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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

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Orbits 101

Satellites Escape and Capture (Comets, Meteorites)

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Two-Body Orbits are Conic Sections

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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

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Orientation of an Elliptical Orbit

First Point of Aries

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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

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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 ⎦⎥ ⎣ ⎦ ⎣ ⎦

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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

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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

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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

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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

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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

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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

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Acceleration Due To Gravity

Gm1m2 F2 = m2a1on2 = 2 “Inverse-square r Law”

Gm1 µ1 a1on2 = 2 ! 2 r r

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 R2 6,378,137m 2 surface ( )

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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

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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

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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

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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 where r t = x2 t + y2 t I ( ) I ( ) I ( ) 18

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Equatorial Orbits Calculated with Inverse-Square-Law Model

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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

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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

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Gravitational Force is Gradient of a Potential, V(r)

Gravity potential, V(r), is a function only of position

µE ∂ ⎛ µE ⎞ ∂ Fg = −m rI = m ! V (r) r 3 ∂r ⎝⎜ r ⎠⎟ ∂r I

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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

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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 ⎜ r ⎟ ⎜ r ⎟ r r ⎝ 2 ⎠ ⎝ 1 ⎠ 2 1

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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

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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

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Interchange Between Potential and Kinetic Energy in a Conservative System

E 2 − E1 = 0 ⎛ µ 1 ⎞ ⎛ µ 1 ⎞ −m + mv2 − −m + mv2 = 0 ⎝⎜ r 2 2 ⎠⎟ ⎝⎜ r 2 1 ⎠⎟ 2 1

P2

⎛ µ µ ⎞ ⎛ 1 2 1 2 ⎞ ⎜ −m + m ⎟ = ⎜ mv2 − mv1 ⎟ ⎝ r2 r1 ⎠ ⎝ 2 2 ⎠

PE2 − PE1 = KE2 − KE1 P1

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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

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Angular Momentum of a Particle (Point Mass)

h = (r × mv) = m(r × v) = m(r × r!) 29

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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

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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 ⎥ ⎢ v ⎥ xvy − yvx ⎣ ⎦ z ⎣⎢ ( ) ⎦⎥ ⎣⎢ ⎦⎥ 31

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Angular Momentum Vector is Perpendicular to Both Moment Arm and Velocity

⎡ yv − zv ⎤ ⎢ ( z y ) ⎥ h = mr × v = m ⎢ zv − xv ⎥ ⎢ ( x z ) ⎥ ⎢ xv − yv ⎥ ⎣⎢ ( y x ) ⎦⎥ ⎡ ⎤ ⎡ 0 −z y ⎤ vx ⎢ ⎥⎢ ⎥ v = m ⎢ z 0 −x ⎥⎢ y ⎥ = mr!v ⎢ ⎥⎢ ⎥ −y x 0 vz ⎣ ⎦⎣⎢ ⎦⎥ 32

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Specific Angular Momentum Vector of a Satellite … is the angular momentum per unit of the satellite’s mass, referenced to the center of attraction m hS = r × v = r × v = r × r! m Perpendicular to the orbital plane

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Equations of Motion for a Particle in an Inverse-Square- Law Field

Acceleration is µ ⎛ r t ⎞ µ a t v t r t I ( ) r t ( ) = ! ( ) = !!( ) = − 2 ⎜ ⎟ = − 3 ( ) r (t) ⎝ r(t) ⎠ r (t)

… or

µ !r!+ 3 r = 0 r 34

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Cross Products of Radius and Radius Rate Then

⎡ µ ⎤ r × !r!+ r = 0 ⎢ 3 ⎥ ⎣ r ⎦

r × r = 0 r! × r! = 0 … because they are parallel

Chain Rule for Differentiation d (r × r!) = (r! × r!) + (r × !r!) = (r × !r!) dt 35

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Specific Angular Momentum

⎡ µ ⎤ µ r × !r!+ r = (r × !r!) + (r × r) ⎣⎢ r 3 ⎦⎥ r 3 0 d dh = (r × r!) = S = 0 dt dt Consequently

hS = Constant h ! h = r × r" (Perpendicular to the plane of motion) S ( ) Orbital plane is fixed in inertial space 36

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Eccentricity Vector is a Constant of Integration

⎡ µ ⎤ µ !r!+ r × h = !r!× h + r × h = 0 ⎢ 3 ⎥ 3 ⎣ r ⎦ r µ µ !r!× h = − 3 r × h = − 3 r × (r × r!) r r With triple vector product identity (see Supplement) µ µ d ⎛ r⎞ !r!× h = − r × (r × r!) = − (r!r − rr!) = µ 3 2 ⎝⎜ ⎠⎟ r r dt r Integrating

⎛ r ⎞ (!r!× h)dt = r! × h = µ + e ∫ ⎝⎜ ⎠⎟ r e = Eccentricity vector (Constant of integration) 37

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Significance of Eccentricity Vector

T ⎡ ⎛ r ⎞ ⎤ ⎡ ⎛ r ⎞ ⎤ r! × h − µ + e h = 0 because r! × h − µ + e = 0 ⎢ ⎝⎜ r ⎠⎟ ⎥ ⎢ ⎝⎜ r ⎠⎟ ⎥ ⎣ ⎦ ⎣ ⎦

T T µr h ∴(r! × h) h − − µeT h = 0 r 0 0

∴−µeT h = 0 § e is perpendicular to angular momentum, § which means it lies in the orbital plane § Its angle provides a reference direction for the perigee 38

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General Polar Equation of a Conic Section

T ⎡ ⎛ r ⎞ ⎤ r r! × h − µ + e = 0 ⎢ ⎝⎜ r ⎠⎟ ⎥ ⎣ ⎦ 1st term is angular momentum squared rT r! × h = hT r × r! = hT h = h2 ( ) ( ) Then

⎛ rT r ⎞ h2 − µ + rT e = 0 h2 = µ(r + rT e) = µ(r + recosθ ) = 0 ⎝⎜ r ⎠⎟

h2 µ r = 1+ ecosθ 39

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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”

p h2 µ r = = , m or km 1+ e cosθ 1+ e cosθ θ : True Anomaly = Angle from perigee direction, deg or rad 40

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Properties of Elliptical Orbits Eccentricity can be determined from apogee and perigee radii

r − r r − r e = a p = a p ra + rp 2a

rp = a /(1− e) ra = a /(1+ e)

Semi-major axis is the average of the two r + r a = a p 2 41

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Properties of Elliptical Orbits § Semi-latus rectum, p, can be expressed as a function of h or a and e p = h2 µ = a(1− e2 ) § Semi-minor axis, b, can be expressed as a function of ra and rp

b = rarp

§ Area of the ellipse, A, is

A = π a b = πa2 1− e2 , m2

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Energy is Inversely Proportional to the Semi-Major Axis

At the periapsis, rp

r! = 0 and v = r θ! 2 2 p p p p = h µ = a(1− e ) 2 1 ! 2 µ 1 h µ E S " E = rpθ p − = 2 − 2 ( ) r 2 r r rp = a(1− e) p p p 1 µ = µp − 2µr = ⎡ 1− e2 − 2 1− e ⎤ E 2 ( p ) 2 ⎣( ) ( )⎦ 2rp 2a(1− e) µ(1− e) = − 2a(1− e)

µ = − E 43 2a 43

Classification of Conic Section Orbits

Orbit Shape Eccentricity, e Energy, E Semi-Major Semi-Latus Axis, a Rectum, p Circle 0 < 0 > 0 a

Ellipse 0 < e < 1 < 0 > 0 a(1– e2)

Parabola 1 0 Undefined 2rp (è∞) Hyperbola >1 > 0 < 0 a(1– e2)

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“Vis Viva (Living Force) Integral”

Velocity is a function of radius and specific energy

1 2 µ ⎛ µ ⎞ v = + E v = 2 + E 2 r ⎝⎜ r ⎠⎟

§ Specific total energy, E, is µ inversely proportional to the E = − semi-major axis 2a § Velocity is a function of ⎛ 2 1⎞ radius and semi-major axis v = µ − ⎝⎜ r a⎠⎟

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Maximum and Minimum Velocities on an Ellipse Velocity at periapsis

⎛ 2 1⎞ ⎛ 2 1⎞ v p = µ⎜ − ⎟ = µ⎜ − ⎟ ⎝ rp a⎠ ⎝ a(1− e) a⎠ µ (1+ e) = a (1− e) Velocity at apoapsis

⎛ 2 1⎞ µ (1− e) va = µ⎜ − ⎟ = ⎝ ra a⎠ a (1+ e)

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Velocities at Periapses and Infinity of Parabola and Hyperbola

Parabola (a è ∞)

2µ vp = v∞ = 0 rp Hyperbola (a < 0)

⎛ 2 1⎞ µ v p = µ⎜ − ⎟ v∞ = − ⎝ rp a⎠ a

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Relating Time and Position in Elliptical Orbit Rearrange and integrate angular momentum over time and angle

2 dθ d( Ellipse Area) h = µp = r = 2 dt dt

t θ θ f f r2 p3 f dθ dt = dθ = ∫ ∫ µp µ ∫ 1 ecos 2 to θo θo ( + θ )

… but difficult to integrate analytically

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Anomalies of an Elliptical Orbit Angles measured from last periapsis

θ (or ν): True Anomaly E (or ψ ): Eccentric Anomaly M: Mean Anomaly

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Kepler’s Equation for the Mean Anomaly M = E − esin E From the diagram, ae + ( p − r) e a − r cosE = = a ae r = a(1− cosE) r! = E! aesin E

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Relationship of Time to Eccentric Anomaly

2 µ 1 ⎛ 2 h ⎞ E = − = r! + 2 − µr 2a 2 ⎝⎜ r ⎠⎟ where h2 = µa(1− e2 )

2 2 ⎡⎛ 2 1⎞ 2 2 ⎤ then r r! = µ ⎢⎜ − ⎟ r − a(1− e )⎥ ⎣⎝ r a⎠ ⎦ 3 2 a 2 ⎛ dE ⎞ leading to (1− ecosE) ⎜ ⎟ = 1 µ ⎝ dt ⎠ or

µ 3 dt = (1− ecosE)dE a 51

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Integrating the Prior Result …

t E 1 µ 1 dt = (1− ecosE)dE ∫ a3 ∫ t0 E0

µ E1 3 (t1 − t0 ) = (E − esin E) = M1 − M 0 a E0 Time is proportional to Mean Anomaly

M1 − M 0 M1 − M 0 (t1 − t0 ) = or t1 = t0 + µ a3 µ a3 Orbital Period, P

µ 2π (P − 0) = (E − esin E) = 2π a3 0 3 P = 2π a µ 52

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Orbital Period

Orbital period is related to the total energy a3 −µ2 P = 2π = π 3 where E < 0 for an ellipse µ 2 E Mean Motion, n, is the inverse of the Period

a3 2π P = 2π ! where n is the Mean Motion µ n 53

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Position and Velocity in Orbit at Time, t

Mean Anomaly, given time from perigee passage, tp

µ M (t) = t − t p a3 ( ) 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)

Eo (t) = M (t) + esin M (t)

Iterate until ΔM i < Tolerance

ΔM i = M (t) − ⎣⎡Ei (t) − esin Ei (t)⎦⎤

ΔEi+1 = ΔM i ⎣⎡1− ecosEi (t)⎦⎤

54 Ei+1 (t) = Ei (t) + ΔEi+1

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Position and Velocity in Orbit at Time, t

Calculate True Anomaly, given Eccentric Anomaly

−1 ⎡ 1+ e E(t)⎤ θ (t) = 2 tan ⎢ tan ⎥ ⎣ 1− e 2 ⎦

Compute magnitude of radius a(1− e2 ) r(t) = 1+ ecosθ (t)

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Position and Velocity in Orbit at Time, t Radius vector, in the orbital plane

⎡ x(t) ⎤ ⎡ r(t)cosθ (t) ⎤ r(t) = ⎢ ⎥ = ⎢ ⎥ ⎢ y t ⎥ ⎢ r t sinθ t ⎥ ⎣ ( ) ⎦ ⎣ ( ) ( ) ⎦ Velocity vector, in the orbital plane

⎡ ⎤ ⎡ ⎤ vx (t) µ −sinθ (t) v(t) = ⎢ ⎥ = ⎢ ⎥ ⎢ vy (t) ⎥ p ⎢ e + cosθ t ⎥ ⎣ ⎦ ⎣ ( ) ⎦ see Weisel, Spaceflight Dynamics, 1997, pp. 64-66

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Next Time: Planetary Defense

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Supplemental Material

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First Point of Aries (Ecliptic Intercept at Right)

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Dimension of energy? Scalar (1 x 1) Dimension of linear momentum? Vector (3 x 1) Dimension of angular momentum? Vector (3 x 1)

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Sub-Orbital (Sounding) Rockets 1945 - Present

NASA Wallops Island Control Center

Canadian LTV Black Brant XII Scout

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MATLAB Code for Flat- Earth Trajectories Script for Script for Numerical Solution Analytic Solution tspan = 40; % Time span, s g = 9.8; xo = [10;100;0;0]; % Init. Cond. t = 0:0.1:40; [t1,x1] = ode45('FlatEarth',tspan,xo);

vx0 = 10; Function for vz0 = 100; Numerical Solution x0 = 0; z0 = 0; function xdot = FlatEarth(t,x) % x(1) = vx vx1 = vx0; % x(2) = vz vz1 = vz0 – g*t; % x(3) = x x1 = x0 + vx0*t; % x(4) = z z1 = z0 + vz0*t - 0.5*g*t.* t; g = 9.8; xdot(1) = 0; xdot(2) = -g; xdot(3) = x(1); xdot(4) = x(2); xdot = xdot'; 62 end

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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

Dynamic Equations Analytic (Closed-Form) Solution v t v x ( ) = x0

v!z (t) = −g (z positive up) v T v x ( ) = x0 x!(t) = vx (t) T

z!(t) = vz (t) vz (T ) = vz − gdt = vz − gT 0 ∫ 0 0 Initial Conditions x T x v T ( ) = 0 + x0 v 0 = v x ( ) x0 T 2 vz (0) = vz 0 z(T ) = z0 + vz T − gt dt = z0 + vz T − gT 2 0 ∫ 0 0 x(0) = x0 63 z(0) = z0

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Trajectories Calculated with Flat-Earth Model

vx (0) = 10m / s

vz (0) = 100,150, 200m / s x(0) = 0 z(0) = 0

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MATLAB Code for Spherical-Earth Trajectories Script for Numerical Solution

R = 6378; % Earth Surface Radius, km tspan = 6000; % seconds options = odeset('MaxStep', 10) xo = [7.5;0;0;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 xdot = RoundEarth(t,x) % x(1) = vx % x(2) = vy % x(3) = x % x(4) = y Function for mu = 3.98*10^5; % km^2/s^2 Numerical Solution 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

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Equations that Describe Ellipses

x2 y2 + = 1 a2 b2 b a : Semi - major axis, m or km y o x b : Semi - minor axis, m or km a

x(θ ) = a cos(θ ) y(θ ) = b sin(θ ) θ : Angle from x - axis (origin at center) rad

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Constructing Ellipses

F1P1 + F2P1 = F1P2 + F2P2 = 2a Foci (from center), ⎡ ⎤ x f ⎡ 2 2 ⎤ ⎡ 2 2 ⎤ ⎢ ⎥ = ⎢ − a − b ⎥, ⎢ a − b ⎥ ⎢ y f ⎥ ⎢ 0 ⎥ ⎢ 0 ⎥ ⎣ ⎦1,2 ⎣ ⎦ ⎣ ⎦

String, Tacks, and Pen Archimedes Trammel Hypotrochoid

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Ellipses

Semi - latus rectum ("The Parameter"), b2 p = , m a

b2 Eccentricity, e = 1− a2 b2 = 1− e; b = a 1− e a2

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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): µ µ V (r) = −m +Vo = −m 1 2 +Vo r (rT r)

⎡ ∂V ∂x ⎤ ⎡ x ⎤ ∂V (r) ⎢ ⎥ µ ⎢ ⎥ = ⎢ ∂V ∂y ⎥ = m 3 y = −Fg ∂r r ⎢ ⎥ ⎢ ∂V ∂z ⎥ ⎢ z ⎥ ⎣ ⎦ ⎣ ⎦ This derivative is also called the gradient of V with respect to r 69

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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

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Examples of Circular Orbit Periods for Earth and Moon

Period, min Altitude above Surface, km Earth Moon 0 84.5 108.5 100 86.5 118 1000 105.1 214.6 10000 347.7 1905

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Typical Satellite Orbits

GPS Constellation 26,600 km Sun- Synchronous Orbit 72

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Geo-Synchronous Ground Track

Geo- Synchronous Ground Track 42, 164 km

Marco Polo- 1 & 2

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Background Math

Triple Vector Product Identity

a × (b × c) ≡ (a i c)b − (a i b)c = aT c b − aT b c ( ) ( ) Dot Product of Radius and Rate dr rT r! = r i r! = r dt 74

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