Free Space Relative Orbital Motion Airless Major Bodies (moons) 1 2016 David L. Akin - All rights reserved http://spacecraft.ssl.umd.edu
Propulsive Motion in Free Space Basic motion governed by Newton s Law ẍ (actually, F = m ) F = ma Over a distance d and time t, assuming the motion is predominately coasting, V = 2 d t (required to accelerate and decelerate) The rocket equation (relates propellant to ΔV) m final m o = e V V exhaust 2
Cost of Propulsive Maneuvering Assuming V Use the Taylor s Series expansion of e m final m o 1 Since m o =m initial =m prop +m final, m prop 2 d m o V exhaust t V exhaust or V V exhaust m prop m o 2 V travel V exhaust 3
Hill s Equations (Proximity Operations) Linearized equations of motion relative to a target in circular orbit in a rotating Cartesian reference frame ẍ = 3n 2 x + 2nẏ + a dx Ref: J. E. Prussing and B. A. Conway, Orbital Mechanics Oxford University Press, 1993 ÿ = z = n = 2nẋ + a dy n 2 z + a dz µ a 3 adx, ady, adz are disturbing accelerations (e.g., thrust, solar pressure) 4
Clohessy-Wiltshire ( CW ) Equations Force-free solutions to Hill s Equations x(t) = [4 3 cos (nt)]x o + y(t) = 6[sin (nt) nt]x o + y o 2 n [1 sin (nt) n x o + 2 n [1 cos (nt)] x o + 4 sin (nt) 3nt n cos (nt)] y o y o ẋ(t) =3n sin (nt)x o + cos(nt)ẋ o +2sin(nt)ẏ o ẏ(t) = 6n [1 cos (nt)] x o 2sin(nt)ẋ o + [4 cos (nt) 3] ẏ o z(t) = z o cos (nt) + z o sin (nt) n ż(t) = z o n sin (nt)+ż o cos (nt) 5
V-Bar Approach 0.01 m/sec 0.0075 m/sec 0.005 m/sec 5$ 0$!120$!100$!80$!60$!40$!20$!5$ 0$!10$ 6
R-Bar Approach Approach from along the radius vector ( Rbar ) Gravity gradients decelerate spacecraft approach velocity - low contamination approach Used for Mir, ISS docking approaches Ref: Collins, Meissinger, and Bell, Small Orbit Transfer Vehicle (OTV) for On-Orbit Satellite Servicing and Resupply, 15th USU Small Satellite Conference, 2001 7
Hopping (Airless Flat Planet) V v, h V V h, d Use F=ma for vertical motion V v = g h = V v t t flt =2V v /g Constant velocity in horizontal direction produces d = V h t flt =2 V hv v g V h = V cos ; V v = V sin d =2 V 2 sin g cos 8 = V 2 g sin (2 ) 1 2 gt2
Hopping (Airless Flat Planet) V v, h V Horizontal distance is maximized when sin (2 )=1 V h, d opt = 2 = 45o d max = V 2 g V = gd V total =2V =2 gd h max = V v V v g h max = V 2 4g = 1 2 g V v g gd2 4g = d 4 9 2 V v = V 2
An Example of Propulsive Gliding 10
Propulsive Gliding (Airless Flat Planet) T = mg V g Assume horizontal velocity is V V h =2V t flt = d/v Total ΔV becomes (includes acceleration and deceleration) V v = gt flt = gd V V total = V v + V h =2V + gd V 11
Propulsive Gliding (Airless Flat Planet) V Want to choose V to minimize 2V + gd V =0 2 gd V 2 =0 V opt = gd 2 gd 2 V total =2 2 + gd gd =2 2 gd 12
Delta-V for Hopping and Gliding Delta-V (m/sec) 2000 1800 1600 1400 1200 1000 800 600 400 200 0 0 200000 400000 600000 800000 1000000 1200000 Distance (m) Ballistic Hop 13 Propulsive Glide
Hopping (Spherical Planet) a = r r = 1 e cos 1 e 2 v =2v o v = µ 2 r p 1+ecos = a(1 e2 ) 1 e cos v = µ 2 r 1 a 1 e 2 r(1 e cos v e =0 r(1 e cos )( 2e) + (1 e2 )r( cos ) r 2 (1 e cos ) 2 =0 14
Hopping (Spherical Planet) 2er 2e 2 r cos r cos + re 2 cos =0 e opt = 2 ± 2 2 4 cos 2 2 cos cos e 2 2e + cos =0 = 1 ± sin cos + produces e>1 (hyperbolic orbit); gives elliptical orbit e opt = 1 sin cos a opt = r 1 e opt cos 1 e 2 opt 15
Propulsive Gliding (Airless Round Planet) 2 r = V 2 r V g Assume horizontal velocity is V V h =2V t flt = d/v Total ΔV becomes (includes acceleration and deceleration) V v = g V 2 r t flt = gd V dv r V total = V v + V h =2V + gd V dv r 16
Propulsive Gliding (Airless Round Planet) V 2V + gd V Want to choose V to minimize dv r =0 2 gd V 2 d r =0 V total =2 gd 2 d r V opt = + gd V total =2 17 gd 2 d r 2 2 d r gd d r d r gd gd 2 d r
Hopping on Flat and Round Bodies 9000 8000 Delta-V (m/sec) 7000 6000 5000 4000 3000 2000 1000 0 0 1000000 2000000 3000000 4000000 5000000 6000000 Distance (m) Ballistic Hop Propulsive Glide Hop on Sphere Glide on Sphere 18
Nondimensional Forms Define V dg d r h max d flat glide =2 2 flat hop =2 = 1 4 spherical glide =2 2 (0 1) 19
Multiple Hops Assume n hops between origin and destination At each intermediate touchdown, v v has to be reversed V total =2V + 2(n 1)V v t peak = V v g t total =2nt peak =2n V v g d = V h t total = 2n g V hv v V v = 2gh max v = 2 n V dg h max d/n V h = dg 2nV v h = 1 2 1 2n 20
Multiple Hop Analysis =2 + 2(n 1) v =2 2 v + 2 h + 2(n 1) v = =2 2n + 1 8n + 2(n 1) 2n 1 2 1 2 n + 1 n 8n 2 +(n 1) 8n 2 n =0 Analytically messy, but note that for n =1 opt = 1 4 (In general, solve numerically) 21
Optimal Solutions for Multiple Hops opt 0.3 3 0.25 2.5 0.2 2 0.15 1.5 0.1 1 0.05 0.5 0 1 10 100 1000 10000 0 1 100 10000 Number of Hops (n) Number of Hops (n) 22
Hopping Between Different Altitudes Relative to starting point, landing elevation h 2 v 1 =(v h,v v1 ) v 2 =(v h,v v2 ) v v1 = v v2 h = v v1 t 1 t peak = v v1 2 gt2 g h peak = 1 2 23 v 2 v1 g v v1 = 2gh peak 1 From peak, v v = gt fall ; h = h peak 2 gt2 fall 1 vv2 2 h 2 = h peak t fall = 2 2 g g (h peak h 2 ) v v2 = 2g(h peak h 2 )
Optimal Hop with Altitude Change v v1 2 d = v h (t peak + t fall )=v h g + g (h peak h 2 ) 2h peak 2 d = v h + g g (h peak h 2 ) d g = v h 2hpeak + 2(h peak h 2 ) v h = v = d g 2h peak + 2(h peak h 2 ) v 2h + v2v1 + vh 2 + v2 v2 24
Nondimensional Form of Equations Remember that = 2 h + v dg ; 2 v1 + h peak 2 h + d ; h 2 d v1 = 2 v2 = 2( ) 1 h = 2 + 2( ) 2 v2 = 2 1 +2 + 2 + 2( ) 2 1 + 2( ) 2 + 2( ) 25
Optimization of Height-Changing Hop This is not going to be one where you can take the derivative and set equal to zero, so use the equation to find a numerical optimization Set =0 to check for plain hop solution =2 1 8 +2 opt = 1 4 26
Trajectory Design for Height Change 3 2.5 2 opt 1.5 1 0.5 0 Height Change -3-2 -1 0 1 2 3 27
Apollo Concept of Lunar Flying Vehicle from Study of One-Man Lunar Flying Vehicle - Final Report Volume 1: Summary North American Rockwell, NASA CR-101922, August 1969 28
Apollo 15 Revisited: LFV Sortie Basic assumptions Vehicle inert mass=300 kg Crew mass=150 kg Science package=100 kg Total propellant=130 kg 29
Apollo 15 Revisited: Leg 1 Base camp to bottom of rille Distance 3 km Altitude change -150 m ΔV=139 m/sec Propellant used=22 kg Collect 20 kg of samples at landing site 30
Apollo 15 Revisited: Leg 2 Propulsive glide along bottom of rille Distance 2 km No net altitude change ΔV=160 m/sec Propellant used=25 kg Collect 20 kg of samples at landing site; leave 25 kg science package 31
Apollo 15 Revisited: Leg 3 Hop to top of mountain Distance 15 km Altitude change 1600 m ΔV=310 m/sec Propellant used=46 kg Collect 30 kg of samples at landing site; leave 50 kg science package 32
Apollo 15 Revisited: Leg 4 Return to base Distance 12 km Altitude change -1450 m ΔV=278 m/sec Propellant used=37 kg Return with 25 kg of science equipment and 70 kg of samples 33
Apollo 15 Revisited: Discussion Current minimum estimates are for 400 kg of residual propellants in Altair at landing - would support three equivalent sorties Presence of water ice or ISRU propellant production at outpost would easily support moderate flier mission requirements Challenges in routine refueling of cryogenic propellants on the lunar surface, reliable flight and landing control system 34
Landing Impact Attenuation Cannot rely on achieving perfect zero velocity at touchdown Specifications for landing conditions Vertical velocity 3 m/sec Horizontal velocity 1 m/sec Kinetic Energy = 1 2 mv2 = 1 2 m(v2 h + v 2 v) Max case 500 kg vehicle = E = 2500Nm 35
Mars Phoenix Lander 36
Apollo Lunar Module 37
Landing Deceleration Look at 3 m/sec vertical velocity Constant force deceleration 1 2 mv2 = Fd t decel = Spring deceleration F = kx k = mv2 v a desired d 2 1 2 v2 = F m d = a desiredd d = 1 2 Fdx = 1 2 mv2 a peak = kd m 38 v 2 a desired a desired d cm t d sec 1/6 g 281 1.88 1/2 g 92 0.61 1g 46 0.31 2g 23 0.15 3g 15 0.10
Effect of Lateral Velocity at Touchdown Resolve torques around landing gear footpad h mg w F h F v = tot I tot = F h h F v w mgw I cg + m 2 Worst cases - hit obstacle (high force), landing downhill Issue: rotational velocity induced is counteracted by vehicle weight Will vehicle rotation stop before overturn limit? 39
Simple Approach to Landing Stability Kinetic energy at landing w K.E. = 1 2 mv2 = 1 2 m(v2 v + v 2 h) h mg F h F v Dissipated by potential energy of raising C.G. by rotation around impact point P.E. = mg h = mg(` h) s v 2 v crit = p 2g(` h) or w req = 2g + h 2 h 2 40