# 1. The orbit of each planet is an ellipse with the Sun at one focus. 2. The line joining the planet to the Sun sweeps out equal areas in equal times.

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1 Appendix A Orbits As discussed in the Introduction, a good first approximation for satellite motion is obtained by assuming the spacecraft is a point mass or spherical body moving in the gravitational field of a spherical planet. This leads to the classical two-body problem. Since we use the term body to refer to a spacecraft of finite size (as in rigid body), it may be more appropriate to call this the two-particle problem, but I will use the term two-body problem in its classical sense. The basic elements of orbital dynamics are captured in Kepler s three laws which he published in the 17th century. His laws were for the orbital motion of the planets about the Sun, but are also applicable to the motion of satellites about planets. The three laws are: 1. The orbit of each planet is an ellipse with the Sun at one focus. 2. The line joining the planet to the Sun sweeps out equal areas in equal times. 3. The square of the period of a planet is proportional to the cube of its mean distance to the sun. The first law applies to most spacecraft, but it is also possible for spacecraft to travel in parabolic and hyperbolic orbits, in which case the period is infinite and the 3rd law does not apply. However, the 2nd law applies to all two-body motion. Newton s 2nd law and his law of universal gravitation provide the tools for generalizing Kepler s laws to non-elliptical orbits, as well as for proving Kepler s laws. A.1 Equations of Motion and Their Solution In the two-body problem, one assumes a system of only two bodies that are both spherically symmetric or are both point masses. In this case the equations of motion may be developed as r + µ r 3 r = 0 (A.1) A-1

2 A-2 APPENDIX A. ORBITS where µ = G(m 1 + m 2 ) and r is the relative position vector as shown in Fig. A.1. The parameter G is Newton s universal gravitational constant, which has the value G = m 3 kg 1 s 2. The scalar r is the magnitude of r, i.e., r = r. Since normally m 1 m 2, µ Gm 1, and this is the value usually given in astrodynamics tables. m 2 r m 1 Figure A.1: Relative position vector in the two-body problem. There are several ways to derive the conserved quantities associated with two-body motion. These constants of the motion are also called first integrals or conservation laws. Here I simply state these conservation laws and interpret them physically and graphically. First is conservation of energy, E = 1 2 v2 µ r = constant (A.2) where v = v = r. Note that E is the energy per unit mass: v 2 /2 is the kinetic energy and µ/r is the potential energy. As I discuss more below, energy is closely related to the size of the orbit. The second constant of motion is the angular momentum vector h = r v (A.3) Again, this is the angular momentum per unit mass. The fact that h is constant in direction means that r and v always lie in the plane to h. This plane is called the orbital plane and it is fixed in inertial space in the two-body problem. It is relatively straightforward to show that conservation of angular momentum is equivalent to

3 A.2. ORBIT DETERMINATION AND PREDICTION A-3 Kepler s second law, and from it can be derived the mathematical version of Kepler s third law for elliptical orbits: a 3 P = 2π (A.4) µ where P is the orbital period and a is the semimajor axis of the ellipse. The third constant is also a vector quantity, sometimes called the eccentricity vector e = 1 [( v 2 µ ) ] r ( r v) v (A.5) µ r Note that e is in the orbital plane, since it is a linear combination of r and v. In fact, e points in the direction of periapsis, the closest point of the orbit to the central body. Also, e = e is the eccentricity of the orbit. Using these constants, it is possible to develop a solution to Eq. (A.1) in the form r = p 1 + e cos ν (A.6) where ν is an angle called the true anomaly, and p = h 2 /µ is the semilatus rectum of the conic section. Equation (A.6) is the mathematical expression of Kepler s second law. This relation describes how the radius changes as the satellite moves but it does not tell how ν varies with time. To determine ν(t), we must solve Kepler s equation M = n(t t 0 ) = E e sin E (A.7) where E is the eccentric anomaly, M is the mean anomaly, n = µ/a 3 is the mean motion, and t 0 is the time of periapsis passage. A satellite s orbit is normally described in terms of its orbital elements. The classical orbital elements are semimajor axis, a, eccentricity, e, inclination, i, right ascension of the ascending node, Ω, argument of periapsis, ω, and true anomaly of epoch, ν 0. The last of these six orbital elements is sometimes replaced by mean anomaly at epoch, M 0, or by the time of periapsis passage, T p. A.2 Orbit Determination and Prediction We often need to be able to determine the orbit of a spacecraft or other space object. The primary motivation for orbit determination is to be able to predict where the object will be at some later time. This problem was posed and solved in 1801 by Carl Friedrich Gauss in order to predict the reappearance of the newly discovered minor planet, Ceres. He also invented the method of least squares which is described below. In practice, orbit determination is accomplished by processing a sequence of observations in an orbit determination algorithm. Since it takes six numbers (elements) to describe an orbit, the algorithm is deterministic if and only if there are six independent numbers in the observation data. If there are fewer than six numbers, then the

4 A-4 APPENDIX A. ORBITS problem is underdetermined and there is an infinite number of solutions. If there are more than six numbers in the observations, then the problem is overdetermined, and in general no solution exists. In the overdetermined case, a statistical method such as least squares is used to find an orbit that is as close as possible to fitting all the observations. In this section, we first present some deterministic orbit determination algorithms, then a statistical algorithm, and finally we discuss the orbit prediction problem. A.2.1 Solving Kepler s Equation function E = EofMe(M,e,tol) This function computes E as a function of M and e. function E = EofMe(M,e,tol) if ( nargin<3 ), tol=1e-11; end En = M; En1 = En - (En-e*sin(En)-M)/(1-e*cos(En)); while ( abs(en1-en) > tol ) En = En1; En1 = En - (En-e*sin(En)-M)/(1-e*cos(En)); end; E = En1; A.2.2 Deterministic Orbit Determination In the best of all possible worlds, we might be able to measure r and v directly, in which case we would have a set of obs comprising six numbers, three in each vector. With this set of observations, it is straightforward to develop an algorithm that computes the orbital elements as a function of r and v. This algorithm is developed in numerous orbit dynamics texts, and here we simply present the mathematical algorithm and a MatLab function that implements the algorithm. rv2oe.m r,v,mu -> orbital elements oe = rv2oe(rv,vv,mu) rv = [X Y Z] vv = [v1 v2 v3] in IJK frame oe = [a e i Omega omega nu0] function oe = rv2oe(rv,vv,mu) first do some preliminary calculations K = [0;0;1];

5 A.3. TWO-LINE ELEMENT SETS A-5 hv = cross(rv,vv); nv = cross(k,hv); n = sqrt(nv *nv); h2 = (hv *hv); v2 = (vv *vv); r = sqrt(rv *rv); ev = 1/mu *( (v2-mu/r)*rv - (rv *vv)*vv ); p = h2/mu; now compute the oe s e = sqrt(ev *ev); eccentricity a = p/(1-e*e); semimajor axis i = acos(hv(3)/sqrt(h2)); inclination Om = acos(nv(1)/n); RAAN if ( nv(2) < 0-eps ) fix quadrant Om = 2*pi-Om end; om = acos(nv *ev/n/e); arg of periapsis if ( ev(3) < 0 ) fix quadrant om = 2*pi-om; end; nu = acos(ev *rv/e/r); true anomaly if ( rv *vv < 0 ) fix quadrant nu = 2*pi-nu; end; oe = [a e i Om om nu]; assemble "vector" A.3 Two-Line Element Sets Once an orbit has been determined for a particular space object, we might want to communicate that information to another person who is interested in tracking the object. The standard format for communicating the orbit of an Earth-orbiting object is the two-line element set, or TLE. An example TLE for a Russian spy satellite is COSMOS U 94023A Note that the TLE actually has three lines. Documentation for the TLE format is available at the website and is reproduced here for your convenience. This website also includes current TLEs for many satellites. Data for each satellite consists of three lines in the following format:

6 A-6 APPENDIX A. ORBITS AAAAAAAAAAAAAAAAAAAAAAAA 1 NNNNNU NNNNNAAA NNNNN.NNNNNNNN +.NNNNNNNN +NNNNN-N +NNNNN-N N NNNNN 2 NNNNN NNN.NNNN NNN.NNNN NNNNNNN NNN.NNNN NNN.NNNN NN.NNNNNNNNNNNNNN Line 0 is a twenty-four character name (to be consistent with the name length in the NORAD SATCAT). Lines 1 and 2 are the standard Two-Line Orbital Element Set Format identical to that used by NORAD and NASA. The format description is: Line 1 Column Description 01 Line Number of Element Data Satellite Number 08 Classification (U=Unclassified) International Designator (Last two digits of launch year) International Designator (Launch number of the year) International Designator (Piece of the launch) Epoch Year (Last two digits of year) Epoch (Day of the year and fractional portion of the day) First Time Derivative of the Mean Motion Second Time Derivative of Mean Motion (decimal point assumed) BSTAR drag term (decimal point assumed) 63 Ephemeris type Element number 69 Checksum (Modulo 10) (Letters, blanks, periods, plus signs = 0; minus signs = 1) Line 2 Column Description 01 Line Number of Element Data Satellite Number Inclination [Degrees] Right Ascension of the Ascending Node [Degrees] Eccentricity (decimal point assumed) Argument of Perigee [Degrees] Mean Anomaly [Degrees] Mean Motion [Revs per day] Revolution number at epoch [Revs] 69 Checksum (Modulo 10) All other columns are blank or fixed. Example:

7 A.3. TWO-LINE ELEMENT SETS A-7 NOAA U 94089A For further information, see "Frequently Asked Questions: Two-Line Element Set Format" in the Computers & Satellites column of Satellite Times, Volume 4 Number 3. A Matlab function to compute the orbital elements given a TLE is provided below. function [oe,epoch,yr,m,e,satname] = TLE2oe(fname); fname is a filename string for a file containing a two-line element set (TLE) oe is a 1/6 matrix containing the orbital elements [a e i Om om nu] yr is the two-digit year M is the mean anomaly at epoch E is the eccentric anomaly at epoch satname is the satellite name Calls Newton iteration function file EofMe.m function [oe,epoch,yr,m,e,satname] = TLE2oe(fname); Open the file up and scan in the elements fid = fopen(fname, rb ); A = fscanf(fid, 13c*s,1); B = fscanf(fid, d6d*c5d*3c2dff5d*c*d5d*c*dd5d,[1,10]); C = fscanf(fid, d6dffffff,[1,8]); fclose(fid); satname=a; The value of mu is for the earth mu = e5; Calculate 2-digit year (Oh no!, look out for Y2K bug!) yr = B(1,4); Calculate epoch in julian days epoch = B(1,5); ndot = B(1,6); n2dot = B(1,7);

8 A-8 APPENDIX A. ORBITS Assign variables to the orbital elements i = C(1,3)*pi/180; inclination Om = C(1,4)*pi/180; Right Ascension of the Ascending Node e = C(1,5)/1e7; Eccentricity om = C(1,6)*pi/180; Argument of periapsis M = C(1,7)*pi/180; Mean anomaly n = C(1,8)*2*pi/(24*3600); Mean motion Calculate the semi-major axis a = (mu/n^2)^(1/3); Calculate the eccentric anomaly using mean anomaly E = EofMe(M,e,1e-10); Calculate true anomaly from eccentric anomaly cosnu = (e-cos(e)) / (e*cos(e)-1); sinnu = ((a*sqrt(1-e*e)) / (a*(1-e*cos(e))))*sin(e); nu = atan2(sinnu,cosnu); if (nu<0), nu=nu+2*pi; end Return the orbital elements in a 1x6 matrix oe = [a e i Om om nu]; A.4 Summary This appendix is intended to give a brief overview of orbital mechanics so that the problems involving two-line element sets can be completed. Here is a summary of some basic formulas and constants. Two-body problem relationships r = a(1 e 2 )/(1 + e cos ν) p = a(1 e 2 ) = h 2 /µ h = r v h = rv cos φ TP = 2π a 3 /µ E = v 2 /2 µ/r v c = µ/r vesc = 2µ/r e = 1 µ [(v2 µ/r) r ( r v) v] e = (r a r p )/(r a + r p ) n = ˆK h

9 A.5. REFERENCES AND FURTHER READING A-9 Orbital elements Perifocal frame a semimajor axis b = ap semiminor axis e eccentricity i inclination h cos i = h ˆK Ω RAAN n cos Ω = n Î check n j ω argument of periapsis ne cos ω = n e check e k ν true anomaly er cos ν = e r check r v Rotation matrices PQW to IJK R = v = µ [ sin ν ˆP + (e + cos ν) ˆQ] p cos Ω cos ω sin Ω sin ω cos i cos Ω sin ω sin Ω cos ω cos i sin Ω sin i sin Ω cos ω + cos Ω sin ω cos i sin Ω sin ω + cos Ω cos ω cos i cos Ω sin i sin ω sin i cos ω sin i cos i Constants and other parameters Heliocentric Geocentric 1 AU (or 1 DU) = km 1 DU (R ) = km 1 TU = sec 1 TU = sec µ = km 3 /s 2 µ = km 3 /s 2 A.5 References and further reading One good reference for this material also has the advantage of being the least expensive. Bate, Mueller, and White 1 is an affordable Dover paperback that includes most of the basic information about orbits that you ll need to know. Another inexpensive book is Thomson s Introduction to Space Dynamics, 2 which covers orbital dynamics as well as attitude dynamics, rocket dynamics, and some advanced topics from analytical dynamics. Other recent texts include Prussing and Conway, 3 Vallado, 4 and Wiesel. 5 Bibliography [1] Roger R. Bate, Donald D. Mueller, and Jerry E. White. Fundamentals of Astrodynamics. Dover, New York, [2] W. T. Thomson. Introduction to Space Dynamics. Dover, New York, [3] John E. Prussing and Bruce A. Conway. Orbital Mechanics. Oxford University Press, Oxford, 1993.

11 A.7. PROBLEMS A Make a plot of the gravitational acceleration magnitude vs. radius for a satellite in Earth orbit. The abscissa should be the orbital radius in units of Earth radii (use 6378 km) from 1 to 10, and the ordinate should be acceleration in units of gees (1 gee = 9.81 m/s 2 ). Your plot should be labeled and should include labeled markers to indicate space station altitude, GlobalStar altitude, GPS altitude and geostationary altitude. Discuss the significance of the term zero-gravity commonly used to describe the near-earth orbital environment. 11. A 100 kg spacecraft is in a circular orbit (e = 0) at 400 km altitude. What are the orbital period [s], total mechanical energy [W], specific mechanical energy [W/kg], and specific angular momentum [km 2 /s 2 ]? What is the instantaneous access area of the satellite [km 2 ]? If the spacecraft is in an equatorial orbit (i = 0) and the position vector is r = 6778 Ĵ [km], then what is the velocity vector [km/s]? Make a supporting sketch. 12. A spacecraft is in a geostationary orbit. Compute the acceleration magnitude [m/s 2 ] due to the gravitational attraction of Earth, Moon, Mars, Sun, and Jupiter. You will need to make some assumptions about the relative positions of the spacecraft and the primary bodies. You should choose the closest approach (worst case) and you can assume circular orbits for all the planets. Discuss the significance of the two-body assumptions used in developing Keplerian motion. A.7 Problems 1. An orbit with parameter p = 2 DU connects two position vectors: r 1 = 3 Î DU r 2 = 4 Ĵ DU Assuming a long-way trajectory, how long does a spacecraft take to get from r 1 to r 2? t = [TU] 2. A mission to Mars begins in a parking orbit at 500 km altitude above the Earth. Assume the parking orbit is in the plane of the ecliptic, and that Earth and Mars are in coplanar circular orbits. (a) 5 points. Determine the angular separation required between Earth and Mars at the time of initiating a heliocentric Hohmann transfer.

12 A-12 BIBLIOGRAPHY θ = [rad] (b) 5 points. Determine the v required to get from the same heliocentric circular orbit that the Earth is on to the heliocentric orbit of Mars, using a Hohmann transfer. v = [km/s] (c) 10 points. Determine the actual v required to get from the Earth parking orbit to the heliocentric transfer orbit. v = [km/s] (d) 5 points. Assuming that the maneuver takes place on the first day of spring, determine the sidereal time of the position at which the actual v takes place. θ v = [degrees] 3. The Mars Global Surveyor used an aerobraking maneuver to reduce the energy in the orbit until it reached a low Mars orbit suitable for its surface-mapping mission (400 km altitude circular orbit). Useful information about Mars is in your textbook on pp The aerobraking maneuver is equivalent to applying a v at each periapsis to reduce the periapsis speed, thereby reducing the energy and lowering the apoapsis altitude, without appreciably affecting periapsis altitude. (a) 20 points. Assume that the periapsis altitude is 110 km. If the apoapsis altitude in one orbit is 20,000 km, and the apoapsis altitude in the next orbit is 19,500 km, determine what the effective v at periapsis was (in km/s). v = [km/s] (b) 5 points. Assuming that the effective v is the same at each periapsis passage, determine the number of orbits it will take to lower the apoapsis to 400 km. N = [orbits]

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