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

Conic section Equations of motion Momentum and energy Kepler’s Equation Position and in

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Orbits 101

Satellites Escape and Capture (, )

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Two-Body Orbits are Conic Sections

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Classical 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., – and or – Earth and or – Earth and : radius and velocity are constant • : 17,000 mph = 24,000 ft/s = 7.3 km/s • Super-circular – 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 , small changes have huge influence on apogee 6

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Newton’s 2nd Law

§ Particle of fixed (also called a point mass) acted upon by a changes velocity with § 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 : (from ), deg

λE : Longitude (from Prime Meridian), deg R : Radius (from Earth's center), deg

Earth's 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

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 , e.g., Earth and Moon Magnitude of gravitational attraction

Gm m F = 1 2 r2

G : = 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 : between centers of mass of m1 and m2 , m

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Acceleration Due To

Gm1m2 F2 = m2a1on2 = 2 “Inverse- 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 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 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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⎡ 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 = (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 Equation of a

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 § of the 2 bodies is at Earth’s center of mass § Center of mass is at one of ’s focal points § Other focal point is “vacant”

p h2 µ r = = , m or km 1+ e cosθ 1+ e cosθ θ : = 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

§ 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∞) >1 > 0 < 0 a(1– e2)

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“Vis Viva (Living Force) Integral”

Velocity is a function of radius and

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 of 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 ): M:

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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 , 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 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, 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 ( 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) 1945 - Present

NASA Wallops Island Control Center

Canadian LTV Black Brant XII Scout

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MATLAB Code for Flat- Earth 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

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 , 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- 72

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Geo-Synchronous

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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