Astrodynamics (AERO0024) L04: Non-Keplerian Motion Gaëtan Kerschen Space Structures & Systems Lab (S3L)
Non-Keplerian Motion 4 Dominant Perturbations Analytic Treatment Numerical Methods Concluding Remarks The Restricted 3-body Problem 2
Summary Mathematical modeling of 4 dominant perturbations. Analytic propagation: Meaning of J2 and its effect on the orbit. Periodic and secular contributions. STK: J2 and SGP4 propagators. 3
STK Propagators 2-body: constant orbital elements. J2: analytic propagator which accounts for secular variations in the orbit elements due to Earth oblateness. HPOP: orbit trajectories are computed using numerical integration of the equations of motion (periodic and secular effects included). 4
Numerical Integration The high accuracy required today for satellite orbits can only be achieved by using numerical integration. Incorporation of any arbitrary disturbing acceleration (versatile). Truncation and round-off may degrade the accuracy for lengthy propagation intervals. New data means new integration (lengthy computing times). 5
Numerical Integration A variety of methods has been applied in astrodynamics. Each of these methods has its own advantages and drawbacks: Accuracy: what is the order of the integration scheme? Efficiency: how many function calls? Versatility: can it be applied to a wide range of problems? Complexity: is it easy to implement and use? Step size: automatic step size control? Four methods are implemented in STK. 6
DLR Example 7
DLR Example 8
Further Reading on the Web Site 9
Envisat Example http://nng.esoc.esa.de/envisat/envpred.html 10
Did you Know? NASA began the first complex numerical integrations during the late 1960s and early 1970s. 1968 1969 12
What is Numerical Integration? Given μ r = r+ a r r( t ), r ( t ) 3 perturbed n Δ t = t t i+ 1 n i Compute r( t ), r ( t ) n+ 1 n+ 1 13
State-Space Formulation μ r = r+ a r 3 perturbed u = f ( u, t) u r = r 6-dimensional state vector 14
How to Perform Numerical Integration? u( t n ) u( ) t n + 1 h h f t h f t hf t f t f t R 2 s! 2 s ( s) ( n + ) = ( n) + '( n) + ''( n) +... + ( n) + s Taylor series expansion 15
First-Order Taylor Approximation (Euler) along the tangent u( t +Δ t) = u( t ) +Δtu ( t ) n n n u = u +Δtf( u, t) n+ 1 n n n Euler step 40 35 Exact solution x(t)=t 2 30 25 20 15 10 5 0 0 1 2 3 4 5 6 Time t (s) The stepsize has to be extremely small for accurate predictions, and it is necessary to develop more effective algorithms. 16
Numerical Integration Methods m u = α u Δt β u n+ 1 j n+ 1 j j n+ 1 j j= 1 j= 0 State vector m β 0 β = 0 0 0 Implicit, the solution method becomes iterative in the nonlinear case Explicit, u n+1 can be deduced directly from the results at the previous time steps α, β =0 j for j > 1 α, β 0 j for j > 1 j j Single-step, the system at time t n+1 only depends on the previous state t n Multi-step, the system at time t n+1 depends several previous states t n,t n-1,etc. 17
Examples: Implicit vs. Explicit Trapezoidal rule (implicit) u n+ 1 un t ( + ) n = +Δ u u 2 n+ 1 r tn t n+ 1 Euler backward (implicit) u = u +Δt u n+ 1 n n+ 1 r tn t n+ 1 Euler forward (explicit) u = u +Δt u + n 1 n n r tn t n+ 1 18
Runge-Kutta Family: Single-Step Perhaps the most well-known numerical integrator. Difference with traditional Taylor series integrators: the RK family only requires the first derivative, but several evaluations are needed to move forward one step in time. Different variants: explicit, embedded, etc. 19
Runge-Kutta Family: Single-Step () t = f( u,) t with u( t 0) = u0 u Slopes at various points within the integration step u = u +Δt bk k n+ 1 n i i i= 1 = f u, t + cδt ( ) 1 n n 1 s i 1 ki = f un +Δ t aijk j, tn + ciδ t, i= 2... s j= 1 20
Runge-Kutta Family: Single-Step The Runge-Kutta methods are fully described by the coefficients: c 1 c 2 c s a 21 a s1 b 1 a s2 b 2 a s,s-1 b s-1 b s s i= 1 c c 1 i b = = i 0 = 1 i 1 j= 1 a ij Butcher Tableau 21
RK4 (Explicit) u + 2 + 2 + = +Δ k k k k 6 n+ 1 un t 1 2 3 4 k 1 = f u, t ( ) n n Δt Δt k2 = f un + k1, tn + 2 2 Δt Δt k3 = f un + k2, tn + 2 2 k = f u + k Δ t, t +Δt ( ) 4 n 3 n Butcher Tableau 22
RK4 (Explicit) k 1 = u f + 2 + 2 + = +Δ k k k k 6 n+ 1 un t u, t ( ) n n Δt Δt k2 = f un + k1, tn + 2 2 Δt Δt k3 = f un + k2, tn + 2 2 k = f u + k Δ t, t +Δt ( ) 4 n 3 n 1 2 3 4 Estimated slope (weighted average) Slope at the beginning Slope at the midpoint (k 1 is used to determine the value of u Euler) Slope at the midpoint (k 2 is now used) Slope at the end 23
RK4 (Explicit) u k 4 k 1 k 3 Estimate at new time k 2 t n tn + Δt/2 tn + Δt 24
RK4 (Explicit) The local truncation error for a 4 th order RK is O(h 5 ). The accuracy is comparable to that of a 4 th order Taylor series, but the Runge-Kutta method avoids the calculation of higher-order derivatives. Easy to use and implement. The step size is fixed. 25
RK4 in STK 26
Embedded Methods They produce an estimate of the local truncation error: adjust the step size to keep local truncation errors within some tolerances. This is done by having two methods in the tableau, one with order p and one with order p+1, with the same set of function evaluations: s ( p) ( p) ( p) un+ 1 = un +Δt bi ki i= 1 s ( p + 1) ( p + 1) ( p 1) n 1 n t b + + = +Δ i i i= 1 u u k 27
Embedded Methods The two different approximations for the solution at each step are compared: If the two answers are in close agreement, the approximation is accepted. If the two answers do not agree to a specified accuracy, the step size is reduced. If the answers agree to more significant digits than required, the step size is increased. 28
Ode45 in Matlab / Simulink Runge-Kutta (4,5) pair of Dormand and Prince: Variable step size. Matlab help: This should be the first solver you try 29
Ode45 in Matlab / Simulink edit ode45 30
Ode45 in Matlab / Simulink Be very careful with the default parameters! options = odeset('reltol',1e-8,'abstol',1e-8); 31
RKF 7(8): Default Method in STK Runge-Kutta-Fehlberg integration method of 7th order with 8th order error control for the integration step size. 32
Multi-Step Methods (Predictor-Corrector) They estimate the state over time using previously determined back values of the solution. Unlike RK methods, they only perform one evaluation for each step forward, but they usually have a predictor and a corrector formula. Adams (*) Bashforth - Moulton, Gauss - Jackson. (*) The first with Le Verrier to predict the existence and position of Neptune 34
Multi-Step Methods: Principle u () t = f( u,) t u t u t f u t dt n+ 1 ( n+ 1) = ( ) + n (, ) tn t unknown u Four function values interpolated by a third-order polynomial Replace it by a ploynomial that interpolates the previous values t 35
Multi-Step Methods: Initiation () t = f( u,) t with u( t 0) = u0 u u t 36
Multi-Step Methods: Initiation Because these methods require back values, they are not self-starting. One may for instance use of a single-step method to compute the first four values. 37
Gauss-Jackson in STK One of the most recommendable fixed-stepsize multistep methods for orbit computations. 38
Extrapolation Methods Not discussed herein. More details in Montenbruck and Gill, Satellite orbits, Springer, 2000. 39
Integrator Selection Montenbruck and Gill, Satellite orbits, Springer, 2000 40
Integrator Selection Multi-step Single step Pros Very fast Pros Plug and play Error control Cons Special starting procedure Fixed time steps Error control Cons Slower 41
Why is the Step Size So Critical? Theoretical arguments: 1. The accuracy and the stability of the algorithm are directly related to the step size. 2. Nonlinear equations of motion. Data for Landsat 4 and 6 in circular orbits around 800km indicates that a one-minute step size yields about 47m error. A three-minute step size produces about a 900m error! 42
Why is the Step Size So Critical? More practical arguments: 1. The computation time is directly related to the step size. 2. The particular choice of step size depends on the most rapidly varying component in the disturbing functions (e.g., 50 x 50 gravity field). 43
Appropriate Step Size The problem of determining an appropriate step size is a challenge in any numerical process. Δ t = T orbit Fixed step size: (rule of thumb for standard applications). 100 But an algorithm with variable step size is really helpful. The step size is chosen in such a way that each step contributes uniformly to the total integration error. 44
Three Examples: XMM / OUFTI-1 / ISS Can you plot the step size vs. true anomaly? 45
XMM: Report in STK 46
XMM (e~0.8) 6000 5000 Postprocessing in Matlab Step size (s) 4000 3000 2000 1000 0 0 50 100 150 200 250 300 350 400 True anomaly (deg) 47
Exercise Session Reproduce this graph! 48
OUFTI-1 (e~0.07) 90 80 Step size (s) 70 60 50 40 30 0 50 100 150 200 250 300 350 400 True anomaly (deg) 49
ISS (e~0) 70 65 60 Step size (s) 55 50 45 40 35 30 0 50 100 150 200 250 300 350 400 True anomaly (deg) 50
Difficult Orbits Automatic time step is especially nice on highly eccentric orbits (Molniya, XMM). These orbits are best computed using variable step sizes to maintain some given level of accuracy: Without this variable step size, we waste a lot of time near apoapsis, when the integration is taking too small a step. Likewise, the integrator may not be using a small enough step size at periapsis, where the satellite is traveling fast. 51
HPOP Propagator: ISS Example 1. J2 only 2. Drag only 3. Sun and moon only 4. SRP only 5. All together. 52
1. J2: Comparison of the Propagators 2-body HPOP J2 53
1. J2: Secular and Periodic Effects HPOP with central body (2,0 + WGS84_EGM96) (without drag/srp/sun and Moon) 54
2. Drag: HPOP 55
2. Drag: Lifetime (Satellite Tools) 56
2. Drag HPOP with drag Harris Priester (without oblateness/srp/sun and Moon) 57
2. Drag: Relationship with Eclipses 58
3. Third-Body Gravity: HPOP 59
3. Third-Body Gravity HPOP with Sun and Moon (without oblateness/srp/drag) 60
4. SRP HPOP with SRP (without oblateness/drag/sun and Moon) 61
4. SRP: Relationship with Eclipses 62
All Perturbations Together 63
2-Body: ISS Example 64
Non-Keplerian Motion 4 Dominant Perturbations Analytic Treatment Numerical Methods Concluding Remarks The Restricted 3-body Problem 65
Motivation For nearly half a century, space missions planners have used tools based largely on a two-body decomposition of the solar system. This is called the patched conic approach (Lecture 06). While that approach remains very valuable for some missions, new trajectory paradigms must be developed to meet the challenges of today. 66
The 3-Body Problem 18 th -order system, but ten known integrals: 6 (linear momentum). 1 (energy). 3 (angular momentum). No closed-form solution. Not integrable. 67
Matlab Example Three identical masses: Two are at rest. The third one has a velocity directed upward to the right making a 45 degrees angle with the X axis. Y v 0 m 1 m 2 m 3 X 68
Why Is the 3-Body Problem So Difficult? Displacement of the first mass (x direction) 16 x 105 Comparison of time series 14 12 10 8 6 4 2 Nominal IC Perturbed IC 0 0 0.5 1 1.5 2 2.5 3 Time x 10 4 2-body problem (0.1% IC perturbation) Displacement of the first mass (x direction) 14 x 106 Comparison of time series 12 10 8 6 4 2 Chaotic by nature! 3-body problem (0.1% IC perturbation) Nominal IC Perturbed IC 0 0 5 10 69 15 Time x 10 4
Why Is the 3-Body Problem So Difficult? x 10 5 8 6 4 Nominal IC m1 m2 m3 Chaotic by nature! 2 0 Y -2-4 -6-8 -10-1 -0.5 0 0.5 1 X x 10 6 8 6 4 2 0 x 10 5 IC perturbed by 0.1% m1 m2 m3 Y -2-4 -6-8 -10 70-1 -0.5 0 0.5 1 X x 10 6
The Planar, Circular, Restricted 3BP z m (x,y,z) m is vanishingly small compared to m 1 and m 2 (no effect on their motions) r 1 r y What is the motion of m? m 1 (x 1,0,0) CM (0,0,0) r 2 r 12 Plane of motion of m 1 and m 2. The orbit around each other is a circle of radius r 12. m 2 (x 2,0,0) x Co-moving xyz frame 71
Position of m 1, m 2, m (co-moving frame) mx 1 1+ mx 2 2= 0 x = x + r 2 1 12 x x = π r 1 2 12 = π r 2 1 12 with m π =, π = m 1 2 1 2 m1+ m2 m1+ m2 r = xˆi+ yˆj+ zkˆ & r = x+ π r ˆi+ yˆj+ zkˆ ( ) 1 2 12 r = x π r ˆi+ yˆj+ zkˆ ( ) 2 1 12 72
Absolute Acceleration of m dω r = acm + arel + 2Ω vrel + r+ Ω Ω r dt ( ) r R ρ Rotating body frame (angular velocity ω) Inertial frame a ω ρ ω ω ρ 2 2 d R d ρ i dρi dω i = + 2 2 2 + + i + i dt dt dt body body dt 73 ( )
Absolute Acceleration of m r = Ω ( Ω r) + 2Ω vrel + arel Inertial angular velocity ˆ μ Ω =Ω k = kˆ r 3 12 Velocity and acceleration of m in the moving frame v = x ˆi+ y ˆj+ z kˆ rel a = xˆi+ yˆj+ zkˆ rel r = x Ωy Ω x ˆi+ y + Ωx Ω y ˆj + zkˆ ( 2 2 ) ( 2 2 ) 74
Newton s Second Law (Secondary Body) μ m μ m m r = F + F = r r 1 2 1 2 3 1 3 2 r1 r2 μ μ r = r r r 1 2 3 1 3 2 1 r2 x Ωy Ω x ˆi+ y+ Ωx Ω y ˆj+ zkˆ = ( 2 ) ( 2 2 2 ) μ ( ) ˆ ˆ ˆ μ x + π ( ) ˆ ˆ ˆ 212 r + y + z x π112 r + y + z 75 r i j k i j k 1 2 3 3 1 r2
Overall: Three Scalar Equations μ μ x 2Ωy Ω x= x+ π r x π r 1 2 3 3 1 r2 ( ) ( ) 2 1 2 3 212 3 112 r1 r2 μ y y+ 2Ωx Ω y = μ z z = r μ y 2 1 2 3 3 r1 r2 μ z 76
Lagrange Points The three scalar equations of motion have no closed form analytical solutions. But the equilibrium points, called the Lagrange points, can be determined. x = y = z = 0, x= y = z = 0 77
Lagrange Points Lie in the Orbital Plane μ μ Ω x = x+ π r + x π r ( ) ( ) 2 1 2 3 212 3 112 r1 r2 μ y Ω y = + μ y 2 1 2 3 3 r1 r2 μ1 μ 2 + z 0 3 3 = r1 r2 78
Some Points Are Collinear μ μ Ω x = x+ π r + x π r ( ) ( ) 2 1 2 3 212 3 112 r1 r2 μ y Ω y = + μ y 2 1 2 3 3 r1 r2 μ1 μ 2 + z 0 3 3 = r1 r2 79
Two Sets of Lagrange Points μ μ Ω x = x+ π r + x π r ( ) ( ) 2 1 2 3 212 3 112 r1 r2 μ y Ω y = + μ y 2 1 2 3 3 r1 r2 y 0, L4,L5 y=0, L1,L2,L3 with π = 1 π 1 2 Ω= μ 3 r 12 μ π =, π = μ μ μ 1 2 1 2 80
L4, L5: Vertices of Equilateral Triangles 1 1 1 π x+ π r + π x+ π r r = ( )( ) ( ) ( ) 2 2 12 3 2 2 12 12 3 3 r1 r2 r12 1 1 1 1 π + π =, if y 0 2 3 2 3 3 r1 r2 r12 x 1 1 1 = = r = r = r r r r 3 3 3 1 2 12 1 2 12 81
Coordinates of L4, L5 (Co-Moving Frame) π = 1 π r = x+ π r ˆi+ yˆj+ zkˆ ( ) 1 2 12 r = x π r ˆi+ yˆj+ zkˆ 1 2 ( ) 2 1 12 with z = 0 r1 = r2 = r12 2 ( ) 2 2 12 = + π 2 12 + r x r y 2 ( ) 2 2 12 = + π 2 12 12 + r x r r y x y r 2 12 = π 212 =± 3 2 r 12 r 82
L1, L2, L3: Collinear Points y = z = 0 r r = x+ π r ( ) 1 2 12 ˆi = x+ π r r ( ) 2 2 12 12 ˆi r = x+ π r 1 2 12 r = x+ π r r 2 2 12 12 with π = 1 π 1 2 Ω= μ 3 r 12 μ π =, π = μ μ μ 1 2 1 2 83
L1, L2, L3: Numerical Technique μ μ Ω x = x+ π r + x π r ( ) ( ) 2 1 2 3 212 3 112 r1 r2 f 1 π π ξ = ξ + π 3 2 + ξ + π 3 2 1 ξ = 0 ξ + π ξ + π 1 ( ) 2 ( ) 2 ( ) 2 2 μ μ π1 = 1 π 2 Ω= π =, π = 3 r 12 μ r1 = x+ π 2r12 x ξ = r = x+ π r r r12 2 2 12 12 μ μ 1 2 1 2 84
Earth - Moon System m m r 1 12 2 24 5.974 10 kg = 22 7.348 10 kg = 5 3.844 10 km = m π = 2 2 m1+ m = 2 0.01215 85
Earth - Moon System 4 3 2 1 L3: 1.005 L2:1.156 f(ksi) 0-1 -2-3 L1: 0.8369-4 -1.5-1 -0.5 0 0.5 1 1.5 ksi 86
Earth - Moon System 87
Lagrange Points: Stability L1,L2,L3 are unstable: the satellite drifts away from the equilibrium point for slight perturbations. L4,L5 are stable: the satellite remains around the equilibrium point for slight perturbations. 88
Lagrange Points: Physical Interpretation Are the Lagrange points such that the gravitational pulls of the Sun and of the Earth are equal? L1 The forces in equilibrium are the gravitational attraction of the massive bodies and the centrifugal force. 89
L1, Physical Interpretation 2-body reasoning: any object going around the Sun in an orbit smaller than Earth's will not keep a fixed station relative to Earth. 3-body reasoning: Earth's gravity cancels some of the gravitational pull of the Sun. With a weaker pull towards the Sun, the spacecraft then needs less speed to maintain its orbit. At L1, the spacecraft will need just one year to go around the Sun. 90
L2, Physical Interpretation Same reasoning! 91
Sun - Earth System 92
Sun - Earth System L1, a solar observatory: ISSE-3, SOHO, ACE L2, a stable environment: Herschel, Planck, JWST, Gaia 93
Sun - Earth System Assuming that the required stationkeeping is performed (unstable equilibrium), is it meaningful to maintain at spacecraft at L1? The tracking antennas are also aimed at the Sun, a source of interfering radio waves! Use of a Halo orbit. 94
Sun Jupiter System: Trojan Asteroids 95
Look for A Constant of the Motion μ μ x 2Ωy Ω x= x+ π r x π r 1 2 3 3 1 r2 ( ) ( ) 2 1 2 3 212 3 112 r1 r2 μ y y+ 2Ωx Ω y = z μ z = r μ y 2 1 2 3 3 r1 r2 μ z x y z Σ d 1 2 1 2( 2 2) μ1 μ 2 v Ω x + y = dt 2 2 r1 r2 0 96
Jacobi Constant 1 1 2 2 ( ) μ r μ r v 2 2 x 2 y 2 1 2 Ω + = C 1 2 Kinetic energy + rotation of the reference frame + gravitational pull of the two primary masses = C v is the speed of the secondary mass relative to the rotating frame 97
Allowed Regions of Motion ( ) 2μ 2μ 0 Ω + + + + 2 0 r r 2 2 2 2 1 2 v x y C 1 2 ( ) 2μ 2μ Ω x + y + + + C = r r 2 2 2 1 2 Boundaries: 2 0 1 2 Zero velocity curves for a given Jacobi constant. 98
Earth - Moon System 99
Exercise A spacecraft has a burnout velocity at a point on the Earth- Moon line with an altitude of 200 km. Find the value of this velocity for the different scenarios discussed. m m r 1 12 2 = 24 5.974 10 kg = 22 7.348 10 kg = 5 3.844 10 km m π = 1 1 m1+ m = 2 0.9878 x1 = π1r12 = 4670.6 km x = 6578 4670.6 = 1907.3km y = 0 Coordinates of the burnout point 101
Solution ( ) 2μ 2μ r r 2 2 2 1 2 v= Ω x + y + + + 2C 1 2 C C C C C C 0 1 2 3 4 5 : v = 10.845km/s bo : v = 10.857 km/s bo : v = 10.858km/s bo : v = 10.866km/s bo : v = 10.867 km/s bo : v = 10.868km/s bo A few m/s can have a significant influence on the regions of Earth- Moon space accessible to the spacecraft! 102
Comparison with Escape Velocity v esc,200 km 2μ 2 398600 = = = 11.01km/s r 6578 C C C C C C 0 1 2 3 4 5 : v = 10.845km/s bo : v = 10.857 km/s bo : v = 10.858km/s bo : v = 10.866km/s bo : v = 10.867 km/s bo : v = 10.868km/s bo 103
Further Reading on the Course Web Site 104
Non-Keplerian Motion 4 Dominant Perturbations Analytic Treatment Numerical Methods Concluding Remarks The Restricted 3-body Problem 105
Perturbations Perturbations are small with respect to the spherical central body attraction. But they must be taken into account for accurate orbit propagation. Realistic models for the 4 dominant perturbations and understanding of their effects. 106
Orbit Propagation Analytical methods Exact solutions to simple approximating problems. Approximate solutions to approximating problems. Semi-analytical methods Better approximate solutions to realistic problems. Numerical methods Best solutions to most realistic problems. 107
STK Propagators 108
STK Integrators and Step Size Importance 109
Practical Example: GEO Satellites Nice illustration of: 1. Perturbations of the 2-body problem. 2. Secular and periodic contributions. 3. Accuracy requires by practical applications. 4. The need for orbit correction and thrust forces. And it is a real-life example (telecommunications, meteorology)! 110
Practical Example: GEO Satellites A station-keeping box is defined by a longitude and a maximum authorized distance for satellite excursions in longitude and latitude. For instance, TC2: -8º ± 0.07º E/W ± 0.05º N/S 111
Practical Example: GEO Satellites A GEO satellite orbit changes over time due to perturbations: 1. Inclination modified by the lunisolar attraction: N/S. 2. The longitude is modified by the tesseral terms of the geopotential: E/W. 112
GEO: Inclination Vector Drift Period HPOP with Sun and Moon (without oblateness/srp/drag) 113
GEO: Inclination Vector Drift Period HPOP with Sun and Moon (without oblateness/srp/drag) 114
GEO: Longitude Drift C 2,2 corresponds to the equatorial ellipticity. 115
GEO: Longitude Drift HPOP with 2,2 (without Sun and moon/srp/drag) 116
GEO: Longitude Drift HPOP with 2,2 (without Sun and moon/srp/drag) 117
GEO: Longitude Drift HPOP with 2,2 (without Sun and moon/srp/drag) 118
Thrust Forces GEO spacecraft require continual stationkeeping to stay within the authorized box using an onboard thruster system. Next lecture: basic orbital maneuvers. 119
Astrodynamics (AERO0024) L04: Non-Keplerian Motion Gaëtan Kerschen Space Structures & Systems Lab (S3L)