ballistic atmospheric entry · let m c da ⇥ ballistic coecient d(v2) d⇤ ... let n ref v2 e h s,...

Ballistic Entry ENAE 791 - Launch and Entry Vehicle Design

U N I V E R S I T Y O FMARYLAND

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

1

© 2016 David L. Akin - All rights reserved http://spacecraft.ssl.umd.edu

http://spacecraft.ssl.umd.edu



Ballistic Entry (no lift)s = distance along the flight path

dv

dt= �g sin � � D

m

dv

dt=

dv

ds

ds

dt= V

dv

ds=

12

d(v2)ds

12

d(v2)ds

= �g sin � � D

mDrag D � 1

2�v2AcD

12

d(v2)ds

= �g sin � � ⇥v2

2mAcD

sin �

2d(v2)dh

= �g sin � � ⇥v2

2mAcD

ds =dh

sin �

Dmg

horizontal

v, s

�

ds dh�

2



Ballistic Entry (2)

d�

�o= e�

hhs

��dh

hs

⇥

dh =�hs

�d�

=�oe

� hhs

�o

��dh

hs

⇥=

�

�o

��dh

hs

⇥

sin �

2d(v2)dh

= �g sin � � ⇥v2

2mAcD

Exponential atmosphere � � = �oe� h

hs

sin �

2d(v2)d⇥

��⇥

hs

⇥= �g sin � � ⇥v2

2AcD

m

d(v2)d⇥

=2ghs

⇥+

hsv2

sin �

AcD

m

3



Ballistic Entry (3)

Let � � m

cDA⇥ Ballistic Coe�cient

d(v2)d⇤

� hs

� sin ⇥v2 =

2ghs

⇤

Assume mg � D to get homogeneous ODE

d(v2)d⇤

� hs

� sin ⇥v2 = 0

d(v2)v2

=hs

� sin ⇥d⇤

Use�v2

⇥as integration variable

� v

ve

d(v2)v2

=hs

� sin ⇥

� �

0d⇤ ve = velocity at entry

4



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

lnv2

v2e

= 2 lnv

ve=

hs⇤

� sin ⇥

v

ve= exp

�hs⇤

2� sin ⇥

⇥

v

ve= exp

�hs⇤o

2� sin ⇥

⇤

⇤o

⇥Check units:

�m kg

m3

kgm2

⇥

5



0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

Beta=100 kg/m^3 300 1000 3000 10000

Earth Entry, γ=-60°

�/�o

v/ve

6



What About Peak Deceleration?

n ⇥ dv

dt= �⇥v2

2�

To find nmax, setd

dt

�dv

dt

⇥=

d2v

dt2= 0

d2v

dt2= � 1

2�

�2⇥v

dv

dt+ v2 d⇥

dt

⇥= 0

d2v

dt2= � 1

2�

��2⇥2v3

2�+ v2 d⇥

dt

⇥= 0

⇥2v3

�= v2 d⇥

dt⇥2v = �

d⇥

dt

7



Peak Deceleration (2)

⇥2v = �d⇥

dt

From exponential atmosphere,d�

dt= ��o

hse�

hhs

dh

dt= � �

hs

dh

dt

d⇥

dt= �⇥v

hssin �

⇤2v = �

��⇤v

hssin ⇥

⇥

Remember that this refers to the conditions at max deceleration

⇤nmax = � �

hssin ⇥

8

From geometry,dh

dt= v sin �



Critical β for Deceleration Before Impact

�crit = �⇤ohs

sin ⇥

At surface, � = �o

� Value of � at which vehicle hitsground at point of maximum deceleration

How large is maximum deceleration?

dv

dt=

⇥v2

2��

��dv

dt

��max

=⇥nmaxv2

2�

��dv

dt

��max

=v2

2�

⇥� �

hssin ⇥

⇤= �1

2v2

hssin �

Note that this value of v is actually vnmax

9



Peak Deceleration (3)

10

vnmax

ve= exp

⇤hs

2� sin ⇥

��

hssin ⇥

⇥⌅= e�

12

v

ve= exp

�hs⇤

2� sin ⇥

⇥

��dv

dt

��max

= �12

⇥vee�

12

⇤2

hssin � = �v2

e sin �

2hse

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 spacecraft parameters (i.e., ballistic coefficient).

From page 14,



Terminal Velocity

11

Full form of ODE -d

�v2

⇥

d⇤� hs

� sin ⇥v2 =

2ghs

⇤

At terminal velocity, v = constant � vT

� hs

� sin ⇥v2

T =2ghs

⇤

v2T =

�

�2g� sin ⇥

⇤



“Cannon Ball” γ=-90° Ballistic Entry6.75” 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

12

*Artifact of assumption that D � mg



Nondimensional Ballistic Coefficient

13

v

ve= exp

�hs⇤o

2� sin ⇥

⇤

⇤o

⇥

�crit = �⇤ohs

sin ⇥

v

ve= exp

�1

2⇤� sin ⇥

⇤

⇤o

⇥

��crit = � 1sin ⇥

not the actual surface pressure.

= exp

✓Po

2�g sin �

⇢

⇢o

◆

Note that we are using the estimated value of Po

= ⇢o

ghs

,

Let

b� ⌘ �

⇢o

hs

=

�g

Po

(Nondimensional form of ballistic coe�cient)



Entry Velocity Trends, γ=-90°

14

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

Velocity Ratio

Den

sity

Rati

o

0.03 0.1 0.3 1 3��



Ballistic Entry, Agains = distance along the flight path

dv

dt= �g sin � � D

m

Drag D � 12�v2AcD

Dmg

horizontal

v, s

�

15

Again assuming D � g,

dv

dt= �D

mdv

dt= ��cDA

2mv2

dv

v2= � ⇥

2�dt

Separating the variables,



Calculating the Entry Velocity Profile

16

� dt =dh

v sin �

dv

v2= � ⇤

2�v sin ⇥dh ⇥ dv

v= � ⇤

2� sin ⇥dh

dv

v= � ⇤o

2� sin ⇥e�

hhs dh

� v

ve

dv

v= � ⇤o

2� sin ⇥

� h

he

e�h

hs dh

lnv

ve=

⇤ohs

2� sin ⇥

�e�

hhs

⇥h

he

=1

2�� sin ⇥

⇥e�

hhs � e�

hehs

⇤

dh

dt= v sin �



Deriving the Entry Velocity Function

17

Remember that e�hehs =

�e

�o� 0

v

ve= exp

�1

2⇤� sin ⇥e�

hhs

⇥

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

ve= exp

�1

2⇤� sin ⇥e�

hhe

hehs

⇥

he

hsand �� are the only variables that relate to a specific planet



Earth Entry, γ=-90°

18

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

Velocity Ratio

Alt

itu

de R

ati

o

0.03 0.1 0.3 1 3��



Deceleration as a Function of Altitude

19

v

ve= exp

�1

2⇤� sin ⇥e�

hhe

hehs

⇥Start with

v

ve= exp

�Be�

hhs

⇥

Let B � 12�� sin ⇥

d

dt

⇤v

ve

⌅= exp

�Be�

hhs

⇥ d

dt

�Be�

hhs

⇥

dv

dt= ve exp

�Be�

hhs

⇥ �B

hs

�e�

hhs

⇥ dh

dt

dh

dt= v sin � = ve sin � expBe�

hhs



Parametric Deceleration

20

dv

dt= ve exp

�Be�

hhs

⇥ �B

hs

�e�

hhs

⇥ve sin � exp

�Be�

hhs

⇥

dv

dt=�Bv2

e

hssin �

�e�

hhs

⇥exp

�2Be�

hhs

⇥

dv

dt=�v2

e

2hs⇧�

�e�

hhs

⇥exp

⇤1

⇧� sin ⇥e�

hhs

⌅

Let nref �v2

e

hs, � � dv/dt

nref, ⇥ � he

hs

⇤ =�12⇧�

�e�� h

he

⇥exp

⇤1

⇧� sin ⇥e�� h

he

⌅



Nondimensional Deceleration, γ=-90°

21

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

Deceleratio

Alt

itu

de R

ati

o h

/h

e

0.03 0.1 0.3 1 3

��



Deceleration Equations

22

⇤ =�12⇧�

�e�� h

he

⇥exp

⇤1

⇧� sin ⇥e�� h

he

⌅

n =�⇤oV 2

e

2�

�e�

hhs

⇥exp

⇤⇤ohs

� sin ⇥e�

hhs

⌅

Nondimensional Form

Dimensional Form

Note that these equations result in values <0 (reflecting deceleration) - graphs are absolute values of deceleration for clarity.



Dimensional Deceleration, γ=-90°

23

0

20

40

60

80

100

120

0 500 1000 1500 2000

Deceleration

Alt

itu

de

277 922 2767 9224 27673

(km

)

� m

sec2

⇥

�

�kg

m2

⇥



Altitude of Maximum Deceleration

24

⇥ = �B sin ��e�

hhs

⇥exp

�2Be�

hhs

⇥Returning to shorthand notation for deceleration

⇥ = �B sin ��e��

⇥exp

�2Be��

⇥Let � � h

hs

d⇤

d⇥= �B sin �

⇤d

d⇥

�e��

⇥exp

�2Be��

⇥+

�e��

⇥ d

d⇥exp

�2Be��

⇥⌅

d⇤

d⇥= �B sin �

⇤�

�e��

⇥exp

�2Be��

⇥+

�e��

⇥ ��2Be��

⇥exp

�2Be��

⇥⌅

d⇤

d⇥= B sin �e�� exp

�2Be��

⇥ ⇤1 +

�2Be��

⇥⌅= 0




25

1 +�2Be��

⇥= 0 ⇥ e� = �2B

�nmax = ln (�2B)

⇤nmax = ln�

�1⇤� sin ⇥

⇥

hnmax = hs ln��⇤ohs

� sin ⇥

⇥

Altitude of maximum deceleration is independent of entry velocity!

Converting from parametric to dimensional form gives




26

0

2

4

6

8

10

12

0 0.2 0.4 0.6 0.8 1

Ballistic Coefficient

Alt

itu

de o

f M

ax D

ece

l h

/h

s

-15 -30 -45 -60 -90

��



Magnitude of Maximum Deceleration

27

and insert the value of � at the point of maximum deceleration

⇤nmax = ln�

�1⇤� sin ⇥

⇥⇥ e�� = �� sin ⇥

⇤nmax =�12⇧�

��⇧� sin ⇥

⇥exp

⇤�⇧� sin ⇥⇧� sin ⇥

⌅� ⇥nmax =

sin �

2e

Start with the equation for acceleration -

⇤ =�12⇤�

e�� exp�

1⇤� sin ⇥

e��

⇥

nmax =v2

e

hs

sin �

2e

Maximum deceleration is not a function of ballistic coe�cient!



Peak Ballistic Deceleration for Earth Entry

28

0

1000

2000

3000

4000

5000

6000

0 5 10 15

Entry Velocity (km/sec)

Peak D

ece

lera

tio

n (

m/

sec^

2)

-15 -30 -45 -60 -90�



Velocity at Maximum Deceleration

29

Start with the equation for velocity

and insert the value of � at the point of maximum deceleration

⇤nmax = ln�

�1⇤� sin ⇥

⇥⇥ e�� = �� sin ⇥

v

ve= exp

�1

2⇤� sin ⇥e��

⇥

v

ve= exp

��⇤� sin ⇥

2⇤� sin ⇥

⇥⇥ vnmax =

ve⇤e�= 0.606ve

Velocity at maximum deceleration is independent of everything except ve



Planetary Entry - Physical Data

30

Radius (km)

(km

(kg/m

h(km)

v(km/sec)

Earth 6378 398,604 1.225 7.524 11.18

Mars 3393 42,840 0.0993 27.7 5.025

Venus 6052 325,600 16.02 6.227 10.37

µ �o



Comparison of Planetary Atmospheres

31

1E-20

1E-18

1E-16

1E-14

1E-12

1E-10

1E-08

1E-06

1E-04

0.01

1

100

0 50 100 150 200 250 300

Altitude (km)

Atm

osp

heri

c D

en

sity

(kg

/m

3)

Earth Mars Venus



Planetary Entry Profiles

32

V

Ve

� = �15o

ve = 10km

sec

� = 300kg

m2



Planetary Entry Deceleration Comparison

33

� = �15o

ve = 10km

sec

� = 300kg

m2



Check on Approximation Formulas

34

� = �15o

ve = 10km

sec

� = 300kg

m2

ballistic atmospheric entry · let m c da ⇥ ballistic coecient d(v2) d⇤ ... let n ref v2 e h s,...

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