dynamics chapter 5

7/24/2019 Dynamics Chapter 5

1/18

84

Chapter 5. REENTRY DYNAMICS

5.1 Motion with Velocity Dependent Forces (VDF)

Spacecraft and satellite have tight mass budget. By the time they

need to land, very little fuel is left for active manouvre (such as

backward jet propulsion as shown at right). Atmospheric drag

provides an important means of achieving the brake from a speed of

around 8 km/s in orbit, or 10 km/s from the moon or other planets.

We shall start with a simplistic case of a sphere falling in a uniform

fluid. Drag on a sphere depends on velocity in a complex manner.When fluid creeps, the flow is well organized layer-by-layer (laminar

flow) and the drag is proportional to speed. At higher speed, the flow

loses stability and becomes turbulent (see Fluid Mechanics in years

2/3). The drag is then proportional to the square of velocity, V2 (the

coefficient in this case is one of the best guarded secrets of nature)

The governing equation of motion for a falling object is

mdV/dt = mg Drag Eq (1)

When drag balances the weight, mg=drag, there is no more

acceleration, and the object is said to have reached a

terminal velocity of falling, VT.
http://commons.wikimedia.org/wiki/Image:Vortex-street-animation.gifhttp://commons.wikimedia.org/wiki/Image:Vortex-street-animation.gif


2/18

85

5.1 Motion with VDF, Falling Sphere

Von-Karm

on

vortexstreet

2

3 laminar

Drag= , 3 for a perfect sphere of diameter

At terminal veloity /

1 Eq 2

6 18

f f

f T T f

sphere

sphere T

C V C d d

C V mg V mg C

gdm d V

For laminar flow

Example: calculate the terminal velocity of a spherical rain-drop of 0.2 mm in diameter

Ans: With =1.5x10-5kg/m/s, Eq (2) gives 1.45 m/s

Note that, for large raindrops, such as 2 mm in diameter, drag

is proportional to V 2 and the above result is not suitable

The full equation is a standard, inhomogeneous, first-order, ordinary

differential equation (ODE)

/ /f fdV

m C V mg dV dt C m V g dt


3/18

86

5.1 Motion with VDF, general laminar/turbulent solution

/ / / /

00 00

For ODE of the following standard form

Hence

Denote , . Hence

/ con

1 1 ex

stant

p

f f f TC t m C t m C t m gt VT

f

T TtT T

p

T

t pt

mgV t e ge dt C Ce V CeC

V t V gtV V V C C V V

V V

dy dt py q y t e qe d

V

t

Eq 3

0 1 2 3 4 50

0.5

1

1.5

2

Laminar

Laminar

Turbulent

Turbulent

Time gt/VT

V/V

T

General results forlaminar and turbulent

flow (next page) are

given here for initial

V0=0 and V0=2VT.

Turbulent flow reachesterminal velocity more

quickly than laminar

Most drastic

acceleration happens at

t=0: velocity and drag

are highest.


4/18

87

5.1 Motion with VDF, Turbulent flow and Parachute

Theparachute areaA is designed in

such a way that the terminal velocity

is what the skydiver can tolerate,

normally VT =(3~5) m/s.

V

2

2turbulent

2

0

0

Turbulent flow drag= , where 0,2 is dimensionless and is frontal area.

Terminal v from drag: and 1 Eq 5

1 1Soln ln

Eq 4

ln2 2

D

T

D T

TT

T T T

DC V A A

mg dV V

mg V g C A dt V

V VV Vgt

V V V V V

C

2 /

0

2 /

0

1Eq 6 and , Eq 7

1

T

T

gt V

T

gt V

T T

V VV CeC

V Ce V V

Example: a spacecraft of m=5000 kg, CD=1, =1 kg/m

3,

frontal areaA=20 m2. Eq (4)

gives VT=50m/s. Reduction

to 5m/s requires 100 times

increase in frontal area A !


5/18

88

5.1 Motion with VDF, Example 1

Question (B&F Q13.49) A sky diver jumps from a helicopter and is falling straight down at

30 m/s when her parachute opens. From then on, her downward acceleration is

approximated by g-cv2 where c is a constant. After an initial transient period of time,

she descends at a nearly constant velocity of 5m/s.

(a) How much is the constant c?

(b) What maximum acceleration or deceleration does she experience?

(c) What is her downward speed when she has fallen 2m from the point when the parachute

opens?

Solution: known VT= 5m/s, V0=30 m/s,s=2m

0

0

2 2 -1

2 2 2

max 0

2 2 2 0

2 22 2

2 2

0

a : 0 at terminal velocity , / 0.3924m

b : 9.81 0.3924 30 343.35 m/s or 35

2

c : , , 2

ln 2 ,

T T

v s

vT

v vT

Tv v

T

a g cv V c g V

a g cv g

vdv vdv

vdv ads ds cdsg cv v v

v vv v cs

v v

2 2 2 2 2

0

2 2 2 2 0.3924 2

,

5 30 5 14.4 m/s

cs cs

T Te v v v v e

v e

In reality, parachute opens gradually andmax deceleration is not as high as 35g


6/18

89

5.1 Motion with VDF, Example 2

Given that the drag coefficient for small particulates is 3Vd, where =1.5x10-5

kgm-1s-1 is the air viscosity, and d is the diameter, estimate the time it takes for a

water droplets of 1m and 10m in diameter to fall by 1m.

Solution: terminal velocity given by Eq. (2) , VT=spheregd2/(18),

Drag

mg

2

3 6

5

5

5

3

10 9.81 101m, 3.6333 10 m/s

18 1.5 101

27523 sec= 7 hrs 39 min3.6333 10

10m, 3.6333 10 m/s, 275 sec=4m35s

T

T

T

d V

st

V

d V t

Comments: when a person with SARS virus (typical size


7/18

90

If a spacecraft has an estimated terminal velocity of 50m/s, it is too fast to

handle by increasing frontal area (by 100 times, parachuting), but it is not a huge

speed compared with its initial outer space speed of 8km/s. What is the biggest

technical challenge for re-entry ?

The answer is that the time to slow down from 8km/s to 50m/s is too short due

to the huge drag proportional to V2

. To estimate the time, we set VT=50m/s,V0=8000m/s=160VT, and ask how much time it takes to slow down from V0 to

2VT=100m/s (since theoretically it takes forever to slow down to VT).

00

0

/ 1/ 11Eq 7 : ln ln

2 / 1 / 1

1 2 1 160 1 1ln ln ln3

2 2 1 160 1 2

0.55, 0.55 2.8 sec

Too short a time for braking, will smash like a comet

TT

T T T

T

T

T

V VV Vgt

V V V V V

gt

V

Vgtt

V g


5.2 Ballistic Estimate and Atmospheric Model

Initial braking by

CDAV02 is too fast.

It takes no time!
http://www.space.com/php/multimedia/imagedisplay/img_display.php?pic=ig113_ikeya4_02.jpg&cap=Ikeya-Zhang%27s%20head,%20or%20coma%20as%20it%20is%20called,%20lights%20up%20the%20surrounding%20sky.%20Stars%20appear%20as%20multiple%20dots%20due%20to%20multiple%20exposures.%20%20Click%20to%20enlarge.http://www.space.com/php/multimedia/imagedisplay/img_display.php?pic=ig113_ikeya4_02.jpg&cap=Ikeya-Zhang%27s%20head,%20or%20coma%20as%20it%20is%20called,%20lights%20up%20the%20surrounding%20sky.%20Stars%20appear%20as%20multiple%20dots%20due%20to%20multiple%20exposures.%20%20Click%20to%20enlarge.


8/18

91

5.2 Ballistic Estimate and Atmospheric Model

Luckily, the earth atmosphere is not uniform and the initial braking at high sky by the drag

force, CDAV02, is not calculated by the density of air on the earth surface. Air rarefies

with height in an exponential function of y (y=0 being the ground),

300 1.225 kg/m , 7.25 km for the ear exp / tEq 8 hy HH

This is a gross approximation of the atmosphere which

has a complex layer structure (of temperatureseerightwith initial lapse rate of 6.5 deg/km), and

differentHvalue should be used for different

atmospheric layers. Nevertheless, the main feature of

the exponential distribution suffices for our analysis of

spacecraft re-entry!

We are now interested in establishing a more credible

model of satellite braking by the exponential

atmosphere. During most of the drastic slowing down,

gravity and centrifugal forces may be ignored, and the

deceleration by drag is studied in a rectilinear fashion,

shown below for an oblique spacecraft re-entry


9/18

92

Localhorizon

V

mg

mVq2/r

Earth

/

0

/0

2

, and cos , cos

hence / cos

Eq 9cos

H

D

y HD

y

dv dv dy dy dv dvv v

dt dy dt dt dt dy

m dv dy C A e

C Adv e dyv m

v v

y y

y

y

yThis equation can be integrated and the

constant determined by the initial

condition

/2 2

small

0/ cos , where , andy H

D

dvm Drag m g v r Drag

teC v A

d q y

Consider a re-entry with an angle of from the vertical free fall

0, i it y yv v


5.3 Maximum Brake, Oblique Entry

When braking by air resistance really occurs with non-trivial density , both

gravity and centrifugal force (seen from rotating earth) are negligible. Change

the independent variable from time t to vertical coordinatey


10/18

93

Local horizon

V y

Earth

//0

/ 0

/0exp

ln ,cos

if , 0, and ln expcos

co

Eqs

10

i

i

y Hy HD

i

y H Di

i

i

y HD

C AHve e

v m

C AHv yy H e

v m H

C AHev v

m

y

y

y

y

Notice the double exponential function! Time is found by dy/dt= -vcosy,

Deceleration is [-dv/dt]=Drag/m:

20 Ep q 12exDC Adv Drag

dt m

y

Hmv

As y, velocity decreases, v2, but the density factor, exp(-y/H). These two

competing factors lead to a critical point of maximum deceleration as found

analytically by zero derivative: da/dy=0.

Eq 111

cos

iy

y

dyt y

v yy

5.3 Maximum Brake, Competing Factors


11/18

94

y y

2 2

0 0

2

max 0

max

Let , 0 2 0

Obtain from Eq 9 : 1 2 0cos cos

2 cos

ln ,Eq 1 and3 Eqcos 42 1

y y y

y yD D

a D i

y da dy e v e v e vH dy dy

C AH C AH dvv e e

dy m

dv

dy

d

m

y C AH v

aH m e

y

H

v

d

y y

y

y

Deceleration has a peak value due to the two competing factors:

(1) Normally, an exponentially decelerating motion has the strongest braking at

thebeginning, yi(2) In this case, the braking force strengthens gradually as there is little airat

very high altitudes (scaled byH )

The critical altitude and the peak deceleration are found as follows

Normally, the maximum deceleration occurs at the start when both drag and

velocity are highest. In this case, it depends mainly on the initial velocity, angle

of descend, and the atmospheric scale height (H). The max deceleration does

NOT depend on spacecraft details (A, CD, m).

5.3 Maximum Brake, Peak Braking Point


12/18

95

Earth

Both astronauts and mechanical payload are fragile. A peak deceleration has to

be specified, say 10g for astronauts. For a satellite with vi=8km/s, the entrance

angle y is found from Eq (14)

3

1 1

223

2 10 9.81 7.25 102cos cos 86.5363

8 10

10

i

ee Hg

vy

If Y90o, the spacecraft would simply skip out of

the atmosphere without much of a second chance of

attempt (as there is little or no fuel), the range of

angle that can be used for safe return is called the re-

entry corridor. This corridor would be narrowed to1.146o if the maximum brake is limited to 3g.

The corridor may be extended if lifting device is

installed for steering, like the American space shuttle,

which was found to be costly and not entirely safe.

5.3 Maximum Brake, Safe Entry Corridor


13/18

96

The following graphs are based on the data of Apollo command module:

Mass m=5000 kg, CD=1, frontal areaA=20 m2, initial vi=10 km/s

The free fall (y=0) and the 45deg ballistic entry both experience huge deceleration

which the passenger cannot bear. The result for =88o seems to be acceptable with

maximum deceleration below 10g.

5.3 Maximum Brake, Why Oblique?

The time it takes to brake from

initial vi=10 km/s to 5km/s is

found by Eq (11): simple

vertical free fall 7.6 sec; 88deg

entry 147.1sec.

The avoidance of thecrushing brake is attributed

to the thin air in the upper

atmosphere, together with

the near-grazing entry angle.

97


14/18

97

Mach>>1

Despite the slower brake by oblique entry, the amount of energy to be dissipated is

still overwhelming (well, consider how much fuel was burnt to get the satellite up

there in the first place!). The kinetic energy per unit mass to be dissipated isV2/2=80002/2 J/kg =32 MJ/kg. When compared with the heat required to melt

aluminium, 0.4 MJ/kg, we get 32/0.4=80: the amount of energy available can melt

the capsule 80 times over!

In reality, most energy is not absorbed by the deorbiting object. First of all, the

supersonic flow around a blunt object creates a bow-shock in the front. Much ofkinetic energy is converted to heat in the thin layer of shock.

Shock wave

helps insulating

the spacecraft

5.3 Maximum Brake, Shocking Hot

Mach= Velocity / (speed of sound) >>1

Behind the (oblique) shock, the flow is still quite

fast. A super-insulating material is needed to

prevent heat penetration, which also includes theradiation from the shock!

Luckily, the crucial insulating material was

associated with the non-stick frying pan that

existed in France before the Apollo project!

98


15/18

98

Local

horizon

V

mg

mVq2

/r

Earth

yt

V

If V0VT

2

laminar turbul

2

ent

or Terminal occurs when Drag

Hence: or18

Df T

sph

T T

D

dm mg C V V mg

dt

g

V

D mgV V

C

C AV

A

Normally, maximum |a| occurs at t=0, but atmospheric density varies with height:

2

max 0

max

2 cosln , and

cos 2

a D iy C AH vaH m e H

y

y

Why is y corridor so small?

Calculations:

Given size, determine VT

Given amax, determine entry

angle and ya,max

VT

Chapter 5. Reentry Dynamics Summary

99


16/18

99

Chapter 5. Reentry Dynamics Revision Concepts

Key points

1. The concept of drag acting on a moving (falling) object. Mathematical

difference between laminar and turbulent flows. Drag coefficients.

2. Terminal velocity for laminar and turbulent flows, and its applications.

3. When gravity and fluid density are constants, maximum deceleration happens

at the beginning of brake when the velocity is maximum.

4. The reason why a reentry spacecraft cannot withstand a free fall. The

theoretical model for an oblique ballistic entry.

5. Exponential atmospheric model, and the mechanism for the peak deceleration

point during a ballistic reentry. Why do we need an oblique reentry?

Note that you do not have to remember the complex formulations in this topic, butyou are required to understand their meaning when they are presented to you

with certain remarks.

Try this: if the initial velocity is 8km/s, what is the velocity at the point where

maximum deceleration occurs? Ans: v0/exp(0.5)=4.8522km/s

100


17/18

100

Q5.3 A mission to the Mars is to return to earth with a shuttle mass of 10,000 kg. It is

decided that the re-entry into the earth atmosphere should avoid two issues: (a)

maximum deceleration exceeding 15g. (b) maximum deceleration occurring in the D-

layer of the ion-sphere spanning from 50 km to 90 km above the earth surface.

Assuming an exponential atmospheric model with a scaling height of 7.25 km, a

turbulent drag coefficient of CD

=1, and a distant entry flight velocity of vi=10 km/s.

(a) Determine the entry angle y relative to the vertical line, and the equivalent frontal area

of the shuttle for braking purpose. Note that Eqs (15) and (16) may be used in the

design.

21 max

max 2

31

24

3 maxmax

0

4

3

2cos cos

2

2 15 9.81 7.25 10cos 86.675

10

cos 50 10 , exp

2

10 cos 86.675 exp

2 1 1.225 7.2

An

5 10 7.2

s

5

:

50

i

i

aa

D

v e a H a

e H v

e

ymy A

C H H

A

yy

y

y

232.3 m

Earth

Avoid max

deceleration within

ion-sphere 50-

90km

Slow down to


18/18

101

(b) Find the velocity at the maximum deceleration height.

(c) If, in addition to the above requirement of avoiding ionosphere, the return trip needs to

spend some time in the stratosphere to collect air samples for the analysis of ozone

recuperation, the maximum flight speed entering the upper stratosphere at 32km should

be less than 100m/s. Check whether the design in (a) can satisfy this additional

requirement. If not, adjust the design to suit both requirement.

(d) Comment on whether the formulation would still hold under such velocity at such

height. Hint: compare gravity with the turbulent drag force.

/050

350/7.25

4

expcos

1 1.225 32.3 7.25 10

a

10 km/s exp 6.1 km/s10 cos86.675

ctually = /iy HD

km i

C AHv v e

m

e

v e

y

3

4 32/7.25

32 4

1 1.225 32.3 7.25 1010 m/s exp 25.1 m/s

10 cos86.675kmv e

/ 2 32/7.25 2 40

The ratio of drag to gravity forces is

/ 1 1.225 25.1 32.3/ 10 9.81

0.0031 1

Therefore the theoretic

y H

DC e V A mg e

al model is no longer valid.

dynamics chapter 5

Documents