TEST BANK FOR Fundamentals of Dynamics and Control of Space Systems By Krishna Dev Kumar

Preface v
2 Kinematics, Momentum and Energy 1
3 Forces and Torques 33
4 Dynamics I 39
5 Dynamics II 45
6 Mathematical and Numerical Simulation 65
7 Control System 85
8 Formation Flying 115
Index 121
vii
Chapter 2
Kinematics, Momentum
and Energy
Problem Set 2
2.1 The coordinate frames used in studying the dynamics of a spacecraft
are as follows:
a) Inertial reference frame,
b) Orbital reference frame,
c) Perifocal reference frame,
c) Satellite body-fixed reference frame.
2.2 The inertial frames are those coordinate frames that are nonrotating
and nonaccelerating frames. The inertial frames are relevant because
in applying the Newton’s second law of motion
~F = m
d~V
dt
(2.1)
to derive the equation of motion of a system, the velocity ~V and the
corresponding acceleration d~V /dt in the right-hand side of the above
equation are to measured with respect to an inertial frame of reference.
An Earth-fixed frame is not an inertial frame as it is spinning about
its axis with a period of 24 hour. When viewed from space, the point
on the surface of the earth moves in a circle as the earth spins on its
axis. Thus, it is accelerating with an centripetal acceleration of r!2,
2 CHAPTER 2. KINEMATICS, MOMENTUM AND ENERGY
where r is the position of the point of the Earth center of mass and !
is the rate of spin of the Earth. With the earth a point on its surface
also orbits the Sun. With the solar system, it orbits the center of the
galaxy. Thus, the Earth-fixed frame is an accelerating frame and not
an inertial frame.
We consider just the effect of the spinning motion of the Earth and
therefore the inertial acceleration can be written as
d~V
dt

inertial
=
d~V
dt

body
+ ~! × ~Vbody (2.2)
The corresponding error in considering an Earth-fixed frame as an inertial
frame is
Error =
d~V
dt

inertial
−
d~V
dt

body
= ~! × ~Vbody (2.3)
The Earth’s spin rate ! is
~! = !kˆk = −
2
T
ˆk
= −
2
24 × 3600
ˆk = 7.275 × 10−5ˆk (2.4)
where ˆk is a unit vector along the z-direction as taken for the aircraft
body-fixed frame.
The order of magnitude error would be 10−4×Vbody. As this magnitude
is usually very small when compared to the magnitude of other relevant
accelerations like the gravitational acceleration, which is 9.81 m/s2, and
we often treat the Earth-fixed frame as an inertial frame. when solving
problems.
2.3 The inertial position vectors for spacecraft m1 and m2 are
~ R1 =
~R
− ~L (2.5)
~ R2 =
~R
+ (1 − )~L (2.6)
where  = m2/(m1 + m2). The corresponding magnitudes are
R1 = [R2 + 2L2 − 2 ~R · ~ L]1/2 (2.7)
R2 = [R2 + (1 − )2L2 + 2(1 − )~R · ~ L]1/2 (2.8)
where L = L0 + vt. The nomenclature Lo defines the initial length of
the cable while v is the speed by which the length of the cable varies.
3
Z
Orbit
E
q
b
Y
X
S
L
m2
m1
R
Figure 2.1: A dumbbell satellite system undergoing in-plane libration.
Expressing ~R and ~L in terms of unit vectors of the respective coordinate
frames as
~R
= Rˆio, ~L = Lˆi (2.9)
Applying the transformation between coordinate frames S−xoyozo and
S − xyz, we get
ˆi
o ·ˆi = cos (2.10)
and using this relation in Eqs. (2.7) and (2.8), we have the inertial
positions of the spacecraft m1 and m2 as
R1 = [R2 + 2L2 − 2 RLcos]1/2 (2.11)
R2 = [R2 + (1 − )2L2 + 2(1 − )RLcos]1/2 (2.12)
The inertial velocity vectors for spacecraft m1 and m2 are
~V1 =
˙~ R1 =
˙~R
−
˙~L
(2.13)
~V2 =
˙~ R2 =
˙~R
+ (1 − )
˙~L
(2.14)
The corresponding magnitudes are
V1 = [
˙~R
2 + 2 ˙~L2 − 2
˙~R
·
˙~L
]1/2 (2.15)
V2 = [
˙~R
2 + (1 − )2 ˙~L2 + 2(1 − )
˙~R
·
˙~L
]1/2 (2.16)
4 CHAPTER 2. KINEMATICS, MOMENTUM AND ENERGY
˙~R
and
˙~L
can be written as
˙~R
=

˙~R

xoyozo
+ ~!o × ~R (2.17)
˙~L
=

˙~L

xyz
+ ~! × ~L (2.18)
Knowing the system is orbiting in a circular orbit (i.e., ˙R = 0), and the
cable connecting the two spacecraft is moving with a constant speed of
v, we get

˙~R

xoyozo
= 0,

˙~L

xyz
= ~v (2.19)
where ~v is in the direction of ~ L. Substituting the above relations in
Eqs. (2.17)-(2.18), we obtain
˙~R
= ~!o × ~R (2.20)
˙~L
= ~v + ~! × ~L (2.21)
The terms
˙~R
2 and
˙~L
2 can be written as
˙~R
2 = (~!o × ~R )2 (2.22)
˙~L
2 = v2 + (~! × ~ L)2 + 2~v · (~! × ~ L) (2.23)
Writing ~!o, ~R, ~!, ~ L, and ~v in terms of the unit vectors of the respective
coordinate frames, we have
!o = ˙ ˆko, ~R = Rˆio, ~! = ( ˙ + ˙)ˆk, ~L = Lˆi, ~v = vˆi (2.24)
Inserting these expressions into Eqs. (2.22)-(2.23) and solving, we have
˙~R
2 = ˙2R2 (2.25)
˙~L
2 = v2 + ( ˙ + ˙)2L2 (2.26)
Note that as ~v ? (~! × ~L), ~v · (~! × ~L) = 0.
Next we derive
˙~R ·
˙~L
. Using Eqs. (2.20)-(2.21), we can write
˙~R
·
˙~L
= (~!o × ~R) · (~v + ~! × ~L)
= (~!o × ~R) · ~v + (~!o × ~R) · (~! × ~L)
= ˙ Rv(ˆjo · ˆi) + ˙ ( ˙ + ˙)RL(ˆjo · ˆj )