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Home , Azimuth, Circular orbit, Hohmann transfer orbit, Hyperbolic trajectory, Mean anomaly, Mean motion

Space Engineering Astrodynamics Where are we?

Where are we going? How do we get there? Early Nicolaus Copernicus 1473-1543

De revolutionibus orbium coelestium (1543)

• The center of the is not the center of the universe, but only of and of the lunar sphere.

• All the spheres revolve about the as their midpoint, and therefore the sun is the center of the universe.

• Whatever motion appears in the firmament arises not from any motion of the firmament, but from the earth's motion. The earth together with its circumjacent elements performs a complete on its fixed poles in a daily motion, while the firmament and highest heaven abide unchanged.

• What appear to us as motions of the sun arise not from its motion but from the motion of the earth and our sphere, with which we revolve about the sun like any other .

• The apparent retrograde and direct motion of the arises not from their motion but from the earth's. The motion of the earth alone, therefore, suffices to explain so many apparent inequalities in the heavens. Tycho Brahe 1546-1601

Tycho's geo-heliocentric • Danish Nobleman, Astronomer, Astrologer, Alchemist • Built two observatories – Hven, Prague • Accurate and Comprehensive Astronomical Observations • Published - De nova stella (1573) • Combined geometric benefits of Copernican system with philosophical benefits of the Ptolemaic system • Assisted by Johannes Kepler 1571-1630 Issac Newton 1642-1727

. Principia Mathematica (1687)

. Newton derived Kepler’s laws of planetary motion from his mathematical description of gravity, removing the last doubts about the validity of the heliocentric model of the cosmos. Richard Battin 1925-2014 Astronauts & Rendezvous EVA Rescue EVA Rescue (2)

Elements of Astrodynamics Launch into desired . , Inclination . LEO/GEO/Departure Orbital Maneuvers . Feasible Trajectories/Orbit Types . Minimize Propulsion Required . Orbit/Plane Changes Interplanetary Transfers . Hyperbolic . Changing Reference Frames . Orbital Insertion Rendezvous/Proximity Operations . Relative Motion Observations/Targeting/Entry/Landing . Ground Coverage (/swath) . Deorbit Burn Mission to (Spirit & Opportunity) Astrodynamics Reference Frames Coordinate Systems

(X,Y,Z) Origin? . Center of Earth . Sun or a . Center of a planetary body . Others…. Reference Axes . Axis of rotation or revolution (R,q,l) . Earth spin axis (R,j,l) • Equatorial Plane . Plane of the Earth’s orbit around the Sun • Plane . Need to pick two axes and then 3rd one is determined Ecliptic and Equatorial Planes

Obliquity of the Ecliptic = 23.44 ° Vernal Equinox vector - Earth to Sun on March 21st - Planes intersect @ Equinox Inertial Coordinates Relationship between Coordinate Frames Solar and

The Sun Drifts east in the ~1° per . Rises 0.066 later each day. (because the earth is orbiting)

The Earth… Rotates 360° in 23.934 hours (Celestial or “Sidereal” Day) Rotates ~361° in 24.000 hours (Noon to Noon or “Solar” Day)

Satellites orbits are aligned to the Sidereal day – not the solar day Astrodynamics Properties of Orbits • a is the semimajor axis; • b is the semiminor axis;

• rMAX = ra, rMIN = rp are the maximum and minimum -vectors; • c is the distance between the focus and the center of the ; • e = c/a is eccentricity 2 b 2  = 1 - e a

• 2p is the latus rectum (latus = side, rectum = straight) p — semilatus rectum or semiparameter • A = ab is the of the ellipse

p = 1 + e cosq r Orbital Elements e 120° 150° 90° Eccentricity V= (0.0 to 1.0) j ()

Apogee a Perigee 180° Semi-major 0° axis

e=0.8 vs e=0.0 e defines ellipse shape a defines ellipse size v/j defines angle from perigee Inclination i

Intersection of the equatorial and Inclination orbital planes (above) (angle) i

(below) Ascending Node Equatorial Plane ( defined by Earth’s )

Ascending Node is where a satellite crosses the equatorial plane moving south to of the ascending node Ω and Argument of perigee ω

Ω = angle from vernal equinox to ascending node on the equatorial plane Perigee Direction

ω = angle from ascending node to perigee on the orbital plane ω Ω Ascending Node

Vernal Equinox Orbital Elements

a - semi-major axis Ω - right ascension of ascending node e - eccentricity  - argument of perigee i - inclination  - true anomaly (also j) The Six Orbital Elements a = Semi-major axis (usually in kilometers or nautical miles) e = Eccentricity (of the elliptical orbit) v/j = True anomaly The angle between perigee and satellite in the orbital plane at a specific time i = Inclination The angle between the orbital and equatorial planes Ω = Right Ascension () of the ascending node The angle from the Vernal Equinox vector to the ascending node on the equatorial plane

 = Argument of perigee The Shape, Size, angle measured between the Orientation, and Satellite ascending node and perigee Location. Two Line Orbital Elements N ASA and NORAD Standard for specifying orbits of Earth-orbiting ISS (ZARYA) 1 25544U 98067A 08264.51782528 −.00002182 00000-0 -11606-4 0 2927 2 25544 51.6416 247.4627 0006703 130.5360 325.0288 15.72125391563537

Ref: http://en.wikipedia.org/wiki/Two-line_element_set Astrodynamics Integrating Multi-Body Dynamics Equations of Motion – The 2-Body Problem Equations of Motion (2) Equations of Motion - Energy Equations of Motion – Equations of Motion - Eccentricity Vector Equations of Motion – Possible Orbital Trajectories

• e=0 -- circle • e<1 -- ellipse • e=1 -- parabola • e>1 -- hyperbola

e < 1 Orbit is ‘closed’ – recurring path (elliptical) e > 1 Not an orbit – passing trajectory (hyperbolic) Conic Section Other Useful Properties Equations of Motion – Kepler’s 3rd Law vs. Altitude

a3 T  2 

h = 160 n.mi T = 90 minutes h = 19,324 n.mi “High” Earth Orbit T = 23 h 56 m 4 s h = 3444 n.mi T = 4 hours Orbital Velocity vs. Altitude

 21 v    ra

h = 160n.mi v = 25,300 ft/s Geosynchronous Orbit h = 19,324 n.mi “High” Earth Orbit v = 10,087 ft/s h = 3444 n.mi v = 18,341 ft/s Orbital Velocity vs. Altitude (2) (Elliptical Orbits)  21 v    ra

h = 160 n.mi h = 19,324 n.mi v = 33,320 ft/s v = 5,273 ft/s

Geosynchronous a = 13,186 n.mi e = 0.726 Parabolic Trajectories Total Energy = 0 Hyperbolic Trajectories Total Energy > 0  21 v    ra As Example 1 – A satellite in a circular orbit with an altitude of 274.6 km passes over USYD at time t=0, when is the next fly-over? Assumptions: 32 REE6,378 km, E = 398,600.441 km / s ,  360 / 24 hr s Example 1 – Circular Orbit A satellite in a polar circular orbit with an altitude of 274.6 km passes over USYD at time t=0, when is the next fly-over?

Orbit Period: h  274.6 km, R E  6,378 km  r a 6,652. 6 a36652.6 3 km 3 T 2   2 32  5400 sec = 90 min E 398,600.441 km / s When does USYD cross the orbit plane? 360  15 / hr  not exactly right as we shall see E 24 hrs It happens twice - near the ascending and decending nodes. 180 360 12 hrs, 24 hrs 15 / hr 15 / hr Where is the satellite in 12 and 24 hours? 360 Since Tn 90 min,   4 / min T In 12 hrs, j  12  60  4 / min  2880  8.0 orbits The satellite is back near the ascending node, but USYD is on the descending node side... In 24 hrs, j  24  60  4 / min  5760  16.0 orbits The satellite and USYD meet again near the ascending node! Example 2 – Elliptical Orbit A giant space station is placed in a permanent elliptical orbit for transporting crew and cargo from the Earth to the . This ‘Cycler’ space station has an apogee at the same radius as the Moon’s orbit. Describe/draw the orbit that cycles twice per (i.e. has a lunar rendezvous every other cycle)? What is the semi-major axis and the eccentricity? For a vehicle launching from the surface of the Earth, what is the rendezvous altitude to dock with the cycler ? How fast does a vehicle need to be going to rendezvous?

Assumptions: 32 E =398,600.441 km / s , is circular with T  27.3 days Example 2 – Elliptical Orbit A giant space station is placed in a permanent elliptical orbit for transporting crew and cargo from the Earth to the Moon. This ‘Cycler’ space station has an apogee at the same radius as the Moon’s orbit. Describe the orbit that cycles twice per month (i.e. has a lunar rendezvous every other cycle)? What is the semi-major axis and the eccentricity? For a vehicle launching from the surface of the Earth, what is the rendezvous altitude to dock with the cycler spacecraft? How fast does a vehicle need to be going to rendezvous?

27.3 km aa333 T  * 24*360022  32 2398,600.441 km / sE  a=241,263 km

raereaMoon(1) 0.5933 384,403 km

raep = (1- )98,122 km (rendezvous altitude) 21 va rv ap paErv , and 410 km/s raa

vraa v p 1606 km/s rp Example 3 – A from the has entered the inner and has a 20% probability of crashing into the Earth. If it misses the Earth, it will pass within 10 million kilometres of the Sun and then crash into with 100% probability on it’s way back out. In order to avert disaster, and for scientific exploration, we would like to capture this comet and place it into Earth orbit. What change in velocity is required to capture the comet?

Assumptions:

3 2 3 2 E =398,600.441 km / s , Sun = 132,712,440,018 km / s Planned Earth orbit is circular with h  1,000 km

RrE 6,378 km, Distance to the Sun =  149,597,870.7 km Example 3 – Parabolic Trajectory A comet from the Oort Cloud has entered the inner solar system and has a 20% probability of crashing into the Earth. If it misses the Earth, it will pass within 10 million kilometres of the Sun and then crash into Mercury with 100% probability on it’s way back out. In order to avert disaster, and for scientific exploration, we would like to capture this comet and place it into Earth orbit. What change in velocity is required to capture the comet? 12 E 0  v2 Sun  v  Sun 2 rr

At Earth's orbit intersection, r  149,597,870.7 km The comet's velocity would be,

2Sun vcomet  r A satellite in circular Earth orbit at h  1000 km

E has velocity vorbit  RE 1000 The change in velocity is

E2 Sun v  vorbit  v comet    34.77 km/s RrE 1000  Example 4 – An interplanetary probe needs to depart Earth with a velocity of 10 km/s (relative to the Earth). The last engine firing occurs at the perigee point with an altitude of 1000 km. Find the velocity needed at perigee and the parameters of the orbit (a, e)? What is the asymptotic departure angle  ?

v  10 km/s Assumptions: 32 E =398,600.441 km / s , RE  6,378 km  Example 4 – Hyperbolic Trajectory An interplanetary probe needs to depart Earth with a velocity of 10 km/s (relative to the Earth). The last engine firing occurs at the perigee point with an altitude of 1000 km. Find the velocity needed at perigee and the parameters of the orbit (a, e)? What is the asymptotic departure angle  ?

 21 E Using, v   , for r   , v  10 km/s=  r a a  a   3986 km  21 Therefore at perigee, v pE    14.42 km/s  RaE 1000  22 Now h rp v p and h/  a (1 e )

2 2 h RvE 1000 p  So solving for e  1  = 12   .85 aaE E p From r , r   as 1  e cosj  0 1 e cosj

1  1  This occurs at j cos  110.5 ,   180  j  69.5  2.85  Astrodynamics Kepler’s Equation When Will You Be Where? What do we know... a3  Given T , or a , we know the energy of an orbit T 2 , E    2a h2 Given e , we also know the angular momentum  p  a (1  e2 ) 

2 and everything else about the orbit's shape... b a 1 e , rpa a (1  e ), r  a (1  e ) h2 /  Given a true anomaly we can compute r  1 e cosj  21 and with position we know velocity v    ra What don’t we know? - WHEN? - All of this is in the orbit plane – we need 3D? Position and Time Kepler’s Equation  1eejj 12 sin From M 2tan1  tan  e  E - e sin E Kepler’s 1ee 2 1 cosj   Equation It is convenient to define, EE in which case, sin (2) 1 e2 sinj EEsin 1 ecosj 1 e2 sinj tan   22 2 1 cos E 2  1 e2 sinj  1e cosj  1  e cos j  1  e2 sin j Trig Identity 11      1 e cosj  1 e2 sinj  1e cosj  1  2 e cos j  e2 cos 2 j  1  e 2 1  cos 2 j  (1ee )(1 ) sinj  1e cosj  1  2 e cos j  e2 cos 2 j  1  e 2  e 2 cos 2 j  cos 2 j (1ee )(1 ) sinj (1e )(1  e ) sin j (1  e )(1  e ) sin j   1e cosj  2 e cos j  e2  cos 2 j 1  e cos j  ( e  cos j ) 2 1ee cosjj   cos (1ee )(1 ) sinj (1e )(1  e ) sinjj 1  e    tan (1e )(1  cosjj ) (1  e )2 1  cos 1  e 2 Ee1 j  1 e j  Therefore, tan tan and E 2tan1  tan 2 1 e 2  12 e 

Kepler’s Equation (3) What does this mean? Now we can relate position and time! For a given orbit (ae , ) and position (i.e. true anomaly) j ,  1 e j  Compute the '' E  2tan1  tan   12 e  Compute the M E - e sin E Then using the period T , the M time since periapsis passage is t t T 0 2 Note: From j  t is easy! From tj is not so easy (cannot write E f ( M , e ))

Physical Interpretation of the Eccentric Anomaly E Computing Orbital Position

Mean Anomaly

Eccentric Anomaly

True Anomaly

Mean Motion Solving Kepler’s Equation

In the case of finding position as a function of time j (t ), r ( t ) From, 2 t M: M  t  t  T 0 M E: E  e sin E  M  1 eE E  j: j  2tan1  tan   12 e  E/j  r : r  a (1  e cos E ) or r  a (1  e2 ) / (1  e cos j ) So the only problem is finding EM from Fortunately, the function M ( E ) E  sin E is monotonic ( if E  then M  ) Solving Kepler’s Equation (2) Method 1: Simple Iteration For small eccentricity e , the values of M and E are close Initially guess that EM

Then calculate a new estimate E using Eii1  M e sin E Solving Kepler’s Equation (3) Method 2: Newton's Method To find the root of fx ( ) 0, start with an estimate xi and update the estimate based on a simple slope calculation:

f()() xii1  f x fx (i )  xxii1 

Choose xi1 such that fx (i1 ) 0,

fx()i that is... xxii1  fx()i In this case, f ( E ) E  e sin E  M f ( E ) 1 e cos E

Eii esin E M So the iteration would be... EEii1  1 eE cos i

fx()i Stop when the update is smaller than a desired tolerance  errordesired fx()i Solving Kepler’s Equation (4) 350 of searching for the fastest method (least number of computations) Newton (1686) Lagrange (1771) fx( ) 0 kk1 Euler (1740), EM0  1dM (sin ) fx() E M ek xx k k 1 kk1 Ekk1  M esin E k 1 k! dM fx()k Levi-Civita (1904)E L ( M ) zk Bessel (1817) k k 1

2 2 eeexp 1 E M Jk ( ke )sin( kM ) k 1 k ze() 2 1 11e ij Jk ( x ) cos x sin k d 0 Lk( M ) p c i cos M , s j sin M Halley (1742) f( Ek ) E k e sin E k M 2f ( Ekk ) f ( E ) Ek1 E k2 f( E k ) 1 e cos E k 2f ( Ek ) f ( E k ) f ( E k ) f( Ekk ) e sin E Position and Time for Parabolic Trajectories hd3 j j Beginning again with t  2  2  0 1 e cosj  For e  1, the result is less complicated than for the ellipse, j h3 dj h 3 11 j j  h 3 tM  tan  tan3  2 2 2 2 p 0 1 cosj   2 2 6 2   1 jj1 h3 From j tM : tan tan3 , tM p 2 2 6 2  2 p  2 From tj : M t p h3 3 11 jj j = root of  tan tan  M p  0 6 2 2 2 The only real root is,

1/3 1/3 1 22 j  2tan 3M 9MMM1  3  9  1  p p  p p  Position and Time for Hyperbolic Trajectories Analogous to the Elliptical Solution, but with hyperbolic functions... Recall: sinhx ex  e x /2, cosh x  e x  e x /2

Kepler's Equations becomes: Mh  e sinh F F (hyperbolic eccentric anomaly) The relation between j and F becomes, Fe1 j tanh tan 2e  1 2 From j  t :

3 12 e 1 j  h 2/3 F 2tanh tan , Mh  e sinh F F , t 2 Mh  e  1  e 12  From t  j :

2  213/2   eF1  Meh 3   1 t , e sinh F F  M h (Iteration), j  2tan  tanh  h  e 12 Astrodynamics Orbital Maneuvers Changing Orbits - The Effects of Burns Posigrade & Retrograde

Initial Orbit h Final Orbit V V h Typical for v of 1 ft / s   h  0.5 nmi A posi-grade burn will RAISE orbital altitude. A retro-grade burn will LOWER orbital altitude. Note – max effect is at 180° from the burn point. Changing Orbits - The Effect of Burns Radial In & Radial Out

V =  2 1 EXAMPLE: Radial In Burn at Perigee (r a )

Resultant Velocity Initial Velocity Initial Orbit Final Orbit Radial burns shift the argument of perigee without significantly altering other orbital parameters Orbital Transfers - Changing Planes

V2 V

V1 •Burn point must be intersection of two orbits (“nodal crossings”)

•Extremely expensive energy- wise: For 160 nmi circular orbit, V25,600 ft/sec. A plane change of 1° requires a V of 470 ft/sec. Plane Change Maneuver The components of a v determine how the orbit is affected. - In-plane v can change the parameters ae , ,j ,  vh - Out-of-plane v can change the parameters ( ,i ). v can be expressed as: vorbit v   v   v  v rˆ   v radial orth radial orth v vradial cannot change the orbit plane radial

h  r   vradial  r  v radial rˆ  0

vorth can have components in and out of plane

vorth  vvorbit   h  v orbitφvˆ    h where φˆ r ˆ and φ ˆ h

vhorbit can only change to magnitude of (not direction)

vh is the only component that can change the orbit plane Plane Change Maneuver (2) Hohmann Transfer

A Hohmann Transfer is an that transfers a satellite from one circular orbit to another. It was invented by Walter Hohmann, a German scientist, in 1925.

A Hohmann Transfer is the fuel efficient way to get from one circular orbit to another circular orbit. Hohmann Transfer (2) • For Example, if we want to move a vc2 spacecraft from LEO → GEO and assuming both orbits are in the GEO same plane

• Initial LEO orbit has radius r1 and velocity vc1 LEO  vc1  r1 r1

• Desired GEO orbit has radius r2 and velocity v c2 vc1 r2 • At LEO (r1), vc1 = 7,724 m/s

• At GEO (r2), vc2 = 3,074 m/s

• Could accomplish this in many ways Hohmann Transfer (3) • For Example, if we want to move a vc2 spacecraft from LEO → GEO and assuming both orbits are in the GEO same plane

• Initial LEO orbit has radius r1 and velocity vc1 LEO  vc1  r1 r1

• Desired GEO orbit has radius r2 and velocity v c2 vc1 r2 • At LEO (r1), vc1 = 7,724 m/s

• At GEO (r2), vc2 = 3,074 m/s

• Could accomplish this in many ways Hohmann Transfer (4) • For Example, if we want to move a vc2 spacecraft from LEO → GEO and assuming both orbits are in the GEO same plane

• Initial LEO orbit has radius r1 and velocity vc1 LEO  vc1  r1 r1

• Desired GEO orbit has radius r2 and velocity v c2 vc1 r2 • At LEO (r1), vc1 = 7,724 m/s

• At GEO (r2), vc2 = 3,074 m/s

• Could accomplish this in many ways Hohmann Transfer (5) • For Example, if we want to move a vc2 spacecraft from LEO → GEO and assuming both orbits are in the GEO same plane

• Initial LEO orbit has radius r1 and velocity vc1 LEO  vc1  r1 r1

• Desired GEO orbit has radius r2 and velocity v c2 vc1 r2 • At LEO (r1), vc1 = 7,724 m/s

• At GEO (r2), vc2 = 3,074 m/s

• The Hohmann transfer is the most efficient path Hohmann Transfer (7) vc2

GEO

• Impulsive V1 is applied to get on at perigee:

v1  vtransfer  v LEO LEO

 21  r    1    rtransfer a transfer  r1 v  21  vc1 1 v     1 1  r2 v  r12  r 1 r 2  r 1 1

• Leave LEO (r1) with a total velocity of v1 Hohmann Transfer (8) v Apogee c2 V2 GEO • Transfer orbit is an ellipse with

– Perigee located at r1

– Apogee located at r2 LEO

• Arrive at GEO (apogee) r1

with v2 v v1  21 c1 v    r2 2  v1  ra2 transfer 

Perigee Hohmann Transfer (9) • When arriving at GEO, which is at vc2 apogee of elliptical transfer orbit, must v v2 apply some v in order to circularize: 2 2 GEO

v2  vGEO  v transfer

  21 =    r22 r atransfer  LEO 22   =  r1 r2 r 2 r 1 r 2 • This is exactly the v that should be v applied to circularize the orbit at GEO vc1 1 (r2) r2 v1 vC 2 v 2   v 2

• If this V is not applied, spacecraft will continue on dashed elliptical trajectory Hohmann Transfer (10) vc2 Hohmann Transfer Summary v2 v2 • Initial LEO orbit has radius r1 and GEO velocity vc1  v  c1 r

• Desired GEO orbit has radius r2 and velocity vc2 LEO • Impulsive V1 is applied to get on Hohmann transfer orbit at perigee: r1 22   v    1 r r r r 1 1 2 1 v vc1 1 • Coast to apogee and apply r2 v impulsive V2: 1 22   v2    r2 r 2 r 1 r 2 Example 1: Planar Hohmann Transfer

From an initial orbit with radius r1  14,000 km compute the

Hohmann transfer orbit to achieve an orbit with r2  28,000 km . What are the initial, intermediate and final ? What is the total v ?

Useful information: 32 E  398,600.441 km / s

   21 vc1  vc2  v  r  r1 2  ra Example 1: Planar Hohmann Transfer(2)

  v 5.336 v 3.773 c1 c2 r r1 2

 11  v1 2     0.825, v 1  vC 1   v 1  6.161  r1 r 1 r 2 r 1

 11 v2 2    3.08  r2 r 1 r 2 

22   v2  vC 2  v 2     0.693 r2 r 2 r 1 r 2

vtot   v12   v  1.5197 km / sec Geosynchronous Orbit Geosynchronous Orbit

Coverage from GEO

TDRSS Comm Coverage Geosynchronous Transfer So, how can we get a spacecraft from a non-equatorial orbit into a geosynchronous one? • Launching into a LEO orbit will have an inclination greater than or equal to the of the launch site. • Need to do a plane change as well as raising the orbital altitude. • Solution - Do the transfer orbit first and do the plane change and circularization burn at apogee! • Rationale – For the same plane change angle, v is less where v is less. Example 2: Geosynchronous Transfer For launch from Cape Canaveral the initial orbit has an inclinition of i 28 deg and an altitude of h1 300 km . We are targeting a geosynchronous orbit with r2  42,186 km , design the two v burns, taking into account the plane change. Is it possible that there is a more efficient maneuver?

Useful information: 32 E  398,600.441 km / s

   21 vc1  vc2  v  r  r1 2  ra 2 2 2 vv1 v 1 v 2 v 22 v 1 v 2 v 1 v 2 2 v 1 v 2 cos q Example 2: Geosynchronous Transfer (2)

 rr   11 12 vC 2 vC11, a  , v  2    r12  r 1 r 1 r 2  i  11  v v v1  v 1  vC 1 2     2.426 km/s 2 2  r1 r 1 r 2 r 1 Determine velocity needed at apogeeof transfer orbit   11 vvC 22 3.074 km/s,  2    1.607 km/s r2 r 2 r 1 r 2  2 2 2 v2  vCC 2  v 2 2 v 2 v 2 cos i   v 2  1.819km/s

vtot   v 1   v 2  4.245 km/s Bi-Elliptic Transfer The Hohmann transfer is the most efficient 2-burn solution. The Bi-Elliptic transfer can be more efficient with 3-burns when the final orbit has a much greater radius than the initial orbit. Bi-Elliptic sequence:

(1) Circular Orbit 1 with radius rA

(2) vAAB to Orbit 2 with radii r , r

(3) vBBC to Orbit 3 with radii r , r

(4) vC to Circularize to Orbit 4 Notes:

- As rvBB  ,   0

- vC will be retrograde (slowdown) 1 1 1  aa33 - Long transfer time , TTT   2 23  2   Manueuver 23 2 2 2   Bi-Elliptic Transfer (2)

The Hohmann transfer is more efficient if rrCA / <11.94

The Bi-Elliptic transfer is more efficient if rrCA / >15.58

Otherwise, it depends on rrBA / as shown below. Super Geosynchronous Transfer (Super GTO – Launch to GEO) 3. Hohmann burn circularizes at GEO GEO Initial transfer orbit has LEO Target greater apogee than v3 Initial Orbit standard GTO. Orbit v Plane change at much 1 higher altitude requires T2 far less ΔV. PRO: Less overall ΔV from higher inclination launch sites. CON: Takes longer to T1 establish the final orbit.

1. LEO (or 2. Plane change Launch) to plus Hohmann v ‘Super GTO’ burn 2 Phasing Maneuvers Phasing orbits are used to change a spacecraft's position in it's orbit (i.e. for rendezvous, targeting, timing). Prograde Burn: Additional v New orbit is slower, fall back " Up To Slow Down " Retrograde Burn: Negative v Smaller orbit is faster, move ahead "Slow Down To Speed Up "  aa33 t 2  N Phase  , where N  # of cycles on phasing orbit    Launch Angle to Orbit Plane The lowest attainable orbit inclination matches the latitude of the launch site. Launch to Orbit Plane (2) Relationship between Launch Azimuth Angle and as Function of Latitude Propulsion Requirements Balance of momentum d mV  mv mv 0 dt mve mv vm ffdm dv v e m vmii  m  v  v  v   vln m  ln m  v ln i f i e f i e    m f  T mv v Defining I ee   v  gI spmg mg g e sp

 v     mm gI v  gIlnii   e sp  Example: sp    mmff For an engine using LH22 / LO (I sp =450), Since m m m i prop f what is the fraction of propellant  v   v       mi  gIsp   gIsp  to obtain a v 10 km / s ?  e , mi e = m i m prop mm  10,000  i prop m  prop 1 e 9.8 450   0.896  v     mi Finally, m m 1 e gIsp  prop i  That is, 90% Propellant  Astrodynamics Orbital Perturbations Orbital Perturbations

- Nonspherical Planet - Gravitational Anomalies - Atmospheric Density - Magnetic Fields

- Solar Pressure - Thruster Firings - Other Celestial Bodies - Phaser Blasts Perturbations - J2000 Inertial Frame The Earth’s U

where Spherical Term  = Universal x Mass of Earth r = Spacecraft Radius Vector from Center of Earth

ae = Earth Equatorial Radius P() = Legendre Polynomial Functions f = Spacecraft Latitude l = Spacecraft Longitude J = Zonal Harmonic Constants Zonal Harmonics n Cn,m,Sn,m = Tesseral & Sectorial Harmonic Coefficients

Notes: J2=0.001082635 has 1/1000th the effect of the spherical term. All other terms start at 1/1000th of J2’s effect. Tesseral Harmonics Sectorial Harmonics Only the first 4x4 (n=4, m=4) nm nm Typically used for S/C software. The J2 Effect

The Earth’s oblateness causes the most significant of any of horbit the nonspherical terms. h J2 Right Ascension of the Ascending Node (Ω), Argument of Perigee (ω) and the (n) are affected.

- 1 0 100 -9 200 -8 300 Typical Shuttle -7 Nodal Regression is the most important -6.7 500 Orbits operationally. -6 -5 Magnitude depends on orbit size (a), 1000 shape (e) and inclination (i). -4 -3

N o d al S h ift, d eg rees/d ay Posigrade orbits’ nodes regress -2

Westward (0° < i < 90°) -1

Retrograde orbits’ nodes regress 0 0 10 20 30 3940 50 60 70 80 90 180 170 160 150 140 130 120 110 100 Eastward (90° < i <180°) Inclination, Degrees Nodal Regression

Orbital planes rotate westward over time. (above)

Ascending Node (below)

Nodal Regression can be used to advantage (such as assuring desired lighting conditions) Ground Track

Ground tracks drift westward as the Earth rotates below. 360 deg / 24 hrs = 15 deg/hr + Nodal Regression

Molniya - 12hr Period Sets inclination such that argument of perigee regression is zero. This enables a long loitering time over the apogee position. Used by USA and USSR for spy satellites Sun-Synchronous Orbits (2)

A Sun- has a shift in ascending node ~1° per day.

Scans the same path under the same lighting conditions each day.

Requires a slightly retrograde orbit (For example: I = 97.56° for a 550km SSO).

Used for reconnaissance, terrain mapping, etc.. Sunsynchronous ()

i = 98.8° Period 98 min 700 km Atmospheric Drag Perturbations

Atmospheric density is a function of altitude, latitude, solar heating, season, time of day, land mass vs water, etc.

Drag depends on atmospheric density, but also spacecraft speed, attitude, frontal area, material properties, etc. Effect is to lower the apogee of an elliptical orbit

Perigee remains relatively constant Relative Magnitude of Perturbations Relative Magnitude of Perturbations (2) Astrodynamics Relative Motion Relative Motion, Rendezvous & Proximity Operations (Prox Ops) Relative Motion Consider two spacecraft flying in proximity to one another. Target vehicle AB is passive. All maneuvers are performed by chase vehicle . We are interested in the motion of BA relative to - as viewed from vehicle A .

Inertial Perspective Relative Perspective from Vehicle A

Astrodynamics Targeting Maneuvers Targeting Maneuvers Space Shuttle Orbital Targeting Display

Initial and desired position and times

Current State Vector

Computing the v for targeted maneuvers What's under the hood? Lambert Targeting! Lambert’s Problem Given r 1, r 2 and the flight time  t from P 1 to P 2, find the orbital trajectory from PP 1 to 2.

Note: Considered an problem, but commonly used as a rendezvous and intercept technique. The trajectory from P 1 to P 2 also determines the  v required at P 1. A sequence of desired relative points can be accomplished with a series of Lambert targeted burns. To Solve this we're going to need Lagrange Coefficients and Universal Variables! Astrodynamics Shuttle/ISS Rendezvous & Prox Ops Space Shuttle Maneuver Targeting

Astrodynamics Interplanetary Trajectories Interplanetary Trajectories (Patched Conic Approximation) Sphere of Influence Interplanetary Trajectory has 3 regimes: - 1st Planet’s gravity field is dominant R - Sun’s gravity field is dominant  r + nd - 2 Planet’s gravity field is dominant 0 mp mS Sphere of influence - The sphere about a body in which its gravity field is dominant.

Approximately coincides with the definition of the L1 Lagrangian point.

Consider a spacecraft in orbit about planet with mass mrp at radius , and the planet is in orbit about the Sun () mass mS at radius R . L1 is located where forces balance. GmGm Gm Gm Gm Gm SSSS pp22R r  R  r 0 since  R r22r2 R r r 2 R 3 R 3

2 2  r  Generallyr R , R  r R  1  2  therefore,  R  1/3 mSSS rmp m r m m p r  m p  21  2   2  2  1    0  3 3r  2    R R r R R R r R 3 mS  2/5 r  mp  (Lagrange - more precise analysis - see Curtis) SOI   Rm S  Interplanetary Hohmann Transfer The most efficient inter-planetary trajectory occurs when the departure and arrival velocities are tangent to the planetary orbits. Assumptions: - Both planets orbit the Sun in the same plane - The planetary orbits are circular - The transfer ellipse is only affected by the Sun's gravity

 2 1  2 1  V     D Sun    r a  R1 R 1 R 2  /2

Sun 2R2 VD  RRR1 1 2

Sun V1  R1

Sun2R2 Sun VVVDD  1   RRRR1 1 2 1   Sun 2R2 VD  1 RRR1 1 2    Sun 2R1 Likewise, VA  1   RRR2 1 2  Hyperbolic Departure

Example: Earth-to-Mars Hohmann Transfer VE - Departure is a hyperbolic trajectory in the Earth's SOI - Departure is parallel to the Earth's velocity

Define VP  Required Perihelion Velocity,

Then VPE V v Circular = v - Transfer to Planet 2 is Elliptical around Sun c - Arrival is hyperbolic relative to Planet 2 - Trajectory begins in a circular Earth orbit Define the following terms:

RVEE, - Heliocentric and velocity rv, - Geocentric spacecraft position and velocity RV, - Heliocentric spacecraft position and velocity

v - Hyperbolic excess velocity at infinity from Earth

r0 - radius of circular Earth

vrc - speed in Earth circular orbit of radius 0

vr10 - speed at perigee of Earth hyperbolic trajectory (at )

VVAP, - speed at aphelion and perihelion of heliocentric transfer ellipse Hyperbolic Departure (2)

First compute the transfer orbit velocity at perihelion: RR The Earth to Mars transfer orbit has a  EM tr 2 The equation of energy applied to perihelion V 2    PSSS      sphere of influence 22REEM atr R R So,

VE (Earth orbit) 2 2SMR VP  RRREEM   r0  At sphere of influence R R r d E  vv1 c  v (Patch conditions) VPE V v v

departure hyperbola Hyperbolic Departure (3)

At the SOI the velocity is vvVVPE    Comparing energy between perigee and the SOI, v2 v 2 v 2 1 EE  SOI   2rr 2 2 0 SOI sphere of influence which gives the velocity required at perigee,  vv2 2 E 1  V (Earth orbit) r0 E

Therefore, the required v1 is r 2 EE 0  v11  v  vc  v  2  rr00 d

To determine the position for the burn, vv1 c  v  v use, a E and r  a (1  e ),  hyperbola 2 0 r compute, e  1 0  1, ahyperbola p a(1 e2 ) departure hyperbola then from r  1ee cos j 1  cos j

1  1  Note that as r rSOI  ,  1  e cos j  0 and j    cos   e  Hyperbolic Departure (4)

Any departure velocity parallel to VE with magnitude v is valid, so this results in a surface of possible departure hyperbolas.

- Each hyperbolic plane includes the Earth center and has a perigee r0 , while prescribes a cone of departure points. Hyperbolic Arrival Arrival trajectories similarly fall on a surface of possible hyperbolas. For planetary capture (outer planet): - a positive v is required - rendevous must occur from ahead of the planet! - a burn is required at periapsis for capture - The v required depends

on the periapsis radius rp

Gravity Assist Maneuvers V Trailing side flyby increases velocity P v2 V Leading side flyby decreases velocity 2 VVVV vv and   1PP 1 2 2  VP vv12

Consider the energy gain VP 11 22SSV v (before) 1VV 1  , (after) 2  2  1 1 22RR12

RR12

112 2 2 2 2 2 V2  V 1 [( Vp  2VV pvv 2 v 2  ) ( V p 2 p 1  v 1  )] 22 22 vv1 2  vv 1   2  The energy gained through gravity assist is:

 Vpp vv21    2Vv  cos(    )  true anomaly at SOI (or at  ), as   ,   Gravity Assist Maneuvers (2) Voyager 2

Cassini Farewell