Orbital Mechanics! Space System Design, MAE 342, Princeton University! Robert Stengel

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1 Orbital Mechanics Space System Design, MAE 342, Princeton University Robert Stengel Conic section orbits Equations of motion Momentum and energy Kepler s Equation Position and velocity in orbit Copyright 216 by Robert Stengel. All rights reserved. For educational use only. 1 Orbits 11 Satellites Escape and Capture (Comets, Meteorites) 2

2 Two-Body Orbits are Conic Sections 3 Classical Orbital Elements Dimension and Time a : Semi-major axis e : Eccentricity t p : Time of perigee passage Orientation :Longitude of the Ascending/Descending Node i : Inclination of the Orbital Plane : Argument of Perigee 4

3 Orientation of an Elliptical Orbit irst Poin of Aries 5 Orbits 12 (2-Body Problem) e.g., Sun and Earth or Earth and Moon or Earth and Satellite Circular orbit: radius and velocity are constant Low Earth orbit: 17, mph = 24, ft/s = 7.3 km/s Super-circular velocities Earth to Moon: 24,55 mph = 36, ft/s = 11.1 km/s Escape: 25, mph = 36,6 ft/s = 11.3 km/s Near escape velocity, small changes have huge influence on apogee 6

4 Newton s 2 nd Law Particle of fixed mass (also called a point mass) acted upon by a force changes velocity with acceleration proportional to and in direction of force Inertial reference frame Ratio of force to acceleration is the mass of the particle: F = m a d mv( t) dt = m dv t dt f x F = f y f z = ma( t) = F = force vector m d dt ( t) ( t) = ( t) v x v y v z f x f y f z 7 Equations of Motion for a Particle Integrating the acceleration (Newton s 2 nd Law) allows us to solve for the velocity of the particle f dv( t) = v ( t) = 1 x m dt m F = f y m f z m = a x a y a z v( T ) = T dv t T T 1 dt + v( ) = a( t)dt dt + v( ) = m Fdt + v 3 components of velocity v x v y v z ( T ) T ( T ) a x t T = a y ( t) dt + a z ( t) v x v y v z m f x t T = f y ( t) m dt + f z ( t) m v x v y v z 8

5 Equations of Motion for a Particle Integrating the velocity allows us to solve for the position of the particle dr( t) = r ( t) = v( t) = dt r( T ) = T dr t dt + r dt x t y t z t = T = v dt v x v y v z ( t) ( t) ( t) + r 3 components of position x T y T z T v x t T = v y ( t) dt + v z ( t) x y z 9 Spherical Model of the Rotating Earth Spherical model of earth s surface, earth-fixed (rotating) coordinates R E = x o y o z o E = cos L E cos E cos L E sin E sin L E R L E : Latitude (from Equator), deg E : Longitude (from Prime Meridian), deg R : Radius (from Earths center), deg Earths rotation rate,, is 15.4 deg/hr 1

6 Non-Rotating (Inertial) Reference Frame for the Earth Celestial longitude, C, measured from First Point of Aries on the Celestial Sphere at Vernal Equinox C = E + ( t t epoch ) = E + t 11 Transformation Effects of Rotation R E = Transformation from inertial frame, I, to Earth s rotating frame, E cost sint sint cost 1 ) R ) I = () cost sint sint cost 1 ) ) () x o y o z o ) ) ) ( I Location of satellite, rotating and inertial frames r E = cos L E cos E cos L E sin E sin L E ( R + Altitude); r I = cos L E cos C cos L E sin C sin L E R + Altitude Orbital calculations generally are made in an inertial frame of reference 12

7 Gravity Force Between Two Point Masses, e.g., Earth and Moon Magnitude of gravitational attraction F = Gm 1m 2 r 2 G : Gravitational constant = Nm 2 /kg 2 m 1 : Mass of 1 st body = kg for Earth m 2 : Mass of 2 nd body = kg for Moon r : Distance between centers of mass of m 1 and m 2, m 13 Acceleration Due To Gravity F 2 = m 2 a 1on2 = Gm 1 m 2 r 2 Inverse-square Law a 1on2 = Gm 1 µ 1 r 2 r 2 µ 1 = Gm 1 Gravitational parameter of 1 st mass At Earth s surface, acceleration due to gravity is a g g oearth = µ E = m 3 s 2 2 R surface 6, 378,137m 2 = 9.798m s 2 14

8 Gravitational Force Vector of the Spherical Earth Force always directed toward the Earth s center F g = m µ E r I 2 r I r I = m µ E 3 r I ( * * ) * x y z (vector),, - I as r I = r I (x, y, z) establishes the direction of the local vertical r I r I = x y z I x 2 I + y 2 2 I + z = I cos L E cos I cos L E sin I sin L E 15 Equations of Motion for a Particle in an Inverse-Square-Law Field Integrating the acceleration (Newton s 2 nd Law) allows us to solve for the velocity of the particle v( T ) = dv( t) = v ( t) = 1 dt m F = µ E g 2 r I T dv t dt + v dt v x v y v z T = a t r I r I dt + v = µ ( E * 3 r * I ) * = 3 components of velocity ( T ) 3 x r I v x T 3 ( T ) = µ E ( y r I dt + v y 3 ( T ) z r I v z x y z + - -, - T 1 m Fdt + v 16

9 Equations of Motion for a Particle in an Inverse-Square-Law Field As before; Integrating the velocity allows us to solve for the position of the particle dr( t) = r ( t) = v( t) = dt x t y t z t = v x v y v z ( t) ( t) ( t) r( T ) = T dr t dt + r dt T = v dt + r 3 components of position x T y T z T v x t T = v y ( t) dt + v z ( t) x y z 17 v x t Dynamic Model with Inverse- Square-Law Gravity No aerodynamic or thrust force Neglect motions in the z direction m satellite << m Earth = µ E x I ( t) r 3 I ( t) = µ E y I ( t) r 3 I ( t) x I ( t) = v x ( t) y I ( t) = v y ( t) v y t where r I Dynamic Equations ( t) = x 2 I ( t) + y 2 I ( t) Example: Initial Conditions at Equator = 7.5, 8, 8.5 km/s = = = 6,378 km = R v x v y x y 18

10 Equatorial Orbits Calculated with Inverse-Square-Law Model 19 Work Work is a scalar measure of change in energy With constant force, In one dimension W 12 = F r 2 r 1 = Fr In three dimensions W 12 = F T r 2 r 1 = F T r With varying force, work is the integral r 2 r 2 W 12 = F T dr = ( f x dx + f y dy + f z dz), dr = r 1 r 1 dx dy dz 2

11 Conservative Force Assume that the 3-D force field is a function of position F = F( r) The force field is conservative if Iron Filings Around Magnet Force Emanating from Source r 2 F T r + F T r dr r 1 dr = r 1 r 2 for any path between r 1 and r 2 and back 21 Gravitational Force is Gradient of a Potential, V(r) Gravity potential, V(r), is a function only of position F g = m µ E r r = 3 I I r m µ E r ( r V r 22

12 Gravitational Force Field Gravitational force field F g = m µ E r I 3 r I Gravitational force field is conservative because r 2 r V ( r)dr I r 1 r 1 r V ( r)dr I = r 2 r 2 m µ r E 3 r r dr 2 I I + m µ E 3 r 1 I r r dr I I = r 1 I for any path between r 1 and r 2 and back 23 Potential Energy in Gravitational Force Field Potential energy, V or PE, is defined with respect to a reference point, r PE( r ) = V ( r ) = V ( U ) PE V ( r 2 ) V ( r 1 ) = m µ + V r 2 ( + m µ r + V ( = m µ + m µ r 1 2 r 1 24

13 Kinetic Energy Apply Newton s 2 nd Law to the definition of Work dr dt = v; dr = vdt F = m dv dt r 2 t 2 W 12 = F T dr = m dv dt t 1 r 1 t 1 T v dt = 1 t 2 m 2 d ( v T v)dt = 1 t dt 2 m 2 d ( v 2 )dt dt t 1 = 1 2 mv2 t 1 t 2 = 1 2 m ( v2 t 2 ) v 2 ( t 1 ) = 1 2 m v v 1 T 2 T 1 KE Work = integral from 1 st to 2 nd time T (= KE) is the kinetic energy of the point mass, m 25 Total Energy in Point-Mass Gravitational Field Potential energy of mass, m, depends only on the gravitational force field Kinetic energy of mass, m, depends only on the velocity magnitude measured in an inertial frame of reference Total energy, E, is the sum of the two: V = PE = m µ r T KE = 1 2 mv2 E = PE + KE = m µ r mv2 = Constant 26

14 Interchange Between Potential and Kinetic Energy in a Conservative System E 2 E 1 = m µ + 1 r 2 2 mv 2 2 m µ + 1 r 1 2 mv 2 1 = P 2 m µ + m µ r 2 r 1 = 1 2 mv mv 2 1 PE 2 PE 1 = KE 2 KE 1 P 1 27 Specific Energy Energy per unit of the satellite s mass E S = PE S + KE S = 1 m mµ r mv2 = µ r v2 P 1 P 2 28

15 Angular Momentum of a Particle (Point Mass) h = ( r mv) = m( r v) = m( r r ) 29 Angular Momentum of a Particle Moment of linear momentum of a particle Mass times components of the velocity that are perpendicular to the moment arm = m( r v) h = r mv Cross Product: Evaluation of a determinant with unit vectors (i, j, k) along axes, (x, y, z) and (v x, v y, v z ) projections on to axes r v = i j k x y z = ( yv z zv y )i + zv x xv z v x v y v z k j + xv y yv x 3

16 Cross Product in Column Notation Cross product identifies perpendicular components of r and v r v = i j k x y z = ( yv z zv y )i + zv x xv z v x v y v z k j + xv y yv x Column notation r v = ( yv z zv y ) ( zv x xv z ) ( xv y yv x ) ( ( ( ( ( = z y z x y x v x v y v z 31 Angular Momentum Vector is Perpendicular to Both Moment Arm and Velocity h = mr v = m = m z y z x y x yv z zv ( y ) ( ( zv x xv z ) ( ( ( xv y yv x ) ( ( ( ( ( v x v y v z ( ( = mrv ( ( 32

17 Specific Angular Momentum Vector of a Satellite is the angular momentum per unit of the satellite s mass, referenced to the center of attraction h S = m m r v = r v = r r Perpendicular to the orbital plane 33 Equations of Motion for a Particle in an Inverse-Square-Law Field Acceleration is a( t) = v ( t) = r ( t) = µ r 2 ( t) or ( t) r( t) = µ r 3 t r I r( t) r + µ r 3 r = 34

18 Cross Products of Radius and Radius Rate Then r r + µ r r 3 = r r = r r = because they are parallel d dt Chain Rule for Differentiation ( r r ) = ( r r ) + ( r r ) = ( r r ) 35 Specific Angular Momentum r r + µ r r 3 = r r = d dt ( r r ) = dh S dt Consequently h S = Constant + µ ( r r) r 3 = h S h = ( r r ) (Perpendicular to the plane of motion) Orbital plane is fixed in inertial space 36

19 Eccentricity Vector is a Constant of Integration r + µ r r 3 h = r h + µ r r h = 3 r h = µ r r h = µ 3 r r ( r r ) 3 With triple vector product identity (see Supplement) r h = µ r r ( r r ) = µ ( rr rr ) = µ d 3 r 2 dt e = Eccentricity vector Integrating (r h)dt = r h = µ r r + e ( r r ( ( Constant of integration) 37 Significance of Eccentricity Vector ) r h µ r r + e, + (. * - T ) h = because r h µ r r + e, + (. * - = ( r h) T h µrt h µe T h = r µe T h = e is perpendicular to angular momentum, which means it lies in the orbital plane Its angle provides a reference direction for the perigee 38

20 General Polar Equation of a Conic Section r T ) r h µ r r + e, + (. * - = 1 st term is angular momentum squared r T ( r h) = h T ( r r ) = h T h = h 2 Then h 2 µ rt r r + rt e = h2 = µ ( r + r T e) = µ ( r + recos ) = r = h 2 µ 1+ ecos 39 Elliptical Planetary Orbits Assume satellite mass is negligible compared to Earth s mass Then Center of mass of the 2 bodies is at Earth s center of mass Center of mass is at one of ellipse s focal points Other focal point is vacant r = p 1+ e cos = h 2 µ, m or km 1+ e cos : True Anomaly = Angle from perigee direction, deg or rad 4

21 Properties of Elliptical Orbits Eccentricity can be determined from apogee and perigee radii e = r r a p = r r a p r a + r p 2a r p = a / ( 1 e) r a = a / ( 1+ e) Semi-major axis is the average of the two a = r a + r p 2 41 Properties of Elliptical Orbits Semi-latus rectum, p, can be expressed as a function of h or a and e p = h 2 µ = a 1 e 2 b = r a r p Semi-minor axis, b, can be expressed as a function of r a and r p Area of the ellipse, A, is A = a b = a 2 1 e 2, m 2 42

22 Energy is Inversely Proportional to the Semi-Major Axis r p = and v = r p p At the periapsis, r p E S E = 1 ( 2 r p p ) 2 µ = 1 h 2 r p 2 r µ 2 p r p E = 1 2r p 2 ( µ p 2µr p ) = µ 2 1 e 2 2a 1 e = µ 1 e 2a 1 e p = h 2 µ = a( 1 e 2 ) r p = a( 1 e) 2 1 e E = µ 2a 43 Classification of Conic Section Orbits Orbit Shape Eccentricity, e Energy, E Semi-Major Axis, a Semi-Latus Rectum, p Circle < > a Ellipse < e < 1 < > a(1 e 2 ) Parabola 1 Undefined () 2r p Hyperbola >1 > < a(1 e 2 ) 44

23 Vis Viva (Living Force) Integral Velocity is a function of radius and specific energy 1 2 v2 = µ r + E v = 2 µ r + E Specific total energy, E, is inversely proportional to the semi-major axis E = µ 2a Velocity is a function of radius and semi-major axis v = µ 2 r 1 a 45 Maximum and Minimum Velocities on an Ellipse Velocity at periapsis v p = µ 2 1 r p a = µ 2 a 1 e = µ ( 1+ e) a 1 e v a = µ 2 1 r a a = 1 a Velocity at apoapsis µ ( 1 e) a 1+ e 46

24 Velocities at Periapses and Infinity of Parabola and Hyperbola Parabola (a ) v p = 2µ r p v = Hyperbola (a < ) v p = µ 2 1 r p a v = µ a 47 Relating Time and Position in Elliptical Orbit Rearrange and integrate angular momentum over time and angle d d Ellipse Area h = µ p = r 2 = 2 dt dt t f dt = t o f r 2 µ p d = o p 3 µ f o d ( 1+ ecos ) 2 but difficult to integrate analytically 48

25 Anomalies of an Elliptical Orbit Angles measured from last periapsis (or ): True Anomaly E (or ): Eccentric Anomaly M: Mean Anomaly 49 Kepler s Equation for the Mean Anomaly M = E esin E e cos E = ae + p r a r = a( 1 cos E) r = E aesin E From the diagram, = a r ae 5

26 Relationship of Time to Eccentric Anomaly E = µ 2a = 1 2 r 2 + h2 r 2 where h 2 = µa 1 e 2 µr ( then r 2 r 2 2 = µ r 1 + * a r 2 a( 1 e 2 )- ), leading to a 3 de ( 1 ecos E)2 µ dt or 2 = 1 µ dt = ( 1 ecos E 3 a )de 51 Integrating the Prior Result t 1 E µ 1 a dt = ( 1 ecos E)dE 3 t E µ E ( t a 3 1 t ) = ( E esin E) 1 E = M1 M Time is proportional to Mean Anomaly ( t 1 t ) = M 1 M or t µ a 3 1 = t + M 1 M µ a 3 Orbital Period, P µ 2 ( P ) = ( E esin E) a 3 = 2 P = 2 a 3 µ 52

27 Orbital Period P = 2 Orbital period is related to the total energy a3 µ = µ2 2E 3 where E < for an ellipse Mean Motion, n, is the inverse of the Period P = 2 a3 µ 2 n where n is the Mean Motion 53 Position and Velocity in Orbit at Time, t Mean Anomaly, given time from perigee passage, t p M ( t) = µ a 3 ( t t p ) Eccentric Anomaly, E(t), given Mean Anomaly, M(t) E( t) esin E( t) = M ( t) Newton s method of successive approximation, using M(t) as starting guess for E(t) E o ( t) = M t + esin M ( t) Iterate until M i < Tolerance M i = M t E i ( t) esin E i ( t) ( t) = E i ( t) + E i+1 E i+1 = M i 1 ecos E i t E i+1 54

28 Position and Velocity in Orbit at Time, t Calculate True Anomaly, given Eccentric Anomaly t = 2 tan 1 1+ e 1 e tan E t 2 Compute magnitude of radius r( t) = a 1 e2 1+ ecos t ( 55 Position and Velocity in Orbit at Time, t Radius vector, in the orbital plane r( t) = x t y t = r( t)cos t r( t)sin t Velocity vector, in the orbital plane v( t) = v x v y ( t) ( t) = µ p sin( t e + cos( t see Weisel, Spaceflight Dynamics, 1997, pp

29 ext Time: lanetary Defense 57 upplemental Marial 58

30 First Point of Aries (Ecliptic Intercept at Right) 59 Dimension of energy? Scalar (1 x 1) Dimension of linear momentum? Vector (3 x 1) Dimension of angular momentum? Vector (3 x 1) 6

31 Sub-Orbital (Sounding) Rockets Present NASA Wallops Island Control Center Canadian Black Brant XII LTV Scout 61 MATLAB Code for Flat- Earth Trajectories Script for Analytic Solution g = 9.8; t = :.1:4; Script for Numerical Solution tspan = 4; Time span, s xo = [1;1;;]; Init. Cond. [t1,x1] = ode45(flatearth,tspan,xo); vx = 1; vz = 1; x = ; z = ; vx1 = vx; vz1 = vz g*t; x1 = x + vx*t; z1 = z + vz*t -.5*g*t.* t; Function for Numerical Solution function xdot = FlatEarth(t,x) x(1) = vx x(2) = vz x(3) = x x(4) = z g = 9.8; xdot(1) = ; xdot(2) = -g; xdot(3) = x(1); xdot(4) = x(2); xdot = xdot; end 62

32 Trajectories Calculated with Flat-Earth Model Constant gravity, g, is the only force in the model, i.e., no aerodynamic or thrust force Can neglect motions in the y direction v x Dynamic Equations ( t) = v x = g ( z positive up) = v x ( t) = v z ( t) v z t x t z t Initial Conditions = v x = v z = x = z v x v z x z v x v z Analytic (Closed-Form) Solution ( T ) = v x T ( T ) = v z gdt = v z gt x T = x + v x T = z + v z T gt dt z T T = z + v z T gt Trajectories Calculated with Flat-Earth Model = 1m / s = 1,15, 2m / s = = v x v z x z 64

33 MATLAB Code for Spherical-Earth Trajectories Script for Numerical Solution R = 6378; Earth Surface Radius, km tspan = 6; seconds options = odeset(maxstep, 1) xo = [7.5;;;R]; [t1,x1] = ode15s(roundearth,tspan,xo,options); for i = 1:length(t1) v1(i) = sqrt(x1(i,1)*x1(i,1) + x1(i,2)*x1(i,2)); r1(i) = sqrt(x1(i,3)*x1(i,3) + x1(i,4)*x1(i,4)); end Function for Numerical Solution function xdot = RoundEarth(t,x) x(1) = vx x(2) = vy x(3) = x x(4) = y mu = 3.98*1^5; km^2/s^2 r = sqrt(x(3)^2 + x(4)^2); xdot(1) = -mu * x(3) / r^3; xdot(2) = -mu * x(4) / r^3; xdot(3) = x(1); xdot(4) = x(2); xdot = xdot; end 65 Equations that Describe Ellipses x 2 a + y2 2 b = 1 2 a : Semi - major axis, m or km b : Semi - minor axis, m or km a b y o x = a cos ( ) = b sin ( ) x y : Angle from x - axis (origin at center) rad 66

34 Constructing Ellipses F 1 P 1 + F 2 P 1 = F 1 P 2 + F 2 P 2 = 2a Foci (from center), x f y f 1,2 = a 2 b 2, a 2 b 2 String, Tacks, and Pen Archimedes Trammel Hypotrochoid 67 Ellipses Semi - latus rectum (The Parameter), p = b2 a, m Eccentricity, e = b 2 = 1 e; 2 a b 1 b2 a 2 = a 1 e 68

35 How Do We Know that Gravitational Force is Conservative? Because the force is the derivative (with respect to r) of a scalar function of r called the potential, V(r): = m µ r + V o = m µ V r V ( r) = r V x V y V z ( r T r) + V 1 2 o = m µ x y r 3 z = (F g This derivative is also called the gradient of V with respect to r 69 Conservation of Energy Energy is conserved in an elastic collision, i.e. no losses due to friction, air drag, etc. Newton s Cradle illustrates interchange of potential and kinetic energy in a gravitational field 7

36 Examples of Circular Orbit Periods for Earth and Moon Period, min Altitude above Surface, km Earth Moon Typical Satellite Orbits GPS Constellation 26,6 km Sun- Synchronous Orbit 72

37 Geo-Synchronous Ground Track Geo- Synchronous Ground Track 42, 164 km Marco Polo Background Math Triple Vector Product Identity a ( b c) ( a i c)b ( a i b)c = ( a T c)b ( a T b)c Dot Product of Radius and Rate r T r = r i r = r dr dt 74