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Satellite communication - Dennis Roddy notes chapter 2

Satellite communication - Dennis Roddy notes chapter 2

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• 1. Chapter 2 Orbits and Launching MethodsIntroduction Satellites (spacecraft) orbiting the earth follow the same laws thatgovern the motion of the planets around the sun. From early times muchhas been learned about planetary motion through careful observations. Johannes Kepler derived empirically three laws describingplanetary motion. Later Sir Isaac Newton derived Kepler’s laws from hisown laws of mechanics and theory of gravitation. Kepler’s laws apply quite generally to any two bodies in space whichinteract through gravitation. The more massive of the two bodies isreferred to as the primary, the other, the secondary or satellite.Kepler’s three laws of planetary motion Kepler’s laws apply to any two bodies interacting in spacethrough gravitation, the more massive of the two being the primary andthe other the secondary or the satellite.KepIer’s First Law Keplers first law states that the path followed by a satellite aroundthe primary will be an ellipse.➔An ellipse has two focal points shown as F1 and F2 in Fig. 2.1. Thecenter of mass of the two-body system, termed as barycenter is alwayscentered on one of the foci.
• 2. ➔ln our specific case, because of the enormous difference between themasses of the earth and the satellite, the center of mass coincides with thecenter of the earth, which is therefore always at one of the foci.➔The semimajor axis of the ellipse is denoted by a, and the semiminoraxis, by b. The eccentricity e is given by➔ Two of the orbital parameters specified for satellites (spacecraft) orbiting theearth are the eccentricity e and the semimajor axis, a . For an elliptical orbit, 0<e< 1. When e = 0, the orbit becomes circular. P H Y S IC A L L A W S K e p le r’s 1 s t L a w : L a w o f E llip s e s T h e o rb its o f th e p la n e ts a re e llip s e s w ith th e s u n a t o n e fo c u sKepler’s Second LawKepler’s second law states that, for equal time intervals, a satellite willsweep out equal areas in its orbital plane, focused at the barycenter.
• 3. ➔Assume the satellite travels distances S1 and S2 meters in 1 s, then theareas A1 and A2 will be equal.➔ The average velocity in each case is S1 and S2 m/s, and because of theequal area law, it follows that the velocity at S2 is less than that at S1.➔An important consequence of this is that the satellite takes longer to travel agiven distance when it is farther away from earth. This property can be madeuse to increase the length of time a satellite can be seen from particulargeographic regions of the earth. PHYSICAL LAWS Kepler’s 2nd Law: Law of Equal Areas The line joining the planet to the center of the sun sweeps out equal areas in equal times T4 T3 T5 T2 A4 A3 A5 A2 T1 A1 T6 A6
• 4. Kepler’s Third LawKepler’s third law states that the square of the periodic time of orbit isproportional to the cube of the mean distance between the two bodies.➔The orbital period in seconds is P and the mean distance is equal to thesemimajor axis a. P 2  a3➔The orbital period P is defined as the time, the orbiting body takes to returnto the same reference point in space with respect to the inertial space.➔P can be written as➔where n is the mean motion of the satellite in radians per second.➔Then 2  2 3 or   a n➔For the artificial satellites orbiting the earth, Kepler’s third law can be writtenin the formwhere n is the mean motion of the satellite in radians per second and is the earth’s geocentric gravitational constant or Keplers constant. Itsvalue is =3.986005×10 14 m3 / s 2➔This equation applies only to the ideal situation of a satellite orbiting aperfectly spherical earth of uniform mass, with no perturbing forces acting, suchas atmospheric drag➔The importance of Kepler’s third law is that it shows there is afixed relationship between period and semimajor axis. One very importantorbit , known as the geostationary orbit, is determined by the rotationalperiod of the earth➔another importance of this law is that it is used to find the period of theorbit of any satellite and is used in GPS receivers to calculate positions ofGPS satellite.
• 5. Explain the three features the orbit of a satellite needs to have to be perfectlygeostationary.1.It must be exactly circular (ie e=0)2.it must be at the correct altitude (ie have the correct period)3.it must be in the plane of equator (ie have a zero inclination with respect tothe equator).If the inclination is note zero and/ or if the eccentricity is not zero, but theorbital period is correct then the satellite will be in geosynchronous orbit. P H Y S IC A L L A W S K e p le r’s 3 rd L a w : L a w o f H a rm o n ic s T h e s q u a re s o f th e p e rio d s o f tw o p la n e ts ’ o rb its a re p ro p o rtio n a l to e a c h o th e r a s th e c u b e s o f th e ir s e m i- m a jo r a x e s : T 12/T 22 = a 13/a 23 In E n g lis h : O rb its w ith th e s a m e s e m i- m a jo r a x is w ill h a v e th e s a m e p e rio dDefinitions of Terms for Earth-Orbiting SatellitesORTerms used to describe the position of the orbit with respect tothe earth.1. Subsatellite path. This is the path traced out on the earth’s surfacedirectly below the satellite.
• 6. 2. Apogee. The point farthest from earth. Apogee height is shown as hain Fig. 2.3.3. Perigee. The point of closest approach to earth. The perigee height isshown as hp in Fig. 2.3. ORBIT CLASSIFICATIONS Orbit Geometry Eccentricity = c/a a Perigee Apogee c c4. Line of apsides. The line joining the perigee and apogee through thecenter of the earth.
• 7. 5. Ascending node. The point where the orbit crosses the equatorialplane going from south to north.6. Descending node. The point where the orbit crosses the equatorialplane going from north to south.7. Line of nodes. The line joining the ascending and descending nodesthrough the center of the earth.8. Inclination. The angle between the orbital plane and the earth’sequatorial plane. It is measured at the ascending node from the equator tothe orbit, going from east to north. The inclination is shown as i in Fig.2.3. It will be seen that the greatest latitude, north or south, reached bythe subsatellite path is equal to the inclination. COORDINATE SYSTEMS Geocentric Inertial Inclination9. Prograde orbit. An orbit in which the satellite moves in the samedirection as the earth’s rotation, as shown in Fig. 2.4. The prograde orbitis also known as a direct orbit. The inclination of a prograde orbit alwayslies between 0° and 90°.Most satellites are launched in a prograde orbit because the earth’srotational velocity provides part of the orbital velocity with a consequentsaving in launch energy.10. Retrograde orbit. An orbit in which the satellite moves in a directioncounter to the earth’s rotation, as shown in Fig. 2.4. The inclination of aretrograde orbit always lies between 90° and 180°
• 8. Argument of perigee. The angle from ascending node to perigee,measured in the orbital plane at the earth’s center, in the direction ofsatellite motion. The argument of perigee is shown as w in Fig. 2.5.
• 9. Right ascension of the ascending node. To define completely theposition of the orbit in space, the position of the ascending node isspecified. However, because the earth spins, while the orbital planeremains stationary the longitude of the ascending node is not fixed, and itcannot be used as an absolute reference. For the practical determinationof an orbit, the longitude and time of crossing of the ascending node arefrequently used. However,for an absolute measurement, a fixed referencein space is required. The reference chosen is the first point of Aries,otherwise known as the vernal, or spring, equinox. The vernal equinox occurs when the sun crosses the equator goingfrom south to north, and an imaginary line drawn from this equatorialcrossing through the center of the sun points to the first point of Aries(symbol ϒ). This is the line of Aries. The right ascension of the ascendingnode is then the angle measured eastward, in the equatorial plane, fromthe ϒ line to the ascending node, shown as Ω in Fig. 2.5. COORDINATE SYSTEMS Geocentric Inertial Direction of Satellite motion Vernal Equinox Ascending Node Right AscensionMean anomaly. Mean anomaly M gives an average value of the angularposition of the satellite with reference to the perigee. For a circular orbit,M gives the angular position of the satellite in the orbit. For ellipticalorbit, the position is much more difficult to calculate.True anomaly. The true anomaly is the angle from perigee to thesatellite position, measured at the earth’s center. This gives the trueangular position of the satellite in the orbit as a function of time. satellite True anomaly
• 10. Orbital ElementsEarth-orbiting artificial satellites are defined by six orbital elementsreferred to as the keplerian element set.1.Semimajor axis a2.Eccentricity ea and e give the shape of the ellipse.3.Mean anomaly M0, gives the position of the satellite in its orbit at areference time known as the epoch.4.Argument of perigee w, gives the rotation of the orbit’s perigee pointrelative to the orbit’s line of nodes in the earth’s equatorial plane. It isshown in figure 2.5.5.Inclination i6.Right ascension of the ascending node Ωi and Ω relate the orbital plane’s position to the earth.[Note: write the definition of these terms.]Because the equatorial bulge causes slow variations in w and Ω, andbecause other perturbing forces may alter the orbital elements slightly,the values are specified for the reference time or epoch, and thus theepoch also must be specified.Two-line elements Two line elements contains a great deal of general information onpolar orbiting satellites as well as weather satellites in the geostationaryorbit.Figure 2.6 shows how to interpret the NASA (National Aeronautics andSpace Administration) two-line elements.
• 11. Apogee and Perigee Heights Although not specified as orbital elements, the apogee height andperigee height are often required.The length of the radius vectors at apogee and perigee can be obtainedfrom the geometry of the ellipse:ra = a(1+e)rp = a(1-e)In order to find the apogee and perigee heights, the radius of the earthmust be subtracted from the radii lengths. h a =r a−R ; h p=r p−RConic section
• 12. SN =ZN −ZS SN =QP−WM SP SM =e , =e PQ MW r p SN = − e e SN =r cos  r p So ,r cos= − e e 1 −p rcos − = e e r e cos−1=− p p r= 1−ecos  p r= 1ecos v p when =0 ; r= =SA 1−e pwhen =1800 ; r= =SA 1e
• 13. AA a= =semimajor axis 2 SASA = AA p p 2a=  1−e 1e 2p 2a= 1−e 2 so p=a 1−e 2  1−e2  SA =a =a 1e=apogee ,r a 1−e 1−e 2  SA=a =a 1−e= perigee ,r p 1eOrbit Perturbations The keplerian orbit is ideal in the sense that it assumes that theearth is a uniform spherical mass and that the only force acting is thecentrifugal force resulting from satellite motion balancing thegravitational pull of the earth. In practice, other forces which can be significant are a) asymmetry of earths gravitational fields( due to nonshericalearth) b)the gravitational forces of the sun and the moon c) solar radiation pressured)atmospheric drag.➔All these interfering forces cause the true orbit to be different fromKeplerian ellipse.➔The gravitational pulls of sun and moon have negligible effect on low-orbiting satellites, but they do affect satellites in the geostationary orbit.➔Atmospheric drag, on the other hand, has negligible effect ongeostationary satellites but does affect low orbiting earth satellites belowabout 1000 km.➔External influences causes changes in longitude and inclination.a) Effects of a nonspherical earth➔The earth is neither a perfect sphere nor a perfect ellipse.➔However, there being an equatorial bulge and a flattening at the poles,a shape described as an oblate spheroid.➔There are regions where the average density of the earth appears to behigher. These are called as regions of mass concentration or Mascons.➔The nonshericity of earth , the noncircularity of the equitorial radius,
• 14. and the Mascons lead to a nonuniform gravitational field around theearth. The force on an orbiting satellite will therefore vary with position.➔For a spherical earth of uniform mass, Kepler’s third law gives thenominal mean motion n0 as➔The 0 subscript is included for a perfectly spherical earth of uniformmass.➔When the earth’s oblateness is taken into account, the mean motion,denoted as n, is modified to➔K1 is a constant which evaluates to 66,063.1704 km2.➔The earth’s oblateness has negligible effect on the semimajor axis a, andif a is known, the mean motion is readily calculated.➔The orbital period taking into account the earth’s oblateness is termedthe anomalistic period (e.g., from perigee to perigee). Because satellitedoes not return to the same point in space once per revolution.➔ The mean motion specified in the NASA bulletins is the reciprocal ofthe anomalistic period.➔The anomalistic period is➔where n is in radians per second.➔Steps for finding a is➔If the known quantity is n one can solve for a, by finding the root of thefollowing equation:➔The oblateness of the earth also produces two rotations of the orbitalplane.➔1. regression of the nodes,2. rotation of line of apsides in the orbital plane
• 15. ➔1. Regression of the nodes is where the nodes appear to slide along theequator.➔In effect, the line of nodes, which is in the equatorial plane, rotatesabout the center of the earth. Thus Ω, the right ascension of the ascendingnode, shifts its position.➔If the orbit is prograde , the nodes slide westward, and if retrograde,they slide eastward.➔As seen from the ascending node, a satellite in prograde orbit moveseastward, and in a retrograde orbit, westward.➔The nodes therefore move in a direction opposite to the direction ofsatellite motion, hence the term regression of the nodes.➔For a polar orbit (i = 90°), the regression is zero.➔Both effects (1 and 2) depend on the mean motion n, the semimajor axisa, and the eccentricity e. These factors can be grouped into one factor Kgiven by➔K will have the same units as n. Thus, with n in rad/day, K will be inrad/day, and with n in degrees/day, K will be in degrees/day. An
• 16. approximate expression for the rate of change of Ω with respect to time is➔where i is the inclination. The rate of regression of the nodes will havethe same units as n.➔When the rate of change given by the above equation is negative, theregression is westward, and when the rate is positive, the regression iseastward. d Ω/dt => positive => eastwardd Ω/dt => negative => westward➔It will be seen, therefore that for eastward regression, i must be greaterthan 90o, or the orbit must be retrograde.Cos(0 to 90 ) is positive and cos( 90 to 180) is negative➔It is possible to choose values of a, e, and i such that the rate of rotationis 0.9856°/day eastward. Such an orbit is said to be sun synchronous➔2. Rotation of line of apsides in the orbital plane,➔Line of apsides. The line joining the perigee and apogee through thecenter of the earth.➔The other major effect produced by the equatorial bulge is a rotation ofthe line of apsides.➔This line rotates in the orbital plane, resulting in the argument ofperigee changing with time. The rate of change is given by➔The units for the rate of rotation of the line of apsides will be the sameas those for n➔When the inclination i is equal to 63.435°, the term within theparentheses is equal to zero, and hence no rotation takes place. d 2 =k2−2.5 sin i dt d =0 ; dt 2 2−2.5 sin i=0 2 2.5 sin i=2 1.10714×180 sin2 i=0.8 ; i=1.10714c = =63.4349 o ➔Denoting the epoch time by t0, the right ascension of the ascending
• 17. node by Ω0, and the argument of perigee by w0 at epoch gives the newvalues for Ω and w at time t as➔The orbit is not a physical entity, and it is the forces resulting from anoblate earth, which act on the satellite to produce the changes in theorbital parameters.➔Thus, rather than follow a closed elliptical path in a fixed plane, thesatellite drifts as a result of the regression of the nodes, and the latitudeof the point of closest approach (the perigee) changes as a result of therotation of the line of apsides.➔With this in mind, it is permissible to visualize the satellite as followinga closed elliptical orbit but with the orbit itself moving relative to theearth as a result of the changes in Ω and w.➔ So, The period PA is the time required to go around the orbital pathfrom perigee to perigee, even though the perigee has moved relative to theearth.➔If the inclination is 90° ,➔the regression of the nodes is zero ,➔the rate of rotation of the line of apsides is dω/dt = −K/2➔ Imagine the situation where the perigee at the start of observations isexactly over the ascending node. One period later the perigee would be atan angle −KPA/2 relative to the ascending node or, in other words, wouldbe south of the equator.
• 18. i=900 sin 90=1  d =−k×0.5 dt➔ −k For 1sec .... 2 −k P A For P A sec .... 2➔The time between crossings at the ascending node would be PA(1+ K/2n), which would be the period observed from the earth.➔In addition to the equatorial bulge, the earth is not perfectly circularin the equatorial plane; it has a small eccentricity of the order of 10−5.➔This is referred to as the equatorial ellipticity.➔The effect of the equatorial ellipticity is to set up a gravity gradient,which has a pronounced effect on satellites in geostationary orbit .➔ A satellite in geostationary orbit ideally should remain fixed relative tothe earth.➔The gravity gradient causes the satellites in geostationary orbit to driftto one of two stable points, which coincide with the minor axis of theequatorial ellipse.➔Due to the positions of mascons and equitorial bulges there are fourequilibrium points in the geostationary orbit.➔Two are stable and the other two are unstable.➔These two stable points are separated by 180° on the equator and are atapproximately 75° E longitude and 105° W longitude.➔Satellites in service are prevented from drifting to these points throughstation-keeping maneuvers.➔Because old, out-of-service satellites eventually do drift to these points,they are referred to as “satellite graveyards.”➔The effect of equatorial ellipticity is negligible on most other satelliteorbits.d) Atmospheric drag➔For near-earth satellites, below about 1000 km, the effects ofatmospheric drag are significant.➔Because the drag is greatest at the perigee, the drag acts to reduce thevelocity at this point, with the result that the satellite does not reach thesame apogee height on successive revolutions.
• 19. ➔The result is that the semimajor axis and the eccentricity are bothreduced.➔Drag does not noticeably change the other orbital parameters, includingperigee height.➔In the program used for generating the orbital elements given in theNASA bulletins, a pseudo-drag term is generated, which is equal to one-half the rate of change of mean motion.➔ An approximate expression for the change of major axis can be derivedas follows➔The change in mean motion,n is dn n=n 0  t−t0  dt n=n 0 n t−t0   a 3= 2 0 n0  a3 = 2 n 2 a 3 n0   =  a0 n 2 n0 3 a=a0   n   2 n0 3 a=a 0 n 0 n t−t0 ➔where the “0” subscripts denote values at the reference time t0, and➔ n 0 is the first derivative of the mean motion.➔The mean anomaly is also changed, an approximate value for thechange being:➔The changes resulting from the drag term will be significant only forlong time intervals, and for present purposes it will be ignored.
• 20. Inclined Orbits➔The orbital elements are defined with reference to the plane of theorbit.➔The position of the plane of the orbit is fixed (or slowly varying) inspace.➔The location of the earth station is usually given in terms of the localgeographic coordinates which rotate with the earth. And the earth stationquantities may be the azimuth and elevation angles and range.➔In calculations of satellite position and velocity in space, rectangularcoordinate systems are generally used .➔So transformations between coordinate systems are therefore required.➔Calculation for elliptical inclined orbits:- the first step is to find theearth station look angles and range➔The look angles for the ground station antenna are the azimuth andelevation angles required at the antenna so that it points directly at thesatellite.➔Elevation is measured upward from local horizontal plane COORDINATE SYSTEMS Topocentric Elevation➔Azimuth is measured from north eastward to the projection of thesatellite path onto the local horizontal plane
• 21. COORDINATE SYSTEMS Topocentric Azimuth Range Origin: Antenna FP: Local Horizon PD: True North➔Determination of the look angles and range:The quantities used are1. The orbital elements,2. Various measures of time3. The perifocal coordinate system, which is based on the orbital plane4. The geocentric-equatorial coordinate system, which is based on theearth’s equatorial plane5. The topocentric-horizon coordinate system, which is based on theobserver’s horizon plane.➔The satellites with inclined orbits are not geostationary, and therefore,the required look angles and range will change with time.➔The two major coordinate transformations needed are:➔■ The satellite position measured in the perifocal system is transformedto the geocentric-horizon system in which the earth’s rotation ismeasured, thus enabling the satellite position and the earth stationlocation to be coordinated.➔■ The satellite-to-earth station position vector is transformed to thetopocentric-horizon system, which enables the look angles and range to becalculated.Calendars➔A calendar is a time-keeping device in which the year is divided intomonths, weeks, and days.➔Calendar days are units of time based on the earth’s motion relative tothe sun.➔It is more convenient to think of the sun moving relative to the earth.➔But this motion is not uniform, and so a fictitious sun, termed the meansun, is introduced.➔The mean sun does move at a uniform speed but requires the same timeas the real sun to complete one orbit of the earth.
• 22. ➔Ie period of mean sun = period of real sun.➔This time being the tropical year.➔A day measured relative to this mean sun is termed a mean solar day.➔Calendar days are mean solar days.➔A tropical year contains 365.2422 days.➔In order to make the calendar year, also referred to as the civil year, itis normally divided into 365 days.➔The extra 0.2422 of a day is significant, and for example, after 100years, there would be a discrepancy of 24 days between the calendar yearand the tropical year.➔ 0.2422×100=24.22 days➔Julius Caesar made the first attempt to correct the discrepancy byintroducing the leap year, in which an extra day is added to Februarywhenever the year number is divisible by 4.➔This gave the Julian calendar, in which the civil year was 365.25 dayson average, a reasonable approximation to the tropical year.➔Again a difference of 0.25−0.2422 days per year exists.➔ 0.25−0.2422×400 years=3.12days➔ie On every 400 years we are actually adding an extra 3 days.➔By the year 1582, an appreciable discrepancy once again existedbetween the civil and tropical years. Pope Gregory XIII took matters inhand by abolishing the days October 5 through October 14, 1582, to bringthe civil and tropical years into line and by placing an additionalconstraint on the leap year in that years ending in two zeros must bedivisible by 400 without remainder to be reckoned as leap years.➔This dodge was used to miss out 3 days every 400 years.➔To see this, let the year be written as X00 where X stands for thehundreds. For example, for 1900, X = 19. For X00 to be divisible by 400, Xmust be divisible by 4. Now a succession of 400 years can be written as X+n, X+ (n+ 1), X + (n + 2), X + (n + 3),X + (n + 4),where n is any integerfrom 0 to 9. If X+ n is evenly divisible by 4, then the adjoining threenumbers are not, so these three years would have to be omitted.➔For example let us take X= 20 and n=0; then in the years 2000,2100,2200, 2300, 2400, 0nly 2000 and 2400 area leap year and theremaining are not even though it is divisible by 4.➔ The resulting calendar is the Gregorian calendar, which is the one inuse today.Universal time➔Universal time coordinated (UTC) is the time used for all civil time–keeping purposes.
• 23. ➔And it is the time reference which is broadcast by the National Bureauof Standards as a standard for setting clocks.➔It is based on an atomic time-frequency standard.➔The fundamental unit for UTC is the mean solar day.➔In terms of “clock time,” the mean solar day is divided into 24 h, anhour into 60 min, and a minute into 60 s. Thus there are 86,400 “clockseconds” in a mean solar day.➔Satellite-orbit epoch time is given in terms of UTC.➔Universal time is equivalent to Greenwich mean time (GMT), as well asZulu (Z) time.➔ For computations, UT will be required in two forms: as a fraction of aday and in degrees.➔Given UT in the normal form of hours, minutes, and seconds, it isconverted to fractional days as UT day= 1 24[ hours 1 60 minutes seconds 60 ] or➔In turn, this may be converted to degrees asSidereal time➔Sidereal time is time measured relative to the fixed stars.
• 24. ➔One complete rotation of the earth relative to the fixed stars is not acomplete rotation relative to the sun.➔This is because the earth moves in its orbit around the sun.➔The sidereal day is defined as one complete rotation of the earthrelative to the fixed stars.➔One sidereal day has 24 sidereal hours, 1 sidereal hour has 60 siderealminutes, and 1 sidereal minute has 60 sidereal seconds.➔ But sidereal times and mean solar times, are different even thoughboth use the same basic subdivisions.➔The relationships between the two systems,➔1 mean solar day= 1.0027379093 mean sidereal days= 24h and (0.0027379093 x 24 )hrs= 24 hrs 0.065709823 hrs= 24 hrs (0. 065709823 x 60) min= 24 hrs 3.942589392 min= 24 hrs 3 m (0.942589392 x 60) sec= 24 h 3 m 56.55536 s sidereal time= 56.55536 + 3 x 60 + 24 x 60 x 60= 86,636.55536 mean sidereal seconds➔1 mean sidereal day = (1/1.0027379093 ) mean solar days
• 25. = 0.9972695664 mean solar days= 23 h 56 m 04.09054 s mean solar time= 86,164.09054 mean solar secondsThe orbital planeor perifocal coordinate system➔In the orbital plane, the position vector r and the velocity vector vspecify the motion of the satellite.➔Determination of position vector r : From the geometry of the ellipse p r= 1ecosv and p=a 1−e 2 ➔r can also be calculated using
• 26. ➔Determination of the true anomaly , v can be done in 2 stagesStage 1 : Find the mean anomaly M .Stage 2 : Solve Kepler’s equation.➔Stage 1: The mean anomaly M at time t can be found as➔Here, n is the mean motion, and Tp is the time of perigee passage.➔ For the NASA elements,➔Therefore,➔Substitute for Tp gives➔Stage 2 : Solve Kepler’s equation.➔Kepler’s equation is formulated as follows➔[.........
• 27. In satellite orbital calculations, time is often measured from the instant ofperigee passage. Denote the time of perigee passage as T and any instantof time after perigee passage as t. Then the time interval of significance ist - T. Let A be the area swept out in this time interval, and let Tp be theperiodic time. Then, from Kepler’s second law,The mean motion is n = 2π/Tp and the mean anomaly is M = n (t - T).Combining these
• 28. Combining
• 29. c= ae is the perigee, position of earth. So area is swept for the time t- T isAEarlier we got the area ascombining these equationsThis is the Kepler’s equation....................................]➔Kepler’s equation isE is the eccentric anomaly.➔Once E is found, v can be found from an equation known as Gauss’equation, which is➔For near-circular orbits where the eccentricity is small, anapproximation for v directly in terms of M is➔r may be expressed in vector form in the perifocal coordinate system (0rPQW frame) as r = (r cos v )P + (r sin v)Q
• 30. Local Mean Solar Time andSun-Synchronous Orbits➔The celestial sphere is an imaginary sphere of infinite radius, where thepoints on the surface of the sphere represent stars or other celestialobjects.➔The celestial equatorial plane coincides with the earth’s equatorialplane, and the direction of the north celestial pole coincides with theearth’s polar axis.➔The the sun’s meridian is shown in figure .➔line of Aries is ➔The angular distance along the celestial equator, measured eastwardfrom the point of Aries to the sun’s meridian is the right ascension of thesun, denoted by S.➔In general, the right ascension of a point P, is the angle, measuredeastward along the celestial equator from the point of Aries to themeridian passing through P. This is shown as P.➔ The hour angle of a star is the angle measured westward along thecelestial equator from the meridian to meridian of the star. Thus for point
• 31. P the hour angle of the sun is (P − S) measured westward➔In astronomy, an objects hour angle (HA) is defined as the differencebetween the current local sidereal time (LST) and the right ascension (α)of the object.hour angle = local sidereal time (LST) - right ascension➔1 HA = 15 degrees➔ie right ascension of the sun → Sright ascension of a point P → Pfor point P the hour angle of the sun is (P − S)➔The apparent solar time of point P is the local hour angle of the sun,expressed in hours, plus 12 h.Ie apparent solar time of point P =LHA of the sun+12 h➔The 12 h is added because zero hour angle corresponds to midday, whenthe P meridian coincides with the sun’s meridian.➔Because the earth’s path around the sun is elliptical rather thancircular, and also because the plane containing the path of the earth’sorbit around the sun (the ecliptic plane) is inclined at an angle ofapproximately 23.44°, the apparent solar time does not measure outuniform intervals along the celestial equator, in other words, the length ofa solar day depends on the position of the earth relative to the sun.➔To overcome this difficulty a fictitious mean sun is introduced, whichtravels in uniform circular motion around the sun.➔The time determined in this way is the mean solar time.
• 32. ➔Figure shows the trace of a satellite orbit on the celestial sphere.➔ Point A corresponds to the ascending node.➔The hour angle of the sun from the ascending node of the satellite is Ω -S measured westward.➔The hour angle of the sun from the satellite (projected to S on thecelestial sphere) is Ω - S +β .➔thus the local mean (solar) time issince 1 degree = 1/15 HA (hour angle)➔ To find β:- Consider the spherical triangle defined by the points ASB.➔This is a right spherical triangle because the angle between themeridian plane through S ( projection of satellite) and the equatorialplane is a right angle.➔The triangle contains the inclination i and latitude λ➔Inclination i is the angle between the orbital plane and the equatorialplane.➔Latitude λ is the angle measured at the center of the sphere goingnorth along the meridian through S.➔ The solution of the right spherical triangle gives β SB tan = OB SB tan i= AB tan  AB = tan i OB AB sin = OB =sin   −1 tan  tan i =arcsin  tan  tan i➔The local mean (solar) time for the satellite is therefore (by substituting thevalue of β)
• 33. ➔As the inclination i approaches 90° angle β approaches zero. i 90o ; tan i ∞ ;  0➔The right ascension of the sun S can be calculated as follows,Consider the earth completes a 360° orbit around the sun in 365.24 days. 365.24 days≡3600  d days≡ s  d days  s= ×360 0 365.24 days➔where Δd is the time in days from the vernal equinox (or line of Aries  ,.➔For a sun-synchronous orbit the local mean time must remain constant.➔The advantage of a sun-synchronous orbit is that the each time thesatellite passes over a given latitude, the lighting conditions will beapproximately the same.➔So it can be used for placing weather satellites and environmentalsatellites.➔Eq. for tSAT shows that for a given latitude λ and fixed inclination i , theonly variables are S and Ω.➔In effect, the angle (Ω- S ) must be constant for a constant local meantime.➔Let Ω0 represent the right ascension of the ascending node at the vernalequinox and Ω the time rate of change of Ω. =0  t−t0  t−t0=d➔For this to be constant the coefficient of Δd must be zero, or
• 34. ➔The orbital elements a, e, and i can be selected to give the requiredregression of 0.9856° east per day.➔These satellites follow near-circular, near-polar orbits.