Solo Hermelin, profile picture
Uploaded bySolo Hermelin
PPT, PDF2,087 views

Keplerian trajectories

The document discusses Kepler's laws of planetary motion and Newton's laws of motion and universal gravitation, providing insights on the two-body problem and its mathematical implications. It elaborates on various concepts including conservation of angular momentum, mechanical energy, and the integration of equations of motion relevant to orbital trajectories. The document serves as a comprehensive guide to understanding the dynamics of celestial bodies and their orbits.

Embed presentation

Downloaded 55 times
1 KEPLERIAN TRAJECTORIES SOLO HERMELIN
2 SOLO Table of Contents 2. Newton’s Laws 3. Newton’s Law of Universal Gravitation 4. Two Body Problem 4.1 Polar Representation 4.2 Conservation of Angular Momentum KEPLERIAN TRAJECTORIES 1. Kepler’s Laws 4.3 Conservation of Mechanical Energy 4.4 Integration of Equations of Motion 4.5 Velocity on the Trajectory 4.6 Specific Mechanic Energy 4.7 Eccentricity Vector 4.8 Orbit Determination from Initial Conditions 4.9 Flat Earth Approximation 4.10 Orbital Trajectories 4.11 Time of Flight on an Elliptic Orbit 5. References
3 SOLO Kepler’s Laws KEPLERIAN TRAJECTORIES Johannes Kepler 1571 - 1630 Tycho Brache 1546 - 1601 From 1601 to 1606 Kepler tried to fit various geometrical curves to Tycho Brache’s data on Mars orbit.
4 SOLO Kepler’s Laws KEPLERIAN TRAJECTORIES 1. The orbit of each planet is an ellipse, with the sun at a focus. 2. The line joining the planet to the sun sweeps out equal area in equal times. 3. The square of the period of a planet orbit is equal to the cube of its mean distance to the sun. b a a dA h dt 2 = b a 2/322 a GMGM ab TP ππ == b
5 SOLO Newton’s Laws of Motion KEPLERIAN TRAJECTORIES . Every body continues in its state of rest or of uniform motion in straight line unless it is compelled to change that state by forces impressed upon it. . The rate of change of momentum of a given body is proportional to the force impressed on the body and is in the same direction as the force. . To every action there is always an equal reaction “THE MATHEMATICAL PRINCIPLES OF NATURAL PHILOSOPHY”1687 Isaac Newton 1643-1727
6 SOLO Newton’s Law of Universal Gravitation KEPLERIAN TRAJECTORIES “THE MATHEMATICAL PRINCIPLES OF NATURAL PHILOSOPHY”1687 Isaac Newton 1643-1727 Any two body attract one on other with a force proportional to the product of the masses and inversely proportional to the square of the distance between them.    EQPOISSON G G r GM r GM g gm r GM mr r mM GF ρπϕϕϕ 4&& 1 2 2 =∇=−∇=      −∇= −=      ∇=−= 228 /1067.6 gmcmdyneG − ×= The Universal Gravitational Constant Instantaneous propagation of the force along the direction between the masses (“Action at a Distance”) GF  GF  M m
7 SOLO Two Body Problem KEPLERIAN TRAJECTORIES Assumptions 1. The bodies are spherical symmetric. 2. The gravitational forces are the only forces acting on the bodies. Mr  mr  r  M m Cr  Let Mr  -the position vector of the mass M relative to an inertial point. mr  -the position vector of the mass m relative to an inertial point. Mm rrr  −= -the relative position vector of the mass m relative to M. r r r  = → :1 -unit vector in direction.r  Let apply the Newton’s Law of Universal Gravitation to describe the forces each mass applies on the other. → → =→= −=→−= r r mG r r r r MmG rM r r MG r r r r MmG rm MM mm 1 1 22 22      
8 SOLO Two Body Problem (continue – 1) KEPLERIAN TRAJECTORIES → → =→= −=→−= r r mG r r r r MmG rM r r MG r r r r MmG rm MM mm 1 1 22 22       Mm rMrm r Mm C + + =   The Center of Mass of those masses is ( ) ( )MmGr r r r MmG rrr Mm +=←−= + −=−= →→ :11 22 µ µ From those equations we obtain: and 0   = + + = Mm rMrm r Mm C The Center of Mass is not accelerating
9 SOLO Two Body Problem (continue – 2) KEPLERIAN TRAJECTORIES Polar Representation Define ω  r  - Angular Velocity Vector →→ Θ+= nrR 11  ωω Decompose in and perpendicular to direction.r  r  ω  Since we can define a right hand Cartesian system using vector product: 011 =⋅ →→ rn →→→ ×= rnt 11:1          −=×      Θ+=×==      +Θ−=×      Θ+=×==      Θ=×      Θ+=×==      →→→→→→→ →→→→→→→→ →→→→→→→ tnnrnnn td d nrtnrttt td d trnrrrr td d RR RR R 1111111 11111111 1111111 ωωω ωωω ωω       Differentiating we obtain:
10 SOLO Two Body Problem (continue – 3) KEPLERIAN TRAJECTORIES Polar Representation (continue – 1) or                               − Θ− Θ =               → → → → → → n t r n t r td d R R 1 1 1 00 0 00 1 1 1 ω ω  → = rrr 1  Therefore →→→→ Θ+=+= trrrrrrrr 1111   ( ) ( ) ( ) →→→ →→→→→ Θ+Θ+Θ+Θ−=       +Θ−Θ+Θ+Θ+Θ+= nrtrrrrr nrrtrrtrrrr R R 1121 11111 2   ω ω → = rrr 1  vtrrrr  =Θ+= →→ 11 ( ) ( ) →→→→ −=Θ+Θ+Θ+Θ−= r r nrtrrrrrr R 11121 2 2 µ ω 
11 SOLO Two Body Problem (continue – 4) KEPLERIAN TRAJECTORIES Conservation of Angular Momentum Vector Solution Cross-multiply the equation of motion by → −= r r r 12 µ r  03 =×−=× rr r rr  µ Define the specific angular momentum as:h  vrrrh  ×=×=: Differentiate h  0    =×+×= rrrr td hd Therefore is a constant vector in space that is perpendicular to the trajectory vectors and , therefore those vectors (and the trajectory) remain in the plane defined by the initial conditions vrh  ×= r  v  ( ) ( ) ( )000 =×=== tvtrth 
12 SOLO Two Body Problem (continue – 5) KEPLERIAN TRAJECTORIES Conservation of Angular Momentum (continue – 1) Polar Coordinate Solution →→→→ Θ=      Θ+×=×= nrtrrrrrvrh 1111 2   The equation of motion in polar coordinates is ( ) ( ) →→→→ −=Θ+Θ+Θ+Θ−= r r nrtrrrrrr R 11121 2 2 µ ω  from which 02 =Θ+Θ  rr and r rdd 2−= Θ Θ   Integrated both sides ( ) ConstrhConstrConstr =Θ=→=Θ→+−=Θ  222 lnlnlnlnln 1 000 ≠Θ←=→=Θ  rr RR ωω2 →→→→ =→=−=      constntn td d R 1011  ωalso →→ =Θ=×= constnrvrh 12  therefore
13 SOLO Two Body Problem (continue – 6) KEPLERIAN TRAJECTORIES Conservation of Mechanical Energy Vector Solution Dot-multiply the equation of motion by → −= r r r 12 µ r  rr r rr  ⋅−=⋅ 3 µ Use vrvr  == & 03 =⋅+⋅ rr r vv  µ Use also the identity aaaaaaaaaa aa    ⋅=→→→ →→ =      +⋅=⋅ 011 111 0 22 0 22 3 =      −=      −+      →=+ r v td d rtd dv td d rr r vv µµµ    const r v E energy potential specific energy kinetic specific =−= µ 2 : 2 We can see that E is the specific mechanical energy.
14 SOLO Two Body Problem (continue – 7) KEPLERIAN TRAJECTORIES Integration of Equations of Motion Vector Solution Cross-multiply the equation of motion with the specific angular momentum → −= r r r 12 µ h  hr r hr   ×−=× 3 µ ( ) ( ) ( )rr td d rhrh td d rhrhr td d      ××+×=×+×=× The left side is ( ) ( ) hrr r rrhrrrrrrrhr           ×=            −××+×=××+××+×= 0 3 0 µ The right side can be written ( ) ( ) ( )[ ] ( )       =−=−=⋅−⋅=××−=×− r r td d r r r v r rrrrv r vrrrrv r vrr r hr r     µ µµµµµµ 2 2 3333 Equaling both sides gives ( )       =× r r td d hr td d   µ Integrating both sides B r r hr   +=× µ where is the constant of integrationB 
15 SOLO Two Body Problem (continue – 8) KEPLERIAN TRAJECTORIES Integration of Equations of Motion (continue – 1) Vector Solution (continue – 1) Dot-multiply this equation by : B r r hr   +=× µ r  ( ) Br r r rhrr     ⋅+⋅=×⋅ µ Use ( ) ( ) 2 hhhhrrhrr =⋅=⋅×⋅=×⋅    r r r r µµ =⋅   ( )pBrBr Θ−Θ=⋅ cos  to obtain ( )pBrrh Θ−Θ+= cos2 µ or ( ) ( ) ( )pp e p B h r Θ−Θ+ = Θ−Θ+ = cos1cos/1 /2 µ µ where µ/: 2 hp = µ/: Be = pΘ parameter (semi-latus rectum) eccentricity periapsis angle
16 SOLO Two Body Problem (continue – 9) KEPLERIAN TRAJECTORIES Integration of Equations of Motion (continue – 2) Polar Coordinate Solution The equations of motion in polar coordinates are: 2 2 r rr µ −=Θ−  Consthr ==Θ2 Let define a new variable: r u 1 := from which urrr r u 2 2 1 −=→−=  If , we can write( )00 ≠≠Θ h Θ =  h r2 and 0≠ Θ −= Θ −= h d ud hu h r    Differentiating again we obtain 02 2 22 22 2 2 2 ≠ Θ −= Θ −=Θ Θ −= h d ud uh r h d ud h d ud hr  We also have 32 4 2 2 uh r h rr ==Θ The first equation of motion becomes 0232 2 2 22 ≠=+ Θ huuh d ud uh µ
17 SOLO Two Body Problem (continue – 10) KEPLERIAN TRAJECTORIES Integration of Equations of Motion (continue – 3) Polar Coordinate Solution (continue – 1) Therefore 32 4 2 2 uh r h rr ==Θ The solution of this equation is 0232 2 2 22 ≠=+ Θ huuh d ud uh µ 02 3 2 2 ≠=+ Θ h h u d ud µ ( ) 221 cos 1 h CC r u µ +−Θ== where C1 and C2 are constant of integration pC p e C h p Θ=== 212 ,:,: µ Define to obtain ( )pe p r Θ−Θ+ = cos1 This is the equation of a conic section in polar coordinates.
18 SOLO A right circular cone is a cone obtained by generators (straight lines) passing through a circle, and the apex C that is situated on the normal to the circle plane and passing trough the center of the circle. β is the angle between the cone axis and the generators. CONIC SECTIONS Cutting Plane generating a "hyperbola" Right Circular Cone Cone Apex Conical Section C Cone Axis Cutting Plane generating a "parabola" Cutting Plane generating a "ellipse" Cutting Plane generating a "circle" Cutting Plane generating two "lines" α β       −= β π α 2       −< β π α 2       −> β π α 2 ( )0=α                   −> −= −< lines line po 2 2 1 2 int 2 β π α β π α β π α P F F * Cutting Plane (Hyperbola ) Right Circular Cone Hyperbola 2 Branches C β α Ellipse Parabola Cutting Plane (Ellipse) Cutting Plane (Circle) Cutting Plane (Parabola) By cutting the right circular conic by a plane we obtain different conic sections, as a function of the inclination angle α of the plane relative to the base of the conic section and the angle β between the generators and the base. The discovery of the Conical Sections is attributed to the greek Menachmus who lived around 350 B.C..
19 SOLO The conical sections are: CONIC SECTIONS 1. Circle if the cutting plane is normal to the cone axis (α=0) and is above or bellow the apex. 2. Ellipse if the cutting plane is inclined to the basis at an angle that falls short of the angle between generators to the base (α<π/2-β) (in greek word elleipsis means falls, short or leaves out. 3. Hyperbola if the cutting plane is inclined to the basis at an angle that exceeds of the angle between generators to the base (α>π/2-β)(in greek word hyperbole means excess. 4. Parabola if the cutting plane is parallel to a generator of the right circular cone (α=π/2-β) (in greek word parabole is the origin of the words parabola and parallel. 5. A point- apex (α<π/2-β), one straight line (α=π/2-β), two straight lines (α>π/2-β), if the cutting plane passes through the apex and intersects the cone basis.
20 SOLO Two Body Problem (continue – 13) KEPLERIAN TRAJECTORIES Integration of Equations of Motion (continue – 6) ( )pe p r Θ−Θ+ = cos1 Conic Section (continue – 2) Θ pΘpΘ−Θ r d directrixfocus conic section x y → P1 → Q1 → r1 → t1 periapsis Let define e p d =: and rewrite ( )pe ed r Θ−Θ+ = cos1 as follows ( )prd r e Θ−Θ− = cos From which we obtain the following definition of conic sections: Conic Section A conic section is a planar curve, such that the ratio of distances, for any point on the curve, to a fixed point F* (focus) and to a line directrix is constant and equal to e.
21 SOLO Two Body Problem (continue – 14) KEPLERIAN TRAJECTORIES Integration of Equations of Motion (continue – 7) Conic Section (continue – 3) Let write the conic section equation in cartesian coordinates using →→ QP 1,1 ( ) ( ) ( ) ( ) →→→ →→→ Θ−Θ+Θ−Θ−= Θ−Θ+Θ−Θ= QPt QPr pp pp 1cos1sin1 1sin1cos1 and ( ) ( )p p ry rx Θ−Θ= Θ−Θ= sin cos ( )pe p r Θ−Θ+ = cos1 Substitute those in the equation rewritten as ( ) prer p =Θ−Θ+ cos xepyx −=+ 22 to obtain Squaring both sides gives 22222 2 xexpepyx +−=+ or ( ) 2222 21 pyxpexe =++−
22 SOLO Two Body Problem (continue – 15) KEPLERIAN TRAJECTORIES Integration of Equations of Motion (continue – 8) Conic Section (continue – 4) For e ≠ 1 can we can write ( ) 2222 21 pyxpexe =++− ( ) 2 2 2 22 22 2 2 2 111 1 e p e ep py e ep xe − = − +=+      − +− We see that for p ≠ 0 ( h ≠ 0 ) 1 0=e 222 pyx =+ circle 2 10 << e 1 11 1 2 2 2 2 2 =             − +             − − + e p y e p e ep x ellipse 3 1=e pe yp x 2 22 − = parabola 4 1>e hyperbola1 11 1 2 2 2 2 2 =             − −             − − − e p y e p e ep x
23 SOLO Two Body Problem (continue – 16) KEPLERIAN TRAJECTORIES Velocity on the Trajectory Let find the velocity on the trajectory in and coordinates →→ tr 1,1 →→ QP 1,1 →→→→ +=Θ+== QyPxtrrrrv 1111  ( ) ( )[ ] ( ) ( )[ ] →→ Θ−ΘΘ+Θ−Θ+Θ−ΘΘ−Θ−Θ= QrrPrr pppp 1cossin1sincos  From the equation of motion ( )pe p r Θ−Θ+ = cos1 hr =Θ2 µ/: 2 hp = we obtain ( ) ( )[ ] ( ) ( )p p p p r e p p eh r h e ep td d d dr rv Θ−Θ= Θ−Θ = Θ−Θ+ Θ−Θ = Θ Θ == sin sin cos1 sin 22 µ  ( )[ ]pt e pr p r h rrv Θ−Θ+===Θ= cos12 µµ
24 SOLO Two Body Problem (continue – 17) KEPLERIAN TRAJECTORIES Velocity on the Trajectory (continue – 1) Let substitute those results in the velocity equation to obtain the components in coordinates. →→ QP 1,1 ( ) ( )[ ] →→→→ Θ−Θ++Θ−Θ−=+= Qe p P p QyPxv pp 1cos1sin11 µµ   ( ) ( ) ( ) ( )[ ]{ } 2/12222/1222/122 cos1sin pptr ee p yxvvv Θ−Θ++Θ−Θ=+=+= µ  ( )[ ] 2/12 cos21 pee p v Θ−Θ++= µ Θ pΘ pΘ−Θ r d directrix focus conic section x y → P1 → Q1 → r1 → t1 v  rv tv periapsis The velocities at the periapsis and apoapsis are ( )pΘ=Θ ( )π+Θ=Θ p ( )e p va −= 1 µ ( )e p vp += 1 µ
25 SOLO Two Body Problem (continue – 18) KEPLERIAN TRAJECTORIES Specific Mechanic Energy The specific mechanic energy is ( ) ( ) ( ) ap e p e p e r v r v E a a 22 11 2 1 22 : 2222 µµµµµµ −= − −= − − − =−=−= or ar v E 22 : 2 µµ −=−= where we used ( )2 1 eap −= From the last equation we get a p e −= 1 Substitute and to obtainµ/: 2 hp = ( )Ea 2/µ−= 2 2 2 1 µ hE e +=
26 SOLO Two Body Problem (continue – 19) KEPLERIAN TRAJECTORIES Eccentricity Vector We defined the eccentricity as a scalar µ/: Be = Let define the eccentricity vector as µ/: Be  = where is defined asB  B r r hr   +=× µ Therefore ( ) ( ) ( ) → = ⋅−      − =− ⋅− =− ×× =− × = Pe vvr r vr r rvvrvr r rvrv r rhv e 1 2 2 µ µ µµµ    The last equation is obtained from the definition of , and , and can be checked by substituting the range and velocity vectors in coordinates, in previous equation. B  e  ( )pΘ−Θ→→ QP 1,1
27 SOLO Two Body Problem (continue – 20) KEPLERIAN TRAJECTORIES Orbit Determination from Initial Conditions Given the initial position and velocity vectors , find the orbital parameters00 ,vr  Θ pΘr d directrix focus conic section x y → P1 → Q1 → r1 → t1 v rv tv( )pΘ−Θ γ 1 From the specific angular momentum of the orbit we can find00 vrh  ×= 01 00 ≠ × = → h h vr R  µ 2 h p = 2 From the specific mechanical energy on an elliptic orbit ar vv E 22 0 00 µµ −=− ⋅ =  we obtain 00 0 2 vv r a  ⋅− = µ µ
28 SOLO Two Body Problem (continue – 21) KEPLERIAN TRAJECTORIES Orbit Determination from Initial Conditions (continue – 1) Θ pΘr d directrix focus conic section x y → P1 → Q1 → r1 → t1 v rv tv( )pΘ−Θ γ 3 The eccentrity vector of a Keplerian trajectory is given by from which ( ) → =      ⋅−      −⋅= Pevvrr r vve 1 1 0000 0 00  µ µ ee  = 01 ≠= → e e e P  →→→ ×= PRQ 111 ( )pe p r Θ−Θ+ = cos1 The position and velocity vectors are given by ( ) ( )       Θ−Θ+Θ−Θ= →→ QPrr pp 1sin1cos  ( ) ( )( )     Θ−Θ++Θ−Θ−=+= →→→→ QeP p QyPxv pp 1cos1sin11 µ  
29 SOLO Two Body Problem (continue – 22) KEPLERIAN TRAJECTORIES Orbit Determination from Initial Conditions (continue – 2) Θ pΘr d directrix focus conic section x y → P1 → Q1 → r1 → t1 v rv tv( )pΘ−Θ γ 4 Initial trajectory angle , on the conic section, from the periapsis is given by pΘ−Θ=Θ 01       Θ Θ =Θ ⋅         ×=Θ ⋅=Θ − → → → → → 1 11 1 0 0 1 0 0 1 cos sin tan 11sin 1cos R r r P r r P One other way to find is by using →→→ RQP 1,1,1 ( )       Θ++Θ−= Θ+Θ= →→ →→→ QePv p QPr 1cos1sin 1sin1cos1 110 110  µ or ( ) ( ) ( )       Θ+Θ=Θ+ Θ−Θ+=Θ+ →→ →→ 01011 01011 cos1sin1cos1 sin1cos1cos1 v p rQe v p rePe   µ µ ( ) 1 1 sin cos Θ− Θ+e 1 1 cos sin Θ Θ
30 SOLO Two Body Problem (continue – 23) KEPLERIAN TRAJECTORIES Orbit Determination from Initial Conditions (continue – 3) Θ pΘr d directrix focus conic section x y → P1 → Q1 → r1 → t1 v rv tv( )pΘ−Θ γ Using 1 0 cos1 Θ+ = e p r we finally obtain ( )        Θ+ Θ = Θ− Θ+ = →→ →→ 01 0 0 1 01 0 0 1 cos sin 1 sin cos 1 v p r r p Q v p r r p e P   µ µ ( )[ ] ( ) ( )[ ] ( ) ( )[ ] ( ) 01 0 01 01 0 01 011 0 011 sincos11 sincos1cos1 sincoscossinsinsincoscos 1sin1cos v p rr r p r v p rr re p r v p rr re p r QPrr     Θ−Θ+       Θ−Θ+−= Θ−Θ+Θ−Θ+−Θ+= ΘΘ−ΘΘ+ΘΘ+Θ+Θ=       Θ+Θ= → → → →→ µ µ µ Therefore
31 SOLO Two Body Problem (continue – 24) KEPLERIAN TRAJECTORIES Orbit Determination from Initial Conditions (continue – 4) Θ pΘr d directrix focus conic section x y → P1 → Q1 → r1 → t1 v rv tv( )pΘ−Θ γ where we used ( ) ( ) ( ) ( )[ ] ( ) ( )[ ] 011 0 0 11 011 0 0 11 coscos sinsinsin coscossinsin sincoscossin 1cos1sin ve p r r p ee p ve p r r p ee p QeP p v    Θ+Θ+Θ+ Θ−Θ−Θ−Θ = ΘΘ++ΘΘ+ ΘΘ+−Θ+Θ −=     Θ++Θ−= → → →→ µ µ µ ( ) ( )[ ] ( ) ( )[ ] ( )[ ] ( )[ ] ( )[ ] ( )[ ] ( ) ( )[ ] 01 0 01 0 1 0 00 01 0 0 1111 011 0 0 111111 cos11sin 1 cos1 cos11 cos1sincos1sin cos1cos1 sinsincoscossinsin v p r r pr e pr vr v p r r p eee p ve p r r p ee p           Θ+Θ−−+         Θ−Θ−Θ−Θ− ⋅ =       Θ+Θ−−+ Θ+Θ−Θ−Θ−Θ−Θ = Θ++Θ+Θ+−+ Θ−Θ−Θ−ΘΘ+Θ−ΘΘ−Θ = → → → → µ µ µ ( ) 10 1111000 sin 1cos1sin1sin1cos Θ=     Θ++Θ−⋅      Θ+Θ=⋅ →→→→ e p r QeP p QPrvr µ µ
32 SOLO Two Body Problem (continue – 25) KEPLERIAN TRAJECTORIES Orbit Determination from Initial Conditions (continue – 5) Θ pΘr d directrix focus conic section x y → P1 → Q1 → r1 → t1 v rv tv( )pΘ−Θ γ Summarize ( )[ ] ( ) 01 0 01 sincos11 v p rr r p r r  Θ−Θ+       Θ−Θ+−= → µ ( )[ ] ( ) ( )[ ] 01 0 01 0 1 0 00 cos11sin 1 cos1 v p r r pr e pr vr v          Θ+Θ−−+         Θ−Θ−Θ−Θ− ⋅ = → → µ
33 SOLO Two Body Problem (continue – 26) KEPLERIAN TRAJECTORIES Orbit Determination from Initial Conditions (Second Method) Start from the solution for Keplerian orbit ( ) 221 cos 1 h CC r µ +−Θ= where C1 and C2 must be determined from initial conditions 0r - initial range from the center of mass M 0v - initial velocity magnitude 00000 2 cosγvrvrrvrh =×=Θ=×=  0γ - initial trajectory angle (between velocity vector and the horizon plane)
34 SOLO Two Body Problem (continue – 27) KEPLERIAN TRAJECTORIES Orbit Determination from Initial Conditions (Second Method - 1) We have ( ) ( ) 0 2 0 21 0 22 0 2 0 21 cos 1 cos cos cos 1 γλγ µ r CC vr CC r +−Θ=+−Θ= where 0 2 0 / : r v µ λ = Let differentiate ( ) ( ) ( )2121 2 212 sinsinsin 1 CChCCrrCC r r rdt d −Θ=−ΘΘ=→−ΘΘ−=−=       or ( )21000 sincos CCvrr −Θ−= γ Let use now the initial conditions to find C1, C2. ( ) 0 2 0 201 0 cos 1 cos 1 γλ r CC r +−Θ= ( ) 002010000 sinsincos γγ vCCvrr =−Θ= or ( ) 0 2 00 201 cos 11 cos γλ rr CC −=−Θ ( ) 0 0 201 tan 1 sin γ r CC =−Θ
35 SOLO Two Body Problem (continue – 28) KEPLERIAN TRAJECTORIES Orbit Determination from Initial Conditions (Second Method - 2) We have ( ) 0 2 00 201 cos 11 cos γλ rr CC −=−Θ ( ) 0 0 201 tan 1 sin γ r CC =−Θ ( ) ( )[ ] ( ) ( ) ( ) ( ) ( ) ( ) 0 2 0 00 0 0 0 2 00 0 2 0 02010201 0 2 0 2001 cos 1 sintan 1 cos cos 11 cos 1 sinsincoscos cos 1 cos 1 γλ γ γλ γλ γλ rrrr r CCCC r CC r +Θ−Θ−Θ−Θ      −= +Θ−Θ−Θ−Θ−Θ−Θ= +−Θ+Θ−Θ= ( ) ( ) ( ) ( ) ( ) 0 00 0 2 0 0 0000 0 2 00 cos cos cos cos1 cos sinsincoscos cos cos1 γ γ γλ γ γγ γλ +Θ−Θ + Θ−Θ− = Θ−Θ−Θ−Θ + Θ−Θ− = r r which, when developed further, gives
36 SOLO Two Body Problem (continue – 29) KEPLERIAN TRAJECTORIES Orbit Determination from Initial Conditions (Second Method - 2) ( ) ( ) 0 00 0 2 00 cos cos cos cos1 γ γ γλ +Θ−Θ + Θ−Θ− = r r Finally ( ) ( )0000 0 2 0 coscoscos1 cos γγλ γλ +Θ−Θ+Θ−Θ− = r r
37 SOLO Two Body Problem (continue – 30) KEPLERIAN TRAJECTORIES Orbit Determination from Initial Conditions (Second Method - 3) ( ) ( )0000 0 2 0 coscoscos1 cos γγλ γλ +Θ−Θ+Θ−Θ− = r r ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] ( )2 0 2 0 2 0 22 00 2 0002 0 2 0 2 0 22 00 2 000 0000 1coscossin cos1cossincossin 1coscossin1 cos1cossincossin1 coscoscos1 −+ Θ−Θ−+Θ−Θ− −++= Θ−Θ−+Θ−Θ−= +Θ−Θ+Θ−Θ− γλγγλ γλγγλ γλγγλ γλγγλ γγλ The denominator can be expressed as 0 2 0 cos: γλrp = ( ) ( ) 1cos21cos2cos1coscossin: 0 2 0 2 0 222 0 2 0 2 0 22 +−=+−=−+= γλλγλγλγλγγλe We define       − =Θ−Θ − 1cos cossin tan: 0 2 001 0 γλ γγλ p ( ) ( ) ( ) ( )[ ] ( )pp ee Θ−Θ+=Θ−Θ+Θ−Θ+= +Θ−Θ+Θ−Θ− cos1cos1 coscoscos1 00 0000 γγλto obtain
38 SOLO Two Body Problem (continue – 31) KEPLERIAN TRAJECTORIES Orbit Determination from Initial Conditions (Second Method - 4) ( ) ( )0000 0 2 0 coscoscos1 cos γγλ γλ +Θ−Θ+Θ−Θ− = r r The denominator can be expressed as ( ) ( ) ( ) ( )[ ] ( )pp ee Θ−Θ+=Θ−Θ+Θ−Θ+=+Θ−Θ+Θ−Θ− cos1cos1coscoscos1 000000 γγλ We obtain again the fact that the trajectory is a conic section given by ( )pe p r Θ−Θ+ = cos1 We can also write ( )[ ] λγλλ γλ − = +−− = − = 21cos21 cos 1 : 0 0 2 0 2 0 2 rr e p a and 0 2 0 cos: γλrp = ( ) 1cos2: 0 2 +−= γλλe
39 SOLO Two Body Problem (continue – 32) KEPLERIAN TRAJECTORIES Flat Earth Approximation We assume that , so we can use10 <<Θ−Θ ( ) ( ) ( )2 0000 2 1 1cos&sin Θ−Θ−Θ−ΘΘ−Θ≈Θ−Θ ( ) ( ) ( ) ( ) ( ) ( ) ( )2 0 0 200 0 000 2 00 22 0 0 2 0 0000 0 2 0 cos 1 1 2 1 tan1 cossin 2 1 1cos 2 1 cos coscoscos1 cos Θ−Θ      −−Θ−Θ− = Θ−Θ−      Θ−Θ−+Θ−Θ ≈ +Θ−Θ+Θ−Θ− = γλ γ γγλγλ γλ γγλ γλ r r r r and where gr v r r v r v 0 2 0 2 0 0 2 0 0 2 0 ≈== µµ λ The order of magnitude of λ is ( ) 64 000,1 10000,400,61 22 0 0 = ⋅ == v gr λ
40 SOLO Two Body Problem (continue – 33) KEPLERIAN TRAJECTORIES Flat Earth approximation (continue – 1) Therefore if γ is not close to 90° , we can assume that , and then1 cos 1 0 2 >> γλ ( ) ( ) ( ) ( ) ( ) ( ) 0 22 0 2 0 2 00000 0 22 0 02 0000 0 22 0 02 000 0 cos2 1 tan cos2 1 tan1 cos2 1 tan1 γ γ γ γ γ γ v g rrr v gr r v gr r r Θ−Θ−Θ−Θ+=       Θ−Θ−Θ−Θ+≈ Θ−Θ+Θ−Θ− ≈ Use to obtain ( )0000 && Θ−Θ=+=+= rxzRrzRr ee 0 22 0 2 00 cos2 1 tan γ γ v g xxzz −+=
41 SOLO Two Body Problem (continue – 34) KEPLERIAN TRAJECTORIES Flat Earth approximation (continue – 2) We obtained the Flat Earth approximation of the trajectory, that is a parabola. 0 22 0 2 00 cos2 1 tan γ γ v g xxzz −+= 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 Vacuum Ballistic Trajectories g V 2 2 0 g V 2 2 0 0 22 0 2 00 cos2 γ γ V g xtgxzz −+= g V 2 0 0V 0γ0z
42 SOLO Two Body Problem (continue – 35) KEPLERIAN TRAJECTORIES Orbital Trajectories Let discuss in more detail the Keplerian trajectories. Ellipse (e < 1) ( Kepler’s First Law ) and Circle ( e = 0 ) x y ( )2 1 eap −= eac = a a ( ) 2/12 1 eab −= ( )eara += 1 ( )earp −= 1 r ΘFOCUS EMPTY FOCUS Apoapsis Radius Periapsis Radius Apoapsis PeriapsisGeometric Center Directrix Directrix c epd /= Semimajor axis Semiminor axis v  nv rv γ → P1 → Q1 1 11 1 2 2 2 2 2 =             − +             − − + e p y e p e ep x We found that the ellipse equation in cartesian coordinates is given by:
43 SOLO Two Body Problem (continue – 36) KEPLERIAN TRAJECTORIES Orbital Trajectories (continue – 1) Ellipse (e < 1) ( Kepler’s First Law ) and Circle ( e = 0 ) x y ( )2 1 eap −= eac = a a ( ) 2/12 1 eab −= ( )eara += 1 ( )earp −= 1 r ΘFOCUS EMPTY FOCUS Apoapsis Radius Periapsis Radius Apoapsis PeriapsisGeometric Center Directrix Directrix c epd /= Semimajor axis Semiminor axis v  nv rv γ → P1 → Q1Define: 2 1 : e p a − = semimajor axis 2 1 : e p b − = - semiminor axis ea e ep c = − = 2 1 : - distance of geometrical center from focus e p d =: - distance of directrix from focus ( )eara += 1: - apoapsis radius ( )earp −= 1: - periapsis radius Θ - true anomaly γ - flight-path angle ( ) 2/pa rra += ( ) ( )papa rrrre +−= / ( ) ( ) ( )ererabeap ap −=+==−= 11/1 22 From those relations we get: 1 22 =      +      + b y a cx and the ellipse equation will be:
44 SOLO Two Body Problem (continue – 37) KEPLERIAN TRAJECTORIES Time of Flight on an Elliptic Orbit From the equation Θ= 2 rh we can write h Ad h dr dt 2 2 = Θ = where is the area defined by the radius vector as it moves through an angle 2 2 Θ = dr Ad Θd Θ pΘ pΘ−Θ r focus conic section x y → P1 → Q1 → r1 → t1 v  rv tv Θd Θ= drAd 2 2 1 periapsis This proves the 2nd Kepler’s Law that equal area are swept out equal in equal times by the radius vector.
45 SOLO Two Body Problem (continue – 38) KEPLERIAN TRAJECTORIES Time of Flight on an Elliptic Orbit (continue – 1) The period of the orbit depends only on the major axis of the ellipse a. ( ) ( ) p pa h eaa h ea h ba T eap ph µ ππππ µ 2/3122/322 2 1 2 1 22 2 = −= = − = − == or 2/3 2 aT µ π = The period of an elliptical orbit T is obtained by integrating from Θ= 0 to Θ=2π , and the radius vector sweeps the area of the ellipse A = π a b. This proves the Kepler’s third law: “the square of the period of a planet orbit is equal To the cube of its mean distance to the sun”.
46 SOLO Two Body Problem (continue – 39) KEPLERIAN TRAJECTORIES Time of Flight on an Elliptic Orbit (continue – 2) Let draw an auxiliary circle of radius a, and the same center O as the geometric center of the ellipse. x y eac = a a ( ) 2/12 1 eab −= r Θ FOCUS EMPTY FOCUS c → P1 → Q1 a F Q O VS E P Let take any point P on the ellipse with polar coordinates r,Θ and define the point Q on the circle at the same coordinate x as P. Eeary r a x ea a x by Ea a x ay ellipse ellipse circle sin1sin sin111 sin1 2 2 2 2 2 2 2 2 −=Θ=→        Θ=−−=−= =−= The angle E of OQ with x axis is called the eccentric anomaly. aeEarxellipse −=Θ= coscos
47 SOLO Two Body Problem (continue – 40) KEPLERIAN TRAJECTORIES Time of Flight on an Elliptic Orbit (continue – 3) Let compute x y eac = a a ( ) 2/12 1 eab −= r Θ FOCUS EMPTY FOCUS c → P1 → Q1 a F Q O VS E P ( ) ( )( ) ( )( ) ( ) 0cos11 sin1sincos1cos 1 2 22 2 >←−−= −+−−= −=×=−= EEEeea EEeaEaEEeaaeEa xyyxvreah ellipseellipseellipseellpse     µ We obtain ( ) n a EEe ==− :cos1 3 µ ( ) ( ) ( )pttntEetE −=− sin Integrating this equation gives Kepler’s Equation where tp is the time of periapsis ( E (tp) = 0 )
48 SOLO Two Body Problem (continue – 41) KEPLERIAN TRAJECTORIES Time of Flight on an Elliptic Orbit (continue – 4) From x y eac = a a ( ) 2/12 1 eab −= rΘ FOCUS EMPTY FOCUS c → P1 → Q1 a F Q O VS E P Eeary aeEarx ellipse ellipse sin1sin coscos 2 −=Θ= −=Θ= we have ( ) ( )[ ] [ ] ( )Eea EeEeaEeaaeEar cos1 coscos21sin1cos 2/1222/12222 −= +−=−+−= Therefore Θ+ Θ− = − − =Θ Θ+ Θ+ = − − =Θ cos1 sin1 sin cos1 sin1 sin cos1 cos cos cos1 cos cos 22 e e E Ee Ee e e E Ee eE ( )( ) Ee Ee Ee eEEe sin1 cos11 sin1 coscos1 sin cos1 2 tan 22 − −+ = − +−− = Θ Θ− = Θ From 2 tan 1 1 2 tan E e e − + = Θ or and are always in the same quadrant.2 Θ 2 E
49 SOLO References 2. Battin, R.H., “An Introduction to the Mathematics and Methods of Astrodynamics”, AIAA Education Series, AIAA, Washington. DV., 1987 KEPLERIAN TRAJECTORIES 3. Battin, R.H., “Astronautical Guidance”, McGraw Hill, 1964 1. Bate, R.R., Mueller, D.D., White, J.E., “Fundamentals of Astrodynamics”, Dover Publications, New York, 1971 4. Chobotov, V.A., Editor, “Orbital Mechanics”, AIAA Education Series, , AIAA, 2nd Edition, 1996 5. Reagan, F.J., “Re-Entry Vehicle Dynamics”, AIAA Education Series, , AIAA, 1984 6. Reagan, F.J. and Anandakrishnan, S.M., “Dynamics of Atmospheric Re-Entry”, AIAA Education Series, , AIAA, 1993 7. Montenbruck, O. and Eberhard, G., “Satellite Orbits”, Springer-Verlag, 2000 8. Kaplan, M.H., “Modern Spacecraft Dynamics and Control”, John Wiley & Sons, 1976
August 13, 2015 50 SOLO Technion Israeli Institute of Technology 1964 – 1968 BSc EE 1968 – 1971 MSc EE Israeli Air Force 1970 – 1974 RAFAEL Israeli Armament Development Authority 1974 – 2013 Stanford University 1983 – 1986 PhD AA

More Related Content

PPTX
Materi Presentasi - Vektor untuk SMA dan kuliah
byAgungPermanaMukti
 
PPT
Geometría Circunferencia y Círculo PAES.
byFaviolaAlarcnCornejo
 
PDF
soldadura y técnicas de union.pdf
byHeberArcana1
 
PPT
Anti ballistic missiles ii
bySolo Hermelin
 
PPT
Anti ballistic missiles i
bySolo Hermelin
 
PPT
Reduced order observers
bySolo Hermelin
 
PPT
11 fighter aircraft avionics - part iv
bySolo Hermelin
 
PPT
Sliding Mode Observers
bySolo Hermelin
 
Materi Presentasi - Vektor untuk SMA dan kuliah
byAgungPermanaMukti
 
Geometría Circunferencia y Círculo PAES.
byFaviolaAlarcnCornejo
 
soldadura y técnicas de union.pdf
byHeberArcana1
 
Anti ballistic missiles ii
bySolo Hermelin
 
Anti ballistic missiles i
bySolo Hermelin
 
Reduced order observers
bySolo Hermelin
 
11 fighter aircraft avionics - part iv
bySolo Hermelin
 
Sliding Mode Observers
bySolo Hermelin
 

Viewers also liked

PPT
6 computing gunsight, hud and hms
bySolo Hermelin
 
PPT
Stabilization of linear time invariant systems, Factorization Approach
bySolo Hermelin
 
PPT
10 fighter aircraft avionics - part iii
bySolo Hermelin
 
PPT
8 fighter aircraft avionics-part i
bySolo Hermelin
 
PPT
3 modern aircraft cutaway
bySolo Hermelin
 
PPT
9 fighter aircraft avionics-part ii
bySolo Hermelin
 
PPT
Slide Mode Control (S.M.C.)
bySolo Hermelin
 
PPT
12 performance of an aircraft with parabolic polar
bySolo Hermelin
 
PPT
2 aircraft flight instruments
bySolo Hermelin
 
PPT
4 navigation systems
bySolo Hermelin
 
PPT
3 earth atmosphere
bySolo Hermelin
 
PPT
5 introduction to quantum mechanics
bySolo Hermelin
 
PPT
Inner outer and spectral factorizations
bySolo Hermelin
 
PPT
2 estimators
bySolo Hermelin
 
PPT
2 classical field theories
bySolo Hermelin
 
PPT
6 radar range-doppler-angular loops
bySolo Hermelin
 
PPT
Introduction to elasticity, part i
bySolo Hermelin
 
PDF
Bgrv and indian missiles arun vishwakarma, rev 1 c
byRamana Brf
 
PPT
Optical aberrations
bySolo Hermelin
 
PPT
5 cramer-rao lower bound
bySolo Hermelin
 
6 computing gunsight, hud and hms
bySolo Hermelin
 
Stabilization of linear time invariant systems, Factorization Approach
bySolo Hermelin
 
10 fighter aircraft avionics - part iii
bySolo Hermelin
 
8 fighter aircraft avionics-part i
bySolo Hermelin
 
3 modern aircraft cutaway
bySolo Hermelin
 
9 fighter aircraft avionics-part ii
bySolo Hermelin
 
Slide Mode Control (S.M.C.)
bySolo Hermelin
 
12 performance of an aircraft with parabolic polar
bySolo Hermelin
 
2 aircraft flight instruments
bySolo Hermelin
 
4 navigation systems
bySolo Hermelin
 
3 earth atmosphere
bySolo Hermelin
 
5 introduction to quantum mechanics
bySolo Hermelin
 
Inner outer and spectral factorizations
bySolo Hermelin
 
2 estimators
bySolo Hermelin
 
2 classical field theories
bySolo Hermelin
 
6 radar range-doppler-angular loops
bySolo Hermelin
 
Introduction to elasticity, part i
bySolo Hermelin
 
Bgrv and indian missiles arun vishwakarma, rev 1 c
byRamana Brf
 
Optical aberrations
bySolo Hermelin
 
5 cramer-rao lower bound
bySolo Hermelin
 

Similar to Keplerian trajectories

PDF
41050EPathshala_071720200709PM.pdf
byAmritSingh610022
 
PPT
physics430_lecture06. center of mass, angular momentum
byMughuAlain
 
PPTX
Planetary Motion- The simple Physics Behind the heavenly bodies
byNISER-sac
 
PPTX
Dynamics slideshare
byMalarMohana
 
PPTX
Dynamics problems
byMalathiNagarajan20
 
PPT
Basic Concept of physics430_lecture06.ppt
byhamidkhan26620
 
PPT
physics430_lecture06.Engineering Physics
byAnithaRaghavan3
 
PPT
physics430_lecture06.ppt-Engineering Phy
byAnithaRaghavan3
 
PPT
Circular and gavitational force
byeshwar360
 
PDF
Solutions for Problems from Celestial and Stellar Dynamics by Barbara Ryden
bynazrinajeeb73
 
PPTX
PHYSICS CLASS XI Chapter 5 - gravitation
byPooja M
 
PDF
chapter5-gravitationppt-copy-211229151431 (2).pdf
byRavindraWaykole
 
PDF
Modelling and Simulations
byMinjie Lu
 
PDF
Astrodynamics_Note.pdf
byfurexpose
 
PDF
phys4550lecture notes Lecture-Note-6-1.pdf
bySuLaiting
 
PDF
Assignment_1_solutions.pdf
byAbhayRupareliya1
 
PDF
Comp3 gravitation
byLulu Delgado
 
PPT
Phys111_lecture13.ppt
byAnshulNayak6
 
PDF
A New Solution To The Lagrange S Case Of The Three Body Problem
bySophia Diaz
 
PPT
Uniform_Circ_Motion_and_UG.ppt
byLeahMaeGarcia1
 
41050EPathshala_071720200709PM.pdf
byAmritSingh610022
 
physics430_lecture06. center of mass, angular momentum
byMughuAlain
 
Planetary Motion- The simple Physics Behind the heavenly bodies
byNISER-sac
 
Dynamics slideshare
byMalarMohana
 
Dynamics problems
byMalathiNagarajan20
 
Basic Concept of physics430_lecture06.ppt
byhamidkhan26620
 
physics430_lecture06.Engineering Physics
byAnithaRaghavan3
 
physics430_lecture06.ppt-Engineering Phy
byAnithaRaghavan3
 
Circular and gavitational force
byeshwar360
 
Solutions for Problems from Celestial and Stellar Dynamics by Barbara Ryden
bynazrinajeeb73
 
PHYSICS CLASS XI Chapter 5 - gravitation
byPooja M
 
chapter5-gravitationppt-copy-211229151431 (2).pdf
byRavindraWaykole
 
Modelling and Simulations
byMinjie Lu
 
Astrodynamics_Note.pdf
byfurexpose
 
phys4550lecture notes Lecture-Note-6-1.pdf
bySuLaiting
 
Assignment_1_solutions.pdf
byAbhayRupareliya1
 
Comp3 gravitation
byLulu Delgado
 
Phys111_lecture13.ppt
byAnshulNayak6
 
A New Solution To The Lagrange S Case Of The Three Body Problem
bySophia Diaz
 
Uniform_Circ_Motion_and_UG.ppt
byLeahMaeGarcia1
 

More from Solo Hermelin

PPT
Analytic dynamics
bySolo Hermelin
 
PPT
2Anti-aircraft Warhead
bySolo Hermelin
 
PPT
1 susceptibility vulnerability
bySolo Hermelin
 
PPT
15 sky cars
bySolo Hermelin
 
PPT
14 fixed wing fighter aircraft- flight performance - ii
bySolo Hermelin
 
PPT
13 fixed wing fighter aircraft- flight performance - i
bySolo Hermelin
 
PPT
Calculus of variation problems
bySolo Hermelin
 
PPT
Calculus of variations
bySolo Hermelin
 
PPT
7 air-to-air combat
bySolo Hermelin
 
PPTX
Aerodynamics part iii
bySolo Hermelin
 
PPT
Aerodynamics part ii
bySolo Hermelin
 
PPT
Aerodynamics part i
bySolo Hermelin
 
Analytic dynamics
bySolo Hermelin
 
2Anti-aircraft Warhead
bySolo Hermelin
 
1 susceptibility vulnerability
bySolo Hermelin
 
15 sky cars
bySolo Hermelin
 
14 fixed wing fighter aircraft- flight performance - ii
bySolo Hermelin
 
13 fixed wing fighter aircraft- flight performance - i
bySolo Hermelin
 
Calculus of variation problems
bySolo Hermelin
 
Calculus of variations
bySolo Hermelin
 
7 air-to-air combat
bySolo Hermelin
 
Aerodynamics part iii
bySolo Hermelin
 
Aerodynamics part ii
bySolo Hermelin
 
Aerodynamics part i
bySolo Hermelin
 

Recently uploaded

PPTX
PPT-Template-Seminar Proposal Skripsi POLINEMA 2025.pptx
by9arifiansyah
 
PDF
Machine Learning Lectures -- Loss Functions (2)
byZahra Sadeghi
 
PDF
이영욱 교수님의 물리학회 발표자료 전채_20251120.pdf
bysciencepeople
 
PDF
Machine Learning Lectures - Loss Functions (1)
byZahra Sadeghi
 
PPT
Grade 8 - Chemistry - Acids and Alkalis.ppt
byEnas813292
 
PDF
Periodic Table ( आवर्त सारणी ) Chemistry Notes PDF @irfanullah_mehar @world_o...
byWorld of Wisdom
 
PDF
Physics: Ultrasound and Applications ks3 ppt
byfizzyrao14
 
PDF
f - Block Handwritten Notes PDF Download - Irfanullah Mehar - Hindi Medium.pdf
byWorld of Wisdom
 
PPTX
Celebrating Breakthroughs in Metal Organic Frameworks - A Poster Showcasing t...
byRunjhunDutta
 
PPTX
Warming and cooling the Earth and the Atmosphere
byJohnLoydCorros
 
PDF
Chemical Bonding ( रासायनिक बंधन ) | Hybridization ,VSPER Theory , MOT, Fazan...
byWorld of Wisdom
 
PPTX
Second Law of Thermodynamics - General Chemistry 2
byjohndaenielverzosa2
 
PPTX
Pharmaceutical Calculation Techniques and Terminologies (1).pptx
bymuhammedsajjadhussei
 
PPTX
SCIENCE 9 Q4 Module2 MOMENTUM AND IMPULSE
byMarkjohnBello4
 
PPT
ch7_Acids&Bases.ppt. acid and bases , reactions of acids and bases, equilibri...
bykhadijaHijazi1
 
PPTX
PPT-Template-Seminar Hasil Skripsi POLINEMA 2025.pptx
by9arifiansyah
 
PPTX
chapter 3: Classical Encryption Techniques
byEbtihalNoor
 
PPTX
adrenal gland and its hormone functions
byanooper
 
PPTX
Acute kidney injury asu teaching lectures
byfathyradwan1991
 
PPTX
THE FUTURE OF ROBOTICS PowerPoint presentation.pptx
by0001aditibasu
 
PPT-Template-Seminar Proposal Skripsi POLINEMA 2025.pptx
by9arifiansyah
 
Machine Learning Lectures -- Loss Functions (2)
byZahra Sadeghi
 
이영욱 교수님의 물리학회 발표자료 전채_20251120.pdf
bysciencepeople
 
Machine Learning Lectures - Loss Functions (1)
byZahra Sadeghi
 
Grade 8 - Chemistry - Acids and Alkalis.ppt
byEnas813292
 
Periodic Table ( आवर्त सारणी ) Chemistry Notes PDF @irfanullah_mehar @world_o...
byWorld of Wisdom
 
Physics: Ultrasound and Applications ks3 ppt
byfizzyrao14
 
f - Block Handwritten Notes PDF Download - Irfanullah Mehar - Hindi Medium.pdf
byWorld of Wisdom
 
Celebrating Breakthroughs in Metal Organic Frameworks - A Poster Showcasing t...
byRunjhunDutta
 
Warming and cooling the Earth and the Atmosphere
byJohnLoydCorros
 
Chemical Bonding ( रासायनिक बंधन ) | Hybridization ,VSPER Theory , MOT, Fazan...
byWorld of Wisdom
 
Second Law of Thermodynamics - General Chemistry 2
byjohndaenielverzosa2
 
Pharmaceutical Calculation Techniques and Terminologies (1).pptx
bymuhammedsajjadhussei
 
SCIENCE 9 Q4 Module2 MOMENTUM AND IMPULSE
byMarkjohnBello4
 
ch7_Acids&Bases.ppt. acid and bases , reactions of acids and bases, equilibri...
bykhadijaHijazi1
 
PPT-Template-Seminar Hasil Skripsi POLINEMA 2025.pptx
by9arifiansyah
 
chapter 3: Classical Encryption Techniques
byEbtihalNoor
 
adrenal gland and its hormone functions
byanooper
 
Acute kidney injury asu teaching lectures
byfathyradwan1991
 
THE FUTURE OF ROBOTICS PowerPoint presentation.pptx
by0001aditibasu
 

Keplerian trajectories

  • 1.
  • 2.
    2 SOLO Table of Contents 2.Newton’s Laws 3. Newton’s Law of Universal Gravitation 4. Two Body Problem 4.1 Polar Representation 4.2 Conservation of Angular Momentum KEPLERIAN TRAJECTORIES 1. Kepler’s Laws 4.3 Conservation of Mechanical Energy 4.4 Integration of Equations of Motion 4.5 Velocity on the Trajectory 4.6 Specific Mechanic Energy 4.7 Eccentricity Vector 4.8 Orbit Determination from Initial Conditions 4.9 Flat Earth Approximation 4.10 Orbital Trajectories 4.11 Time of Flight on an Elliptic Orbit 5. References
  • 3.
    3 SOLO Kepler’s Laws KEPLERIAN TRAJECTORIES JohannesKepler 1571 - 1630 Tycho Brache 1546 - 1601 From 1601 to 1606 Kepler tried to fit various geometrical curves to Tycho Brache’s data on Mars orbit.
  • 4.
    4 SOLO Kepler’s Laws KEPLERIAN TRAJECTORIES 1.The orbit of each planet is an ellipse, with the sun at a focus. 2. The line joining the planet to the sun sweeps out equal area in equal times. 3. The square of the period of a planet orbit is equal to the cube of its mean distance to the sun. b a a dA h dt 2 = b a 2/322 a GMGM ab TP ππ == b
  • 5.
    5 SOLO Newton’s Laws ofMotion KEPLERIAN TRAJECTORIES . Every body continues in its state of rest or of uniform motion in straight line unless it is compelled to change that state by forces impressed upon it. . The rate of change of momentum of a given body is proportional to the force impressed on the body and is in the same direction as the force. . To every action there is always an equal reaction “THE MATHEMATICAL PRINCIPLES OF NATURAL PHILOSOPHY”1687 Isaac Newton 1643-1727
  • 6.
    6 SOLO Newton’s Law ofUniversal Gravitation KEPLERIAN TRAJECTORIES “THE MATHEMATICAL PRINCIPLES OF NATURAL PHILOSOPHY”1687 Isaac Newton 1643-1727 Any two body attract one on other with a force proportional to the product of the masses and inversely proportional to the square of the distance between them.    EQPOISSON G G r GM r GM g gm r GM mr r mM GF ρπϕϕϕ 4&& 1 2 2 =∇=−∇=      −∇= −=      ∇=−= 228 /1067.6 gmcmdyneG − ×= The Universal Gravitational Constant Instantaneous propagation of the force along the direction between the masses (“Action at a Distance”) GF  GF  M m
  • 7.
    7 SOLO Two Body Problem KEPLERIANTRAJECTORIES Assumptions 1. The bodies are spherical symmetric. 2. The gravitational forces are the only forces acting on the bodies. Mr  mr  r  M m Cr  Let Mr  -the position vector of the mass M relative to an inertial point. mr  -the position vector of the mass m relative to an inertial point. Mm rrr  −= -the relative position vector of the mass m relative to M. r r r  = → :1 -unit vector in direction.r  Let apply the Newton’s Law of Universal Gravitation to describe the forces each mass applies on the other. → → =→= −=→−= r r mG r r r r MmG rM r r MG r r r r MmG rm MM mm 1 1 22 22      
  • 8.
    8 SOLO Two Body Problem(continue – 1) KEPLERIAN TRAJECTORIES → → =→= −=→−= r r mG r r r r MmG rM r r MG r r r r MmG rm MM mm 1 1 22 22       Mm rMrm r Mm C + + =   The Center of Mass of those masses is ( ) ( )MmGr r r r MmG rrr Mm +=←−= + −=−= →→ :11 22 µ µ From those equations we obtain: and 0   = + + = Mm rMrm r Mm C The Center of Mass is not accelerating
  • 9.
    9 SOLO Two Body Problem(continue – 2) KEPLERIAN TRAJECTORIES Polar Representation Define ω  r  - Angular Velocity Vector →→ Θ+= nrR 11  ωω Decompose in and perpendicular to direction.r  r  ω  Since we can define a right hand Cartesian system using vector product: 011 =⋅ →→ rn →→→ ×= rnt 11:1          −=×      Θ+=×==      +Θ−=×      Θ+=×==      Θ=×      Θ+=×==      →→→→→→→ →→→→→→→→ →→→→→→→ tnnrnnn td d nrtnrttt td d trnrrrr td d RR RR R 1111111 11111111 1111111 ωωω ωωω ωω       Differentiating we obtain:
  • 10.
    10 SOLO Two Body Problem(continue – 3) KEPLERIAN TRAJECTORIES Polar Representation (continue – 1) or                               − Θ− Θ =               → → → → → → n t r n t r td d R R 1 1 1 00 0 00 1 1 1 ω ω  → = rrr 1  Therefore →→→→ Θ+=+= trrrrrrrr 1111   ( ) ( ) ( ) →→→ →→→→→ Θ+Θ+Θ+Θ−=       +Θ−Θ+Θ+Θ+Θ+= nrtrrrrr nrrtrrtrrrr R R 1121 11111 2   ω ω → = rrr 1  vtrrrr  =Θ+= →→ 11 ( ) ( ) →→→→ −=Θ+Θ+Θ+Θ−= r r nrtrrrrrr R 11121 2 2 µ ω 
  • 11.
    11 SOLO Two Body Problem(continue – 4) KEPLERIAN TRAJECTORIES Conservation of Angular Momentum Vector Solution Cross-multiply the equation of motion by → −= r r r 12 µ r  03 =×−=× rr r rr  µ Define the specific angular momentum as:h  vrrrh  ×=×=: Differentiate h  0    =×+×= rrrr td hd Therefore is a constant vector in space that is perpendicular to the trajectory vectors and , therefore those vectors (and the trajectory) remain in the plane defined by the initial conditions vrh  ×= r  v  ( ) ( ) ( )000 =×=== tvtrth 
  • 12.
    12 SOLO Two Body Problem(continue – 5) KEPLERIAN TRAJECTORIES Conservation of Angular Momentum (continue – 1) Polar Coordinate Solution →→→→ Θ=      Θ+×=×= nrtrrrrrvrh 1111 2   The equation of motion in polar coordinates is ( ) ( ) →→→→ −=Θ+Θ+Θ+Θ−= r r nrtrrrrrr R 11121 2 2 µ ω  from which 02 =Θ+Θ  rr and r rdd 2−= Θ Θ   Integrated both sides ( ) ConstrhConstrConstr =Θ=→=Θ→+−=Θ  222 lnlnlnlnln 1 000 ≠Θ←=→=Θ  rr RR ωω2 →→→→ =→=−=      constntn td d R 1011  ωalso →→ =Θ=×= constnrvrh 12  therefore
  • 13.
    13 SOLO Two Body Problem(continue – 6) KEPLERIAN TRAJECTORIES Conservation of Mechanical Energy Vector Solution Dot-multiply the equation of motion by → −= r r r 12 µ r  rr r rr  ⋅−=⋅ 3 µ Use vrvr  == & 03 =⋅+⋅ rr r vv  µ Use also the identity aaaaaaaaaa aa    ⋅=→→→ →→ =      +⋅=⋅ 011 111 0 22 0 22 3 =      −=      −+      →=+ r v td d rtd dv td d rr r vv µµµ    const r v E energy potential specific energy kinetic specific =−= µ 2 : 2 We can see that E is the specific mechanical energy.
  • 14.
    14 SOLO Two Body Problem(continue – 7) KEPLERIAN TRAJECTORIES Integration of Equations of Motion Vector Solution Cross-multiply the equation of motion with the specific angular momentum → −= r r r 12 µ h  hr r hr   ×−=× 3 µ ( ) ( ) ( )rr td d rhrh td d rhrhr td d      ××+×=×+×=× The left side is ( ) ( ) hrr r rrhrrrrrrrhr           ×=            −××+×=××+××+×= 0 3 0 µ The right side can be written ( ) ( ) ( )[ ] ( )       =−=−=⋅−⋅=××−=×− r r td d r r r v r rrrrv r vrrrrv r vrr r hr r     µ µµµµµµ 2 2 3333 Equaling both sides gives ( )       =× r r td d hr td d   µ Integrating both sides B r r hr   +=× µ where is the constant of integrationB 
  • 15.
    15 SOLO Two Body Problem(continue – 8) KEPLERIAN TRAJECTORIES Integration of Equations of Motion (continue – 1) Vector Solution (continue – 1) Dot-multiply this equation by : B r r hr   +=× µ r  ( ) Br r r rhrr     ⋅+⋅=×⋅ µ Use ( ) ( ) 2 hhhhrrhrr =⋅=⋅×⋅=×⋅    r r r r µµ =⋅   ( )pBrBr Θ−Θ=⋅ cos  to obtain ( )pBrrh Θ−Θ+= cos2 µ or ( ) ( ) ( )pp e p B h r Θ−Θ+ = Θ−Θ+ = cos1cos/1 /2 µ µ where µ/: 2 hp = µ/: Be = pΘ parameter (semi-latus rectum) eccentricity periapsis angle
  • 16.
    16 SOLO Two Body Problem(continue – 9) KEPLERIAN TRAJECTORIES Integration of Equations of Motion (continue – 2) Polar Coordinate Solution The equations of motion in polar coordinates are: 2 2 r rr µ −=Θ−  Consthr ==Θ2 Let define a new variable: r u 1 := from which urrr r u 2 2 1 −=→−=  If , we can write( )00 ≠≠Θ h Θ =  h r2 and 0≠ Θ −= Θ −= h d ud hu h r    Differentiating again we obtain 02 2 22 22 2 2 2 ≠ Θ −= Θ −=Θ Θ −= h d ud uh r h d ud h d ud hr  We also have 32 4 2 2 uh r h rr ==Θ The first equation of motion becomes 0232 2 2 22 ≠=+ Θ huuh d ud uh µ
  • 17.
    17 SOLO Two Body Problem(continue – 10) KEPLERIAN TRAJECTORIES Integration of Equations of Motion (continue – 3) Polar Coordinate Solution (continue – 1) Therefore 32 4 2 2 uh r h rr ==Θ The solution of this equation is 0232 2 2 22 ≠=+ Θ huuh d ud uh µ 02 3 2 2 ≠=+ Θ h h u d ud µ ( ) 221 cos 1 h CC r u µ +−Θ== where C1 and C2 are constant of integration pC p e C h p Θ=== 212 ,:,: µ Define to obtain ( )pe p r Θ−Θ+ = cos1 This is the equation of a conic section in polar coordinates.
  • 18.
    18 SOLO A right circularcone is a cone obtained by generators (straight lines) passing through a circle, and the apex C that is situated on the normal to the circle plane and passing trough the center of the circle. β is the angle between the cone axis and the generators. CONIC SECTIONS Cutting Plane generating a "hyperbola" Right Circular Cone Cone Apex Conical Section C Cone Axis Cutting Plane generating a "parabola" Cutting Plane generating a "ellipse" Cutting Plane generating a "circle" Cutting Plane generating two "lines" α β       −= β π α 2       −< β π α 2       −> β π α 2 ( )0=α                   −> −= −< lines line po 2 2 1 2 int 2 β π α β π α β π α P F F * Cutting Plane (Hyperbola ) Right Circular Cone Hyperbola 2 Branches C β α Ellipse Parabola Cutting Plane (Ellipse) Cutting Plane (Circle) Cutting Plane (Parabola) By cutting the right circular conic by a plane we obtain different conic sections, as a function of the inclination angle α of the plane relative to the base of the conic section and the angle β between the generators and the base. The discovery of the Conical Sections is attributed to the greek Menachmus who lived around 350 B.C..
  • 19.
    19 SOLO The conical sectionsare: CONIC SECTIONS 1. Circle if the cutting plane is normal to the cone axis (α=0) and is above or bellow the apex. 2. Ellipse if the cutting plane is inclined to the basis at an angle that falls short of the angle between generators to the base (α<π/2-β) (in greek word elleipsis means falls, short or leaves out. 3. Hyperbola if the cutting plane is inclined to the basis at an angle that exceeds of the angle between generators to the base (α>π/2-β)(in greek word hyperbole means excess. 4. Parabola if the cutting plane is parallel to a generator of the right circular cone (α=π/2-β) (in greek word parabole is the origin of the words parabola and parallel. 5. A point- apex (α<π/2-β), one straight line (α=π/2-β), two straight lines (α>π/2-β), if the cutting plane passes through the apex and intersects the cone basis.
  • 20.
    20 SOLO Two Body Problem(continue – 13) KEPLERIAN TRAJECTORIES Integration of Equations of Motion (continue – 6) ( )pe p r Θ−Θ+ = cos1 Conic Section (continue – 2) Θ pΘpΘ−Θ r d directrixfocus conic section x y → P1 → Q1 → r1 → t1 periapsis Let define e p d =: and rewrite ( )pe ed r Θ−Θ+ = cos1 as follows ( )prd r e Θ−Θ− = cos From which we obtain the following definition of conic sections: Conic Section A conic section is a planar curve, such that the ratio of distances, for any point on the curve, to a fixed point F* (focus) and to a line directrix is constant and equal to e.
  • 21.
    21 SOLO Two Body Problem(continue – 14) KEPLERIAN TRAJECTORIES Integration of Equations of Motion (continue – 7) Conic Section (continue – 3) Let write the conic section equation in cartesian coordinates using →→ QP 1,1 ( ) ( ) ( ) ( ) →→→ →→→ Θ−Θ+Θ−Θ−= Θ−Θ+Θ−Θ= QPt QPr pp pp 1cos1sin1 1sin1cos1 and ( ) ( )p p ry rx Θ−Θ= Θ−Θ= sin cos ( )pe p r Θ−Θ+ = cos1 Substitute those in the equation rewritten as ( ) prer p =Θ−Θ+ cos xepyx −=+ 22 to obtain Squaring both sides gives 22222 2 xexpepyx +−=+ or ( ) 2222 21 pyxpexe =++−
  • 22.
    22 SOLO Two Body Problem(continue – 15) KEPLERIAN TRAJECTORIES Integration of Equations of Motion (continue – 8) Conic Section (continue – 4) For e ≠ 1 can we can write ( ) 2222 21 pyxpexe =++− ( ) 2 2 2 22 22 2 2 2 111 1 e p e ep py e ep xe − = − +=+      − +− We see that for p ≠ 0 ( h ≠ 0 ) 1 0=e 222 pyx =+ circle 2 10 << e 1 11 1 2 2 2 2 2 =             − +             − − + e p y e p e ep x ellipse 3 1=e pe yp x 2 22 − = parabola 4 1>e hyperbola1 11 1 2 2 2 2 2 =             − −             − − − e p y e p e ep x
  • 23.
    23 SOLO Two Body Problem(continue – 16) KEPLERIAN TRAJECTORIES Velocity on the Trajectory Let find the velocity on the trajectory in and coordinates →→ tr 1,1 →→ QP 1,1 →→→→ +=Θ+== QyPxtrrrrv 1111  ( ) ( )[ ] ( ) ( )[ ] →→ Θ−ΘΘ+Θ−Θ+Θ−ΘΘ−Θ−Θ= QrrPrr pppp 1cossin1sincos  From the equation of motion ( )pe p r Θ−Θ+ = cos1 hr =Θ2 µ/: 2 hp = we obtain ( ) ( )[ ] ( ) ( )p p p p r e p p eh r h e ep td d d dr rv Θ−Θ= Θ−Θ = Θ−Θ+ Θ−Θ = Θ Θ == sin sin cos1 sin 22 µ  ( )[ ]pt e pr p r h rrv Θ−Θ+===Θ= cos12 µµ
  • 24.
    24 SOLO Two Body Problem(continue – 17) KEPLERIAN TRAJECTORIES Velocity on the Trajectory (continue – 1) Let substitute those results in the velocity equation to obtain the components in coordinates. →→ QP 1,1 ( ) ( )[ ] →→→→ Θ−Θ++Θ−Θ−=+= Qe p P p QyPxv pp 1cos1sin11 µµ   ( ) ( ) ( ) ( )[ ]{ } 2/12222/1222/122 cos1sin pptr ee p yxvvv Θ−Θ++Θ−Θ=+=+= µ  ( )[ ] 2/12 cos21 pee p v Θ−Θ++= µ Θ pΘ pΘ−Θ r d directrix focus conic section x y → P1 → Q1 → r1 → t1 v  rv tv periapsis The velocities at the periapsis and apoapsis are ( )pΘ=Θ ( )π+Θ=Θ p ( )e p va −= 1 µ ( )e p vp += 1 µ
  • 25.
    25 SOLO Two Body Problem(continue – 18) KEPLERIAN TRAJECTORIES Specific Mechanic Energy The specific mechanic energy is ( ) ( ) ( ) ap e p e p e r v r v E a a 22 11 2 1 22 : 2222 µµµµµµ −= − −= − − − =−=−= or ar v E 22 : 2 µµ −=−= where we used ( )2 1 eap −= From the last equation we get a p e −= 1 Substitute and to obtainµ/: 2 hp = ( )Ea 2/µ−= 2 2 2 1 µ hE e +=
  • 26.
    26 SOLO Two Body Problem(continue – 19) KEPLERIAN TRAJECTORIES Eccentricity Vector We defined the eccentricity as a scalar µ/: Be = Let define the eccentricity vector as µ/: Be  = where is defined asB  B r r hr   +=× µ Therefore ( ) ( ) ( ) → = ⋅−      − =− ⋅− =− ×× =− × = Pe vvr r vr r rvvrvr r rvrv r rhv e 1 2 2 µ µ µµµ    The last equation is obtained from the definition of , and , and can be checked by substituting the range and velocity vectors in coordinates, in previous equation. B  e  ( )pΘ−Θ→→ QP 1,1
  • 27.
    27 SOLO Two Body Problem(continue – 20) KEPLERIAN TRAJECTORIES Orbit Determination from Initial Conditions Given the initial position and velocity vectors , find the orbital parameters00 ,vr  Θ pΘr d directrix focus conic section x y → P1 → Q1 → r1 → t1 v rv tv( )pΘ−Θ γ 1 From the specific angular momentum of the orbit we can find00 vrh  ×= 01 00 ≠ × = → h h vr R  µ 2 h p = 2 From the specific mechanical energy on an elliptic orbit ar vv E 22 0 00 µµ −=− ⋅ =  we obtain 00 0 2 vv r a  ⋅− = µ µ
  • 28.
    28 SOLO Two Body Problem(continue – 21) KEPLERIAN TRAJECTORIES Orbit Determination from Initial Conditions (continue – 1) Θ pΘr d directrix focus conic section x y → P1 → Q1 → r1 → t1 v rv tv( )pΘ−Θ γ 3 The eccentrity vector of a Keplerian trajectory is given by from which ( ) → =      ⋅−      −⋅= Pevvrr r vve 1 1 0000 0 00  µ µ ee  = 01 ≠= → e e e P  →→→ ×= PRQ 111 ( )pe p r Θ−Θ+ = cos1 The position and velocity vectors are given by ( ) ( )       Θ−Θ+Θ−Θ= →→ QPrr pp 1sin1cos  ( ) ( )( )     Θ−Θ++Θ−Θ−=+= →→→→ QeP p QyPxv pp 1cos1sin11 µ  
  • 29.
    29 SOLO Two Body Problem(continue – 22) KEPLERIAN TRAJECTORIES Orbit Determination from Initial Conditions (continue – 2) Θ pΘr d directrix focus conic section x y → P1 → Q1 → r1 → t1 v rv tv( )pΘ−Θ γ 4 Initial trajectory angle , on the conic section, from the periapsis is given by pΘ−Θ=Θ 01       Θ Θ =Θ ⋅         ×=Θ ⋅=Θ − → → → → → 1 11 1 0 0 1 0 0 1 cos sin tan 11sin 1cos R r r P r r P One other way to find is by using →→→ RQP 1,1,1 ( )       Θ++Θ−= Θ+Θ= →→ →→→ QePv p QPr 1cos1sin 1sin1cos1 110 110  µ or ( ) ( ) ( )       Θ+Θ=Θ+ Θ−Θ+=Θ+ →→ →→ 01011 01011 cos1sin1cos1 sin1cos1cos1 v p rQe v p rePe   µ µ ( ) 1 1 sin cos Θ− Θ+e 1 1 cos sin Θ Θ
  • 30.
    30 SOLO Two Body Problem(continue – 23) KEPLERIAN TRAJECTORIES Orbit Determination from Initial Conditions (continue – 3) Θ pΘr d directrix focus conic section x y → P1 → Q1 → r1 → t1 v rv tv( )pΘ−Θ γ Using 1 0 cos1 Θ+ = e p r we finally obtain ( )        Θ+ Θ = Θ− Θ+ = →→ →→ 01 0 0 1 01 0 0 1 cos sin 1 sin cos 1 v p r r p Q v p r r p e P   µ µ ( )[ ] ( ) ( )[ ] ( ) ( )[ ] ( ) 01 0 01 01 0 01 011 0 011 sincos11 sincos1cos1 sincoscossinsinsincoscos 1sin1cos v p rr r p r v p rr re p r v p rr re p r QPrr     Θ−Θ+       Θ−Θ+−= Θ−Θ+Θ−Θ+−Θ+= ΘΘ−ΘΘ+ΘΘ+Θ+Θ=       Θ+Θ= → → → →→ µ µ µ Therefore
  • 31.
    31 SOLO Two Body Problem(continue – 24) KEPLERIAN TRAJECTORIES Orbit Determination from Initial Conditions (continue – 4) Θ pΘr d directrix focus conic section x y → P1 → Q1 → r1 → t1 v rv tv( )pΘ−Θ γ where we used ( ) ( ) ( ) ( )[ ] ( ) ( )[ ] 011 0 0 11 011 0 0 11 coscos sinsinsin coscossinsin sincoscossin 1cos1sin ve p r r p ee p ve p r r p ee p QeP p v    Θ+Θ+Θ+ Θ−Θ−Θ−Θ = ΘΘ++ΘΘ+ ΘΘ+−Θ+Θ −=     Θ++Θ−= → → →→ µ µ µ ( ) ( )[ ] ( ) ( )[ ] ( )[ ] ( )[ ] ( )[ ] ( )[ ] ( ) ( )[ ] 01 0 01 0 1 0 00 01 0 0 1111 011 0 0 111111 cos11sin 1 cos1 cos11 cos1sincos1sin cos1cos1 sinsincoscossinsin v p r r pr e pr vr v p r r p eee p ve p r r p ee p           Θ+Θ−−+         Θ−Θ−Θ−Θ− ⋅ =       Θ+Θ−−+ Θ+Θ−Θ−Θ−Θ−Θ = Θ++Θ+Θ+−+ Θ−Θ−Θ−ΘΘ+Θ−ΘΘ−Θ = → → → → µ µ µ ( ) 10 1111000 sin 1cos1sin1sin1cos Θ=     Θ++Θ−⋅      Θ+Θ=⋅ →→→→ e p r QeP p QPrvr µ µ
  • 32.
    32 SOLO Two Body Problem(continue – 25) KEPLERIAN TRAJECTORIES Orbit Determination from Initial Conditions (continue – 5) Θ pΘr d directrix focus conic section x y → P1 → Q1 → r1 → t1 v rv tv( )pΘ−Θ γ Summarize ( )[ ] ( ) 01 0 01 sincos11 v p rr r p r r  Θ−Θ+       Θ−Θ+−= → µ ( )[ ] ( ) ( )[ ] 01 0 01 0 1 0 00 cos11sin 1 cos1 v p r r pr e pr vr v          Θ+Θ−−+         Θ−Θ−Θ−Θ− ⋅ = → → µ
  • 33.
    33 SOLO Two Body Problem(continue – 26) KEPLERIAN TRAJECTORIES Orbit Determination from Initial Conditions (Second Method) Start from the solution for Keplerian orbit ( ) 221 cos 1 h CC r µ +−Θ= where C1 and C2 must be determined from initial conditions 0r - initial range from the center of mass M 0v - initial velocity magnitude 00000 2 cosγvrvrrvrh =×=Θ=×=  0γ - initial trajectory angle (between velocity vector and the horizon plane)
  • 34.
    34 SOLO Two Body Problem(continue – 27) KEPLERIAN TRAJECTORIES Orbit Determination from Initial Conditions (Second Method - 1) We have ( ) ( ) 0 2 0 21 0 22 0 2 0 21 cos 1 cos cos cos 1 γλγ µ r CC vr CC r +−Θ=+−Θ= where 0 2 0 / : r v µ λ = Let differentiate ( ) ( ) ( )2121 2 212 sinsinsin 1 CChCCrrCC r r rdt d −Θ=−ΘΘ=→−ΘΘ−=−=       or ( )21000 sincos CCvrr −Θ−= γ Let use now the initial conditions to find C1, C2. ( ) 0 2 0 201 0 cos 1 cos 1 γλ r CC r +−Θ= ( ) 002010000 sinsincos γγ vCCvrr =−Θ= or ( ) 0 2 00 201 cos 11 cos γλ rr CC −=−Θ ( ) 0 0 201 tan 1 sin γ r CC =−Θ
  • 35.
    35 SOLO Two Body Problem(continue – 28) KEPLERIAN TRAJECTORIES Orbit Determination from Initial Conditions (Second Method - 2) We have ( ) 0 2 00 201 cos 11 cos γλ rr CC −=−Θ ( ) 0 0 201 tan 1 sin γ r CC =−Θ ( ) ( )[ ] ( ) ( ) ( ) ( ) ( ) ( ) 0 2 0 00 0 0 0 2 00 0 2 0 02010201 0 2 0 2001 cos 1 sintan 1 cos cos 11 cos 1 sinsincoscos cos 1 cos 1 γλ γ γλ γλ γλ rrrr r CCCC r CC r +Θ−Θ−Θ−Θ      −= +Θ−Θ−Θ−Θ−Θ−Θ= +−Θ+Θ−Θ= ( ) ( ) ( ) ( ) ( ) 0 00 0 2 0 0 0000 0 2 00 cos cos cos cos1 cos sinsincoscos cos cos1 γ γ γλ γ γγ γλ +Θ−Θ + Θ−Θ− = Θ−Θ−Θ−Θ + Θ−Θ− = r r which, when developed further, gives
  • 36.
    36 SOLO Two Body Problem(continue – 29) KEPLERIAN TRAJECTORIES Orbit Determination from Initial Conditions (Second Method - 2) ( ) ( ) 0 00 0 2 00 cos cos cos cos1 γ γ γλ +Θ−Θ + Θ−Θ− = r r Finally ( ) ( )0000 0 2 0 coscoscos1 cos γγλ γλ +Θ−Θ+Θ−Θ− = r r
  • 37.
    37 SOLO Two Body Problem(continue – 30) KEPLERIAN TRAJECTORIES Orbit Determination from Initial Conditions (Second Method - 3) ( ) ( )0000 0 2 0 coscoscos1 cos γγλ γλ +Θ−Θ+Θ−Θ− = r r ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] ( )2 0 2 0 2 0 22 00 2 0002 0 2 0 2 0 22 00 2 000 0000 1coscossin cos1cossincossin 1coscossin1 cos1cossincossin1 coscoscos1 −+ Θ−Θ−+Θ−Θ− −++= Θ−Θ−+Θ−Θ−= +Θ−Θ+Θ−Θ− γλγγλ γλγγλ γλγγλ γλγγλ γγλ The denominator can be expressed as 0 2 0 cos: γλrp = ( ) ( ) 1cos21cos2cos1coscossin: 0 2 0 2 0 222 0 2 0 2 0 22 +−=+−=−+= γλλγλγλγλγγλe We define       − =Θ−Θ − 1cos cossin tan: 0 2 001 0 γλ γγλ p ( ) ( ) ( ) ( )[ ] ( )pp ee Θ−Θ+=Θ−Θ+Θ−Θ+= +Θ−Θ+Θ−Θ− cos1cos1 coscoscos1 00 0000 γγλto obtain
  • 38.
    38 SOLO Two Body Problem(continue – 31) KEPLERIAN TRAJECTORIES Orbit Determination from Initial Conditions (Second Method - 4) ( ) ( )0000 0 2 0 coscoscos1 cos γγλ γλ +Θ−Θ+Θ−Θ− = r r The denominator can be expressed as ( ) ( ) ( ) ( )[ ] ( )pp ee Θ−Θ+=Θ−Θ+Θ−Θ+=+Θ−Θ+Θ−Θ− cos1cos1coscoscos1 000000 γγλ We obtain again the fact that the trajectory is a conic section given by ( )pe p r Θ−Θ+ = cos1 We can also write ( )[ ] λγλλ γλ − = +−− = − = 21cos21 cos 1 : 0 0 2 0 2 0 2 rr e p a and 0 2 0 cos: γλrp = ( ) 1cos2: 0 2 +−= γλλe
  • 39.
    39 SOLO Two Body Problem(continue – 32) KEPLERIAN TRAJECTORIES Flat Earth Approximation We assume that , so we can use10 <<Θ−Θ ( ) ( ) ( )2 0000 2 1 1cos&sin Θ−Θ−Θ−ΘΘ−Θ≈Θ−Θ ( ) ( ) ( ) ( ) ( ) ( ) ( )2 0 0 200 0 000 2 00 22 0 0 2 0 0000 0 2 0 cos 1 1 2 1 tan1 cossin 2 1 1cos 2 1 cos coscoscos1 cos Θ−Θ      −−Θ−Θ− = Θ−Θ−      Θ−Θ−+Θ−Θ ≈ +Θ−Θ+Θ−Θ− = γλ γ γγλγλ γλ γγλ γλ r r r r and where gr v r r v r v 0 2 0 2 0 0 2 0 0 2 0 ≈== µµ λ The order of magnitude of λ is ( ) 64 000,1 10000,400,61 22 0 0 = ⋅ == v gr λ
  • 40.
    40 SOLO Two Body Problem(continue – 33) KEPLERIAN TRAJECTORIES Flat Earth approximation (continue – 1) Therefore if γ is not close to 90° , we can assume that , and then1 cos 1 0 2 >> γλ ( ) ( ) ( ) ( ) ( ) ( ) 0 22 0 2 0 2 00000 0 22 0 02 0000 0 22 0 02 000 0 cos2 1 tan cos2 1 tan1 cos2 1 tan1 γ γ γ γ γ γ v g rrr v gr r v gr r r Θ−Θ−Θ−Θ+=       Θ−Θ−Θ−Θ+≈ Θ−Θ+Θ−Θ− ≈ Use to obtain ( )0000 && Θ−Θ=+=+= rxzRrzRr ee 0 22 0 2 00 cos2 1 tan γ γ v g xxzz −+=
  • 41.
    41 SOLO Two Body Problem(continue – 34) KEPLERIAN TRAJECTORIES Flat Earth approximation (continue – 2) We obtained the Flat Earth approximation of the trajectory, that is a parabola. 0 22 0 2 00 cos2 1 tan γ γ v g xxzz −+= 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 Vacuum Ballistic Trajectories g V 2 2 0 g V 2 2 0 0 22 0 2 00 cos2 γ γ V g xtgxzz −+= g V 2 0 0V 0γ0z
  • 42.
    42 SOLO Two Body Problem(continue – 35) KEPLERIAN TRAJECTORIES Orbital Trajectories Let discuss in more detail the Keplerian trajectories. Ellipse (e < 1) ( Kepler’s First Law ) and Circle ( e = 0 ) x y ( )2 1 eap −= eac = a a ( ) 2/12 1 eab −= ( )eara += 1 ( )earp −= 1 r ΘFOCUS EMPTY FOCUS Apoapsis Radius Periapsis Radius Apoapsis PeriapsisGeometric Center Directrix Directrix c epd /= Semimajor axis Semiminor axis v  nv rv γ → P1 → Q1 1 11 1 2 2 2 2 2 =             − +             − − + e p y e p e ep x We found that the ellipse equation in cartesian coordinates is given by:
  • 43.
    43 SOLO Two Body Problem(continue – 36) KEPLERIAN TRAJECTORIES Orbital Trajectories (continue – 1) Ellipse (e < 1) ( Kepler’s First Law ) and Circle ( e = 0 ) x y ( )2 1 eap −= eac = a a ( ) 2/12 1 eab −= ( )eara += 1 ( )earp −= 1 r ΘFOCUS EMPTY FOCUS Apoapsis Radius Periapsis Radius Apoapsis PeriapsisGeometric Center Directrix Directrix c epd /= Semimajor axis Semiminor axis v  nv rv γ → P1 → Q1Define: 2 1 : e p a − = semimajor axis 2 1 : e p b − = - semiminor axis ea e ep c = − = 2 1 : - distance of geometrical center from focus e p d =: - distance of directrix from focus ( )eara += 1: - apoapsis radius ( )earp −= 1: - periapsis radius Θ - true anomaly γ - flight-path angle ( ) 2/pa rra += ( ) ( )papa rrrre +−= / ( ) ( ) ( )ererabeap ap −=+==−= 11/1 22 From those relations we get: 1 22 =      +      + b y a cx and the ellipse equation will be:
  • 44.
    44 SOLO Two Body Problem(continue – 37) KEPLERIAN TRAJECTORIES Time of Flight on an Elliptic Orbit From the equation Θ= 2 rh we can write h Ad h dr dt 2 2 = Θ = where is the area defined by the radius vector as it moves through an angle 2 2 Θ = dr Ad Θd Θ pΘ pΘ−Θ r focus conic section x y → P1 → Q1 → r1 → t1 v  rv tv Θd Θ= drAd 2 2 1 periapsis This proves the 2nd Kepler’s Law that equal area are swept out equal in equal times by the radius vector.
  • 45.
    45 SOLO Two Body Problem(continue – 38) KEPLERIAN TRAJECTORIES Time of Flight on an Elliptic Orbit (continue – 1) The period of the orbit depends only on the major axis of the ellipse a. ( ) ( ) p pa h eaa h ea h ba T eap ph µ ππππ µ 2/3122/322 2 1 2 1 22 2 = −= = − = − == or 2/3 2 aT µ π = The period of an elliptical orbit T is obtained by integrating from Θ= 0 to Θ=2π , and the radius vector sweeps the area of the ellipse A = π a b. This proves the Kepler’s third law: “the square of the period of a planet orbit is equal To the cube of its mean distance to the sun”.
  • 46.
    46 SOLO Two Body Problem(continue – 39) KEPLERIAN TRAJECTORIES Time of Flight on an Elliptic Orbit (continue – 2) Let draw an auxiliary circle of radius a, and the same center O as the geometric center of the ellipse. x y eac = a a ( ) 2/12 1 eab −= r Θ FOCUS EMPTY FOCUS c → P1 → Q1 a F Q O VS E P Let take any point P on the ellipse with polar coordinates r,Θ and define the point Q on the circle at the same coordinate x as P. Eeary r a x ea a x by Ea a x ay ellipse ellipse circle sin1sin sin111 sin1 2 2 2 2 2 2 2 2 −=Θ=→        Θ=−−=−= =−= The angle E of OQ with x axis is called the eccentric anomaly. aeEarxellipse −=Θ= coscos
  • 47.
    47 SOLO Two Body Problem(continue – 40) KEPLERIAN TRAJECTORIES Time of Flight on an Elliptic Orbit (continue – 3) Let compute x y eac = a a ( ) 2/12 1 eab −= r Θ FOCUS EMPTY FOCUS c → P1 → Q1 a F Q O VS E P ( ) ( )( ) ( )( ) ( ) 0cos11 sin1sincos1cos 1 2 22 2 >←−−= −+−−= −=×=−= EEEeea EEeaEaEEeaaeEa xyyxvreah ellipseellipseellipseellpse     µ We obtain ( ) n a EEe ==− :cos1 3 µ ( ) ( ) ( )pttntEetE −=− sin Integrating this equation gives Kepler’s Equation where tp is the time of periapsis ( E (tp) = 0 )
  • 48.
    48 SOLO Two Body Problem(continue – 41) KEPLERIAN TRAJECTORIES Time of Flight on an Elliptic Orbit (continue – 4) From x y eac = a a ( ) 2/12 1 eab −= rΘ FOCUS EMPTY FOCUS c → P1 → Q1 a F Q O VS E P Eeary aeEarx ellipse ellipse sin1sin coscos 2 −=Θ= −=Θ= we have ( ) ( )[ ] [ ] ( )Eea EeEeaEeaaeEar cos1 coscos21sin1cos 2/1222/12222 −= +−=−+−= Therefore Θ+ Θ− = − − =Θ Θ+ Θ+ = − − =Θ cos1 sin1 sin cos1 sin1 sin cos1 cos cos cos1 cos cos 22 e e E Ee Ee e e E Ee eE ( )( ) Ee Ee Ee eEEe sin1 cos11 sin1 coscos1 sin cos1 2 tan 22 − −+ = − +−− = Θ Θ− = Θ From 2 tan 1 1 2 tan E e e − + = Θ or and are always in the same quadrant.2 Θ 2 E
  • 49.
    49 SOLO References 2. Battin, R.H.,“An Introduction to the Mathematics and Methods of Astrodynamics”, AIAA Education Series, AIAA, Washington. DV., 1987 KEPLERIAN TRAJECTORIES 3. Battin, R.H., “Astronautical Guidance”, McGraw Hill, 1964 1. Bate, R.R., Mueller, D.D., White, J.E., “Fundamentals of Astrodynamics”, Dover Publications, New York, 1971 4. Chobotov, V.A., Editor, “Orbital Mechanics”, AIAA Education Series, , AIAA, 2nd Edition, 1996 5. Reagan, F.J., “Re-Entry Vehicle Dynamics”, AIAA Education Series, , AIAA, 1984 6. Reagan, F.J. and Anandakrishnan, S.M., “Dynamics of Atmospheric Re-Entry”, AIAA Education Series, , AIAA, 1993 7. Montenbruck, O. and Eberhard, G., “Satellite Orbits”, Springer-Verlag, 2000 8. Kaplan, M.H., “Modern Spacecraft Dynamics and Control”, John Wiley & Sons, 1976
  • 50.
    August 13, 201550 SOLO Technion Israeli Institute of Technology 1964 – 1968 BSc EE 1968 – 1971 MSc EE Israeli Air Force 1970 – 1974 RAFAEL Israeli Armament Development Authority 1974 – 2013 Stanford University 1983 – 1986 PhD AA