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Orbital Mechanics: 11. Orbital Systems


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  • 1. 11 Orbital Systems11.1 Launch Window ConsiderationsLaunch Azimuth The launch of a satellite into an orbit with inclination i requires launching in anazimuth direction Az defined by the formula cos i sin Az - (11.1) COS q~Owhere ~b0is the launch site latitude. For a due east launch (Az = 90 deg), i = q~0,orthe orbital inclination is equal to the launch site latitude. Differentiating Eq. (11.1)and solving for di, one obtains - cos 4)o cos A z d A z (di) = (11.2) sin iwhich shows the sensitivity of inclination error di due to an error in azimuth dAz.For range safety reasons, the azimuth may be constrained between specified limits,as is shown in Fig. 11.1, for example.M-Plane or Out-of-Plane Ascent Using a Launch AzimuthConstrained to the Nominal Value The most efficient launch is an in-plane launch ascent characterized by a waiton the ground until the launch site lies in the mission orbit plane, which is fixedin inertial space. No plane change maneuver is then required unless the orbitinclination is less than the launch site latitude or the launch azimuth constraintis such that a "dog-leg" maneuver is needed. A direct ascent to a parking ormission orbit may be made with or without a phasing orbit to place the satellitein a specified geographic location. The phasing problem where the parking andmission orbits are not coplanar may involve excessive waiting times, which canbe somewhat reduced by "lofted" or other trajectories. The out-of-plane launch is a form of direct ascent when the phasing problemcan be solved by a wait on the ground. The launch window geometry is shown inFig. 11.2. The launch window can be defined as the time, L.W. = 2 Afa/coe (11.3)where Af2 = Wet is the maximum permissible nodal increment consistent withthe plane change angle e required to transfer from the launch orbit plane to themission orbit plane. The latter can be determined from the equation S = COS-1(COS2 i + sin 2 i COS Aria) (11.4) 241
  • 2. 242 ORBITAL M E C H A N I C S :: .... : USSR~2~45o" 2B- (470 inclination) ,L..-,/ /" /" _ ;,~ ,,/--,,: "~7.~/7"~51° 3"05o3o9o~ A ~ USSR ::::: i~ : i: :/:: o 300 ---..~jA PAN ! ] .LoL"2{< I 172° (83 ...inclination) Fig. 11.1 Geographic launch constraints.where i is the inclination of the orbit and the corresponding launch azimuthis constrained to the nominal value provided by Eq. (11.1). The longitudinalseparation Af2 is the same at all latitudes. In general, a graph of e vs time is as shown in Fig. 11.3, where the plane changecapability e of the launch vehicle defines the launch window.Launch Azimuth Constrained to Off-Nominal Value If the launch azimuth is constrained to an off-nominal value, then Fig. 11.4depicts the launch geometry at the time t = 0 the launch site lies in the missionorbit plane. The launch geometry at some later time, t > 0, is shown in Fig. 11.5.Applying spherical trigonometry to the triangle bounded by the launch orbit,mission orbit, and equator results in the equation cos s = cos i cos i + sin i sin i cos(cOet -I- A~0) (11.5)where A g20 is the difference in right ascension of the ascending node of the launchand mission orbits at t = 0.
  • 3. ORBITAL SYSTEMS 243 Z. N. P. C ~ we / ~ ORBIT PLANE ~ / INTERSECTION ON LAUNCHSITE t AT : 0 -~xl/ ~ I/ " ", ~ / EARTH SURFACE x ~ ~/~~"~ E UT R QAO COS i = sin AZ cos LAUNCH WINDOW : 2T Fig. 11.2 Launch window geometry. Launch Vehicle Plane Change Capabil~tyW In-Plane Launch Opportunity Time (t) Launch Window (L.W.) -l Fig. 11.3 Launch window.
  • 4. 244 O BT L E H N S R I AM C A I C ~ e ~ plane AT L L u csite a nh ~ latitude E~aS qh~ r Fig. 11.4 Geometry for fixed launch azimuth at t = 0. Milstead 8 showed that the condition for minimum plane change angle couldbe found by simply differentiating Eq. (11.5). The resulting condition,(09et "b A ~ 0 ) = 0, implies that g r o i n : li - i[ which, in turn, requires that thelaunch and mission orbit planes intersect on the equator. Figure 11.6 from Ref. 8 describes plane change angle vs time for several fixedvalues of launch azimuth and for a launch latitude of 28.34 deg and a missionorbit inclination of 30 deg. For these conditions, a launch azimuth of 79.722 degsatisfies Eq. (11.1), and the satellite can be launched directly into the mission orbitat t -- 0, with no plane change required. Another opportunity to launch directlyinto the mission orbit with no plane change required occurs about 168 min laterwhen the launch site rotates into the mission orbit plane again. For a launch azimuth of 90 deg, the inclination of the launch orbit equals28.34 deg. The minimum plane change angle Smin equals [28.34 deg - 30 degl,which equals 1.66 deg, as shown on Fig. 11.6.Launch Azimuth Continuously Optimized to Minimize Plane Change If the capability exists to vary the launch azimuth with time, then an optimumazimuth solution to minimize the plane change angle can be found. Reference 8derives this solution, again by differentiation. The intuitively appealing solutionis that the minimum plane change angle is achieved by launching the vehicle insuch a manner that it will intercept the target plane 90 deg downrange from the " Fig. 11.5 Geometry for fixed launch azimuth at t > 0.
  • 5. ORBITAL SYSTEMS 245 "10 ~3 g i o "/, t 4 = 79.7 ~ i 10 = Target -- -- :: 32)1v /Vo plane g inclination A L = Fixed -- launch 1 -- -- 4° ~ = Launch s i t e .,,= latitude 0 I - 120 -60 0 60 120 180 240 300 Time from nominal launch time, t, min Fig. 11.6 L a u n c h w i n d o w s for fixed launch azimuth.position of the launch site at the time of launch. This launch geometry is illustratedin Fig. 11.7. Reference 8 derives the following equation for the minimum planechange angle as a function of time: sin emin = sin i cos ~b sin(wet + A L ) - cos i sin ~b (11.6)where i is the inclination of the target orbit, ~b is the launch site latitude, and tan ~b sin A L -- (11.7) tan i Figure 11.8 from Ref. 8 illustrates emin as a function of time for several valuesof target orbit inclination and for a launch site latitude of 28.34 deg. Note thesecond opportunities to launch directly into the mission orbit with no plane changerequired when the launch site again rotates into the mission orbit plane. The optimum launch azimuth to achieve Emin is given by ~ cos i cos ~b + sin i sin(wet + A L ) sin ¢ (11.8) tan Aopt = sin i cos(wet + AL)where 0 < Aopt < 180 deg, and is illustrated in Fig. 11.9.Sun-Orbit Orientation Frequently, thermal or power considerations of spacecraft require that the anglebetween the sun and the orbit plane be maintained within specified bounds for the
  • 6. 246 ORBITAL M E C H A N I C S pole Target plane , Launch plane 90 ° Launch site eJet Equator Fig. 11.7 Launch geometry. 5 ¢9 "t3 ;-4 3 2 O.. 1 0 -240 -120 0 120 240 360 Time from nominal launch time, t, min Fig. 11.8 Launch windows for variable optimum launch azimuths.
  • 7. ORBITAL SYSTEMS 247 120 J f i=2834o I I 29.00 ° 10 30.00 ° ID 32~"xK 31.00 ° / ~. 100 o J~ / / E 90 g c -- 8G E E i = Target plane inclination q~ = Launch site latitude = 28.34 ° O 7O 6_.:,P. 40 -120 I 0 I 120 I 240 360 Time from nominal launch time, t, min Fig. 11.9 O p t i m u m launch azimuth.duration of the mission. This angle, conventionally known as the beta angle/3, isillustrated in Fig. 11.10. Since/3 is the complement of the angle between the sunvector ~ and the positive normal to the orbit fi, it follows from their scalar productthat /3 = sin-l(,~ • fi) (11.9)and ultimately that 1 /3 = sin -1 [cos 3s sin i sin(f2 - RATS) + sin 6s cos i] (11.10)where/3 is defined to lie in the range from - 9 0 to + 9 0 deg. Equation (11.10) reveals that beta angle depends on solar declination 8s, orbitinclination i, and the difference in right ascensions of the true sun and the ascendingnode (~2 - RATS). The first of these quantities, gs, depends on the date duringthe mission. The second quantity, i, is essentially constant during the mission.The last quantity (f2 - RATS) changes because of nodal regression (induced byEarths oblateness perturbations, as described in Chapter 8) and seasonal variationin the right ascension of the true sun. In light of the variability of the terms on the right side of Eq. (11.10), it is clearthat the beta angle cannot be held constant throughout a mission. However, it isgenerally possible to select conditions at the start of the mission so that the betaangle will stay within some prescribed tolerable range of values for that portionof the mission during which/3 is essential to performance.
  • 8. 248 ORBITAL MECHANICS Positive Normal 1 to t h e O r b i t Orbit Plane Fig. 11.10 Definition of sun-orbit angle, beta.Earth Eclipsing of a Circular Orbit It may be important to determine those occasions during its mission whena satellite is eclipsed by the Earth. Such eclipsing occurs whenever the satellitepasses through the Earths shadow, which is assumed cylindrical in this discussion.*Figure 11.11 shows that the Earths shadow intersects the orbital sphere of asatellite at altitude h in a minor circle whose Earth-central-angular radius is/3%where I /3 = sin -1 [R/(R + h)], 0 deg < fl* < 90 deg (11.11)View A-A in Fig. 11.11 reveals that the orbit intersects the perimeter of theshadow circle at points El and E2. Note that the length of the eclipsed orbital arcE1 E2 is just twice arc CE1, where C is the point on the orbit of closest approachto the shadow axis A. That is, the length of arc AC is just the magnitude of ft.Hence, it follows from the right spherical triangle ACE~ that Au = cos -1 (cos/3*/cos/3) (1 1.12)When Eqs. (11.11) and (11.12) are combined, the eclipsed fraction of the circularorbit is found to be 2Au 1 [ ~/h 2 + 2Rh ] - - -- cos -1 (11.13) fE-- 27r 7r / (RT~ cos/3 J *Such an assumption is valid at low satellite altitudes, where there is no appreciable differencebetween the umbral and penumbral regions of total and partial eclipsing, respectively.
  • 9. ORBITAL SYSTEMS 249 " ~ Orbit re A Sunlight Earths Shadow (cylindrical) A -~-- Orbit Eclipsed Orbital Arc. View A-A E1 E2 = 2Au Fig. 11.11 Eclipsing of a circular orbit (from Ref. 1). Figure 11.12 shows the variation of eclipse fraction with/%angle magnitudefor several low-altitude circular orbits. Note that the eclipse fraction diminisheswith an increase in either altitude or magnitude of/~. One should also observethat eclipsing can occur only if I¢~1 is less than the critical angle/~* for the givenorbit altitude. For example, a satellite in a 200-n.mi.- altitude circular orbit will beeclipsed through some portion of its orbit if and only if 1/31 < 70.93 deg. For a circular orbit, the time duration of the eclipse interval is directly propor-tional to its angular extent; in other words, the duration of the eclipse on a givensatellite revolution is just the product of the eclipse fraction fe and the orbitalperiod. However, this is not true for eccentric orbits. The duration and angularextent of eclipsing of a satellite in an eccentric orbit depends on the eccentricityand the relative orientation of sun and perigee, as well as the sun-orbit angle ¢~. Figure 11.13 presents the Earth, its shadow, and the circular synchronous equa-torial orbit at the time of the vernal equinox. The circular synchronous orbit radius
  • 10. 250 ORBITAL M E C H A N I C S.~ 0.4- L Orbit Altitude (h) n mi,..-t o.3-u ~ 0.2"r~ o.z- m 0 I0 20 30 40 50 60 70 80 Magnitude of the Sun-Orbit Angle, 161(deg)Fig. 11.12 Eclipse fraction vs beta-angle magnitude and circular orbit altitude (fromRef. 1).of 6.61 Earth radii produces an orbit period of 24 sidereal hours. A satellite in thisorbit revolves with an angular rate that exactly equals the angular rate of rotationof the Earth on its axis. An Earth-centered inertial (ECI) coordinate system whose origin is at the centerof the Earth is shown in this figure. The X axis points to the vernal equinox, the Z axis points to the celestial North Pole, and the Y axis forms a right-hand triadwith X and Z. In Fig. 11.13a, the sun is along the X axis, and the shadow axis intersects thesynchronous equatorial orbit. The Y axis points 23.45 deg below the ecliptic plane. This angular difference of 23.45 deg is important at the summer solstice(Fig. 11.13b) because the Earth casts a shadow that passes beneath the circu-lar, synchronous equatorial orbit. And so no eclipsing occurs at this time. Figure 11.14 shows the geosynchronous eclipse geometry. For geometricalpurposes, consider the relative motion of the Earths shadow as it moves along theecliptic plane. At the distance of the synchronous orbit, the radius of the disk of theumbra is 8.44 deg. Therefore, 21.6 deg or 22 days before the autumnal equinox,the umbral disk becomes tangent to the equatorial plane. This is the beginning of theeclipse season because this is the first time a satellite in the synchronous equatorialorbit could experience an umbral eclipse, although it would be vanishingly short.At the time of the autumnal equinox, the center of the umbral disk would lie inthe equatorial plane. This geometry produces a maximum time of 67.3 min. Twenty-two days later, the eclipse season ends when the umbral disk is againtangent to the equatorial plane. And so, the eclipse season is 44 days, centered onthe autumnal equinox. A similar eclipse season is centered on the vernal equinox.Sun Synchronism of an Orbit In Chapter 8, one of the principal perturbations of an orbit caused by Earthsoblateness was identified as nodal regression (2. Recall that Eq. (8.3) showed that
  • 11. ORBITAL SYSTEMS 251 N RH OT •C L S I L EE T A Z IPL OE INTERSECTS ~ ~ , : "--.. I - v o COUATORIAt. ~X ~o % SUN % ~. b) N RH OT / C L SI L EE TA / ~ O J . E Y "-.3," BEffEATH ORBIT XFig. 11.13 Geosynchronous orbit geometry: a) vernal equinox shadow; b) summersolstice shadow.~2 depends as follows on orbit inclination i, average altitude h, and eccentricity e: -9.9639 ( R ~3.5 . deg (11.14) -- ~ - ~ - e ~ ~ - h ] c°S t mean solar daywhere h = (ha q- h p ) / 2 e = (ha - h p ) / ( 2 R + ha + hp)and R = the equatorial radius of the Earth. An orbit is said to be sun-synchronous if its line of nodes rotates eastwardat exactly the orbital angular velocity of the mean sun. Since the mean sun
  • 12. 252 ORBITAL MECHANICS END OF BEGINNING ECLIPSE EQUINOX OF ECLIPSE SEASON c,~rONG-° L_ E SEASON ..,.- ~ - 22 DAYS ECLIPSE 22 DAYS AFTER AE = 67 3 MtN BEFORE AE L~.~.~"-RADIUS OF ~ . UMDRALDISK ~ _...--f~o ~ At D,StANCEOF / / ~ / ~.~ . SYNC,RONOUS ( 1 ( _ ( / o.,T DIRECTION OF MOTIONOF TIIE SIIADOI,f DISK WITH RESPECTTO THE EQUATOR Fig. 11.14 Geosynchronous orbit eclipse at vernal or autumnal equinox.moves uniformly eastward along the celestial equator through 360 deg in a trop-ical year (about 365.242 mean solar days), the required rate of nodal regres-sion is 360/365.242, or 0.985647 deg/day. Substituting that numerical value inEq. (11.14) and solving for the sun-synchronous inclination yields iss ----cos -1 - 0.098922(1 - e2) 2 1 + (11.15)Note that, since the cosine of iss is always negative, the inclination of a sun-synchronous orbit must be greater than 90 deg; that is, sun-synchronous orbits arenecessarily retrograde. For circular orbits, e is zero, and h is the constant altitude above a sphericalEarth with radius R. In that case, the variation of sun-synchronous inclinationwith altitude is shown in Fig. 11.15. Note that sun synchronism is possible forretrograde circular orbits up to an altitude of about 3226 n.mi. or a radius of6670 n.mi. (for R = 3444 n.mi.), at which point the orbit inclination reaches itsgreatest possible value, 180 deg. Figure 11.16 shows examples of noon-midnightand twilight orbits.Launch Window to Satisfy Beta-Angle Constraints The term launch window, as used here, will mean the time interval (or intervals)on a given date within which a satellite can be launched into a prescribed orbitwith the subsequent satisfaction of sun-orbit angle constraints (upper and lowerlimits on fl) for the specified mission duration. If the mission duration were zero days, one would simply determine the timeinterval(s) on the given launch date within which the combination of solar decli-nation 3s, orbit inclination i, and sun-node orientation (O~s - ~2) yields acceptablevalues of/3, as shown in Eq. (11.10). However, if the mission has any extent, the determination of launch windowinvolves more effort. In such a case, one must account for the changes that occurin 3s and in (O~s - g2) during the mission and select the reduced launch windowaccordingly. As was mentioned earlier, the seasonal change in 8s is inevitable.However, the initial value of (O~s - ~2) can easily be controlled by varying the
  • 13. ORBITAL SYSTEMS 253 / "~ 180 15o o o 120 f to J I J J 90 0 500 1000 1500 2000 2500 3000 Alltltude I ( n mi I • l ~ I I I I I I I 0 i000 2000 3000 4000 5000 6000 Altitude (kilometers) Fig. 11.15 Sun-synchronous inclination vs circular orbit altitude (from Ref. 1).launch time; a 1-h delay in launch moves the line of nodes 15 deg eastward andthus reduces (Ors - f2) by 15 deg on any given launch date. Furthermore, the rateof change of (O~s - f2) can be adjusted by varying the orbit inclination or, to alesser degree, the altitude. In particular, the average rate of change of (as - f2) canbe made to vanish by employing a sun-synchronous combination of altitude andinclination. In this way, a preferred constant value of (Oes - f2) and an associatedacceptable range of/3 values can often be maintained throughout the mission.11.2 Time of Event Occurrence If an event such as a perigee passage by a satellite occurs over a certain longitudei , then the time t of the event can be specified unambiguously from the equation ~ , = G H A + i + coet (11.16) = right ascension of the eventwhere GHA = Greenwich hour angle (at Greenwich midnight of date) t = longitude of the event We = 15.041067 deg/h = Earth rotation rate t = time from Greenwich midnight (GMT) The values of G H A can be found in standard reference works for each date. Forexample, on January 1, 2000, G H A = 99.97 deg. If the event is a perigee passageby a satellite, then, O/, = ~ ÷ ~pp (11.17)where ~pp = tan -1 (tan co cos i)
  • 14. 254 ORBITAL MECHANICS ~ ""A -EH R T NM OB OG R OH T NT I D-I N I ( TWILIGHT ORBITFig. 11.16 Sun-synchronous orbits- a) noon-midnight orbit (F~ - A = 0); b) tvdlightorbit ( a - ,~ = 90 deg).and w = argument of perigee (< 180 deg)The time of perigee passage tpp is then given by /pp - = ( f ] ~ - / J , pp - - GHA - )~pp)/(o e (11.18)where ~.pp = longitude of perigee passage11.3 Ground-Trace ConsiderationsGeneral Characteristics Circular figure eight, or "eggbeater," types of ground traces can be obtained byusing satellites with 12- or 24-h periods at different inclinations to the equator.The use of several orbital planes equally spaced in node can result in severalsatellites moving in the same ground trace. Two examples of this are shown in
  • 15. ORBITAL SYSTEMS 255 TRACK • I; o" ", • o° ~ ", Fig. 11.17 Regional navigation satellite system cencept.Figs. 11.17 and 11.18, which show the circular and eggbeater types of groundtraces with several satellites in each ground trace. Figure 11.17 is for a regionalsystem with four satellites, whereas Fig. 11.18 is for a possible global navigationsystem employing twenty satellites. Other orbital systems involving up to four 12-h satellites in each of six equallyspaced orbit planes have been found useful for global navigation purposes, as inthe global positioning system (GPS), for example. One measure of performance of a navigation satellite system is the performancefactor called geometric dilution of precision (GDOP), which is a measure of howsatellite geometry degrades accuracy. The magnitude of the ranging errors to aminimum of four selected satellites, combined with the geometry of the satellites,determines the magnitude of the user position errors in the GPS navigation fix.The four "best" visible satellites are those with the lowest GDOP. 2 Thus, GDOP = ~/(PDOP) 2 + (TDOP) 2 (11.19)where PDOP = ratio of radial error in user position 1 ~ in three dimensions to range error 1 ~r. TDOP = ratio of error 1 cr in the range equivalent of the user clock offset to range error 1 a.
  • 16. 256 ORBITAL MECHANICS Z/Perigee 5 Equator 3 4 3 4 Z 5 Global System (Z0 satellites} 4 satellites in each of g o r b i t planes 180° apart In longitude w i t h perigee in southern hemlllphere o 4 satellites in each o f Z o r b i t planes 180 apart in longitude with perigee in northern hemisphere 4 s a t e l l i t e s i n synchronous equatorial orbit, 90° apart In longitude. All inclined planes (i = 60 ° ) with e c c e n t r i c i t y e = . 25 a r e separated 90o in longitude relative to each o t h e r . Fig. 11.18 Global navigation satellite system concept.Perturbation Effects Earths nonsphericity, solar-radiation pressure, and solar/lunar gravitational at-traction tend to alter the satellites orbital elements in time. An example of howthe ground trace can change as a result of the Earths gravitational harmonics isshown in Fig. 11.19, where the initial and a four-year ground trace are illustrated. The results are for a Q = 1 eccentric (e = 0.27) orbit, whose period is23h55"~59.3 S, with an initial inclination i0 = 28.5 deg. The perturbations are dueto the principal gravitational harmonics and the solar/lunar gravitational accel-erations. The satellite trace repetition parameter Q is the number of satelliterevolutions that occur during one rotation of the Earth relative to the osculatingorbit plane. Q is approximately the number of satellite revolutions per day.11.4 Highly Eccentric, Critically Inclined Q = 2 Orbits (Molniya) This discussion is devoted to examining the characteristics of a very specializedorbit: the Q = 2 orbit, which is highly eccentric and critically inclined, withapogee located over the northern hemisphere. This type of orbit has the usefulcharacteristic of enabling observation of vast areas of the northern hemispherefor extended periods of time each day. Typically, two properly phased spacecraftlocated in two ideal ground track locations (groundtracks repeat day after day)will view continuously 55-60% of the northern hemisphere, centered at the NorthPole, as illustrated in Fig. 11.20.Orbital Geometry The following paragraphs examine each of the classical orbital elements, andoptimal values are assigned to those elements for which such assignment hasmeaning.Semimajor Axis The semimajor axis a is determined by calculation to be approximately 26,554km. This value assures that the groundtrack of the orbit will remain fixed relative
  • 17. ORBITAL SYSTEMS 257 90 80 70 60 50 40 30 20 10 N 0 2 -10 -20 -30 -40 -50 60 -70 . -80 -90 190 200 210 220 230 24,0 250 260 270 280 290 300 310 320 330 LONGITUDE(degE)Fig. 11.19 Four-year ground trace change due to Earths gravitational the Earth and will repeat itself every day (Q = 2, an integer, assures this).Operationally, there are forces that tend to alter a (drag, solar pressure, tesseralharmonics, etc.), and fuel must be expended by the spacecraft to make periodiccorrections to a. When a ~ 26,554 kin, the Keplerian period is 11.967 mean solar hours, whichis one-half of the quantity 360 deg (one complete Earth rotation on its axis relativeto the stars) divided by 15.041067 deg/h (the Earths rotation rate). And so, to bea repeating groundtrack, Q = 2 orbit, the period is not 12 h (or one-half a meansolar day), but 11.967 h, which is one-half of a mean sidereal day.Eccentricity The value of eccentricity e will vary over a typical mission lifetime, largelybecause of solar and lunar gravitational perturbations. Eccentricities ranging from0.69 to 0.74 are typical of this orbit type. The initial mission value of e is dictatedby launch date, mission duration, minimum acceptable value of hp, groundtracklocation, and initial right ascension of ascending node (f2). An adequately accuratecomputer program must be utilized to assure that the spacecraft will not re-enterthe atmosphere at a premature time.
  • 18. 258 ORBITAL MECHANICS < t- in II Fig. 11.20 Q = 2 (Molniya) type of orbit.
  • 19. ORBITAL SYSTEMS 259Inclination For observation of northern hemisphere regions, the inclination should be high.In order that apogee be maintained at the northernmost point for long periods oftime, the inclination must be approximately 63.44 deg (the critical inclination).Inclinations higher than 63.44 deg will force the line of apsides to rotate in adirection opposite the direction of satellite motion. Inclinations lower than 63.44deg will rotate the line of apsides in the same direction as the satellite motion.These statements are valid for direct orbits only. The arguments are reversed forretrograde motion.Right Ascension of Ascending Node The node of the orbit can take any value, 0-360 deg. The value chosen essentiallypinpoints the time of day of launch. At launch, the node (or time of day) must becarefully chosen so that lunisolar perturbations do not act to reduce h e below anacceptable level during the nominal mission duration.Argument of Perigee In order that optimal visibility of the northern hemisphere be maintained, ~oshould be held as close to 270 deg as possible. Selecting i = 63.44 deg assuresthat ~o will drive away from 270 deg by less than +5 deg for mission lifetimes onthe order of 5 yr.11.5 Frozen OrbitsIntroduction The term frozen orbit first appeared in the literature in 1978 in a paper byCutting et al. entitled "Orbit Analysis for SEASAT-A. 3 A frozen orbit is onewhose mean elements, specifically, eccentricity e and argument of perigee ~o, havebeen selected to produce constant values, or nearly so, of e and co with time. Thus,perigee rotation is stopped, and the argument of perigee is frozen at 90 deg. Frozenorbits maintain a constant altitude profile over the oblate Earth from revolution torevolution, with very small altitude variations over the northern hemisphere. Thisprofile appears to decay uniformly because of atmospheric drag. Thus, frozenorbits are particularly useful for low-altitude missions that require eccentricitycontrol. If radiation pressure and, perhaps, drag are not too influential, frozen orbiteccentricity will remain nearly constant for years. Otherwise, active maneuveringto maintain the frozen orbit is feasible and has been demonstrated (see Ref. 4).Frozen orbits are necessary for missions that require tight longitudinal control ofthe ground trace because there are no longitudinal variations of the ground tracedue to perigee rotation. The orbit will remain frozen in time if only zonal harmonics are assumed in thegeopotential, i.e., no tesseral harmonics, and if atmospheric drag, solar-radiationpressure, and third-body gravitational perturbations are ignored. Geometric interpretations of the first two zonal harmonics, ,/2 and ,/3, are shownin Fig. 11.21. The Earths oblateness is caused by a combination of J2 and therotation of the Earth. The difference between the polar and equatorial radii of theEarth is 21.4 km. The pear shape of the Earth is caused by J3. The height of the
  • 20. 260 ORBITAL MECHANICS SPHERICAL EARTH + ROTATION + J2 f ( SPHERICAL EARTH = J3 %: REQ -- RpOLE = 2 1 . 3 7 k M J2 = 1 . 0 8 2 6 3 x 1 0 - 3 POLAR BULGE 16.5 M J3 = - 2 . 5 3 2 1 5 × 10 - 6 Fig. 11.21 Shape of the Earth.bulge at the North Pole is about 16.5 m. For the Earth, the value of ,/3 is threeorders of magnitude less than the value of J2.Variational Rate Equations Equations (11.20) and (11.21) are the averaged variational rate equations foreccentricity and argument of perigee for an Earth model consisting of a rotatingEarth plus zonal harmonics ,/2 and ,/3 (see Chapter 9). All elements are meanelements: de 3J3n ( R ) 3 ( 5 sin2 i ) cost ° (11.20) dt - 2(i ---72) 2 sin/ 1 - J3 (11.21) dt - (1~)2 1 - 2J2(1 - e 2) l )sini: in ]where n = the satellite mean motion R = the Earths mean equatorial radiusMean elements reflect secular and long-period perturbative variations. Mean ele-ments can be obtained from osculating elements by averaging out the short-periodoscillations.
  • 21. ORBITAL SYSTEMS 261 Note that there is no J2 term in Eq. (11.20) but that Eq. (11.21) containsboth J2 and ./3. Also, note that both equations contain the well-known expression(1 - 5/4 sin 2 i). These equations equal zero when this expression equals zero, i.e.,when i = 63.4 deg or 116.6 deg, the critical inclination. These are frozen-orbitsolutions. This solution was probably first implemented in 1964 with the adventof the Molniya satellite orbit. However, the frozen-orbit solutions of interest are implemented by setting co =90 deg, so that de/dt = 0, and by setting the bracketed term in Eq. (11.21) equalto zero, so that dco/dt = 0. From this term, the frozen mean eccentricity can besolved for given values of the mean semimajor axis and mean inclination. Notethat the eccentricity [obtained as a solution to Eq. (11.22)] is J3 ( R ) sini (11.22) e ~ -2J2It is of the order of 10 -3 because J3 is three orders of magnitude less than J2 (seeFig. 11.21).Frozen-Orbit Solutions Results from Eq. (11.22) are presented in Fig. 11.22 as the dashed line. Ex-tensive numerical investigations of frozen-eccentricity solutions have revealedthe importance of higher-order zonal harmonics, especially for inclinations nearthe critical inclination. The solid line in Fig. 11.22 describes frozen-eccentricitysolutions obtained by using the J z - J 1 2 zonal harmonic terms in the WGS-84geopotential model. There are distinct behaviors of these solutions in three regions of inclination:i < 63.4 deg, 63.4 deg < i < 66.9 deg, and 66.9 deg < i < 90 deg. At i = 10deg, the frozen eccentricity is 0.132 x 10 -3. As inclination increases, the frozeneccentricity increases in an oscillatory manner as shown in Fig. 11.22. The frozen-eccentricity increases only slightly between i = 35 deg and i = 40 deg. Thefrozen-eccentricity increases dramatically, however, as the inclination approachesthe critical value, i.e., i = 63.435 deg. A frozen solution, e = 10.1 x 10 -3, wasfound at i = 63.0 deg. Presumably, frozen orbits with larger eccentricities can befound in the inclination region between 63.0 and 63.435 deg. Inverted frozen orbits were discovered for inclinations slightly greater than63.435 deg. Inverted frozen-orbit solutions occur at co = 270 deg rather thanco = 90 deg. Thus, perigee is at the southernmost point in the orbit rather than at thenorthernmost point. The initial discovery of inverted frozen orbits was publishedby J. C. Smith in 1986. 6 At i = 63.6 deg, the frozen eccentricity was foundto be 23.5 x 10 -3, or 0.0235. As inclination increases, the frozen eccentricitydecreases dramatically. At i = 66.6 deg, the frozen solution is e = 0.124 x 10 -3and, at i = 66.8 deg, e = 0.041 x 10 -3. Then, as inclination increases slightly, a transition in frozen solutions fromco = 270 deg to co = 90 deg occurs, so that, for i = 67.1 deg, e = 0.068 x 10 -3at co = 90 deg. And, for a narrow range of inclinations, no frozen-orbit solutionscould be found. As inclination increases from i = 67.1 deg, eccentricity increasesrapidly to e = 0.65 x 10 -3 at i = 70.0 deg. Then, as Fig. 11.22 shows, the frozeneccentricity increases less rapidly with increasing inclination until e = 1.26 x
  • 22. 262 ORBITAL MECHANICS MEAN a ° = 7041.1 K M .... J2 - J3 ZONAL FIELD - - J2 - J12 ZONAL FIELD 40 30. 6. T 4 3 " 2__ M E A N = 270 DEG 2 CRITICAL _~_~__ o INCLINATION = ~. o 63.435 DEE ~ ~ ...... i / -, // 0.8 0.6 0.5 -MEAN ~ = 9o DEC 0.4 k MEAN tO = 90 DEG 0.3 0.2 0.I i0 20 30 40 50 60 70 80 90 MEAN I N C L I N A T I O N ~ D E G R E E S Fig. 11.22 Frozen-orbit solutions.10 -3 at i = 90.0 deg. Thus, the curve corresponding to the J2 - J12 zonal field isquite different from the curve corresponding to the J2 and J3 zonal field, especiallyfor inclinations between 50 and 75 deg. In the vicinity of the critical inclination,the values ofde/dt and dco/dt from Eqs. (11.20) and (11.21) are near zero. Termsinvolving higher zonal harmonics than J2 and J3 then become relatively moreimportant.Circulations About the Frozen-Orbit Conditions It is useful and interesting to examine the behavior of mean e and mean co forinitial values that are near the frozen-orbit values. For nearby initial values, e andco will circulate about the frozen-orbit conditions as illustrated in Fig. 11.23. Thisfigure is from Ref. 5. For i = 98.7 deg and a semimajor axis value of 7198.7 km,the frozen eccentricity is 1.15 x 10 -3 at co = 90 deg. Although the semimajoraxis is slightly different, by about 160 km, the value of frozen e is very nearly thevalue obtained from Fig. 11.22 at an inclination of 81.3 deg, i.e., the supplementof 98.7 deg. For initial conditions that are near, but not at, the frozen point, e and co willmove counterclockwise in closed contours. For inclinations less than 63.435 deg
  • 23. ORBITAL SYSTEMS 263 PHASE SPACE DIAGRAM WITH TIME TICS C~ 35.0 - F R ~= 98.7* O 3 0 . 0 - 25.0 -- ~-- 2:0.0 -- c_) Z (-) 15.0 -- 10.0 - ~,.0 O.O o 90 ~8o z~o 36o ARGUMENT OF PERIGEE (degrees) Fig. 11.23 Phase space diagram marked at 13-day interval (from Ref. 5).or greater than 116.565 deg, the motion is clockwise. The period of circulation is360 deg/&, where & is the familiar J2 secular apsidal rate. A -/3 - J21 zonal fieldwas used to generate the circulation curves, and 13-day intervals in time weremarked on the curves. The distribution of time tics on the larger closed curves isinteresting. More time tics are clustered at the higher than the lower eccentricities.Thus, during one circulation, the eccentricity is larger than the frozen value for alonger period of time than it is smaller than the frozen value. For larger deviationsfrom the frozen point, the contours do not close. In Ref. 7, Rosborough and Ocampo derive and present frozen-orbit eccen-tricity for an Earth gravity field complete to degree 50. Their results show thatinverted frozen orbits, i.e., co = 270 deg, occur in a low-altitude (a < 1.09 Earthradii), low-inclination (i < 10 deg) region as well as in a region of inclination63.4 deg < i < 66.9 deg. References IGinsberg, L. J., and Luders, R. D., Orbit Planners Handbook, The Aerospace Corp.,E1 Segundo, CA, Technical Memorandum, 1976. 2Milliken, R. J., and Zoller, C. J., "Principle of Operation of Navstar and System Char-acteristics," Navigation: Journal of the Institute of Navigation, Vol. 25, Summer, 1978,pp. 95-106. 3Cutting, E., Born, G. H., and Frantnick, J. C., "Orbit Analysis for SEASAT-A," Journalof the Astronautical Sciences, Vol. XXVI, Oct.-Dec. 1978, pp. 315-342. 4McClain, W. D., "Eccentricity Control and the Frozen Orbit Concept for the NavyRemote Ocean Sensing System (NROSS) Mission," AAS Paper 87-516, Aug. 1987.
  • 24. 264 ORBITAL MECHANICS 5Nickerson, K. G. et al., "Application of Altitude Control Techniques for Low AltitudeEarth Satellites," Journal of the Astronautical Sciences, Vol. XXVI, April-June 1978,pp. 129-148. 6Smith, J. C., "Analysis and Application of Frozen Orbits for the Topex Mission," AIAAPaper 86-2069-CP, Aug. 1986. 7Rosborough, G. W., and Ocampo, C. A., "Influence of Higher Degree Zonals on theFrozen Orbit Eccentricity," AAS Paper 91-428, Aug. 1991. 8Milstead, A. H., "Launch Windows for Orbital Missions," The Aerospace Corp., E1Segundo, CA, Report No. TDR-269(4550-10)-6, April 1, 1964. Problem11.1. On 19 June, 1995 the Space Shuttle Atlantis was scheduled to launch andrendezvous with the Russian Space Station Mir. Cape Canaveral is the launchsite, ~ = 28.34 deg. Nominally, Atlantis would launch directly into Mirs orbitplane whose inclination was 51.6 deg. For performance reasons Atlantis has aplane change capability of only 52.4 m/sec. If the plane change occurs at an altitudeof 350 km, what was the launch window associated with Atlantis plane changecapability? What was Atlantis launch azimuth from the Cape? Selected Solution11.1. Launch window = + / - 2 min Launch azimuth = 44.9 deg