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Teresa J. Stuchi, IF-UFRJ Teresa J. Stuchi, IF-UFRJ IF-Junho/2015 Space elevators 1. Torre de Babel (Moises, Genesis (11:1-9) 2. Escada de Jacob (Genesis (28:12)) Space elevators 3. K.E.Tsiolkovski Speculation about Earth and Sky and on Vesta (1895) 4. Yuri Artsutanov (1960) Komosomolskaya Pravda 5. Arthur C. Clarke The fountains of Paradise (tower in Sri Lanka) Yuri Artsutanov and Jerome Pearson 35.800 km K.E.Tsiolkovski 35.800 km, GEO: órbita geoestacionária AS THE CAR CLIMBS, THE ELEVATOR TAKES ON A 1 DEGREE LEAN, DUE TO THE TOP OF THE ELEVATOR TRAVELING FASTER THAN THE BOTTOM AROUND THE EARTH (CORIOLIS FORCE). THIS DIAGRAM IS NOT TO SCALE. 35.800 km VARIAÇÃO DA VELOCIDADE COM A ALTITUDE The velocity at the top of the tower is so great (10.93 km/s) that a payload released from there would escape the Earth without rocket propulsion. The Space Elevator will succeed 50 years after everyone has stopped laughing. -Arthur C. Clarke NANO TUBOS DERAM VIDA NOVA AO SONHO DA BABEL Lagrangian Points Earth‐Moon L4 64,000 km 58,000 km L2 L 1 380,000 km L3 L5 LUNAR SPACE ELEVATOR There are two lunar-synchronous points where an elevator could be placed that would be stable: the libration points L1 and L2 Anchored Lunar Satellites Compared to the Anchored Earth Satellite ( Jerome Pearson) What are tethered satellites? Space missions TSS 1992 NASA TSS-1R 1996 NASA TIPS 1996 US Navy YES2 2007 ESA MAST 2007 JAXA Artistic representation (NASA) YES2 (2007) - ESA TSS-1 (1992) - NASA Tether mudando a órbita de satélite Momentum-exchange tethers (nonconductive tethers representing passive propulsion). They allow momentum to be transferred between objects in space, such as two spacecraft (tethers may redistribute momentum of a system from one body to another, but overall momentum is always conserved). The principle is based on the gravity gradient force. Two objects, separated by a distance but tied together by a tether, are “pulled” apart by the gravity gradient force [this causes vertical (radial) alignment between the two objects]. Due to irregularities in the central body's gravitational field, the nearly radially aligned tether system actually librates, or oscillates, in a pendulum-like motion, about the system's center of mass. This swinging motion may be used to raise or lower the orbit of a tandem system without using any propellant. TIPS - A bolo tether rotates end-over-end in the orbit plane. This system could propel a payload attached to one end into a different orbit. The bolo could conceivably catch a payload. - A stationary tether refers to two end- bodies connected by a tether of constant length. The system may be used to drag the lower end-body payload through the higher atmosphere (for sampling) and simultaneously lowering the system's orbit. - A tethered system release of an end- body from the remaining end-body and tether causes a momentum gain for the released end-body, resulting in a higher orbit for the released end-body orbit and a lower orbit for the remaining end-body and tether. The analytical model for the dynamics of a tether in the vicinity of collinear points Determination of the equilibrium points Lyapunov periodic orbits in the vicinity of the equilibrium points Application to the Earth-Moon L2 Primary bodies as point masses on circular orbits around the CM Cable length much smaller than the distance between the primary bodies Planar motion Ideal rigid cable x, yx,θ, y,θ System description Synodic system with origin in L2 of the RTBP Equations of Euler-Lagrange m 2 generalized coordinates : x, y,θ θ (holonomic constraints) CM X Y m ( , ) 1 L(q,q) = T(q,q) −V (q) d ∂L ∂L = 0 − dt ∂q ∂q Kinetic energy as a function of the generalized coordinates 1 T = (m + m ){x 2 + y 2 − 2nx(Y + y) + 2ny(X + x) + 2 1 2 o o (quadratic) (Coriolis) 2 2 2 + n [( X o + x) + (Yo + y) ]+ (centrifugal) n 2 l 2 l2 + β1β2[( +θ) + ]} (rotation/constrain) Potential energy - V m 2 m 1 d d l 1 + 2 = R R1 2 M1 M2 2 3 1 1 d d 1 2 d 1 3 1 1 u .t 1 3 u .t 1 1 5 u .t 3 u .t = + ()1 + ⋅ ()()1 − + ⋅ []()()1 − 1 + R d t R R R 2 R 2 1 − 1 1 1 1 1 GM m GM m GM m GM m GM M Gm m V = − 1 1 − 2 1 − 1 2 − 2 2 − 1 2 − 1 2 R d t R d t R d t R + d t D l 1 − 1 2 − 1 1 + 2 2 2 Lagrangian function 1 = + 2 + 2 + + + + 2 2 2 L (m1 m2 ) x y − 2nx(Yo y) 2ny(X o x) n [( X o + x) + (Yo + y) ]+ 2 n 2 l 2 l2 β1β2[( +θ) + ] + 2 2 1 1 β1β2l 3( cos θ − )1 G()m1 + m2 M1 + + 2 2 1 2 + + 2 + 2 3 2 [(X 0 + D1 + x) + y ] 2 [(X 0 D1 x) y ] 3 3 4 4 2 1 β1β2 1( −3β1β2 )l (35cos θ −30 cos θ + )3 1 β1β2 (β2 − β1)l 5( cos θ − 3cosθ) + 2 2 5 2 2 2 2 8 [(X 0 + D1 + x) + y ] 2 [(X 0 + D1 + x) + y ] 1 1 β β l 2 3( cos2 θ − )1 M + 1 2 + 2 2 2 1 2 2 [(X − D + x)2 + y2 ]3 2 [(X 0 − D2 + x) + y ] 0 2 4 4 2 1 β β (β − β )l 3 5( cos3 θ − 3cosθ) 1 β β 1( −3β β )l (35cos θ −30 cos θ + )3 1 2 2 1 + 1 2 1 2 2 [(X − D + x)2 + y2] 2 2 2 5 2 0 2 8 [(X 0 − D2 + x) + y ] Expansion up to fourth order Equations of motion y x x y = l = l τ = nt dimensionless coordinates : = D D D l,θ l,θ 1−µ l,θ µ l,θ Λ1 Λ2 xˆ ′′ =+2 yˆ′− Λ + Λ −1(Xˆ + xˆ)−µ 1( −µ) − ˆ 3 1 ˆ 3 2 0 ˆ 3 ˆ 3 R1 R2 R1 R2 1−µ l,θ µ l,θ yˆ ′′ = −2xˆ′− Λ + Λ −1 yˆ ˆ 3 1 ˆ 3 2 R1 R2 ˆ l ′ 1− µ l µ l θ ′′ = −2 (θ′ + )1 − Ω ,θ − Ω ,θ lˆ Rˆ 3 1 R3 2 , where: 1 2 2 3 l 3 l l Λ ,θ =1+ β β 3( cos2 θ − )1 + 2β β (β − β ) 5( cos3 θ −3cosθ) + i 1 2 R 1 2 2 1 R 2 1 1 4 5 l β β 1( −3β β ) 35 cos4 θ −30 cos2 θ + 3 1 2 1 2 R () 8 1 2 l 3 3 l 5 l Ω ,θ = sen 2( θ) − (β − β ) 1( −5cos2 θ)senθ − 1( −3β β ) 3( − 7cos2 θ)cosθsenθ i 2 1 R 1 2 R 2 2 i 2 i i∈{},1 2 Constant of motion n ∂L E = q − L (Jacobian constant) ∑ i i=1 ∂qi * 2 2 2 ˆ 2 ˆ 2 E = 2Ω − (xˆ $ + yˆ $ ) − β1β2[(θ $ −1)l − l $ ] , where: (Xˆ + xˆ)2 yˆ 2 1− µ µ Ω = 0 + + Σ + Σ ˆ 1 ˆ 2 2 2 R1 R2 2 3 1 l 1 l Σ =1+ β β (3cos2 θ − )1 + β β (β − β ) (5cos3 θ −3cosθ) + i 1 2 R 1 2 2 1 R 2 i 2 i 4 1 l β β 1( −3β β ) (35cos4 θ −30 cos2 θ + 3) 1 2 1 2 R 8 i i∈{},1 2 If l = 0 ⇒ RTBP Equations of motion 1−µ µ 1 1 Xˆ ′′ = +2Yˆ′− + −1 Xˆ −µ 1( −µ) − ˆ 3 ˆ 3 ˆ 3 ˆ 3 R1 R2 R1 R2 0,0 0,0 Λ1 = Λ2 =1 1−µ µ ˆ ′′ 2 ˆ ′ 1 ˆ Y = − X − ˆ 3 + ˆ 3 − y R1 R2 Jacobian constant E* = C = 2Ω−(Xˆ ′2 +Yˆ′2 ) Σ1 = Σ2 =1 J Xˆ 2 Yˆ 2 1− µ µ Ω = + + + ˆ ˆ 2 2 R1 R2 X 0 + x ↔ X Equilibrium equations Left hand side equals zero and all velocities equal zero l,θ l,θ 1−µ l,θ µ l,θ Λ1 Λ2 xˆ ′′ =+2 yˆ′− Λ + Λ −1(Xˆ + xˆ)−µ 1( −µ) − ˆ 3 1 ˆ 3 2 0 ˆ 3 ˆ 3 R1 R2 R1 R2 1−µ l,θ µ l,θ yˆ ′′ = −2xˆ′− Λ + Λ −1 yˆ ˆ 3 1 ˆ 3 2 R1 R2 ˆ l ′ 1− µ l µ l θ ′′ = −2 (θ′ + )1 − Ω ,θ − Ω ,θ lˆ Rˆ 3 1 R3 2 , where: 1 2 2 3 l 3 l l Λ ,θ =1+ β β 3( cos2 θ − )1 + 2β β (β − β ) 5( cos3 θ −3cosθ) + i 1 2 R 1 2 2 1 R 2 1 1 4 5 l β β 1( −3β β ) 35 cos4 θ −30 cos2 θ + 3 1 2 1 2 R () 8 1 2 l 3 3 l 5 l Ω ,θ = sen 2( θ) − (β − β ) 1( −5cos2 θ)senθ − 1( −3β β ) 3( − 7cos2 θ)cosθsenθ i 2 1 R 1 2 R 2 2 i 2 i i∈{},1 2 ˆ ˆ θ Equilibrium points (xeq, yeq, eq ) 1−µ l µ l xˆ = xˆ Λ ,θ + Λ ,θ = Xˆ + xˆ (degree 7) eq Xˆ x 2 1 Xˆ x 2 2 0 ( 0 + ˆ+µ) ( 0 + ˆ+µ− )1 ˆ =0 yeq m m θ =0 ou θ =π ∀ 1, 2 eq eq yˆ yˆ x x m m ˆ m m ˆ 2 1 1 2 θ eq π m = m θ =0 ou θ =π ou θ = 1 2 eq eq eq yˆ 2 m = m 2 1 xˆ m 1 Equilibrium Points: Earth-Moon 5 45 inclinação: 0 rad inclinação: 0 rad 4.5 inclinação: pi rad 40 inclinação: pi rad 2 4 L 2 99 L 35 9 o β1 = β2 a o β = β 3.5 1 2 o a ã o 30 ç ã a 3 l ç e a r l e 25 r m 2.5 e m m e 20 k 2 m m k e m 1.5 a 15 i e c n a i a t c 1 s n 10 i a t D s i 0.5 D 5 0 0 -0.5 0 1000 20 00 3000 4000 5 000 6000 -5 0 1000 2000 3000 4000 50 00 6000 Comprimento em km Comprimento em km Behavior for different tether length: Earth-Moon 0 16 500 km 500 km initial 800 km 800 km 1000 km 14 10 00 km conditions -5 12 10 x 0 -10 ) 0 = ) m m k k ( 8 ( y x -15 6 x0′ = 0 4 -20 2 y0 = 0 -25 0 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8 dia di a y0′ = 0 7 1.
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