Ballistic Atmospheric Entry

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2 Atmospheric Density with Altitude Ref: V. L. Pisacane and R. C. Moore, Fundamentals of Space Systems Oxford University Press,

3 Energy Loss Due to Atmospheric Drag Drag D 1 2 ρv2 Ac D Drag acceleration a d = D m = ρv2 2 a d = ρv2 2β orbital energy E = µ 2a de dt = µ da 2a 2 dt 3 β m c D A Ac D m <== Ballistic Coefficient

4 Energy Loss Due to Atmospheric Drag Since drag is highest at perigee, the first effect of atmospheric drag is to circularize the orbit (high perigee drag lowers apogee) de drag dt = a d v v 2 circ = µ a de drag dt = ρv2 2β µ a de drag dt µ = a ρ 2β µ a = µ a 3 2 ρ 2β 4

5 Derivation of Orbital Decay Due to Drag Set orbital energy variation equal to energy lost by drag ρ = ρ o e h hs dh dt = µ da 2a 2 dt = ρ 2β da dt = ρ β µ a 3 2 µa a = h + r E = da dt = dh dt µ (h + re ) β ρ o e h hs 5

6 Derivation of Orbital Decay (2) is is a separable differential equation... h 1 re + h e h hs dh = 1 h o re + h e h hs dh = µ β ρ odt µ β ρ o t t o dt Assume r E + h r E for r E h 1 h e h µ hs dh = re h β ρ o (t t o ) o 6

7 Derivation of Orbital Decay (3) h s e h hs re e h hs e h o µre hs = h s β ρ o (t t o ) h(t) = h s ln e h o µre hs h s β ρ o (t t o ) Note that some variables typically use km, and others are in meters - you have to make sure unit conversions are done properly to make this work out correctly! e h o hs 7 = µ β ρ o (t t o )

8 Orbit Decay from Atmospheric Drag Altitude (km) β=500 β=1500 β= Time (sec) 8

9 Time Until Orbital Decay e h hs e h o hs = µre h s β ρ o (t t o ) To find the time remaining (t o =0) until the orbit reaches any given critical altitude, some algebra gives t(h) = h sβ e h o hs µre ρ o e h hs 9

10 Decay Time to r=120 km Altitude (km) β=500 β=1500 β= Decay Time (yrs) 10

11 Ballistic Entry (no lift) s = distance along the flight path dv dt = g D m dv dt = dv ds d(v 2 ) ds d(v 2 ) ds 2 d(v 2 ) dh ds dt = V dv ds = 1 2 = g D m d(v 2 ) ds = g ρv2 2m Ac D ρv2 = g 2m Ac D D Drag D 1 2 ρv2 Ac D ds γ mg dh ds = γ horizontal dh v, s 11

12 Ballistic Entry (2) Exponential atmosphere ρ = ρ o e h hs dρ ρ o = e h hs dh h s = ρ oe h hs ρ o dh h s = ρ ρ o dh h s 2 2 d(v 2 ) dρ dh = h s ρ dρ d(v 2 ) ρv2 = g dh 2m Ac D ρ = g ρv2 2 h s Ac D m d(v 2 ) dρ = 2gh s ρ 12 + h sv 2 Ac D m

13 Ballistic Entry (3) Let β m c D A Ballistic Coefficient d(v 2 ) dρ h s β v2 = 2gh s ρ Assume mg D to get homogeneous ODE d(v 2 ) dρ h s β v2 = 0 Use v 2 as integration variable d(v 2 ) v 2 = h s β dρ v v e d(v 2 ) v 2 = h s β ρ 0 dρ v e = velocity at entry 13

14 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 ln v2 v 2 e v hs ρ o = exp v e 2β = 2 ln v v e = h sρ β v hs ρ = exp v e 2β ρ ρ o Check units: m kg m 3 kg m 2 14

15 Earth Entry, γ=-60 v/v e ρ/ρ o Beta=100 kg/m^

16 What About Peak Deceleration? To find n max, set d dt d 2 v dt 2 = 1 2β d 2 v dt 2 = 1 2β n dv dt = ρv2 2β dv = d2 v dt dt 2 = 0 2ρv dv dt 2ρ2 v 3 2β + v2 dρ dt dρ + v2 dt = 0 = 0 ρ 2 v 3 β dρ = v2 dt ρ 2 v = β dρ dt 16

17 Peak Deceleration (2) From exponential atmosphere, dρ dt = ρ o h s e h hs From geometry, dh dt = v dρ dt = ρv h s ρ 2 v = β dh dt = ρ dh h s dt ρv h s ρ 2 v = β dρ dt Remember that this refers to the conditions at max deceleration ρ nmax = β h s 17

18 Critical β for Deceleration Before Impact At surface, ρ = ρ o β crit = ρ oh s Value of β at which vehicle hits ground at point of maximum deceleration How large is maximum deceleration? dv dt = ρv2 dv 2β dt = ρ n max v 2 max 2β dv dt = v2 β = 1 v 2 max 2β h s 2 h s Note that this value of v is actually v nmax 18

19 Peak Deceleration (3) From page 14, v nmax v e dv dt = 1 max 2 v hs ρ = exp v e 2β h s = exp 2β v e e β h s 2 = e 1 2 = v2 e h s 2h s e 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 (v e and γ) but not spacecra parameters (i.e., ballistic coefficient).

20 Terminal Velocity Full form of ODE - d v 2 dρ h s β v2 = 2gh s ρ At terminal velocity, v = constant v T h s β v2 T = 2gh s ρ v 2 T = 2gβ ρ 20

21 Cannon Ball γ=-90 Ballistic Entry 6.75 diameter sphere, c D =0.2, V E =6000 m/sec Iron Aluminum Balsa Wood Weight 40 lb 15.6 lb 14.5 oz βmd (kg/m 2 ) ρmd (kg/m 3 ) hmd (m) ,300 32,500 Vimpact (m/s) * Vterm (m/sec) *Artifact of assumption that D mg 21

22 Atmospheric Density with Altitude Pressure=the integral of the atmospheric density in the column above the reference area ρ = f(h) P o = o ρgdh = ρ o g o e h hs dh = ρ o gh s e h hs = ρ o gh s [0 1] o Earth: ρ o = kg m 3 ; h s = 7524m; P o = ρ o gh s P o (calc) = 90, 400 Pa; P o (act) = 101, 300 Pa ρ o, P o 22

23 Nondimensional Ballistic Coefficient v hs ρ o = exp v e 2β Let β βg P o ρ ρ o Po =exp 2βg (Nondimensional form of ballistic coefficient) Note that we are using the estimated value of P o = ρ o h s, not the actual surface pressure. v 1 ρ = exp v e 2β ρ o ρ ρ o β crit = ρ oh s β crit = 1 23

24 Entry Velocity Trends, γ=-90 Density Ratio Velocity Ratio β

Rocket Performance MARYLAND U N I V E R S I T Y O F. Ballistic Entry ENAE Launch and Entry Vehicle Design

Rocket Performance MARYLAND U N I V E R S I T Y O F. Ballistic Entry ENAE Launch and Entry Vehicle Design Rocket Performance Parallel staging Modular staging Standard atmospheres Orbital decay due to drag Straight-line (no gravity) entry based on atmospheric density 1 2014 David L. Akin - All rights reserved