Ballistic Atmospheric Entry • Straight-line (no gravity) ballistic entry based on density and altitude • Planetary entries (at least a start) • Basic equations of planar motion © 2016 David L. Akin - All rights reserved http://spacecraft.ssl.umd.edu U N I V E R S I T Y O F Ballistic Entry MARYLAND 1 ENAE 791 - Launch and Entry Vehicle Design Ballistic Entry (no lift) s = distance along the flight path v, s dv D γ = g sin γ dt − − m D horizontal dv dv ds dv 1 d(v2) mg = = V = dt ds dt ds 2 ds 2 1 d(v ) D 1 2 = g sin γ Drag D ρv AcD 2 ds − − m ≡ 2 2 2 1 d(v ) ⇥v ds dh = g sin γ AcD γ 2 ds − − 2m dh ds = sin γ d(v2) ⇥v2 sin γ = g sin γ Ac 2 dh − − 2m D U N I V E R S I T Y O F Ballistic Entry MARYLAND 2 ENAE 791 - Launch and Entry Vehicle Design Ballistic Entry (2) h Exponential atmosphere ρ = ρ e− hs ⇒ o h dρ h dh ρoe− hs dh ρ dh = e− hs − = − = − ρ h ρ h ρ h o s ⇥ o s ⇥ o s ⇥ h dh = − s dρ ρ sin γ d(v2) ⇥v2 = g sin γ Ac 2 dh − − 2m D sin γ d(v2) ⇥ ⇥v2 Ac − = g sin γ D 2 d⇥ h − − 2 m s ⇥ d(v2) 2gh h v2 Ac = s + s D d⇥ ⇥ sin γ m U N I V E R S I T Y O F Ballistic Entry MARYLAND 3 ENAE 791 - Launch and Entry Vehicle Design Ballistic Entry (3) m Let β Ballistic Coefficient ≡ cDA ⇥ d(v2) h 2gh s v2 = s d⇤ − β sin ⇥ ⇤ Assume mg D to get homogeneous ODE d(v2) h d(v2) h s v2 = 0 = s d⇤ d⇤ − β sin ⇥ v2 β sin ⇥ Use v2 as integration variable v d⇥(v2) h ρ = s d⇤ v = velocity at entry v2 β sin ⇥ e ve 0 U N I V E R S I T Y O F Ballistic Entry MARYLAND 4 ENAE 791 - Launch and Entry Vehicle Design Ballistic Entry (4) Note that the effect of ignoring gravity is that there is no force perpendicular to velocity vector ⇒ constant flight path angle γ ⇒ straight line trajectories 2 v v hs⇤ ln 2 = 2 ln = ve ve β sin ⇥ v h ⇤ = exp s v 2β sin ⇥ e ⇥ kg v h ⇤ ⇤ m 3 = exp s o Check units: m v 2β sin ⇥ ⇤ kg e o ⇥ m2 ⇥ U N I V E R S I T Y O F Ballistic Entry MARYLAND 5 ENAE 791 - Launch and Entry Vehicle Design Earth Entry, γ=-60° v/ve 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.2 0.4 0.6 0.8 1 ρ/ρ o Beta=100 kg/m^3 300 1000 3000 10000 U N I V E R S I T Y O F Ballistic Entry MARYLAND 6 ENAE 791 - Launch and Entry Vehicle Design What About Peak Deceleration? dv ⇥v2 n = ⇥ dt − 2β d dv d2v To find n , set = = 0 max dt dt dt2 ⇥ d2v 1 dv d⇥ = 2⇥v + v2 = 0 dt2 −2β dt dt ⇥ d2v 1 2⇥2v3 d⇥ = + v2 = 0 dt2 −2β − 2β dt ⇥ ⇥2v3 d⇥ d⇥ = v2 ⇥2v = β β dt dt U N I V E R S I T Y O F Ballistic Entry MARYLAND 7 ENAE 791 - Launch and Entry Vehicle Design Peak Deceleration (2) From exponential atmosphere, dρ ρo h dh ρ dh = e− hs = dt −hs dt −hs dt dh From geometry, = v sin γ dt d⇥ ⇥v d⇥ = sin γ ⇥2v = β dt − hs dt ⇤v ⇤2v = β sin ⇥ − h s ⇥ Remember that this refers to the conditions at max deceleration β ⇤nmax = sin ⇥ −hs U N I V E R S I T Y O F Ballistic Entry MARYLAND 8 ENAE 791 - Launch and Entry Vehicle Design Critical β for Deceleration Before Impact At surface, ρ = ρo ⇤ h β = o s Value of β at which vehicle hits crit −sin ⇥ ground⇐ at point of maximum deceleration How large is maximum deceleration? 2 2 dv ⇥v dv ⇥n v = = max dt 2β ⇒ dt 2β max 2 dv v β 1 v2 = sin ⇥ = sin γ dt 2β −h max ⇥ s ⇤ −2 hs Note that this value of v is actually vnmax U N I V E R S I T Y O F Ballistic Entry MARYLAND 9 ENAE 791 - Launch and Entry Vehicle Design Peak Deceleration (3) From page 14, v h ⇤ = exp s v 2β sin ⇥ e ⇥ vnmax hs β 1 = exp sin ⇥ = e− 2 v 2β sin ⇥ −h e ⇤ s ⇥⌅ 1 2 dv 1 vee− 2 v2 sin γ = sin γ = e dt −2 ⇥ h ⇤ − 2h e max s s Note that the velocity at which maximum deceleration occurs is always a fixed fraction of the entry velocity - it doesn’t depend on ballistic coefficient, flight path angle, or anything else! Also, the magnitude of the maximum deceleration is not a function of ballistic coefficient - it is dependent on the entry trajectory (ve and γ) but not spacecraf parameters (i.e., ballistic coefficient). U N I V E R S I T Y O F Ballistic Entry MARYLAND 10 ENAE 791 - Launch and Entry Vehicle Design Terminal Velocity Full form of ODE - d v2 h 2gh s v2 = s d⇤ ⇥ − β sin ⇥ ⇤ At terminal velocity, v = constant v ≡ T h 2gh s v2 = s −β sin ⇥ T ⇤ 2 2gβ sin ⇥ vT = − ⇤ U N I V E R S I T Y O F Ballistic Entry MARYLAND 11 ENAE 791 - Launch and Entry Vehicle Design “Cannon Ball” γ=-90° Ballistic Entry 6.75” diameter sphere, cD=0.2, VE=6000 m/sec Iron Aluminum Balsa Wood Weight 40 lb 15.6 lb 14.5 oz β 3938 1532 89 ρ 0.555 0.216 0.0125 h 5600 12,300 32,500 V 1998 355 0* V 251 156 38 *Artifact of assumption that D mg U N I V E R S I T Y O F Ballistic Entry MARYLAND 12 ENAE 791 - Launch and Entry Vehicle Design Nondimensional Ballistic Coefficient v h ⇤ ⇤ P ⇢ = exp s o =exp o v 2β sin ⇥ ⇤ 2βg sin γ ⇢ e o ⇥ ✓ o ◆ β βg Let β = (Nondimensional form of ballistic coefficient) ⌘ ⇢ohs Po Note that we are using the estimated value of P = ⇢ gh , b o o s not the actual surface pressure. v 1 ⇤ = exp v ⇤ e 2β sin ⇥ o ⇥ ⇤ohs ⇤ 1 β = βcrit = crit −sin ⇥ −sin ⇥ U N I V E R S I T Y O F Ballistic Entry MARYLAND 13 ENAE 791 - Launch and Entry Vehicle Design Entry Velocity Trends, γ=-90° Velocity Ratio 0 0.2 0.4 0.6 0.8 1 0 0.1 0.2 0.3 0.4 0.5 0.6 Density Ratio 0.7 0.8 0.9 1 β 0.03 0.1 0.3 1 3 U N I V E R S I T Y O F Ballistic Entry MARYLAND 14 ENAE 791 - Launch and Entry Vehicle Design Ballistic Entry, Again s = distance along the flight path v, s dv D γ = g sin γ dt − − m D horizontal Again assuming D g, mg dv D 1 2 = Drag D ρv AcD dt −m ≡ 2 dv ρc A = D v2 dt − 2m Separating the variables, dv ⇥ = dt v2 −2β U N I V E R S I T Y O F Ballistic Entry MARYLAND 15 ENAE 791 - Launch and Entry Vehicle Design Calculating the Entry Velocity Profile dh dh = v sin γ dt = dt ⇒ v sin γ dv ⇤ dv ⇤ = dh = dh v2 −2βv sin ⇥ ⇥ v −2β sin ⇥ dv ⇤o h = e− hs dh v −2β sin ⇥ v h dv ⇤o h = e− hs dh v −2β sin ⇥ ve he h v ⇤ohs h 1 h he ln = e− hs = e− hs e− hs ve 2β sin ⇥ he 2β sin ⇥ − ⇥ ⇥ ⇤ U N I V E R S I T Y O F Ballistic Entry MARYLAND 16 ENAE 791 - Launch and Entry Vehicle Design Deriving the Entry Velocity Function he ρe Remember that e− hs = 0 ρo ≈ v 1 h = exp e− hs v e 2β sin ⇥ ⇥ We have a parametric entry equation in terms of nondimensional velocity ⇤ ratios, ballistic coefficient, and altitude. To bound the nondimensional altitude variable between 0 and 1, rewrite as v 1 h he = exp e− he hs v e 2β sin ⇥ ⇥ he and β are the⇤ only variables that relate to a specific planet hs U N I V E R S I T Y O F Ballistic Entry MARYLAND 17 ENAE 791 - Launch and Entry Vehicle Design Earth Entry, γ=-90° 1 0.9 0.8 0.7 0.6 0.5 0.4 Altitude Ratio 0.3 0.2 0.1 0 0 0.2 0.4 0.6 0.8 1 Velocity Ratio β 0.03 0.1 0.3 1 3 U N I V E R S I T Y O F Ballistic Entry MARYLAND 18 ENAE 791 - Launch and Entry Vehicle Design Deceleration as a Function of Altitude Start with v 1 h he 1 = exp e− he hs Let B v ≡ e 2β sin ⇥ ⇥ 2β sin ⇥ v h = exp B⇤e− hs ve ⇥ d v h d h = exp Be− hs Be− hs dt ve dt ⇤ ⌅ ⇥ ⇥ dv h B h dh = ve exp Be− hs − e− hs dt hs dt ⇥ ⇥ dh h = v sin γ = v sin γ exp Be− hs dt e U N I V E R S I T Y O F Ballistic Entry MARYLAND 19 ENAE 791 - Launch and Entry Vehicle Design Parametric Deceleration dv h B h h = ve exp Be− hs − e− hs ve sin γ exp Be− hs dt hs ⇥ ⇥ ⇥ 2 dv Bve h h = − sin γ e− hs exp 2Be− hs dt hs ⇥ ⇥ 2 dv ve h 1 h = − e− hs exp e− hs dt 2h β β sin ⇥ s ⇥ ⇤ ⌅ 2 ve dv/dt he ⇧ Let nref ⇧, ν , ⇥ ≡ hs ≡ nref ≡ hs 1 ϕ h 1 ϕ h ⇤ = − e− he exp e− he 2β β sin ⇥ ⇥ ⇤ ⌅ U N I V⇧ E R S I T Y O F ⇧ Ballistic Entry MARYLAND 20 ENAE 791 - Launch and Entry Vehicle Design Nondimensional Deceleration, γ=-90° 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 Altitude Ratio h/he Altitude Ratio 0.2 0.1 0 0 0.05 0.1 0.15 0.2 Deceleratio ν β 0.03 0.1 0.3 1 3 U N I V E R S I T Y O F Ballistic Entry MARYLAND 21 ENAE 791 - Launch and Entry Vehicle Design Deceleration Equations Nondimensional Form 1 ϕ h 1 ϕ h ⇤ = − e− he exp e− he 2β β sin ⇥ ⇥ ⇤ ⌅ Dimensional Form ⇧ 2 ⇧ ⇤oVe h ⇤ohs h n = − e− hs exp e− hs 2β β sin ⇥ ⇥ ⇤ ⌅ Note that these equations result in values <0 (reflecting deceleration) - graphs are absolute values of deceleration for clarity.