Orbital Mechanics - spacecraft.ssl.umd.eduspacecraft.ssl.umd.edu/academics/483F15/483F15L05.orbmech/483F15L...Orbital Mechanics ENAE 483/788D - Principles of Space Systems Design U

Orbital Mechanics ENAE 483/788D - Principles of Space Systems Design

U N I V E R S I T Y O FMARYLAND

Orbital Mechanics

• Lecture #05 – September 15, 2015 • Planetary launch and entry overview • Energy and velocity in orbit • Elliptical orbit parameters • Orbital elements • Coplanar orbital transfers • Noncoplanar transfers • Time in orbit • Interplanetary trajectories • Relative orbital motion (“proximity operations”)

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© 2015 David L. Akin - All rights reserved http://spacecraft.ssl.umd.edu

http://spacecraft.ssl.umd.edu



Space Launch - The Physics

• Minimum orbital altitude is ~200 km !!

• Circular orbital velocity there is 7784 m/sec !!

• Total energy per kg in orbit

Potential Energy

kg in orbit= � µ

rorbit+

µ

rE= 1.9⇥ 106 J

kg

Kinetic Energy

kg in orbit=

12

µ

r2orbit

= 30� 106 J

kg

Total Energy

kg in orbit= KE + PE = 32� 106 J

kg

2



Theoretical Cost to Orbit

• Convert to usual energy units !!!

• Domestic energy costs are ~$0.05/kWhr !!Theoretical cost to orbit $0.44/kg

Total Energy

kg in orbit= 32� 106 J

kg= 8.888

kWhrs

kg

3



Actual Cost to Orbit

• SpaceX Falcon 9 – 13,150 kg to LEO – $65 M per flight – Lowest cost system currently

flying

• $4940/kg of payload • Factor of 11,000x higher

than theoretical energy costs!

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What About Airplanes?

• For an aircraft in level flight, !!

• Energy = force x distance, so !!

• For an airliner (L/D=25) to equal orbital energy, d=81,000 km (2 roundtrips NY-Sydney)

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Total Energykg

=thrust� distance

mass=

Td

m=

gd

L/D

WeightThrust

=LiftDrag

, ormg

T=

L

D



Equivalent Airline Costs?

• Average economy ticket NY-Sydney round-round-trip (Travelocity 9/3/09) ~$1300

• Average passenger (+ luggage) ~100 kg • Two round trips = $26/kg

– Factor of 60x more than electrical energy costs – Factor of 190x less than current launch costs

• But… you get to refuel at each stop!

6



Equivalence to Air Transport

• 81,000 km ~ twice around the world

• Voyager - one of only two aircraft to ever circle the world non-stop, non-refueled - once!

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Orbital Entry - The Physics

• 32 MJ/kg dissipated by friction with atmosphere over ~8 min = 66kW/kg

• Pure graphite (carbon) high-temperature material: cp=709 J/kg°K

• Orbital energy would cause temperature gain of 45,000°K!

• (If you’re interesting in how this works out, you can take ENAE 791 Launch and Entry Vehicle Design next term...)

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Newton’s Law of Gravitation

• Inverse square law !!

• Since it’s easier to remember one number, !

• If you’re looking for local gravitational acceleration,

F =

GMm

r2

µ = GM

g =

µ

r2

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Some Useful Constants

• Gravitation constant µ = GM – Earth: 398,604 km3/sec2

– Moon: 4667.9 km3/sec2

– Mars: 42,970 km3/sec2 – Sun: 1.327x1011 km3/sec2

• Planetary radii – rEarth = 6378 km

– rMoon = 1738 km

– rMars = 3393 km

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• Kinetic Energy

!• Potential Energy!

• Total Energy

Energy in Orbit

<--Vis-Viva Equation

K.E. =1

2mv

2=⇒

K.E.

m=

v2

2

P.E. = −µm

r=⇒

P.E.

m= −

µ

r

Constant =v2

2−

µ

r= −

µ

2a

11

v2 = µ

�2r� 1

a

⇥



Classical Parameters of Elliptical Orbits

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The Classical Orbital Elements

Ref: J. E. Prussing and B. A. Conway, Orbital Mechanics Oxford University Press, 1993

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Implications of Vis-Viva

• Circular orbit (r=a)

!• Parabolic escape orbit (a tends to infinity)

!• Relationship between circular and parabolic orbits

vcircular =

!

µ

r

vescape =

!

2µ

r

vescape =√

2vcircular

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The Hohmann Transfer

vperigee

v1

vapogee

v2r1

r2

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First Maneuver Velocities

• Initial vehicle velocity !!

• Needed final velocity !!

• Delta-V

v1 =

!

µ

r1

vperigee =

!

µ

r1

!

2r2

r1 + r2

∆v1 =

!

µ

r1

"!

2r2

r1 + r2

− 1

#

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Second Maneuver Velocities



• Delta-V

∆v2 =

!

µ

r2

"

1 −

!

2r1

r1 + r2

#

vapogee =

!

µ

r2

!

2r1

r1 + r2

v2 =

!

µ

r2

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Implications of Hohmann Transfers

• Implicit assumption is made that velocity changes instantaneously - “impulsive thrust”

• Decent assumption if acceleration ≥ ~5 m/sec2 (0.5 gEarth)

• Lower accelerations result in altitude change during burn ⇒ lower efficiencies and higher ΔVs

• Worst case is continuous “infinitesimal” thrusting (e.g., ion engines) ⇒ ΔV between circular coplanar orbits r1 and r2 is

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�VLow Thrust

= Vc1 � V

c2 =

rµ

r1�r

µ

r2



Limitations on Launch Inclinations

Equator

19



Differences in Inclination

Line of Nodes

20



Choosing the Wrong Line of Apsides

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Simple Plane Change

vperigee

v1vapogee

v2

Δv2

22



Optimal Plane Change

vperigee v1 vapogee

v2

Δv2Δv1

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First Maneuver with Plane Change Δi1



• Delta-V

v1 =�

µ

r1

vp =�

µ

r1

�2r2

r1 + r2

�v1 =�

v21 + v2

p � 2v1vp cos �i1

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Second Maneuver with Plane Change Δi2



• Delta-V

�v2 =�

v22 + v2

a � 2v2va cos �i2

va =�

µ

r2

�2r1

r1 + r2

v2 =�

µ

r2

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Sample Plane Change Maneuver

Optimum initial plane change = 2.20°

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Calculating Time in Orbit

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Time in Orbit

• Period of an orbit

!• Mean motion (average angular velocity)

!• Time since pericenter passage

!➥M=mean anomaly

P = 2⇥

�a3

µ

n =�

µ

a3

M = nt = E � e sinE

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Dealing with the Eccentric Anomaly

• Relationship to orbit

!• Relationship to true anomaly

!• Calculating M from time interval: iterate

! until it converges

r = a (1� e cos E)

tan�

2=

�1 + e

1� etan

E

2

Ei+1 = nt + e sinEi

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Example: Time in Orbit

• Hohmann transfer from LEO to GEO – h1=300 km --> r1=6378+300=6678 km

– r2=42240 km

• Time of transit (1/2 orbital period)

a =12

(r1 + r2) = 24, 459 km

ttransit =P

2= ⇥

�a3

µ= 19, 034 sec = 5h17m14s

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Example: Time-based Position

Find the spacecraft position 3 hours after perigee !!!!!E=0; 1.783; 2.494; 2.222; 2.361; 2.294; 2.328; 2.311;

2.320; 2.316; 2.318; 2.317; 2.317; 2.317

Ej+1 = nt + e sin Ej = 1.783 + 0.7270 sin Ej

n =�

µ

a3= 1.650x10�4 rad

sec

e = 1� rp

a= 0.7270

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Example: Time-based Position (cont.)

Have to be sure to get the position in the proper quadrant - since the time is less than 1/2 the period, the spacecraft has yet to reach apogee --> 0°<θ<180°

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E = 2.317

tan�

2=

�1 + e

1� etan

E

2=⇥ � = 160 deg

r = a(1� e cosE) = 12, 387 km



Basic Orbital Parameters

• Semi-latus rectum (or parameter) !

• Radial distance as function of orbital position

!• Periapse and apoapse distances !

• Angular momentum

h⃗ = r⃗ × v⃗

p = a(1 − e2)

r =p

1 + e cos θ

rp = a(1 − e) ra = a(1 + e)

h =

√

µp

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Velocity Components in Orbit

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r =p

1 + e cos �

vr =dr

dt=

d

dt

�p

1 + e cos �

⇥=�p(�e sin � d�

dt )(1 + e cos �)2

vr =pe sin �

(1 + e cos �)2d�

dt

1 + e cos � =p

r� vr =

r2 d�dt e sin �

p�⇤h = �⇤r ⇥�⇤v



Velocity Components in Orbit (cont.)

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~h = ~r ⇥ ~v h = rv cos � = r

✓rd✓

dt

◆= r2

d✓

dt



Patched Conics• Simple approximation to multi-body motion (e.g.,

traveling from Earth orbit through solar orbit into Martian orbit)

• Treats multibody problem as “hand-offs” between gravitating bodies --> reduces analysis to sequential two-body problems

• Caveat Emptor: There are a number of formal methods to perform patched conic analysis. The approach presented here is a very simple, convenient, and not altogether accurate method for performing this calculation. Results will be accurate to a few percent, which is adequate at this level of design analysis.

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Example: Lunar Orbit Insertion• v2 is velocity of moon

around Earth • Moon overtakes

spacecraft with velocity of (v2-vapogee)

• This is the velocity of the spacecraft relative to the moon while it is effectively “infinitely” far away (before lunar gravity accelerates it) = “hyperbolic excess velocity”

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Planetary Approach Analysis

• Spacecraft has vh hyperbolic excess velocity, which fixes total energy of approach orbit !

• Vis-viva provides velocity of approach !

• Choose transfer orbit such that approach is tangent to desired final orbit at periapse

v =

!

v2

h+

2µ

r

∆v =

!

v2

h+

2µ

r−

!

µ

r

v2

2� µ

r= � µ

2a=

v2h

2

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Patched Conic - Lunar Approach

• Lunar orbital velocity around the Earth !!

• Apogee velocity of Earth transfer orbit from initial 400 km low Earth orbit !!

• Velocity difference between spacecraft “infinitely” far away and moon (hyperbolic excess velocity)

vm =

!

µ

rm

=

!

398, 604

384, 400= 1.018

km

sec

va = vm

!

2r1

r1 + rm

= 1.018

!

6778

6778 + 384, 400= 0.134

km

sec

vh = vm − va = vm = 1.018 − 0.134 = 0.884km

sec

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Patched Conic - Lunar Orbit Insertion

• The spacecraft is now in a hyperbolic orbit of the moon. The velocity it will have at the perilune point tangent to the desired 100 km low lunar orbit is !

• The required delta-V to slow down into low lunar orbit is

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vpm =

rv2h +

2µm

rLLO=

r0.8842 +

2(4667.9)

1878= 2.398

km

sec

�v = vpm � vcm = 2.398�r

4667.9

1878= 0.822

km

sec

Red text is a correction to original notes



ΔV Requirements for Lunar Missions

To:

From:

Low EarthOrbit

LunarTransferOrbit

Low LunarOrbit

LunarDescentOrbit

LunarLanding

Low EarthOrbit

3.107km/sec

LunarTransferOrbit

3.107km/sec

0.837km/sec

3.140km/sec

Low LunarOrbit

0.837km/sec

0.022km/sec

LunarDescentOrbit

0.022km/sec

2.684km/sec

LunarLanding

2.890km/sec

2.312km/sec

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LOI ΔV Based on Landing Site

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LOI ΔV Including Loiter Effects

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Interplanetary Trajectory Types

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Interplanetary “Pork Chop” Plots• Summarize a number of

critical parameters – Date of departure – Date of arrival – Hyperbolic energy (“C3”) – Transfer geometry

• Launch vehicle determines available C3 based on window, payload mass

• Calculated using Lambert’s Theorem

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C3 for Earth-Mars Transfer 1990-2045

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Earth-Mars Transfer 2033

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Earth-Mars Transfer 2037

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Interplanetary Delta-V

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C3 = V 2h

Hyperbolic excess velocity ⌘ Vh

Vreq =p

V 2esc + C3

�V =pV 2esc + C3 � Vc

2033 Window: �V = 3.55 km/sec

2037 Window: �V = 3.859 km/sec

�V in departure from 300 km LEO



Hill’s Equations (Proximity Operations)

€

˙ ̇ x = 3n2x + 2n˙ y + adx

€

˙ ̇ y = −2n˙ x + ady

€

˙ ̇ z = −n 2z + adz

Ref: J. E. Prussing and B. A. Conway, Orbital Mechanics Oxford University Press, 1993

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Clohessy-Wiltshire (“CW”) Equations

€

x(t) = 4 − 3cos(nt)[ ]xo +sin(nt)

n˙ x o +

2n

1− cos(nt)[ ] ˙ y o

€

y(t) = 6 sin(nt)− nt[ ]xo + yo −2n

1−cos(nt)[ ] ˙ x o +4sin(nt)− 3nt

n˙ y o

€

z( t) = zo cos(nt) +˙ z on

sin(nt)

€

˙ z ( t) = −zonsin(nt) + ˙ z o sin(nt)

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“V-Bar” Approach

Ref: Collins, Meissinger, and Bell, Small Orbit Transfer Vehicle (OTV) for On-Orbit Satellite Servicing and Resupply, 15th USU Small Satellite Conference, 2001

52



“R-Bar” Approach

• Approach from along the radius vector (“R-bar”) • Gravity gradients decelerate spacecraft approach

velocity - low contamination approach • Used for Mir, ISS docking approaches

Ref: Collins, Meissinger, and Bell, Small Orbit Transfer Vehicle (OTV) for On-Orbit Satellite Servicing and Resupply, 15th USU Small Satellite Conference, 2001

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References for This Lecture

• Wernher von Braun, The Mars Project University of Illinois Press, 1962

• William Tyrrell Thomson, Introduction to Space Dynamics Dover Publications, 1986

• Francis J. Hale, Introduction to Space Flight Prentice-Hall, 1994

• William E. Wiesel, Spaceflight Dynamics MacGraw-Hill, 1997

• J. E. Prussing and B. A. Conway, Orbital Mechanics Oxford University Press, 1993

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Orbital Mechanics - spacecraft.ssl.umd.eduspacecraft.ssl.umd.edu/academics/483F15/483F15L05.orbmech/483F15L...Orbital Mechanics ENAE 483/788D - Principles of Space Systems Design U

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