 Ea a Thus a1 ecos E a Thus E ecosEE M3 n 0A Integrating E e sin E nt constant lecmre3 ppt RS Nerem 2004 23 Derivation of Kepler s Equation If reference time is perigee E O 0 0 1th ntp constant gt constant ntp 7 l7 E esinE nt tp E M Kepler s Equation Angle swept out at the M Mean Anomaly mean angular velocity n the mean motion Important Formulae 2 r aI ecosE lecmre3 ppt RS Nerem 2004 24 Derivation of Kepler s Equation cosE e acosE e SO cosv 1ecosE r 2 2 I r 1 ecosE cosEe 1ecosE cosE e tanZZ1COS 1 ecosE 2 1cosv COSE e 1 I ecosE 1eX1 cosE1em 1 eX1cosE 1 e v 1e E tat1 2 tan 6 2 lecmre3 ppt RS Nerem 2004 1 ecosEcosE e 2E 2 Derivation of Kepler s Equation If t is given M nt tp Solve E e sinE M forE r aI ecosE acosE ae and cosv r b sinv 7 sin E r mil 92 sinE If v is given 17 1 ecosv r cosv ae cosE a rsinv san 7 Solve E e sin E M for t or M Another Useful Relation lecmre3 ppt RS Nerem 2004 Solving Kepler s Equation Want to use NewtonRaphson Iteration Assume we want to solve fy 0 for y Assume y x6 where x is an approximate guess and 6 is a small correction Expanding in a Taylor Series 0fyfx6fxf x6 az fx Neglecting 2nd order terms and higher we can solve for 5 ax f39lx lecmre3 ppt RS Nerem 2004 27 Solving Kepler s Equation We need to iterate this since we have neglected higher order terms X Iterate until 6quot is sz Xn 6n Xn acceptably small 5 yx We want to solve M E esinE O fE f E l ecosE So M E 6 sin EH 1 ecos En Careful with e I ie don t use for high e lecmre3 ppt RS Nerem 2004 28 ASEN 5050 SPACEFLIGHT DYNAMICS General Perturbation Techniques Dr Steve Nerem University of Colorado Boulder lecmrel ppt RS Nerem 2004 1 General Perturbation Techniques Note that the Lagrange planetary equations require the partial of the disturbing potential WRT the orbital elements We have R K r need Ra e i 0 9 M Kaula 1966 developed an elegant solution to this problem Eq 928 We won t look at this but let s look at the problem for just J 2 2 Rhee i r r 2 g 3 E1 023 Let s convert this to orbital elements sin pgc sinisinwv 2 R sinzisin2wv l 2r r 3 lecmrel ppt R s Nerem 2004 2 General Perturbation Techniques h Figure 82 Putcnlial Transfurmltinn Convcnmg the potential relicson he remuon of m clas sical artHui clemnm and spherical erdimlcs 5an w r Vauadn t 997 immmppms mm 3 General Perturbation Techniques Eq C12 gives sin2ot cos2a thus R3M2 R78 2 sin2isin2icos2wvl 2r r We want to isolate the secular variation so we ignore long period 00 and short period V terms N 3 2 a sinzi I R 7 2 J 7 7 7 2 2W 2 3 immmppms mm 4 General Perturbation Techniques r changes periodically over each revolution Thus we need to replace ar3 by its average over one period since we want only the secular effect dM1dE a 3 3 3 a I 2 a I 2 a r dE d 7 i all7 i idE 2V r 2nr 2nr a a 1 8 I 2 a 3 r2 I 2 E0 a2 62 dv 2 1eZyZ 1ecosvdv uVJ 2n I a1 e2 I Ieeosv Ieeosv r all ezi lecmrel ppt R s Nerem 2004 General Perturbation Techniques Thus replacing ar3 by its average gives 3 2 I sinzi I lwll 2 393 Substituting into the Lagrange Planetary Equations gives 2 Q mg J2 000108262693 SEC 2172 2 DSEC 3nR 2J2 45Sin2 4p 2 usin 6R 2n 3nR JZwI ez 3sm2l2 g aa n2 da 2 4p 3RAVG anda e39 i 0 lecmrel ppt R s Nerem 2004 M USEC a William K aula 1 92 6 2 000 n man he cnaneo a panel convale by NAsluo lopose malpeouencallylelaleu ploplalns lnal palsy snonlu lnmale and snppomn me neal mule 2 me loaulnap lol Nume y VLEI Doppla selaleo syslelns Lisa lanplnp Leonel Apollo lase allnnelel lean Manuel Nnnonal Acauauy ol selelces m7 Fellow Amalcan Geophyslcal unon 1m Hollolaly Doclol ol Elalce onlo Stale unwesny ms Kuula s Solution conneu polaulal quotem a uqzaluelce on l 5 Mn a uqzaluelce on a ellnm Manu em l I s UM 73 Zn mmqgwmmmwm M Qe Smu C MW m cus72pm72pqMmQrS 5 Ir a Smlemm squammuezwqwmew c M lncllnanonlnncnon Fwl eccamlcnymncuon GWe Substitute into Lagrange Planetary Eq I dalmpq 2ALmeFlmpGlpqaslmpq 01 4 dl nal2 delmpq W FlmpGlpq 2 2 T Wl e 051mpq0M e aslmpqaw l dllmpq WeEmpGlpq fmias awas 09 dt nal3 1 82 sini lmpq lmpq l dglmpq WeaFlrnp alGlpq Slrnpq dl nal3wll e2 sini dwlm AuafZ 1 1 32 coti dl nat3 EmpaGlpq me mempal lpq Slmpq l 2 dM Ma F S 1 8 lm 2 lm lrn 2l1G E7 P Eq 0G we dt nal3 W 8 Pl lecmrels ppL R s Nam 2004 9 Orbit Element Perturbations 2Iw lranlqu 2p LDSlmpq We 12 na 2 12 2 12 FlmpGlpqa E 1 E l 2P q1 217Slmpq Aelmpf We 13 na 21 Ailquot Ma FlmpGlpq U 217COSi mSlmpq I nal3l 829211 A9 l aFlmp aiGlpqSlmpq lmpq We nal31 ez12 Aalmpq sin i1 1 2212211mp0Glpq we cot i1 e2 2aFmp MGM 15qu Awlmpq We Mt31 329211 1 e2y 1aG ae211G F 5 AM 11 1 mp mpq lmpq We nal3w Where 1 l 2pu39 l 217 qM m 2 9 and gram integral ofits argument lecmrels ppL R s Nam 2004 10 Slmpq 51mm C l meven cos1 2pwl 2pqMmg2 9 1 l modd shyly maven Clm l modd 5 sinl 2pw z 2p qM m 2 49 lecturele ppt R s Nerem 2004 Orbit Elements a 52222Aalmpq l m p q e 52222Aelmpq l m p q i Eggggmlmpq 1 mp 617 9 2222A91mpq 1 mp 617 a WEEEEAwlmpq 1 mp q M MEEEEAMMM 1 mp 61 lecturele ppt R s Nerem 2004 General Perturbation Techniques The secular change of the orbital elements due to J2 is given from the Lagrange Planetary Equations as 2 QSEC Cosi 7 2 00m 45sin2i 2 I 2 MUSEC 3nR 22 I e 351712 0 2 anda e i 0 J2 000108262693 lecmrel ppt R s Nam 2004 13 General Perturbation Techniques Which is a secularly precessing ellipse The equatorial bulge introduces a force component toward the equator causing a regression of the node for prograde orbits New Q lt0 for i lt 90 Q 0 for i 90 Qgt0 for igt90 lecmrel ppt R s Nam 2004 14 0blateness Perturbations mmmmw 15 General Perturbation Techniques T mmmmw m General Perturbation Techniques ll 4 7 5 7i 7quot7 1 0 C 2000x2000kme0 5 100 x3000 km e 0 128 7 100 x 1000 km 9 0 65 100 x 60ll km 2 0037 100x400km e0023 100 x 200 km e 0008 10 39 100X100kme0 I I l I I x I I I I l I I I I I I I I I I l I 0 30 60 90 11 0 150 180 Inclination Figure 84 Daily Nodal Regression in lday For the eccentric orbits I used a perigee altitude lecture16 ppt R S Nerem 2004 of 100 km with the apogee values as indicated The 100 X 2000 km orbit shows that the perturbing effect for each of the eccentric orbits would be smaller if the orbit were circular at that apogee altitude 2000 lt 2000 km Vallado 1997 General Perturbation Techniques Perigee also processes 60 0 at the critical inclination 1w 634 1166 lecture16 ppt R S Nerem 2004 Apogee Start Apogee End Perigee End Perigee Start How Apsidal Rotation Works We can see the effect of apsidal rotation after the J2 perturbation effect increases by a factor of 40 for a polar orbit to eliminate the nodal regression Notice how perigee and apogee locations change dramatically in a few 1 t V revo u ions Vallado 1997 Figure 85 18 General Perturbation Techniques 20 l n100X 100 kme0 4 looxsookmeo30 l X 1000 km e 0065 15 i w 74 100 X 2000 km e 0128 4 100 X 3000 km e 0183 10 quotW quot7 100x4mokme023 9 a 4000 x 4000 km 9 3 5 r 39 0 5 l I l t 1 0 30 90 Inclination Figure 86 Daily Apsidal Regression in day As with nodal regression circular orbits at an altitude say 4000 X 4000 would have a smaller daily change than an eccentric orbit with apogee at the same altitude 00 X 4000 km Vallado 1997 lecture16 ppt R s Nerem 2004 19 General Perturbation Techniques Gravity perturbations also affect geosynchronous orbits s 753 E Satellite in geosgnchronous or it Unstable 1653 E Q U 3453OE 6 Circular F about the Earth 2553 E C22 S22 Stable Figure 88 Polar View of an Equatorial Section of the Earth C22 only F is the net tangen tial force on the satellite at the positions shown C22 models a longitudinal asymme try of the Earth Both stable S and unstable U positions are identified Vallado 1997 lecture16 ppt R s Nerem 2004 20 General Perturbation Techniques Application Sun Synchronous orbits Q 3 60 WM 3652421897 098564 7 degday Can adjust a e l39to quot8quotquot accomodate this h 800 km 6 001 986 Figure 109 Geometry for SunSynchronous Orbits We use a nomenclature based on the time of the ascending node to distinguish Sunsynchronous orbits The orbit shown is an 0800 orbit because the satellite reaches the ascending node at 800 AM local time I ve shown the Sun offset from the equatorial plane for generality Vallado 1997 lecturel6 ppt R s Nerem 2004 21 General Perturbation Techniques Sun Synchronous orbits Orbit plane remains at a constant angle 939 with respect to the EarthSun line Orbit plane precession about the Earth is equal to period of Earth s orbit about the Sun Figure 212 Schematic picture of a satellite orbital plane precessing about the Earth with a period equal to the period of the Earth39s orbit about the sun ref ear 28 y lecturel6 ppt R s Nerem 2004 22 General Perturbation Techniques Fig 98 Local time at which a sunsynchronous satellite crosses a given latitude assuming a crossing time of midnight at the ascending node The gure has been calculated for an orbital inclination of99quot but will be similar for any lowaltitude sunsynchronous orbit Relative local crossing time hours lecture16 ppt DR S Nerem 2004 l I 3 0 30 60 9 l Latitude ref year General Perturbation Techniques Semimajor Axis km Figure 1010 lecture16 ppt DR S Nerem 2004 l l l v x r 39110 39114 us Inclination Elements for SunSynchronous Orbits This gure shows that Choosing two orbital elements from i a and e determines the third element needed for a Sunsyn chronous orbit The semimajor axis and radius of perigee must be greater than the Earth s radius to be realistic Vallado 1997 Atmospheric Drag Apogee Figure 810 Contraction of Orbit Under Drag This gure is greatly exaggerated but shows the general effect of drag on the satellite s orbit The radius of perigee tends to remain constant while apogee shrinks and the eccentricity approaches 00 If we assume perigee height remains constant then rPa1e and dr da de P 0 a i I e a7 dt dt dt thus 1e dt a dt lecturel ppt R S Nerem 2004 25 Atmospheric Drag 2 3 i dn 3 2 2 da To determine dadt look at n a i M 2nEa 3n a E 0 da 2a0 n dt 3110 dn 2610 As a first approximation let a do n no a Clo 3 1M1 no 3n 0 a 21 The period also changes P 231 n dn W 2c dP 27Jdn 33M 11 na lecturel ppt R s Nerem 2004 26 Atmospheric Drag Must use Gauss s form of planetary equations to look at whole effect Secular changes in a e i periodic changes in all elements e r an 012 108 12222 011 10639 Perm 2037 010 1041 1352 009 102 1 1667 008 100 1 1432 007 98 1296 006 96 1111 005 94 926 004 92 741 003 90 556 002 00L 370 03901 86 Altitude of perigee 185 000 81 1 1 1 1 1 1 1 1 1 t 1 1 1 1 0 00 Time 100 days Figure 811 Notional Variations of Orbital Properties During a Satellite s Lifetime in gen eral certain orbital parameters tend toward zero and eventually the satellite will reenter and crash The lefthand axes are eccentricity and period min respectively whereas the righthand scale is altitude for perigee and apogee km lecturel ppt R S lvewul 4mm valladoy 1 Solar Radiation Pressure The energy of the orbit is 8 2M a Thus Ag LA 2612 a FSR pSRCRAi A8 F dist Force times the distance work FSR AIZ AI radial displacement of the satellite relative to the sun 2 Aa 2i FSRAI a Thus as the satellite moves towards the Sun the semimajor aXis and energy decrease and as it moves away from the Sun both values increase Thus for an orbit completely in sunlight the net change to the semimajor aXis is 0 lecture16 ppt R S Nerem 2004 Solar Radiation Pressure Let s compare drag and solar radiation pressure l CDAV2 at h800km ad 2 rel pvz amps relz1 p117x1039 kgm3 aSRP M pSR V 74518 ms m Thus radiation pressure exceeds atmospheric drag at altitudes greater than about 800km Affects all orbital elements lecmrel ppt R s Nerem 2004 Propagating the Perturbed Orbit The semimajor aXis mean anomaly and eccentricity may be propagated if it and n are known The mean anomaly is found from a Taylor Series expansion M MU nUAtgAt2 gAt3 Eand gare given in the TLEs no p0 00185 Mg Eg ESI39HE0 2 aa 2a0 At 990 3ngR ezJZCOSQ39Mt 0 3mg 0 2 e M At wu 239245sin2i4t U 371 4P0 i1 MMUnUAtgAt2 At3 lecmrel ppt R s Nerem 2004 Two Line Element Sets Example 1 16609U 86017A 9335253502934 00007889 000000 105293 34 2 16609 516190 133340 0005770 1025680 2575950 1559114070 44786 Epoch Dec 18 1993 12h 50min 265350 sec UTC 1559114070 revday gt L7 I06118087 ER 6768357 km 37889x10395 revclqu 2 00 revday B I0529gtlt10quot1 e 00005770 i 516190 9 133340 w1025680 M 2575950 Errors can be as large as a km or more lecmrels ppL R s Nam 2004 31 ASEN 5050 SPACEFLIGHT DYNAMICS General Perturbation Techniques Dr Steve Nerem University of Colorado Boulder lecturelS ppt R S Nerem 2004 General Perturbation Techniques Perturbations can be categorized as secular short period long period k l Effect of Perturbation Forces on Orbital Elements The change in any orbital ele ment c is illustrated The straight line shows secular effects The large oscillating line shows the secular plus longperiodic effects and the small oscillatory line which combines all three shows the shortperiodic effects It s important to understand which elements are averaged when using mean elements The two examples I39ve shown are greatly exaggerated but representative of possible errors Time Figure 81 Vallado 1997 lecturelS ppt R S Nerem 2004 Perturbations c co e1 I to K1 cos2w K2 sin2v u K3 cos2v secular longperiodic mixedperiodic shortperiodic This equation is known as a Poisson Series K sums and products of polynomials in a e i l lecmrelS ppt R s Nam 2004 3 General Perturbation Techniques Osculating elements orbital elements at an instant in time defined by the position and velocity Osculate is Latin for to kiss The osculating orbitellipse kisses the trajectory at the prescribed instant The osculating orbit is also the path the satellite would follow if the forces were instantaneously removed Mean elements averaged over some period of time Useful for determining the satellite s longterm behavior Method of perturbations a class of mathematical techniques for generating analytical solutions which describe the motion of a satellite subject to disturbing forces Looks at small deviations to the solution for the unperturbed problem One technique is the Variation of Parameters which looks at small variations of the normally constant orbital elements lecmrelS ppt R s Nam 2004 4 Method of Perturbations do i 81fl82gi 81ltltl 82 ltlt1 dt 2 s C CO sla1ta2t 8231l 32I 8 182 2 y2t lecmrelS ppt R s Nam 2004 5 Lagrange Planetary Equations Variation of Parameters VOP Derivation VOP equations of motion are a system of 1 order differential equations describing the rates of change for the timevarying elements A i at Eq 93 dt These elements are osculating elements To derive the osculating element rates we ll use the following notation 17 fcaeio QMt Xeleze3eweje t at y Jaeio QMt Jeleze3edeje t at The siX constants c indicate we can use any set of orbital elements or more generally any siX independent constants of otion lecmrelS ppt R s Nam 2004 6 Lagrange Planetary Equations Variation of Parameters VOP Derivation The Equation of Motion EOM for the unperturbed and perturbed systems respectively can be expressed as MXGJ g A 2a i 0 t 3 m E 94 Mag xc Mat up q Where apert is the disturbing acceleration 5iEt Since velocity is the derivative of the position we can also express the equations for velocity in these two systems at dig Mg Eq 95 d5t axer i 6XEtdex dt at H ac dt at lecmrelS ppt R s Nerem 2004 Lagrange Planetary Equations Variation of Parameters VOP Derivation Differentiate the position vector in the perturbed case Eq 94 twice to get the acceleration using the following constraint for the perturbed system condition of osculation 6 6XEtdex 1 ac dt The second derivative of the position vector gives us d2Et 62XEt 6 6Etdel 72 2 2 dt at H 661 dt 50 Eq96 We can now find the osculating element rates in terms of the original equations by substituting into Eq 94 6226 t 6 615 tde are t 7 7 71 7 a E 9 7 6t2 601 dt 5a pm q lecmrelS ppt R s Nerem 2004 Lagrange Planetary Equations Variation of Parameters VOP Derivation But the unperturbed EOM Eq 94 hold for any time and can be written 2 A A A 6 xct uxct0 at Mair So Eq 97 simplifies to 5 6Etde A 77 E 98 ac dt am q Eq 98 is incomplete because it gives three equations am in the siX element rates dccit ie 3 equations 6 unknowns Remaining equations come from condition of osculation in Eq lecmrelS ppl R s Nam 2004 9 Lagrange Planetary Equations Variation of Parameters VOP Derivation In matriX notation the complete system of equations gives us 6 axerg 1 ac dt 0 54939 11 601 bit The solution to the Lagrange planetary equations of motion begin by taking the dot product of the bottom vector Eq 99 with and the top vector with 6 60 Ck k wherek 12 6 lecmrelS ppl R s Nam 2004 10 Lagrange Planetary Equations Lagrangian VOP Conservative Effects After replacing disturbing acceleration wgradient of disturbing potential function am gt VR the result of that dot product is 2 6261 6Et6f5t6iat H dt 60k 6c 60 60 x k ack Lagrange brackets are typically used for simplification sick Eq910 Using matrix operations and invoking Lagrange bracket notation we can now rewrite the matrix equation for the element rates 612361 6165 6561 6565 dt lecmrelS ppL R s Nam 2004 11 BE 3R Lagrange Planetary Equations Variation of Parameters VOP Derivation Letting L represent the matrix of Lagrange brackets the general form of the Lagrange Planetary equations becomes 1 6R 5 E L Eq9 11 Which reduces to 15 equations using the identities er01 0 0 r Ck Ck 5 Evaluation of the remaining brackets requires the partial derivatives of position and velocity which need to be expressed as functions of the elements M and M wherei126 60 651 x lecmrelS ppL R s Nam 2004 12 Lagrange Planetary Equations Variation of Parameters VOP Derivation From Kaula 1966 we obtain the explicit dependence of position and velocity on the orbital elements using the transformation from the geocentric UK to apsidial PQW frame na sinv r cosv V 1 62 A A nae cosv rPQW rs1nv VPQW 0 e 0 I 2 Remembering that r M I e cosv lecmrelS ppt R s Nam 2004 13 Lagrange Planetary Equations Variation of Parameters VOP Derivation Letting R represent the transformation matriX between PQW and UK the position and velocity vectors are x IJK rPQW x IJK VPQW PQW PQW Previous relations show an important result R is a function of only 9 n and i whereas the position and velocity vectors are functions of a e v Consequently R 6fPQwivPQW 6aeM UK 6a eM PQW lecmrelS ppt R s Nam 2004 14 Lagrange Planetary Equations Variation of Parameters VOP Derivation At periapse the PQW position and velocity vectors become a1 e 0 A A Ie rPQW 0 VPQW na 3 0 0 Three cases arise in the solution 1 C Ck quot02 r62 0le 0R 0R2 0le 11 00 00k 00k 00 3 6R av av m 1 e2 6R 6r 6r 55 a1ei R7PRi 77 R7PRi 2 z k acl 116 12 66k 1e at 116 12 66k 3 Niki szarPavP arP 0VP1IR1R2J139Zr avg arP avgl J 1 11100 00k 00k 00 JT 0 00k 00k BCIJ jarQ av arQ av 2 org 0 arQ avgl R1R21001 ac 00k acijR lacl 00k 00k 001 lecmrelS ppt R s Nam 2004 15 Lagrange Planetary Equations Variation of Parameters VOP Derivation Kaula 1966 shows us an example of this We start with the Lagrange bracket using Case 1 from previous slide mpg Hz j1 The partial derivatives are obtained by differentiating each component of the rotation matrix from PQW9UK Eq 135 91 sinQ cosw cos9cos1smwsm9sm1cosw cos9 cosw sinQ cosz39smwcos9 sinioosw sm9 sinw cosQ cos1coswsm9 sinisinw cos9 sinw sinQ cos1coswcos9 sinisinwna2 W lecmrelS ppt R s Nam 2004 16 Lagrange Planetary Equations Variation of Parameters VOP Derivation Simplification reduces this equation to note the second terms in each brace of the preVious equation cancel 9 i Jr sin2 9 sini eosz a Jposz Q sini eosz a in2 Q sini sin2 a J39gosz Q sinisin2 a hnaz V I e2 Further reduction produces 1 Qj eos2 stinZ eosz Q 1 sin2 stin2 eosz Q Mnaz VI e2 sini l eos2 a m2 00na2wI e2 sini lecmrelS ppl R s Nam 2004 17 Lagrange Planetary Equations Variation of Parameters VOP Derivation After simplifying we find Qi na2Vl e2 sini There are only 12 nonzero results Kaula 1966 though 6 values differ only in sign from the relation Eli iQ mentioned earlier Similarly we obtain the remaining unique Lagrange brackets in terms of classical elements The following are the siX brackets 91 naZV ez sini monkg 9a1wsogtwa1wsogt w e Qe 0030 wecosz WA10 lecmrelS ppl R s Nam 2004 18 Lagrange Planetary Equations Variation of Parameters VOP Derivation From Eq 91 1 we find that L clck reduces to 0 0 0 119 aw aM 0 0 0 e 9 e w 0 0 0 0 13912 0 0 L 911 9e 9139 0 0 0 1 a e 0 0 0 0 M0 at 0 0 0 0 0 Which resembles the following when inverted 0 0 0 0 0 IaM J a L0 0 o o o 1 7 A e Maura e 4 0 0 0 1gi 0 0 L 0 0 1i r2 0 0 0 0 1e w 0 0 0 0 IMwa wa a91ewe91aw 0 0 0 LLM 1e w 1 Q ILLM law lecmrelS ppt R s Nam 2004 19 Lagrange Planetary Equations Variation of Parameters VOP Derivation We can now solve Eq 911 to find the equations for the variation of parameters For example 6R6a 6R6e 6R 639 Q 0 0 71 0 l 139 6R6Q 6R6o 1 1 6R6MU 139 6139 Qi 6139 Q naZVI ez sini 6i Similarly the remaining Lagrange planetary equations can be derived to produce those shown in Eq 912 lecmrelS ppt R s Nam 2004 20 General Perturbation Techniques Lagrangian VOP 9 Lagrange Planetary Equations 9 Only for conservative effects Eq 912 da 2 6R 2 2 E aMu de I e 6R II e 6R 1 mm diw1 82 eoti dt nalel ezsin 600 69 art maze aenazm 61 as 1 6R dMu 1762 i jm2411ezsmig dt naze 6e na 6a Gauss also gives similar equations as a function of the perturbing forces F R FS FW which can also handle nonconservative effects lecmrelS ppt R s Nerem 2004 General Perturbation Techniques Gaussian form of VOP equations 9 Can handle nonconservative effects Eq 924 3 e sinvFR g L sinvFR cosv e WSW dt na 1ecosv dz39 rcosu E ratW W dig rsz39nu F dt nazmsin W L Lg cosvFR sinv1jFS r COIiSinuFW dt nae p h 1 d 2 pcosv 2erFR Cvrsinv lecmrelS ppt RS Nam 200746 8 ASEN 5050 SPACEFLIGHT DYNAMICS TwoBody Motion Dr Steve Nerem University of Colorado Boulder lecmre4 ppt R s Nerem 2004 Example Using MeanEccentric Anomalies For a satellite in an Earth orbit with hA3000 km and hP300 km how long does it take to go from an altitude of 1000 km to one of 2000 km r 6378km gt rp 6678km rA 9378km 39 a A 3quot 8028km e AquotP WWW e 2700 01682 Between and rArP a1eaI e 612 6056 7 r1 7378km gt r1 aI ecosEl gt cosEl 04815 E1 6122 or M r2 8378km gt r2 aI ecosE2 gt cosE2 02593 E2 10503 or Find the time of ight between E16122 to E 21 05 03 MI E esinEl gt M1 52 78 M2 E2 esinE2 M2 9572 n 181 hr M 39860044 kmjsz a nt2 t1M2 M 4294 gt tz tl 142 minutes lecmre4 ppt RS Nerem 2004 Example Using MeanEccentric Anomalies What is the altitude of this same satellite 10 minutes past apogee At apoapse M1180 10 minutes past apoapse M2 M1181 hrhr21016 M221016 E2 esinE2 e01682 Solve forE2 raI ecosE2 a8028km h r 6378km lecmre4 ppt R s Nerem 2004 Review of T woBody Problem Gm m f Gm m 17 A 2 A 2 msazrsaz 2 m m r 2 m r r r r Gm m f M A g r 2 3 r M Gm r r r A A A A A A 2 2 hrgtltV rCMatb hr6rv 2 V Energy Integral 2 r ALKE LPE both relative to one of the bodies r L v 6 w true anomaly 1 ecosv 2 Equatlon of come sectlon p hA semllatus rectum eAh7 M lecmre4 ppt R s Nerem 2004 Review of T woBody Problem Elliptical Orbits O S e lt l 2 2 a semimajor aXis b a 1 e2 6 a b b semiminor aXis a 2 2 p5aI e25 a M rPLal e rALale Ie I e Flight Path Angle h W cos pm 8 i gt V2 g l VisViva Equation 2a r a lecture4 ppt R s Nerem 2004 Review of T woBody Problem gtV le V E e P aI e A ale r VESC 7 aoo r C 23h 2 M P2rrlag t n 6 n2a3y raI ec0sE Also e 1 E esinEnt tpM lecture4 ppt R s Nerem 2004 Example Problems 1 An Earth satellite is observed to have a perigee height of 100 km and an apogee height of 600 km Find the period and e arArP 100637860063786728km 2 2 P2n 549211sec15256h0urs e rA rP 6978 6478 rA rP 6978 6478 00372 Can also nd V M if P given AVPP lecmre4 ppt R s Nerem 2004 7 Example Problems 2 How many days each year is the Earth farther from the Sun than lAU I AUI49597870 km Weneed rgta e00167 err a1 ecosE gt a I ecosE gt1 ecosElt0 cosElt0 90 ltE lt 270 E1 90 M1 E1 esinE1 8904 E 270 M2 E esinEZ 27096 MZ M1 360 n 09856 deg day At 36526 n At 846 days lecmre4 ppt R s Nerem 2004 8 Example Problems 3 Neglecting the eccentricity of Neptune s orbit how many years in each Pluto orbit is Pluto closer to the Sun than Neptune rP aP1 epcosE ap 39544AU eP 0249 rNaN eN0009 aN3011AU Need rp aPI epcosE lt aN gtCosEgti1aiN1 3011 ep gp 0 956077 0249 39544 E lt 1 7 0445 gtlMllt129097 M E ePsinE Required time I 7 7 years 71 p n 360 P 24 6 74 years lecmre4 ppt R s Nerem 2004 ASEN 5050 SPACEFLIGHT DYNAMICS Orbital Elements Dr Steve Nerem University of Colorado Boulder lecmre4 ppt R s Nerem 2004 Canonical Units Reduce size of numbers More mathematically stable Speed up algorithms Allow different orgs to use standard values Reduce maintenance programming 959 De ne distance unit to be one Earth radius 1 ER R9 63 78137km ER3 TU2 R 3 Thus our time unit TU is T U F Me TU is time for satellite to cover 1 radian in a circular orbit of radius R9 U 1 8068 seconds 39860044 Vallado uses canonical units in examples throughout the book We want 91 so de ne Me lecmre4 ppt RS Nerem 2004 1 1 Canonical Units Example 1 Given A geosynchronous orbit Find The semimajor axis a P 24 sidereal hours 861648068 106795869 TU 1067958697 61 1 P 2 13 M 2a 2a a 66107346456378 1363 4216417124 km 2 13 6610734645 ER lecmre4 ppt R s Nerem 2004 12 Orbital Elements mum 5 mm m e323 l n We 95 s Angular mommmm I Wt I I l i l 39 WW Figure 242 Classical ommmmems The 5 classical orbital elmmus are me Mimew axis a eccentricity e incumuwn i langum uaxcemiing node o en referred v v 4 1 u m i n39 n quot scale and sum clemenls a and at because I ve inuudnwd mm in Chm I Vallado i 997 lectuwt ppt R s Nerem 2004 13 Orbital Elements Law of Cosines 23 ABcosoc State Vector 9 Orbital Elements Keplerian Elements From our twobody derivation we have rx uil r Since e the eccentricity vector and y r where id e e points towards periapse lectuwt ppt R s Nerem 2004 14 Orbital Elements Angular momentum in 7 If 39 V t Figure 2712 ClassiCHIOibilalmemenlsi The six Magical arbth element are Ihc sermmmar am a mutt19 e inclimz an r lungllude ujuscending made l oneu referred 4 r I n mu m u l scale and shape elements a and 5 because I ve mzmdums mm m Chap l Vallado1997 12mm ppl R s Nam mm Orbital Elements Now let s de ne our other orbital elements The inclination 139Arefers to the tilt of the orbit plane It is the angle betweenK and h and varies from 01800 0 lt i lt 90 Prograde orbit wEarth s rotation 90 lt i lt 180 Retrograde orbit against Earth rotation i 90 Polar Orbit K cosz A A K kl 1mm ppl R 5 Nisan mm Orbital Elements The right ascension of the ascending node 9 is the angle in the equatorial pla e from to the ascending node The ascending node is the point on the equator where the satellite passes from South to North opposite for the descending node The line of nodes connects the ascending and descending nodes The node vector 72 points towards the ascending node and is denoted nKx The node liesAbetween 0 and 360 cosQIAl If nlt0 then 93600 9 ll l sinQL cos9 hr hsini hsini lecmre4 ppt RS Nerem 2004 17 Orbital Elements The argument at geriagse a measured from the ascending node locates the closest point of the orbit periapse and is the angle between n and E cosw Ifezlt0cu36OO w Md The true anomaly v is the angle between periapse and the satellite position thus A cosv 7 AT sU WltOv3 f v 6 M f 39v is positive going away from periapse negative coming towards periapse lecmre4 ppt R s Nerem 2004 18 Special Cases Elliptical Equatorial Orbits S2 is unde ned so we use trie longitude of periapse a7 lme N I e cos 00m A If e lt 0 then 07m 360 as true For i 0 equivalent to astronomers longitude of periapse 6 where 07 Q 00 Circular Orbits o is unde ned use argument of latitude u Where n cosu If r2 lt 0 then u 3600 u an Circular Equatorial o and 2 unde ned v7 7 1ecture4 ppt R S Nerem 2004 19 COSLme 01339 lt 0 Aime Aime i T we Line Element Sets Designator Epoch Argument Mean Mean Epoch ev Right Inclination of perigee Anomaly Motion node Figure 215 Transmission Format for the 2line Element Set This example 2line element set uses data from the previous example in the text Note the use of implied decimal paints S 18 the ago of the values E 15 the exponent Vallado 1 997 Available on Class web page Can be read by many programs including STK n l3eiQaM r2 n39 CA BgtxltD 2 6 2mp0 c7 17 are Kozai means B lt is a drag parameter UTC 1ecture4 ppt R S Nerem 2004 20 Two Line Element Sets Example 1 16609U 86017A 9335253502934 00007889 000000 105293 34 2 16609 516190 133340 0005770 1025680 2575950 1559114070 44786 Epoch Dec 18 1993 12h 50min 265350 sec UTC 1559114070 revday gt 6 I06118087 ER 6768357 km 3 7889x10 5 revclqu 2 00 revday B I0529gtlt10394 e 00005770 i 516190 9 133340 w1025680 M 2575950 Errors can be as large as a km or more lecmre4 ppt R s Nerem 2004 Orbital Elements from I7 and 17 and t Algorithm 9 in the book First compute the following vectors 7xv 1 x aVXhl M r hl l n e l Compute the energy Vj a 7X r M h2 P Qcosr n a org Agtalez n If nlt0 Q360 Q 12 4171 o e or e1282 w cos E 1fellt0w 360 w 5 h vcos4i If vlt0 v360 v zcos 1 er lecmre4 ppt R s Nerem 2004 Orbital Elements from 1quot and 17 and 2 Test using Example 25 in book Also v tan 7 2 1 n hamAppi RS imam 23 Example 25 Example 25 Finding the Orbital Elements with ELORB GIVEN 7 6524834 I 6862875 J 5442195 K km 1 4901327 I 5533756 1 117634i K kms FIND Clusswil orbital clemcnls Convert umls Io nd 7 1023 I 1076 I 1011 K ER i7 062 I 07 J 025 K ERTU The magnitudes are r 2 1796 225 ER and v 0967 936 ERTU Ecgin by nding he speci c angular momentum i A L023 1076 LDH gt 39 2 ii rXi 062 m 4125 ovamuossz 57 JUU48 98K ER rm 24 mm ppi R s Nenm 2cm Example 25 The magnitude of angular momentum is 1317 297 Find ihe node veclnr not he mean motion using a cross producr 2 xx 70882571 03767J lZl 1316 386 ERln U 1K 22 1t 139 1v 0832 853 E vji 119079363 2 r 2 v 3146 0385 231 O 668 04K m 7 70088 273 ERZITUI 0 a Because 1h arbit isn t parabolic Lhe eccentricity is not 10 nd the semimajor axis by a 7 11 5664 247 ER 3612734 km Then find Lin scmiparumelcr using lhe speci c angular mnmeninm 2 Z L 1735 272 ER 11067790 km 1 The ncxi msk is to demrminc Lhc angles Find lhe incllnuuun wilh a cosine expression No quadrant check is necessary mm was 1 2mm 25 Example 25 0048 98 C05 39 z m kl 1317297 139 87870quot Find the longitude ofthe ascending node Notice Llie quadmnt check here affecLs 1111 nal value quotI 4188257 05 1 1H1 2 quot A 1316386 mu gt 1513 10 lFnJltOTHEN S2 360 XI 22789 Find the argument of perigee similarly but withoul modi cation for quadrants A 1 cos quotJ 2 5333a w 11 1 1316 3860832 853 w 1F2Klt 0TH5N 2 360 7w Finally nd Lhe ime anomaly A 1 ms 1 46 2 92335 W mm 0332 mm 796 226 39 IFgtltOTHEN v 360k 12mm ppl R s Nexem zuuA Example 25 Evalualt the special angles recalling the limjlalions impuscd for planes in each calculalionr Deter man the elc ems relating to the locau un of pcrlgce Notice the dvffcrence in the values of 1115 rue longitude of perigee and 1h lungiludc Dfperigeel comm 3146 a m and1n1tmlly WE 13 112194 lFllt0TllEN 21mm 360 139HW 81quot 247806 11 01 22789C 5338quot 23127 Find Lhc argument of latitullr COSM ln M u COS 1670882 57l023709767 1076 145605 495 1316 3861796 226 Finally delcrmina Lhe Iruc longitude lF1KltOTllL7N u 360 1 1023 C03 1 r 5282587 lt m mm 5 FrjltOTHEN Am 2 EGOLX mu lecmreAppl RS NeanDDA


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