3 Keplerian Orbits3.1 Newtons Universal Law of Gravitation Newtons law of gravitation states that any two particles of mass m 1 and m2,distance r apart, are attracted toward each other with a force Gmlm2 F -- r2 (3.1)where G is a universal constant called the gravitational constant. The value of Gis (6.6726 ± 0.0005) x 10 -11 m3/kg-sZ(lcr), which is ~ o w n to about 1 part in13,000.1 The value of/z = GMe, on the other hand, where Me is the Earth mass,is known to a much higher accuracy, and is therefore used for astrodynamicalcalculation. Accurate measurement of the gravitational attraction between smallmasses is limited by the accuracy to which G is known. There are several laboratory techniques available for the determination of G.One of the most accurate experiments to determine G is the "time of swing method"of Heyl. This approach is illustrated in Fig. 3.1, where two large masses M andtwo small masses m are placed on coaxially suspended torsional springs. When two balances are aligned in parallel, the period of oscillation is less thanwhen they are aligned at right angles. In the former position, the gravitationalattraction adds to the torsional spring whereas, in the latter position, the attractionsubtracts from the spring. The periods are on the order of 0.5 h and can be measuredto 0.1 s of accuracy. A constant torque is thus maintained on the free balance. The angular displace-ment, after many hours, determines G. Other methods for improving the accuracy of G include a dynamically resonanttorsional balance system, with a possible accuracy of 1 part in 106. Miniatureorbital systems with equal or unequal masses have also been proposed that couldbe established in space to determine G to a high accuracy. The gravitational force equation (3.1) can be expressed in terms of the gradientV of a scalar gravitational potential function V of a body. Thus, one can write F = VV (3.2)where V = tz/r for a spherical body. For a nonspherical body, V contains addi-tional terms that can be used to determine F, as will be shown later.3.2 General and Restricted Two-Body Problem A more useful and general expression of the gravitational law is in its vectorformulation. Thus, referring to Fig. 3.2, the masses m and M are moving in aninertial reference frame El, E2, E3. It is desired to determine the motion of mrelative to the larger mass M. 21
22 ORBITAL MECHANICS Torsional Spring M Fig. 3.1 Torsional balance system. The general two-body problem results i f a ~ 0. Then, for the mass M, GMmr M~---- (3.3) F3and, for the mass m, GMmr m/~ r3 (3.4)Subtracting Eq. (3.4) from (3.3) yields G ( M + m)r r3 (3.5) G ( M + m)r Y+ -0 1,.3 In a restricted two-body problem, the principal mass M is assumed fixed ininertial space. This implies that M >> m, so that m does not affect the motion ofM. Letting M be at the origin O (i.e., a = O), then the force on the mass m is GMmr F-- (3.6) r3or, in accordance with Eq. (3.4) with p = r, GMmr F -- - m~ (3.7) r3
KEPLERIAN ORBITS 23 a Fig. 3.2 Two-body system.Therefore, GMr i; + ?.3 -- 0 (3.8)where i~ = d2r/dt 2 = acceleration of mass m relative to the inertial frame. Equation(3.8) represents the motion of mass m in a gravitational field of mass M, assumedspherically symmetric and, therefore, concentrated at the origin of the referencesystem. Equation (3.8) differs from Eq. (3.5) only in the gravitational constant term.The motion of the restricted two-body problem is therefore similar to that of thegeneral two-body system and is affected by the magnitude of the gravitationalterm. The latter is a negligible effect when m << M, which is true for the satellitesof the Earth and other planetary bodies.3.3 Conservation of Mechanical Energy Consider Eq. (3.8) in the form /zr =o (3.9)where/z = GM, and m is assumed negligibly small compared to M. Scalar multiplication of Eq. (3.9) by/" results in ~r • ~.~+ - - --0 (3.10) F3or (3.11) d-t -b ~-~~-~ =0
24 ORBITAL M E C H A N I C Ssince d(r • r)/dt = 2r#, etc. This equation can be integrated to yield (#)2 # --8 2 r = specific mechanical energy (3.12)Here (#)2/2 = v2/2 = specific kinetic energy and -tz/r is the specific potentialenergy of the satellite. The specific potential energy is also equal to the gravita-tional potential function per unit mass.3.4 Conservation of Angular Momentum The specific angular momentum H of a satellite (angular momentum per unitmass) can be obtained by vector-multiplying Eq. (3.9) by r. Then, ]zr rxia+rx~=0 (3.13)which shows that d rx~= ~(rxk) d ~--H dt = 0 (3.14)Consequently, H = const. This means that r and/" are always in the same plane.The actual solution for the satellite motion can be obtained by cross-multiplyingEq. (3.9) by H. Then, i; x H = ~ ( H x r) (3.15)or d /Z @ x H) (r20)r0 dt rj = ~0~ = #d(?) (3.16)where the magnitude of the specific angular momentum H = r20, 0 is a unitvector normal to the unit vector ? along the r vector, and 0 is the angular rate ofthe r vector. Integrating Eq. (3.16), one obtains k x H =/z? +B (3.17)where B is a constant of integration.
KEPLERIAN ORBITS 25 Furthermore, since r • (k x H ) = r • (#~ + B ) = (r x k ) . H =H .H : n 2 H 2 = # r + r B cos 0 (3.18)therefore, H2/l~ r~ (3.19) 1 + (B/lz) cos 0where H 2 / l z = p = semilatus r e c t u m 2 e H 2 ,] 1/2 B/Iz = e = 1 + -7-] = eccentricity 0 = true a n o m a l yThe general e q u a t i o n for the radius r in Eq. (3.9) is therefore of the form P r - (3.20) 1 + ecos0 This is an equation for a conic section, an example of which is the ellipseillustrated in Fig. 188.8.131.52 Orbital P a r a m e t e r s o f a S a t e l l i t e The orbit ellipse geometry is shown in Fig. 3.3. The following notation is used: a s e m i m a j o r axis = (ra + rp)/2 = b = semirninor axis e eccentricity = (ra - r p ) / ( r a -t- rp) = 0 = true a n o m a l y ra = apogee radius = a(1 + e) rp = perigee radius = a(1 - e) p = a(1 - e 2) = b2/a = rp(1 + e) = ra(1 - e) = semilatus r e c t u m y -- flight-path angle = zr/2 - / ~ The radial velocity c o m p o n e n t Vr can be f o u n d as follows: dr d r dO dr • -- 0 vr = ~ - d0dt dO dr H (3.21) dO r2
26 ORBITAL MECHANICS FLIGHT PATH ANGLE VELOCITY J r SEMIMINOR AXIS SEMILATUS RECTUM F 1 APOGEE EMPTY FOCUS GEOMETRIC CENTER GEOMETR F~]CUS ~ERIGEE ~ ae ~a=a(1 =all APGGEE = APOGEE RADIUS ÷e) ;L, R%u, "1" SEMIMAJDR AXIS Fig. 3.3 Ellipse geometry (from Ref. 2).but, since dr d dO = d-O[p(1 + ec°sO)-l] H e sin 0 Ur -- _ _ P = ~ / ~ e sin 0 (3.22)The normal component vn is found as vn = rO =r(H) ~pp ~-~ = (1 + e c o s O ) (3.23)The flight-path angle y is given by V = cos-1 = tan-1 (3.24)where v = (d + @,/2 (3.25) = (1 + e 2 + 2e cos O)
KEPLERIAN ORBITS 27The velocity at perigee Vp is ve = g ~ ( l + e ) (3.26)and, at apogee, va = ~ / - ~ ( 1 - e ) (3.27)These are the m a x i m u m and minimum values of v in orbit, respectively. The velocity v at any position in orbit is found from the energy or vis-vivaequation (3.12) as follows: v2 /z e- (3.28) 2 ras e --+ 1, va --~ 0, ra --+ 2a; therefore, e = - t z / 2 a since the energy e remainsconstant. F r o m Eq. (3.28), (3.29) For a circular orbit r = a and, therefore, the velocity Vc = x / - ~ / r . For the escapetrajectory, a = e~ (parabola) and the velocity Vesc = ~/2vc. The classification ofthe different possible conic sections of orbits in terms of the eccentricity e is asfollows: Orbit 0 Circle (a = r) <1 Ellipse (a > 0) 1 Parabola [a ~ w (undefined)] >1 Hyperbola (a < 0)The period of an elliptic orbit is P = 2 r r / n , where n = v/-~-/a 3 = mean motion. If the position of a satellite is desired at a specified time t, then it can be foundfrom M = n(t - r) (3.30)where r -----time of perigee passage, and M = mean anomaly, and from Keplersequation, M = E - e sinE (3.31)which can be solved for E.
28 ORBITAL MECHANICS F A0×,L,A.Y C,.CLE ~ Fig. 3.4 PERIGEE Definition of eccentric anomaly (from Ref. 2). The definition of the eccentric anomaly E is shown in Fig. 3.4. The true anomaly0 can be determined from 0 I I + e ] 1/2 E = tan -- (3.32) tan ~ L1 - e J 2The trigonometric arguments 0/2 and El2 are not always in the same quadrant. Conversely, if the time t of travel from one point on the ellipse to another pointis desired, then it can be found from Eq. (3.30), where M is given in Eq. (3.31).3.6 Orbital Elements The motion of a satellite around the Earth may be described mathematically bythree scalar second-order differential equations. The integration of these equationsof motion yields six constants of integration. It is these constants of integrationthat are known as the orbital elements. The Keplerian orbital elements are often referred to as classical or conventionalelements and are the simplest and easiest to use. This set of orbital elements can bedivided into two groups: the dimensional elements and the orientation elements. The dimensional elements specify the size and shape of the orbit and relate theposition in the orbit to time (Fig. 3.3). They are as follows: a = semimajor axis, which specifies the size of the orbit. e = eccentricity, which specifies the shape of the orbit. r = time or perigee passage, which relates position in orbit to time (r is oftenreplaced by M, the mean anomaly at some arbitrary time t; the mean anomaly is auniformly varying angle) The orientation elements specify the orientation of the orbit in space (Fig. 3.5).They are as follows: i = inclination of the orbit plane with respect to the reference plane, which istaken to be the Earths equator plane for satellite orbits. 0 deg < i < 180 deg).
KEPLERIAN ORBITS 29 EARTHS o = LONGITUDE OF THE ASCENDING NODE NORTH o= OLE ... . SATE,.+E E r / f . " - - - PERIG E [ .----~-L_~-~--~ :7".4"~=~-/=Z- -.~ECUPTJC //~ ~ ~ "/" ~ ~ PLANE Fig. 3.5 Orientation of orbit in space (from Ref. 2).For 0 deg < i < 90 deg, the motion is "posigrade" or "direct"; for 90 deg < i <180 deg, the motion is termed "retrograde" f2 = right ascension of the ascending node (often shortened to simply "node");f2 is measured counterclockwise in the equator plane, from the direction of thevernal equinox to the point at which the satellite makes its south-to-north crossingof the equator (0 deg < f2 < 360 deg) 09 = argument of perigee; ~o is measured in the orbit plane in the direction ofmotion, from the ascending node to perigee (0 deg < w < 360 deg) The angles i and f2 specify the orientation of the orbit plane in space. The anglew then specifies the orientation of the orbit in its plane. The argument of latitudeu defines the position of the satellite relative to the node line. Still another system of specifying the satellite state vector involves the scalarquantities of v = velocity r ----radius f2 -- node Y = flight-path angle ----geocentric latitude Az = azimuth of v from true north Various sets of elements are used in orbit determination, the inertial rectan-gular (x, y, z, .~, ~, ~) and the spherical (or, ~, fi, Az, r, v), where ot is longitude
KEPLERIAN ORBITS 31 a) b) NP NP E UT R Q AOFig. 3.6 Earth orbits: a) retrograde orbit: 9 0 < i < 1 8 0 deg, 180<Az<360 deg; b)posigrade orbit: 0 < i < 90 deg, 0 < A z < 180 deg.and/3 = rr/2 - y are examples of such elements. Note that both of these setsof elements give the satellites position as a point in space at any specific time.Several orbital element systems are listed in Table 3.1. Posigrade and retrogradeorbits are illustrated in Fig. 3.6. References 1Luther, G. G., and Tower, W. R., "Redetermination of the Newtonian GravitationalConstant G," Physical Review Letters, Vol. 48, No. 3, 18 Jan. 1982, pp. 121-123. 2Cantafio, L. (ed.), Space-Based Radar Handbook, Artech House, Norwood, MA, 1989. Selected Bibliography Battin, R. H., An Introduction to the Mathematics and Methods of Astrodynamics, AIAAEducation Series, AIAA, Washington, DC, 1987. Bate, R. R., Mueller, D., and White, J. E., Fundamentals of Astrodynamics, Dover, NewYork, 1971. Flury, W., Vorlesung Raumfahrtmechanik, Technische Hochschule Damstadt, 1994. Kaplan, M. H., Modern Spacecraft Dynamics and Control, Wiley, New York, 1976. Prussing, J. E., and Conway, B. A., Orbital Mechanics, Oxford, 1993. Roy, A. E., Orbital Motion, 3rd ed., Adam Hilger, Bristol and Philadelphia, 1988. Taft, L. G., Celestial Mechanics, Wiley, New York, 1985. Wertz, J. R., and Larson, W. J., eds., Space Mission Analysis and Design, KluwerAcademic Publishers, 1991. Problems3.1, The period of revolution of a satellite is 106 min. Find the apogee altitudeif the perigee altitude is 200 km. (/z = 3.986 × 105 km3/s 2)3.2. Find the period of revolution of a satellite if the perigee and apogee altitudesare 250 and 300 km, respectively.3,3. Find the maximum and minimum velocity of the Earth if the eccentricityof the Earths orbit about the sun is 1/60. What is the mean velocity if the meandistance to the sun is 149.6 x 10 6 kill? (/zs = 1.327 × 1011 km3/s 3)
32 ORBITAL MECHANICS3.4. Show that for a satellite moving on an elliptical orbit the velocity at thetime of passage through the minor axis is equal in magnitude to the local circularvelocity (vc). (Hint: Vc = ~/-~/r.)3.5. A satellite in a circular orbit at an altitude hc above the Earths surface isgiven a velocity v0 with a flight-path angle y. What is the magnitude of v0 if theperigee altitude of the resultant orbit is to be equal to hp(hp < he)?3.6. A sounding rocket is launched from a planet. Find initial velocity to reachheight H above the surface. Assume a spherical planet of radius R and gravity gat surface.3.7. Find the escape velocity from the moons surface. Assume that 1 gmoon = ~gEarth rmoon = ~rEarth3.8. Assuming that the period of Mars about the sun is 687 Earth days, find themean distance of Mars to sun if the Earth distance to sun is 149.5 x 106 kin.3.9. Find the mass of the sun using Keplers third law.3.10. For a Keplerian orbit with period P = 205 min, eccentricity e = 0.40, andtrue anomaly 0 = 60 deg, find the time t since passage of perigee.3.11. Given the following orbit: hp = 200 km ha = 600 km a) What is the time interval over which the satellite remains above an altitude of400 km? Assume a spherical Earth with radius = 6378 km and/z = 3.986 x 105knl3/s 2 . b) What additional information is needed to solve part a if the Earth is assumedto be nonspherical?3.12. A spaceship is in a 200-km circular orbit above a spherical Earth. At t = 0,it retrofires its engine, reducing its velocity by 600 m/s. How long (in minutes oftime) does it take to impact the Earth?3.13. An object was observed at a distance of (1.05) 2/3 Earth radii from thecenter of the Earth. Sixteen minutes later, the same object was observed at aposition 60 deg (measured at the Earth center) from the original position. Showwhether the object is in a circular orbit. Assume that/z = 0.00553 (ER3/min2).