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2. Orbital Mechanics MAE 342 2016 2/12/20 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 2016 by Robert Stengel. All rights reserved. For educational use only. http://www.princeton.edu/~stengel/MAE342.html 1 1 Orbits 101 Satellites Escape and Capture (Comets, Meteorites) 2 2 1 2/12/20 Two-Body Orbits are Conic Sections 3 3 Classical Orbital Elements Dimension and Time a : Semi-major axis e : Eccentricity t p : Time of perigee passage Orientation Ω :Longitude of the Ascending/Descending Node i : Inclination of the Orbital Plane ω: Argument of Perigee 4 4 2 2/12/20 Orientation of an Elliptical Orbit First Point of Aries 5 5 Orbits 102 (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: 17,000 mph = 24,000 ft/s = 7.3 km/s • Super-circular velocities – Earth to Moon: 24,550 mph = 36,000 ft/s = 11.1 km/s – Escape: 25,000 mph = 36,600 ft/s = 11.3 km/s • Near escape velocity, small changes have huge influence on apogee 6 6 3 2/12/20 Newton’s 2nd 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 dv(t) ⎣⎡mv(t)⎦⎤ = m = ma(t) = F ⎡ ⎤ dt dt vx (t) ⎡ f ⎤ ⎢ ⎥ x ⎡ ⎤ d ⎢ ⎥ fx f ⎢ ⎥ m ⎢ vy (t) ⎥ = ⎢ y ⎥ F = fy = force vector dt ⎢ ⎥ ⎢ ⎥ ⎢ f ⎥ ⎢ f ⎥ ⎢ vz (t) ⎥ ⎢ z ⎥ ⎣⎢ z ⎦⎥ ⎣ ⎦ ⎣ ⎦ 7 7 Equations of Motion for a Particle Integrating the acceleration (Newton’s 2nd Law) allows us to solve for the velocity of the particle ⎡ ⎤ ⎡ ⎤ fx m ax dv(t) 1 ⎢ ⎥ ⎢ ⎥ = v! (t) = F = ⎢ fy m ⎥ = ⎢ ay ⎥ dt m ⎢ ⎥ ⎢ ⎥ fz m az ⎣⎢ ⎦⎥ ⎣⎢ ⎦⎥ T dv(t) T T 1 v(T ) = dt + v(0) = a(t)dt + v(0) = Fdt + v(0) ∫0 dt ∫0 ∫0 m 3 components of velocity ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ vx (T ) ax (t) vx (0) fx (t) m vx (0) ⎢ ⎥ T ⎢ ⎥ ⎢ ⎥ T ⎢ ⎥ ⎢ ⎥ ⎢ vy (T ) ⎥ = ⎢ ay (t) ⎥dt + ⎢ vy (0) ⎥ = ⎢ fy (t) m ⎥dt + ⎢ vy (0) ⎥ ⎢ ⎥ ∫0 ⎢ ⎥ ⎢ ⎥ ∫0 ⎢ ⎥ ⎢ ⎥ v T a t v 0 f t m v 0 ⎣⎢ z ( ) ⎦⎥ ⎣⎢ z ( ) ⎦⎥ ⎣⎢ z ( ) ⎦⎥ ⎣⎢ z ( ) ⎦⎥ ⎣⎢ z ( ) ⎦⎥ 8 8 4 2/12/20 Equations of Motion for a Particle Integrating the velocity allows us to solve for the position of the particle ⎡ ⎤ ⎡ ⎤ x!(t) vx (t) dr(t) ⎢ ⎥ ⎢ ⎥ = r!(t) = v(t) = ⎢ y!(t) ⎥ = ⎢ vy (t) ⎥ dt ⎢ ⎥ ⎢ ⎥ ⎢ z!(t) ⎥ ⎢ vz (t) ⎥ ⎣ ⎦ ⎣ ⎦ T dr(t) T r(T ) = dt + r(0) = v dt + r(0) ∫0 dt ∫0 3 components of position ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ x(T ) vx (t) x(0) ⎢ ⎥ T ⎢ ⎥ ⎢ ⎥ ⎢ y(T ) ⎥ = ⎢ vy (t) ⎥dt + ⎢ y(0) ⎥ ∫0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ z T v t z 0 ⎣⎢ ( ) ⎦⎥ ⎣⎢ z ( ) ⎦⎥ ⎣⎢ ( ) ⎦⎥ 9 9 Spherical Model of the Rotating Earth Spherical model of earth’s surface, earth-fixed (rotating) coordinates ⎡ x ⎤ ⎡ cos L cosλ ⎤ ⎢ o ⎥ ⎢ E E ⎥ RE = ⎢ yo ⎥ = ⎢ cos LE sinλE ⎥ R ⎢ ⎥ ⎢ ⎥ zo sin LE ⎣ ⎦E ⎣ ⎦ LE : Latitude (from Equator), deg λE : Longitude (from Prime Meridian), deg R : Radius (from Earth's center), deg Earth's rotation rate, Ω, is 15.04 deg/hr 10 10 5 2/12/20 Non-Rotating (Inertial) Reference Frame for the Earth Celestial longitude, λC, measured from First Point of Aries on the Celestial Sphere at Vernal Equinox λC = λE + Ω (t − tepoch ) = λE + Ω Δt 11 11 Transformation Effects of Rotation Transformation from inertial frame, I, to Earth’s rotating frame, E ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ xo cosΩΔt sinΩΔt 0 cosΩΔt sinΩΔt 0 ⎢ ⎥ R = ⎢ sin t cos t 0 ⎥R = ⎢ sin t cos t 0 ⎥ y E ⎢ − ΩΔ ΩΔ ⎥ I ⎢ − ΩΔ ΩΔ ⎥⎢ o ⎥ 0 0 1 0 0 1 ⎢ ⎥ ⎣⎢ ⎦⎥ ⎣⎢ ⎦⎥ zo ⎣ ⎦I Location of satellite, rotating and inertial frames ⎡ cos L cosλ ⎤ ⎡ cos L cosλ ⎤ ⎢ E E ⎥ ⎢ E C ⎥ rE = ⎢ cos LE sinλE ⎥(R + Altitude); rI = ⎢ cos LE sinλC ⎥(R + Altitude) ⎢ sin L ⎥ ⎢ sin L ⎥ ⎣ E ⎦ ⎣ E ⎦ Orbital calculations generally are made in an inertial frame of reference 12 12 6 2/12/20 Gravity Force Between Two Point Masses, e.g., Earth and Moon Magnitude of gravitational attraction Gm m F = 1 2 r2 G : Gravitational constant = 6.67 ×10−11Nm2 /kg2 st 24 m1 : Mass of 1 body = 5.98 ×10 kg for Earth nd 22 m2 : Mass of 2 body = 7.35 ×10 kg for Moon r : Distance between centers of mass of m1 and m2 , m 13 13 Acceleration Due To Gravity Gm1m2 F2 = m2a1on2 = 2 “Inverse-square r Law” Gm µ a = 1 ! 1 1on2 r2 r2 µ1 = Gm1 Gravitational parameter of 1st mass At Earth’s surface, acceleration due to gravity is µ 3.98 ×1014 m3 s2 a g E 9.798m s2 g ! oEarth = 2 = 2 = Rsurface (6,378,137m) 14 14 7 2/12/20 Gravitational Force Vector of the Spherical Earth Force always directed toward the Earth’s center ⎡ x ⎤ µE ⎛ rI ⎞ µE ⎢ ⎥ F = −m = −m y (vector), as r = r g 2 ⎜ ⎟ 3 ⎢ ⎥ I I rI ⎝ rI ⎠ rI ⎢ z ⎥ ⎣ ⎦I (x, y, z) establishes the direction of the local vertical ⎡ x ⎤ ⎢ y ⎥ ⎢ ⎥ ⎡ cos L cosλ ⎤ ⎢ z ⎥ E I rI ⎣ ⎦I ⎢ ⎥ = = cos L sinλ 2 2 2 ⎢ E I ⎥ rI x + y + z I I I ⎢ sin L ⎥ ⎣ E ⎦ 15 15 Equations of Motion for a Particle in an Inverse-Square- Law Field Integrating the acceleration (Newton’s 2nd Law) allows us to solve for the velocity of the particle ⎡ x ⎤ dv(t) 1 µE ⎛ rI ⎞ µE ⎢ ⎥ = v! t = F = − = − y ( ) g 2 ⎜ ⎟ 3 ⎢ ⎥ dt m rI ⎝ rI ⎠ rI ⎢ z ⎥ ⎣ ⎦ T dv(t) T T 1 v(T ) = dt + v(0) = a(t)dt + v(0) = Fdt + v(0) ∫0 dt ∫0 ∫0 m 3 components of velocity ⎡ ⎤ ⎡ 3 ⎤ ⎡ ⎤ vx (T ) x rI vx (0) ⎢ ⎥ T ⎢ ⎥ ⎢ ⎥ 3 ⎢ vy (T ) ⎥ = −µE ⎢ y rI ⎥dt + ⎢ vy (0) ⎥ ⎢ ⎥ ∫0 ⎢ ⎥ ⎢ ⎥ v T z r 3 v 0 ⎣⎢ z ( ) ⎦⎥ ⎣⎢ I ⎦⎥ ⎣⎢ z ( ) ⎦⎥ 16 16 8 2/12/20 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 ⎡ ⎤ ⎡ ⎤ x!(t) vx (t) dr(t) ⎢ ⎥ ⎢ ⎥ = r!(t) = v(t) = ⎢ y!(t) ⎥ = ⎢ vy (t) ⎥ dt ⎢ ⎥ ⎢ ⎥ ⎢ z!(t) ⎥ ⎢ vz (t) ⎥ ⎣ ⎦ ⎣ ⎦ T dr(t) T r(T ) = dt + r(0) = v dt + r(0) ∫0 dt ∫0 3 components of position ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ x(T ) vx (t) x(0) ⎢ ⎥ T ⎢ ⎥ ⎢ ⎥ ⎢ y(T ) ⎥ = ⎢ vy (t) ⎥dt + ⎢ y(0) ⎥ ∫0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ z T v t z 0 ⎣⎢ ( ) ⎦⎥ ⎣⎢ z ( ) ⎦⎥ ⎣⎢ ( ) ⎦⎥ 17 17 Dynamic Model with Inverse- Square-Law Gravity No aerodynamic or thrust force Neglect motions in the z direction msatellite << mEarth Dynamic Equations Example: Initial Conditions at Equator 3 v!x (t) = −µE xI (t) rI (t) 3 vx (0) = 7.5, 8, 8.5 km/s v!y (t) = −µE yI (t) rI (t) vy (0) = 0 x!I (t) = vx (t) x(0) = 0 y!I (t) = vy (t) y 0 = 6,378 km = R 2 2 ( ) where rI (t) = xI (t) + yI (t) 18 18 9 2/12/20 Equatorial Orbits Calculated with Inverse-Square-Law Model 19 19 Work “Work” is a scalar measure of change in energy With constant force, In one dimension W12 = F(r2 − r1 ) = FΔr In three dimensions T T W12 = F (r2 − r1 ) = F Δr With varying force, work is the integral ⎡ ⎤ r2 r2 dx T ⎢ ⎥ W = F dr = f dx + f dy + f dz , dr = dy 12 ∫ ∫ ( x y z ) ⎢ ⎥ r1 r1 ⎣⎢ dz ⎦⎥ 20 20 10 2/12/20 Conservative Force Iron Filings § Assume that the 3-D force Around Magnet field is a function of position F = F(r) § The force field is conservative if Force Emanating from Source r2 r1 ∫ FT (r)dr + ∫ FT (r)dr = 0 r1 r2 … for any path between r1 and r2 and back 21 21 Gravitational Force is Gradient of a Potential, V(r) Gravity potential, V(r), is a function only of position µ ∂ ⎛ µ ⎞ ∂ F m E r m E V r g = − 3 I = ⎜ ⎟ ! ( ) rI ∂r ⎝ r ⎠ ∂r 22 22 11 2/12/20 Gravitational Force Field Gravitational force field µE Fg = −m 3 rI rI Gravitational force field is conservative because r r 2 ∂ 1 ∂ V (r)drI − V (r)drI = ∫ ∂r ∫ ∂r r1 r2 r r 2 µ 2 µ − m E r dr + m E r dr = 0 ∫ r 3 I I ∫ r 3 I I r1 I r1 I … for any path between r1 and r2 and back 23 23 Potential Energy in Gravitational Force Field Potential energy, V or PE, is defined with respect to a reference point, r0 PE(r0 ) = V (r0 ) = V0 (≡ −U0 ) ⎛ µ ⎞ ⎛ µ ⎞ µ µ ΔPE ! V (r2 ) −V (r1 ) = −⎜ m +V0 ⎟ + ⎜ m +V0 ⎟ = −m + m ⎝ r2 ⎠ ⎝ r1 ⎠ r2 r1 24 24 12 2/12/20 Kinetic Energy Apply Newton’s 2nd Law to the definition of Work dr dv = v; dr = vdt F = m dt dt r2 t2 T t2 T ⎛ dv ⎞ 1 2 1 2 2 W12 = F dr = m⎜ ⎟ v dt ∫ ∫ ⎝ dt ⎠ = mv = m ⎣⎡v (t2 ) − v (t1 )⎦⎤ r1 t1 2 2 t1 t2 t2 1 d T 1 d 2 1 = m∫ (v v)dt = m∫ (v )dt = m ⎡v2 − v2 ⎤ ! T − T ! ΔKE 2 t dt 2 t dt ⎣ 2 1 ⎦ 2 1 1 1 2 Work = integral from 1st to 2nd time T (= KE) is the kinetic energy of the point mass, m 25 25 Total Energy in Point-Mass Gravitational Field § Potential energy of mass, µ m, depends only on the V = PE = −m gravitational force field r § Kinetic energy of mass, m, 1 depends only on the velocity T ! KE = mv2 magnitude measured in an 2 inertial frame of reference E = PE + KE § Total energy, , is the E µ 1 sum of the two: = −m + mv2 r 2 = Constant 26 26 13 2/12/20 Interchange Between Potential and Kinetic Energy in a Conservative System E 2 − E1 = 0 ⎛ µ 1 2 ⎞ ⎛ µ 1 2 ⎞ ⎜ −m + mv2 ⎟ − ⎜ −m + mv1 ⎟ = 0 ⎝ r2 2 ⎠ ⎝ r1 2 ⎠ P2 ⎛ µ µ ⎞ ⎛ 1 2 1 2 ⎞ ⎜ −m + m ⎟ = ⎜ mv2 − mv1 ⎟ ⎝ r2 r1 ⎠ ⎝ 2 2 ⎠ PE2 − PE1 = KE2 − KE1 P1 27 27 Specific Energy… Energy per unit of the satellite’s mass P E S = PES + KES 1 1 ⎛ mµ 1 2 ⎞ = − + mv m ⎝⎜ r 2 ⎠⎟ µ 1 P2 = − + v2 r 2 28 28 14 2/12/20 Angular Momentum of a Particle (Point Mass) h = (r × mv) = m(r × v) = m(r × r!) 29 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 h = (r × mv) = m(r × v) • Cross Product: Evaluation of a determinant with unit vectors (i, j, k) along axes, (x, y, z) and (vx, vy, vz) projections on to axes i j k x y z r × v = = (yvz − zvy )i + (zvx − xvz ) j + (xvy − yvx )k vx vy vz 30 30 15 2/12/20 Cross Product in Column Notation Cross product identifies perpendicular components of r and v i j k x y z r × v = = (yvz − zvy )i + (zvx − xvz ) j + (xvy − yvx )k vx vy vz ⎡ yv − zv ⎤ ⎡ ⎤ Column ⎢ ( z y ) ⎥ ⎡ 0 −z y ⎤ vx notation ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ r × v = zv − xv = z 0 x vy ⎢ ( x z ) ⎥ ⎢ − ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ −y x 0 ⎥
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