Space Craft Dyanmics

University of Glasgow

Lecture Notes for the Courses

Spaceflight Dynamics 5 (M.En.)Spaceflight Dynamics 2 (M.Sc.)

Academic Year 2009 - 10

Giulio AvanziniDipartimento di Ingegneria Aeronautica e Spaziale

Politecnico di Torino

e-mail: [email protected]

Contents

I Attitude Dynamics and Control 1

1 Rigid Body dynamics 31.1 Frames of reference and transformation matrices . . . . . . . . . . . . . . . 31.2 Euler’s angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2.1 Building the coordinate transformation matrix from elementary ro-tations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2.2 Angular velocity and the evolution of Euler’s angles . . . . . . . . . 111.2.3 The quaternions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.2.4 Evolution of the quaternions . . . . . . . . . . . . . . . . . . . . . . 161.2.5 Quaternions vs Euler angles . . . . . . . . . . . . . . . . . . . . . . 161.2.6 Other attitude representations . . . . . . . . . . . . . . . . . . . . . 16

1.3 Time derivative of vector quantities . . . . . . . . . . . . . . . . . . . . . . 171.4 Euler’s equations of motion of a rigid body . . . . . . . . . . . . . . . . . . 18

1.4.1 The inertia tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.4.2 Rotational kinetic energy . . . . . . . . . . . . . . . . . . . . . . . . 211.4.3 Euler’s equation of motion . . . . . . . . . . . . . . . . . . . . . . . 221.4.4 Conservation of angular momentum . . . . . . . . . . . . . . . . . . 221.4.5 Conservation of kinetic energy . . . . . . . . . . . . . . . . . . . . . 23

1.5 Generalised Euler equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 231.5.1 Derivation of the generalised form of Euler equation . . . . . . . . . 231.5.2 Use of the generalised form of Euler equation . . . . . . . . . . . . 25

2 Dynamics of Spinning Spacecraft 272.1 Torque–free motion of axi–symmetric satellites . . . . . . . . . . . . . . . . 272.2 Torque–free motion of tri–inertial satellites . . . . . . . . . . . . . . . . . . 33

2.2.1 Drawing the polhode curves . . . . . . . . . . . . . . . . . . . . . . 342.2.2 Stability of torque–free motion about principal axes . . . . . . . . . 36

2.3 Nutation damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372.3.1 Effects of energy dissipation on axi–symmetric satellites . . . . . . . 372.3.2 More accurate models . . . . . . . . . . . . . . . . . . . . . . . . . 39

2.4 Attitude Maneuvers of a spinning satellite . . . . . . . . . . . . . . . . . . 40

3 Dual–spin satellites 453.1 Mathematical model of a gyrostat . . . . . . . . . . . . . . . . . . . . . . . 47

3.1.1 Angular momentum formulation . . . . . . . . . . . . . . . . . . . . 473.1.2 Angular velocity formulation . . . . . . . . . . . . . . . . . . . . . . 48

3.2 Simplified models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.2.1 The Kelvin gyrostat . . . . . . . . . . . . . . . . . . . . . . . . . . 493.2.2 The apparent gyrostat . . . . . . . . . . . . . . . . . . . . . . . . . 50

i

ii CONTENTS

3.3 Stability of axial gyrostat . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

4 Active Stabilisation and Control of Spacecraft 53

4.1 Actuators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

4.2 Linear model of rigid satellite attitude motion . . . . . . . . . . . . . . . . 54

4.3 Linear model of gyrostat attitude motion . . . . . . . . . . . . . . . . . . . 55

4.4 Use of thrusters for attitude control . . . . . . . . . . . . . . . . . . . . . . 56

4.4.1 Single axis slews (open loop) . . . . . . . . . . . . . . . . . . . . . . 56

4.4.2 Closed–loop control . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

4.4.3 Fine pointing control . . . . . . . . . . . . . . . . . . . . . . . . . . 65

4.5 Momentum exchange devices for attitude control . . . . . . . . . . . . . . . 69

4.5.1 Open–loop control with RWs . . . . . . . . . . . . . . . . . . . . . 70

4.5.2 Sizing a reaction wheel for single axis slews . . . . . . . . . . . . . . 72

4.5.3 Closed–loop control with RWs for single axis slews . . . . . . . . . 73

4.5.4 Bias torque and reaction wheel saturation . . . . . . . . . . . . . . 73

5 Environmental torques and other disturbances 75

5.1 Environmental torques . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

5.2 Internal disturbances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

5.3 Gravity–gradient stabilization . . . . . . . . . . . . . . . . . . . . . . . . . 77

5.3.1 Origin of the gravity–gradient torque . . . . . . . . . . . . . . . . . 77

5.3.2 Pitch component of the gravity–gradient torque . . . . . . . . . . . 78

5.3.3 Attitude motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

II Advanced Orbital Dynamics: Orbit Control 83

6 Basic concepts 85

6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

6.2 Keplerian motion: a review . . . . . . . . . . . . . . . . . . . . . . . . . . 87

6.2.1 Birth of Astrodynamics: Kepler’s Laws . . . . . . . . . . . . . . . . 87

6.2.2 Newton’s Laws of Motion . . . . . . . . . . . . . . . . . . . . . . . 88

6.2.3 Newton’s Law of Universal Gravitation . . . . . . . . . . . . . . . . 89

6.2.4 Equation of Motion in the Two-Body Problem . . . . . . . . . . . . 89

6.2.5 Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

6.2.6 Constants of the Motion . . . . . . . . . . . . . . . . . . . . . . . . 90

6.2.7 Trajectory Equation . . . . . . . . . . . . . . . . . . . . . . . . . . 91

6.2.8 Relating Energy and Semi-major Axis . . . . . . . . . . . . . . . . 92

6.2.9 Elliptical Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

6.3 Three-Dimensional Analysis of Motion . . . . . . . . . . . . . . . . . . . . 95

6.3.1 Non–rotating Frames . . . . . . . . . . . . . . . . . . . . . . . . . . 95

6.3.2 Classical Orbital Elements . . . . . . . . . . . . . . . . . . . . . . . 96

6.3.3 Modified Equinoctial Orbital Elements . . . . . . . . . . . . . . . . 97

6.3.4 Determining the orbital elements . . . . . . . . . . . . . . . . . . . 97

6.3.5 Orbit propagation: the ideal case . . . . . . . . . . . . . . . . . . . 98

CONTENTS iii

7 Orbit Trim Manoeuvres 1017.1 Variational Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1017.2 Perigee Raise Manoeuvre . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1037.3 Plane Changes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1057.4 Finite Burn Manoeuvres . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

8 Orbit Control 1078.1 Air Drag Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1078.2 Geostationary Orbit Control . . . . . . . . . . . . . . . . . . . . . . . . . . 110

Part I

Attitude Dynamics and Control

1

Chapter 1

Rigid Body dynamics

In order to describe the attitude of a rigid body and to determine its evolution as afunction of its initial angular velocity and applied torques, Euler’s angles and Euler’sequations of motion need to be introduced. The transformation matrix between differentreference frames will be recalled and the concept of inertia tensor will also be brieflydiscussed.

1.1 Frames of reference and transformation matrices

Assuming that a satellite is a rigid body is a reasonable initial model for attitude dy-namics and control. However, in practice, this assumption can only be used as a firstapproximation. For satellites with large deployable solar arrays the structure can bequite flexible. The elastic modes in the structure can be excited through attitude con-trol thrusters firings. This leads to vibrations which reduce the pointing accuracy of thepayload. In addition, fuel consumption and fuel slosh in propellant tanks can cause theinertia properties of the satellite to be time varying, leading to a more complex controlproblem. But if we assume that our spacecraft is a rigid body, we can attach to it a body

frame, FB, described by a set of unit vectors (e1, e2, e3). The position of FB with respectto an inertial reference frame FI , identified by the unit vectors (E1, E2, E3), completelydescribes the attitude of our spacecraft.

Assuming that ~v is a vector quantity, it is possible to write it as

~v = xe1 + ye2 + ze3

or, equivalently, ~v = XE1 + Y E2 + ZE3

The column vectors vB = (x, y, z)T and vI = (X, Y, Z)T provide the component repre-sentations of the same vector quantity ~v in the reference frames FB and FI , respectively.

If we now consider the components eiI = (e1,i, e2,i, e3,i)T of the i–th unit vector ei in

FI , that isei = e1,iE1 + e2,iE2 + e3,iE3

we can write

~v = xe1 + ye2 + ze3

= x(e1,1E1 + e2,1E2 + e3,1E3) +

+ y(e1,2E1 + e2,2E2 + e3,2E3) +

+ z(e1,3E1 + e2,3E2 + e3,3E3)

3

4 G. Avanzini, Spaceflight Dynamics – 1. Rigid Body dynamics

~v = (e1,1x+ e1,2y + e1,3z)E1 +

+ (e2,1x+ e2,2y + e2,3z)E2 +

+ (e3,1x+ e3,2y + e3,3z)E3

This means that the components of ~v in FI can be expressed as a function of those inFB as follows:

X=e1,1x+ e1,2y + e1,3zY=e2,1x+ e2,2y + e2,3zZ=e3,1x+ e3,2y + e3,3z

or, in compact matrix form,

vI = LIBvB

where the transformation matrix LIB is given by

LIB =

e1,1 e1,2 e1,3

e2,1 e2,2 e2,3

e3,1 e3,2 e3,3

=

[

e1I

...e2I

...e3I

]

LIB is made up by the components of the unit vectors ei as expressed in FI . Any matrixmade up by mutually orthogonal row or column unit vectors is an orthogonal matrix andis characterized by several properties, among which we only recall that:

• the inverse of an orthogonal matrix L is given by its transpose: L−1 = LT ;

• the determinant of an orthogonal matrix is det(L) = ±1 (and that of a rotationmatrix is 1);

• if L1 and L2 are orthogonal matrices, their product L1L2 is an orthogonal matrix.

Thanks to the first property, it is possible to write the inverse coordinate transforma-tion as

vB = LBIvI = (LIB)−1vI = (LIB)T vI

which means that it is also

LBI =

eT1I

. . .

eT2I

. . .

eT3I

As an exercise, demonstrate that the dual relations

LBI =

[

E1B

...E2B

...E3B

]

; LIB =

ET

1B

. . .

ET

2B

. . .

ET

3B

also hold.

G. Avanzini, Spaceflight Dynamics – 1. Rigid Body dynamics 5

1.2 Euler’s angles

It is possible to use the coordinate transformation matrix LBI to describe the attitudeof the spacecraft through the unit vectors ei of the body frame attached to it, comingout with a total of 9 parameters. As a matter of fact, these 9 parameters are not free tovary at will, inasmuch as they must satisfy 6 constraints, expressed by the orthonormalitycondition, that is

ei · ej = δi,j =

0 if i 6= j1 if i = j

Roughly speaking, only 9− 6 = 3 parameters should be sufficient to describe the attitudeof FB w.r.t. FI .

One of the set of three parameters most widely used to describe the attitude of a rigidbody (or equivalently the attitude of the body frame attached to it) w.r.t. a fixed frameare the Euler’s angles, a sequence of three rotations that take the fixed frame and makeit coincide with the body frame. The original sequence of rotations proposed by Euler tosuperimpose FI onto FB is the sequences 3-1-3:

1. the first rotation is about the third axis of the initial frame, that is E3, in our case,and takes the first axis E1 to the direction e′

1 perpendicular to the plane determinedby the unit vectors E3 and e3; E2 is rotated onto e′

2; the rotation angle is calledprecession angle Ψ;

2. the second rotation is about the first axis transformed after the first rotation, e′

1,and takes the axis e′

3 into the position of e3; e′

2 is moved onto e′′

2; the rotation angleis called nutation angle Θ;

3. the third and final rotation is about e3 and brings e′′

1 = e′

1 and e′′

2 to their finalpositions, e1 and e2, respectively; the rotation angle is called spin angle Φ.

The three angles, representing the amplitude of the three, successive rotations Ψ,Θ,Φ,respectively about the third, the first, and again the third axis, can be used to representthe attitude of the frame FB: The nutation angle represents the inclination of the thirdbody axis e3 w.r.t. the local vertical E3; The precession angle represents the angle betweenthe first inertial axis E1 and the line of the nodes ξ, i.e. the intersection between theplanes perpendicular to e3 and E3; The spin angle is the rotation about the third bodyaxis.

The transformation matrix LBI can be expressed as a function of these three angles,in terms of three elementary rotation matrices, as will be derived in the sequel.

Other sequences

It must be remembered that the sequence of rotations here described is not the onlypossible choice for rotating FI onto FB. Many other sequences are available and equallyuseful. In atmospheric flight mechanics the most widely used sequence of rotations is the3–2–1, also known as the Bryant’s angles.

In this case the first rotation is about the third axis, E3, and its amplitude is calledyaw angle ψ. The second rotation about the second axis e′

2 is the pitch angle, θ, andtakes the first axis onto its final position. The third rotation about e1 is the roll angle,φ. This set of angles is used also in space flight dynamics, to describe the attitude of a


E1 E2

E3

E1 E2

e′

3 ≡ E3

e′

1

e′

2

Ψ

Ψ

E1 E2

E3

e′

1 ≡ e′′

1

e′

2

Ψ

e′′

3

e′′

2

ΘΘ

E1 E2

E3

e′

1

e′

2

Ψ

e3 ≡ e′′

3

e′′

2

Θ

e1

e2

Φ

Φ

Figure 1.1: Euler’s angles

spacecraft with respect to the Local Horizontal – Local Vertical (LHLV) reference frame.In many textbooks also this latter set of rotations is often referred to as Euler’s angles,and this fact may lead to some confusion.

As a final observation, the order of the rotation sequence is important: Rotations donot commute! This means that the rotation sequence 1–2–3, performed with the sameangles about the same axis, will take the initial frame to another one. The rotationsequence 1–2–3 is known as Cardan angles.

Singularity

There is another problem with the representation of rotations in a three dimensional space,that is the singularity of all the descriptions in terms of three parameters. This meansthat there will always be positions of the two frames that can be described in differentways, once a particular sequence of rotations is chosen. As an example, if the originalEuler’s angle sequence is employed, the case in which Θ = 0 is singular, inasmuch as theprecession and spin rotations will be about the very same axis, E3 ≡ e3. This meansthat all the triplets (Ψ, 0,Φ) for which Ψ + Φ is constant represent the same change ofreference frame.

Similarly, when the Bryant’s angles are used, the case θ = ±π/2 is singular, as in thiscase all the triplets (ψ,±π/2, φ) for which ψ − φ is constant will provide the same finalattitude for FB.

The problem of coordinate transformation singularity has some unpleasant mathemat-


x 1

1y

1x

x

1

y 2

y

2

α

v

Figure 1.2: Planar rotation of amplitude α.

ical consequence that will be underlined in the following paragraphs.

1.2.1 Building the coordinate transformation matrix from ele-mentary rotations

Planar rotations

Consider the sketch of Fig. 1.2, where two planar reference frames F1 and F2, with thesame origin O are represented. The angle α, assumed positive for counter–clockwiserotations, allows one to identify univocally the position of the axes X2–Y2 of F2 w.r.t. theframe F1 defined by the axes X1 and Y1.

Given the components x1 and y1 of a vector ~v expressed in F1, the components x2 ey2 can be expressed as a function of the angle α. The following relations can be easilyinferred from Fig. 1.2:

x2 = x1 cos(α) + y1 sin(α)

y2 = −x1 sin(α) + y1 cos(α)

or, in matrix form,

x2

y2

=

[cos(α) sin(α)− sin(α) cos(α)

]x1

y1

This relation expresses the coordinate transformation that takes the component of avector quantity expressed in F1 into those of a reference frame F2 rotated w.r.t. F1 of anangle α.

In compact notation we can write

v2 = L21v1 = R(α)v1


where the subscript near the vector indicates the frame in which the components of thevector quantity are considered, the matrix L21 is the coordinate transformation matrixfrom F1 to F2 that in the two dimensional case coincides with the elementary rotation

matrix R(α).The inverse transformation from F2 to F1 is given by

v1 = L12v2 = (R(α))−1v2

Recalling the properties of orthogonal matrices, it is

L12 = (R(α))−1 = (R(α))T = R(−α)

Elementary rotations for the sequence 3–1–3

Each one of the Euler’s rotations can be considered an elementary rotation about a givenaxis, that remains unchanged during the transformation. It is still possible to applythe relations derived for the planar case, adding a further equation that states that thecoordinate relative to the rotation axis does not vary.

The coordinate transformation during the first rotation is given by

x′ = X cos(Ψ) + Y sin(Ψ)

y′ = −X sin(Ψ) + Y cos(Ψ)

z′ = Z

that, in matrix form, can be written as:

x′

y′

z′

=

cos(Ψ) sin(Ψ) 0− sin(Ψ) cos(Ψ) 0

0 0 1

XYZ

In an analogous way it is possible to demonstrate that, during the second rotationabout e′

1, the coordinate transformation is given by

x′′ = x′

y′′ = y′ cos(Θ) + z′ sin(Θ)

z′′ = −y′ sin(Θ) + z′ cos(Θ)

that in matrix form becomes:

x′′

y′′

z′′

=

1 0 00 cos(Θ) sin(Θ)0 − sin(Θ) cos(Θ)

x′

y′

z′

Finally, the third rotation about e′′

3 is represented by the transformation

x = x′′ cos(Φ) + y′′ sin(Φ)

y = −x sin(Φ) + y′′ cos(Φ)

z = z′′

or, in matrix form:

xyz

=

cos(Φ) sin(Φ) 0− sin(Φ) cos(Φ) 0

0 0 1

x′′

y′′

z′′


The three elementary rotation matrices of the Euler’s sequence 3–1–3 can thus bedefined as

R3(Ψ) =

cos(Ψ) sin(Ψ) 0− sin(Ψ) cos(Ψ) 0

0 0 1

; R1(Θ) =

1 0 00 cos(Θ) sin(Θ)0 − sin(Θ) cos(Θ)

;

R3(Φ) =

cos(Φ) sin(Φ) 0− sin(Φ) cos(Φ) 0

0 0 1

where the subscript near the rotation matrix symbol R indicates the axis around whichthe rotation is performed, while the argument indicates the amplitude of the rotation.

Summing up...

When passing from the inertial frame FI to the body frame FB using Euler’s sequence, thecoordinate transformation of vector quantities can be obtained combining in the correctorder the elementary rotation matrices, as follows:

v′ = R3(Ψ)vI

v′′ = R1(Θ)v′

vB = R3(Φ)v′′

that is

vB = R3(Φ)R1(Θ)R3(Ψ)vI

This means that

LBI = R3(Φ)R1(Θ)R3(Ψ)

Performing the row–column products, the following expression for LBI is obtained:

LBI =

cos Φ cos Ψ sin Φ cos Θ cos Ψ sin Φ sinΘ− sin Φ cos Θ sinΨ + cos Φ sin Ψ

− cos Φ cos Θ sinΨ cos Φ cos Θ cos Ψ cos Φ sinΘ− sinΦ cos Ψ − sinΦ sin Ψ

sin Θ sinΨ − sin Θ cos Ψ cos Θ

As the product of orthogonal matrices is an orthogonal matrix, the inverse of which isequal to its transpose, the inverse coordinate transformation matrix LIB is simply givenby

LIB = LBI−1 = LBI

T =

cos Φ cos Ψ − cos Φ cos Θ sin Ψ sinΘ sin Ψ− sin Φ cos Θ sinΨ − sin Φ cos Ψ

sin Φ cos Θ cos Ψ cos Φ cos Θ cos Ψ − sinΘ cos Ψ+ cos Φ sin Ψ − sin Φ sinΨ

cos Φ sin Θ cos Φ sin Θ cos Θ


Elementary rotations for the sequence 3–2–1

It is left as an exercise to the reader the composition of elementary rotation matrices forthe sequence 3–2–1. Adopting the same notation used above, it is

R3(ψ) =

cos(ψ) sin(ψ) 0− sin(ψ) cos(ψ) 0

0 0 1

; R2(θ) =

cos(θ) 0 − sin(θ)0 1 0

sin(θ) 0 cos(θ)

;

R1(φ) =

1 0 00 cos(φ) sin(φ)0 − sin(φ) cos(φ)

and the final result is given by

LBI =

cos θ cosψ cos θ sinψ − sin θ

sinφ sin θ cosψ sinφ sin θ sinψ sinφ cos θ− cosφ sinψ + cosφ cosψ

cosφ sin θ cosψ cosφ sin θ sinψ cosφ cos θ+ sinφ sinψ − sinφ cosψ

A first consequence of the Euler’s angle singularity

When Θ = 0, the coordinate transformation matrix does not depend on Ψ and Φ sepa-rately, but only on their sum. In such a case, it is

LBI =

cos Φ cos Ψ sinΦ cos Ψ 0− sinΦ sin Ψ + cos Φ sin Ψ

− cos Φ sin Ψ cos Φ cos Ψ 0− sinΦ cos Ψ − sinΦ sin Ψ

0 0 1

=

cos(Φ + Ψ) sin(Φ + Ψ) 0

− sin(Φ + Ψ) cos(Φ + Ψ) 0

0 0 1

As an exercise, demonstrate that LBI depends on ψ− φ only, when Bryant’s rotationsequence is employed and θ = ±π/2.

How to build elementary rotation matrices

There is a simple way to build mnemonically the elementary rotation matrices. Thematrices are 3 by 3. If a rotation of an angle α about the i-th axis is being considered,place 1 in position i, i, and fill the remaining elements of the i–th row and i–th columnwith zeroes. All the other elements of the principal diagonal are cosα and the last twooutside the diagonal are sinα. The sin element above the row with the 1 must have aminus sign. As an example, let us consider a rotation θ about the second axis (like in thesecond rotation of the Bryant’s sequence). We start filling the matrix with 0s along thesecond row and column, with a one in position 2,2:

· 0 ·0 1 0· 0 ·


E1 E2

E3

e′

1

e′

2

e′′

2

e1

e2

e3

Ψ

Θ

Φ

Figure 1.3: Angular velocity as a funciton of Euler’s angle rates.

Then we fill the diagonal with cos θ:

cos θ 0 ·0 1 0· 0 cos θ

and we put sin θ in the remaining places, with a minus sign in the row above the 1:

cos θ 0 − sin θ0 1 0

sin θ 0 cos θ

This is R2(θ). When a rotation about the first axis is considered, the 1 is on the first rowand apparently there is no row above it. But it is sufficient to cycle and start again fromthe bottom: In this case the minus sign is on the sin in the third row.

1.2.2 Angular velocity and the evolution of Euler’s angles

The angular velocity ~ω is given by

~ω = ω1e1 + ω2e2 + ω3e3

but it is also (see Fig. 1.3)~ω = ΨE3 + Θe′

1 + Φe3

The components of the unit vector E3 in FB are given by the third column of thematrix LBI , that is E3B

= (sin Φ sin Θ, cos Φ sin Θ, cos Θ)T , while the components of e′

1

are (cos Φ,− sin Φ, 0)T . Thus

~ω = Ψ(sin Φ sin Θe1 + cos Φ sin Θe2 + cos Θe3) +

+ Θ(cos Φe1 − sin Φe2) +

+ Φe3

= (Ψ sin Φ sin Θ + Θ cos Φ)e1 +

+ (Ψ cos Φ sin Θ − Θ sin Φ)e2

+ (Ψ cos Θ + Φ)e3


or, in matrix form

ω1

ω2

ω3

=

sin Φ sin Θ cos Φ 0cos Φ sin Θ − sin Φ 0

cos Θ 0 1

Ψ

Θ

Φ

Inverting the 3 × 3 matrix, one obtains the law of evolution of Euler’s angles as afunction of angular velocity components in body axis, that is

Ψ

Θ

Φ

=

sin Φ/ sin Θ cos Φ/ sin Θ 0cos Φ − sin Φ 0

− sin Φ/ tanΘ − cos Φ/ tan Θ 1

ω1

ω2

ω3

or, in explicit form,

Ψ = (ω1 sin Φ + ω2 cos Φ)/ sin Θ

Θ = ω1 cos Φ − ω2 sin Φ

Φ = (−ω1 sin Φ − ω2 cos Φ)/ tanΘ + ω3

These equations can be integrated to obtain the evolution of the Euler angles, if theangular velocity is known. But they also show an unpleasant feature of Euler’s anglesingularity, that is the spin and precession rates go to infinity when Θ approaches 0. Thisfact has some serious consequences on the problem of attitude representation, inasmuchas it is not possible to accept a set of attitude parameters the evolution of which cannotalways be described in an accurate way.

If the Bryant’s angles are used, the reader can demonstrate that

~ω = ψE3 + θe′

2 + φe3

so that

ω1

ω2

ω3

=

1 0 − sin θ0 cosφ cos θ sinφ0 − sin φ cos θ cosφ

φ

θ

ψ

Also demonstrate that the inverse relation is

φ = ω1 + (ω2 sinφ+ ω3 cosφ) tan θ

θ = ω2 cosφ− ω3 sinφ

ψ = (ω2 sin φ+ ω3 cosφ)/ cos θ

Again, in the neighborhood of the singular condition θ = ±π/2 the rate of change of theroll and yaw angles goes to infinity.

1.2.3 The quaternions

Euler’s eigenaxis rotation theorem states that it is possible to rotate a fixed frame FI

onto any arbitrary frame FB with a simple rotation around an axis a that is fixed in bothframes, called the Euler’s rotation axis or eigenaxis, the direction cosines of which arethe same in the two considered frame.

A very simple algebraic demonstration of Euler’s theorem can be obtained from thefollowing considerations:


• the eigenvalues of any (real) orthogonal matrix L have unit modulus; indicatingwith H the Hermitian conjugate, which, for a real matrix is coincident with thetranspose, one has

La = λa ⇒ aHLT La = λλaHa ⇒ (1 − λλ)aHa = 0

that for any nontrivial eigenvector a implies that

λλ = 1 ⇒ |λ| = 1

• (at least) one eigenvalue is λ = 1; any n × n real matrix has at least one realeigenvalue if n is an odd number, which means that a 3× 3 orthogonal matrix musthave at least one eigenvalue which is λ1 = ±1. The other couple of eigenvalues willbe, in the most general case, complex conjugate numbers of unit modulus, whichcan be cast in the form λ2,3 = exp(±iφ). The determinant is equal to the productof the eigenvalues, which is one, for an orthogonal matrix, so that

λ1λ2λ3 = 1 ⇒ λ1 = 1

The eigenvector relative to the first eigenvalue satisfies the relation

La = 1 · a

This means that there is a direction a which is not changed under the action of transfor-mation matrix L. If L represents a coordinate change, the vector a will be representedby the same components in both the considered reference frames,

a = a1e1 + a2e2 + a3e3

= a1E1 + a2E2 + a3E3

For this reason, the transformation that takes the initial frame onto the final one can beconsidered as a single rotation α about the Euler axis a.

In order to express the coordinate transformation matrix LBI as a function of α anda it is sufficient to consider the following sequence of rotations:1

1. take the unit vector E1 onto a, so that the new frame F ′ is given by the unit vectorsa, e′

2, e′

3; call this rotation R;

2. rotate both frames FI and F ′ about the eigenaxis of the rotation angle α; becauseof the definition of Euler rotation, FI goes onto FB, while F ′ will rotate into anew frame F ′′ given by the unit vectors a, e′′

2, e′′

3; this rotation is represented by theelementary rotation matrix R1(α);

3. at this point it should be noted that the rotation ¯R that takes F ′′ onto FB has thesame magnitude of R, but it is performed in the opposite direction so that ¯R = R

T.

Summing up it is

LBI = RTR1(α)R

1This derivation is taken from B. Wie, Space Vehicle Dynamics and Control, AIAA Education Series,Reston (VA), USA, 1998, Chap. 5, pp. 312–315 and 318–320.


(a)

E1 E2

E3

e1

e2

e3a

α

(b)

E1 E2

E3 e′

3

e′

2

a

(c)

E1 E2

E3

e1

e2

e3

e′

3

e′

2

e′′

3

e′′

2a

α

(d)

e1

e2

e3

e′′

3

e′′

2

a

where

R =

a1 a2 a3

R21 R22 R23

R31 R32 R33

Carrying out the calculations, one get, for the components Lij of the coordinate transfor-mation matrix LBI the following expressions:

L11 = a21 + (R2

21 +R231) cosα

L12 = a1a2 + (R21R22 +R31R32) cosα + (R21R32 −R31R22) sinα



L22 = a22 + (R2

22 +R232) cosα





L33 = a23 + (R2

23 +R233) cosα

Taking into account the orthogonality conditions for R, one gets

a21 +R2

21 +R231 = 1 ⇒ R2

21 +R231 = 1 − a2

1

a22 +R2

22 +R232 = 1 ⇒ R2

22 +R232 = 1 − a2

2

a23 +R2

23 +R233 = 1 ⇒ R2

23 +R233 = 1 − a2

3

a1a2 +R21R22 +R31R32 = 0 ⇒ R21R22 +R31R32 = −a1a2

a2a3 +R22R23 +R32R33 = 0 ⇒ R22R23 +R32R33 = −a2a3

a3a1 +R21R23 +R31R33 = 0 ⇒ R21R23 +R31R33 = −a1a2

while remembering that the first row of an orthogonal matrix is given by the cross productof the second and the third ones, it is

a1 = R22R33 − R23R32

a2 = R23R31 − R21R33

a3 = R21R32 − R22R31

Substituting these results into the expressions of the coefficients Lij the followingexpression for LBI is obtained:

LBI =

cosα+ a21(1 − cosα) a1a2(1 − cosα) + a3 sinα a1a3(1 − cosα) − a2 sinα

a2a1(1 − cosα) − a3 sinα cosα + a22(1 − cosα) a2a3(1 − cosα) + a1 sinα

a3a1(1 − cosα) + a2 sinα a3a2(1 − cosα) − a1 sinα cosα + a23(1 − cosα)

or, in compact matrix form

LBI = cosα1 + (1 − cosα)aaT − sinαA

where 1 is the 3 × 3 identity matrix. The tilde symbol, V , is the equivalent matrix formof the cross product, that is, letting

V =

0 −v3 v2

v3 0 −v1

−v2 v1 0

it is v × w = V w.

We now define the Euler parameters or quaternions as

q0 = cos(α/2)

q1 = a1 sin(α/2)

q2 = a2 sin(α/2)

q3 = a3 sin(α/2)

By letting q = a sin(α/2) and rembering that cosα = cos2(α/2)−sin2(α/2) = q20−q·q and

sinα = 2 cos(α/2) sin(α/2) = 2q0 sin(α/2), it is easy to demonstrate that the coordinatetransformation matrix is given by

LBI = (q20 − q · q)1 + 2qqT − 2q0Q

where the ˜ indicates again the cross product matrix equivalent

Q =

0 −q3 q2q3 0 −q1−q2 q1 0


1.2.4 Evolution of the quaternions

The evolution of the quaternions is described by the set of linear differential equations,represented in matrix form as2

q0q1q2q3

=1

2

0 −ω1 −ω2 −ω3

ω1 0 ω3 −ω2

ω2 −ω3 0 ω1

ω3 ω2 −ω1 0

q0q1q2q3

The equivalent matrix form is given by

q0 = −1

2ω · q

q =1

2(q0ω − ω × q)

1.2.5 Quaternions vs Euler angles

The quaternions allow for representing the attitude of a rigid body with several advantagesover Euler’s angles, above all the absence of inherent geometric singularity. Moreover, thelinear equation to be integrated in time in order to determine their evolution as a functionof angular velocity components is less computationally expensive than that derived forthe Euler’s angles. The price to pay is that 4 parameters are used, instead of only three,that are not independent, inasmuch as they must satisfy the constraint

q20 + q · q = 1

Moreover, their geometric interpretation during an evolution is less immediate than thatof the Euler’s angles, the geometric meaning of which is intuitive. For this reason theattitude of a satellite is often integrated in strapdown attitude determination systems interms of quaternions but then represented in terms of Euler angles.

1.2.6 Other attitude representations

The non–minimality of attitude representation in terms of quaternions can be solved forby use of the Gibbs vector, defined as

g = (g1, g2, g3)T = a tan(α/2)

Gibbs parameters, also known as Rodrigues parameters, are strictly related to quaternions,as it is

(g1, g2, g3)T = (q1/q0, q2/q0, q3/q0)

T

The coordinate transformation matrix can also be represented in terms of Gibbs param-eters:

LBI = (1 − G)(1 + G)−1

Being a minimal parametrization of attitudes, Gibbs parameters must present a singu-larity.3 The singular configuration of the Gibbs vector is for any eigenaxis rotation with

2See above, pp. 326–3283Recall that Euler demonstrated that any minimal parametrization of attitudes has at least one

singular configuration


α = ±π, in which case their values diverge towards infinity. To solve for this problem, theso called set of Modified Rodrigues parameters (MRP) was recently introduced, defined as

p = (p1, p2, p3)T = a tan(α/4)

The singularity in the attitude representation is still present, and for a spinning bodythe value of the MRPs will “jump” when α crosses the critical threshol placed at half ofa rotation (that is, α = ±π). Nonetheless the advantage of the MRPs is that they donot diverge, but at the same time, no constraint on their value needs to be enforced inorder to save their meaning. These features make the numerical integration of MRPs lesscritical with respect to both the quaternions case and the Gibbs vector.

1.3 Time derivative of vector quantities

If we consider a vector quantity in an inertially fixed reference frame FI ,

~v = XE1 + Y E2 + ZE3

its time derivative is given simply by

d~v

dt= XE1 + Y E2 + ZE3

that is [d~v

dt

]

I

=

X, Y , ZT

= vI

When the same vector quantity ~v is expressed in terms of components in a moving ref-erence frame FB, rotating with angular velocity ~ωBI = ~ω with respect to FI , the timederivatives of

~v = xe1 + ye2 + ze3

is given by4

d~v

dt= xe1 + ~ω × (xe1) +

+ ye2 + ~ω × (ye2) +

+ ze3 + ~ω × (ze3)

This means that, in terms of vector components in FB, it is[dv

dt

]

B

= vB + ωB × vB

where

vB = x, y, zT

ωB = ω1, ω2, ω3T

4Remeber the Poisson formula for the time derivative of a unit vector,

dei

dt= ~ω × ei


E1 E2

E3

e1

e2

e3

~r0

~r

~v

~h

~ω

dm

CM

Figure 1.4: A rotating rigid body.

1.4 Euler’s equations of motion of a rigid body

1.4.1 The inertia tensor

The angular momentum δ~h of a mass element δm, moving with velocity ~v is

δ~h = ~r × (δm~v)

where ~r is the position vector of the mass, with respect of the pole used for the evaluationof moments of vector quantities.

For an extended rigid body (Fig. 1.4), the total angular momentum is given by

~h =

∫

B

(~r × ~v)δm

If the body is rotating around its center of mass, the velocity of every mass element is

~v = ~ω × ~r

so that~h =

∫

B

[~r × (~ω × ~r)] δm

Expressing the vector quantities in body components as ωB = (ω1, ω2, ω3)T and rB =

(x, y, z)T , the vector product ~ω × ~r is given by

~ω × ~r = (ω2z − ω3y)e1 + (ω3x− ω1z)e2 + (ω1y − ω2x)e3

Carrying on the calculations, the product ~r × (~ω × ~r) is

~r × (~ω × ~r) =[

(y2 + z2)ω1 − (xy)ω2 − (xz)ω3

]e1 +

+[−(xy)ω1 + (x2 + z2)ω2 − (yz)ω3

]e2 +

+[−(xz)ω1 − (yz)ω2 + (x2 + y2)ω2

]e3


The integration over the body B of [~r × (~ω × ~r)] δm is strictly function of the massdistribution only, as angular velocity components are independent of body shape andlocation. This means that, letting

~h = h1e1 + h2e2 + h3e3

it is

h1 = Ixω1 − Ixyω2 − Ixzω3

h2 = −Ixyω1 + Iyω2 − Iyzω3

h3 = −Ixzω1 − Iyzω2 + Izω3

where the moments of inertia Ix, Iy, Iz, and the products of inertia Ixy, Ixz, Iyz, are

Ix =

∫

B

(y2 + z2

)δm ; Iy =

∫

B

(x2 + z2

)δm ; Iz =

∫

B

(x2 + y2

)δm

Ixy =

∫

B

(xy) δm ; Ixz =

∫

B

(xz) δm ; Iyz =

∫

B

(yz) δm

In matrix form the angular momentum components in body axes are given by

hB = IωB

where the symmetric matrix

I =

Ix −Ixy −Ixz

−Ixy Iy −Iyz

−Ixz −Iyz Iz

is the inertia matrix that represent the inertia tensor in body axes.The same relations can be derived directly in a more compact vector form remembering

that, for the double vector product, the following relation holds:

~x × (~y × ~z) = (~x · ~z) ~y − (~x · ~y)~z

so that, in the present case, it is

~r × (~ω × ~r) = (~r · ~r) ~ω − (~r · ~ω)~r

Taking into account the definition of the dyadic tensor

(~x~y)~z = (~y · ~z)~x

and the fact that the angular velocity vector is constant and can be taken out of theintegral, it is

~h =

∫

B

[~r × (~ω × ~r)] δm

=

(∫

B

[(~r · ~r)1 − (~r~r)] δm

)

~ω

= I~ω

where I is the inertia tensor. Expressing ~r in terms of body components and integratingover B the previous expression for the inertia matrix I is obtained.


Principal axes of inertia

The matrix I is real and symmetric, so its eigenvalues are real5 and its eigenvectors aremutually orthogonal.6 This means that there exists a body reference frame FP such thatthe inertia matrix is diagonal,

I =

Jx 0 00 Jy 00 0 Jz

where the principal moment of inertia Jx, Jy, and Jz are the eigenvalues of I. Theeigenvectors are called principal axes.

Symmetries

If the mass distribution of the body B is characterized by symmetries, this propertyreflects onto the inertia matrix I. As an example, if B has a plane of symmetry, one of theprincipal axis will be perpendicular to the plane and the other two will lie on that plane.If this case, the products of inertia relative to the axis perpendicular to the symmetryplane will be zero. This case is typical of fixed wing aircraft, where the longitudinalplane is approximately a symmetry plane. The y body axis, directed perpendicular to thesymmetry plane, is characterized by zero products of inertia, so that the inertia matrix

5From the definition of eigenvalue and eigenvector of a complex matrix A, it is easy to derive thefollowing equation,

Ax = λx ⇒ λ =xHAx

xHx

If A is Hermitian (i.e. the linear operator represented by A is self–adjoint), it is

yHAx = (Ay)Hx

so that, remembering the properties of the hermitian inner product, such that xHy = (yHx), it is

λ =xHAx

xHx=

(Ax)Hx

xHx=

xHAx

xHx= λ

which means that the eigenvalue λ is equal to its conjugate, i.e. it must be real.6Given two distinct eigenvalues λi 6= λj and their respective eigenvectors xi and xj , the following

relations hold:

Axi = λixi

Axj = λjxj

Multiplying the first equation by xHj and the second one by xH

i , taking the complex conjugate of thesecond and subtracting it from the first, one gets

xHj Axi − (xH

i Axj) = λixTj xi − (λjx

Ti xj)

Remembering that the eigenvalues are real, it is

xHj Axi − (Axj)

Hxi = (λi − λj)xHj xi

Taking into account the definition of Hermitian operator the first term of the last equation is zero and sothe Hermitian product xH

j xi is zero if λi 6= λj . Since both xj and xi are real, their Hermitian productcoincides with the scalar product, so that distinct eigenvectors are real and perpendicular.


of an aircraft is typically equal to

I =

Ix 0 −Ixz

0 Iy 0−Ixz 0 Iz

If the body is axi-symmetric (or simply has a regular polygonal mass distributionw.r.t. an axis σ), the symmetry axis σ is a principal axis of inertia, while any couple ofperpendicular axes on the plane Σ normal to σ will complete the set of principal axes.In this case the principal moments of inertia relative to the axes perpendicular to thesymmetry axis will be equal. Assuming that σ = e3 is a symmetry axis, the inertiatensor becomes

I =

Jt 0 00 Jt 00 0 Js

where the subscripts t and s indicate the transverse and spin (or axial) moments ofinertia, respectively.

1.4.2 Rotational kinetic energy

The kinetic energy of a body is given by

T =1

2

∫

B

(~v · ~v) δm

Remembering that for a rotating rigid body it is ~v = ~ω ×~r, the argument of the integralbecomes (~ω × ~r) · (~ω × ~r). Also, taking into account the permutation property of thescalar triple product

~x · (~y × ~z) = ~y · (~z × ~x) = ~z · (~x × ~y)

one obtains the equivalence

(~ω × ~r) · (~ω × ~r) = ~ω · [~r × (~ω × ~r)]

Substituting this expression into the integral, and taking the (constant) angular velocityout of the integration symbol, one gets

T =1

2~ω ·∫

B

[~r × (~ω × ~r)] δm

that, remembering the definition of the inertia tensor, brings

T =1

2~ω ·(

I~ω)

=1

2ωT

B (IωB)

or, equivalently,

T =1

2~ω · ~h =

1

2ωT

BhB


1.4.3 Euler’s equation of motion

The second fundamental law of rigid body dynamics states that the time derivative ofthe angular momentum is equal to the external torque applied to the body B. In vectorform, it is

d~h

dt= ~M

Expressing the vector quantities in body axis components one gets

hB + ωB × hB = MB

If the inertia matrix I is constant, it is

IωB + ωB × (IωB) = MB

When a set of principal axes is chosen as the body axes, the inertia tensor is diagonaland the Euler’s equation of motion for a rigid body are obtained:

Jxω1 + (Jz − Jy)ω2ω3 = M1

Jyω2 + (Jx − Jz)ω3ω1 = M2

Jzω3 + (Jy − Jx)ω1ω2 = M3

These equations can be integrated as a function of the applied torque to obtain thetime history of the angular velocity components. These, in turn, can be used to determinethe variation with time of the Euler’s angles (or of the quanternions), thus describing theevolution of the rigid body attitude.

1.4.4 Conservation of angular momentum

Writing Euler’s equations in a set of principal axes such that (without loss of generality)Jx > Jy > Jz, torque–free motion is described by the following set of ODEs,

Jxω1 + (Jz − Jy)ω2ω3 = 0

Jyω2 + (Jx − Jz)ω3ω1 = 0

Jzω3 + (Jy − Jx)ω1ω2 = 0

It is easy to demonstrate that the magnitude h of the angular momentum vector

~h = h1e1 + h2e2 + h3e3

= Jxω1e1 + Jyω2e2 + Jzω3e3

is constant when a torque–free motion is considered. A first intuitive derivation of thisproperty is that if the applied torque vanishes the angular momentum vector is constant inFI , and its magnitude is independent of the considered reference system. It is also possibleto demonstrate analytically that h = ||h|| is constant, by taking the time derivatives of

h2 = (Jxω1)2 + (Jyω2)

2 + (Jzω3)2

in the hypothesis of torque–free motion (M1 = M2 = M3 = 0),

dh2

dt= 2

(J2

xω1ω1 + J2yω2ω2 + J2

zω3ω3

)

= 2 [Jxω1(Jy − Jz)ω2ω3 + Jyω2(Jz − Jx)ω1ω3 + Jzω3(Jx − Jy)ω1ω2]

= 2 (JxJy − JxJz + JyJz − JyJx + JzJx − JzJy)ω1ω2ω3 = 0


Geometrically, the angular velocity vector must lie on an ellipsoid (the angular momentum

ellipsoid) in FB, the equation of which takes the form

ω21

(h/Jx)2+

ω22

(h/Jy)2+

ω23

(h/Jz)2= 1

1.4.5 Conservation of kinetic energy

In a similar way, it is also possible to demonstrate that the kinetic energy of a rigid bodyis constant if no external torque is applied. Again, taking the time derivative of

T =1

2ωB · h =

1

2

(Jxω

21 + Jyω

22 + Jzω

23

)

it is

dTdt

= Jxω1ω1 + Jyω2ω2 + Jzω3ω3

= ω1(Jy − Jz)ω2ω3 + ω2(Jz − Jx)ω1ω3 + ω3(Jx − Jy)ω1ω2

= 2 (Jy − Jz + Jz − Jx + Jx − Jy)ω1ω2ω3 = 0

This means that the angular velocity vector must satisfy also the equation

ω21

(2T /Jx)+

ω22

(2T /Jy)+

ω23

(2T /Jz)= 1

that is, it must point the surface of the kinetic energy (or Poinsot) ellipsoid in FB. Thecombination of these two last results demonstrate that the angular velocity vector describethe curve given by the intersection of the angular momentum ellipsoid and the kineticenergy ellipsoid, which is called the polhode.

1.5 Generalised Euler equation

In their original formulation, Euler equations are written in a body–fixed reference frameFB with the origin in the center of mass O of the body B. On one side, the expressionemployed for the angular momentum h of B requires that B is rigid, and centering B inO greatly simplify the expression. At the same time, linear and angular momentum con-servation laws do apply to any mechanical system (under sufficiently mild assumptions),so that it is possible to obtain a generalised formulation for the equation of motion in aframe which is non centered in O.

1.5.1 Derivation of the generalised form of Euler equation

The classical equations, referred to the center of mass O, are

m~aO = ~F

d~hO

dt= ~MO

where, m is the mass of B and ~F is the external force, producing an acceleration ~aO ofthe center of mass. Considering an arbitrary point A with arbitrary motion with respect


to the body, such that ~rAO is the position vector of the centre of mass O with respect toA, the torque acting on the body can be referred to A,

~MA = ~MO + ~rAO × ~F

As for the angular momentum relative to A, it is

~hA =

∫

B

[

~rAP × d~rAP

dt

]

δm

where ~rAP is the position vector of the mass element δm with respect to A, while d~rAP/dt

its (relative) velocity. Upon substitution of ~rAP = ~rAO + ~rOP into the expression for ~hA,one obains

~hA =

∫

B

[

(~rAO + ~rOP ) × d

dt(~rAO + ~rOP )

]

δm

=

∫

B

[

~rAO × d~rAO

dt

]

δm+

∫

B

[

~rAO × d~rOP

dt

]

δm+

+

∫

B

[

~rOP × d~rAO

dt

]

δm+

∫

B

[

~rOP × d~rOP

dt

]

δm

The last term is the angular momentum with respect to the centre of mass

~hO =

∫

B

[

~rOP × d~rOP

dt

]

δm

while the second and the third one are both zero, from the definition of centre of mass.7

It is thus possible to write ~hA as

~hA = ~hO +m~rAO × d~rAO

dt

The acceleration of O can be rewritten as a function of the absolute acceleration of A

~aO = ~aA +d2~rA

dt2

By substituting the above expressions in the angular momentum equation, one gets

d

dt

(

~hA −m~rA × d~rA

dt

)

= ~MA − ~rA ×[

m

(

~aA +d2~rA

dt2

)]

which can be rewritten asd~hA

dt+ ~SA × ~aA = ~MA

where ~SA = m~rA is the static moment of the body with respect to A.

7If m is the total mass of the body, the absolute position of the centre of mass is defined as

~rO =1

m

∫

B

~rOP δm

so that, letting the position vector of P with respect to O be defined as ~rOP = ~rP − ~rO, it is∫

B

~rOP δm = 0.

Taking into account that integration is a linear operator, that can be commuted with the time derivative,both the second and the third terms vanishes.


1.5.2 Use of the generalised form of Euler equation

The generalised form of classical Euler equation allows for writing dynamic models ofsystems where the mass distribution is not constant. On one side, the assumption ofrigidity often applies with reasonable accuracy to many spacecraft, at least over a relativelylong time–scale. As an example, the slow consumption of fuel during the whole operationallifetime of the vehicle (years) causes a shift of the centre of mass, but the knowledge ofthe current value is usually sufficient for describing manoeuvers over a faster time–scale(minutes).

On the converse, it is sometimes necessary to take into account phenomena wherevariations of the mass distribution affects significantly the vehicle’s attitude dynamics.Flexible solar panels or other appendages or fuel sloshing in the tanks, the motion ofwhich can be excited by an attitude or an orbit manoeuvre, may cause as a reactionoscillations of the spacecraft with respect to the desired attitude, which may harm overallpointing accuracy, unless properly damped passively by the inherent vehicle stability oractively by its automatic control system. Sometimes, a satellite may feature a nutationdamper, where a mass moving within a viscous liquid induces dissipation in order toasymptotically stabilise a pure spin condition (see Chapter 3).

If the centre of mass O does not maintain a fixed position with respect to the bodyand one must choose whether (i) keeping the reference frame centred in O or (ii) choosinganother reference point, allowing for displacements of O with respect to the origin of theframe. In the second case, the generalised form of Euler equation can be employed forthe angular momentum balance, the current position of the centre of mass and the inertiatensor being usually available from the knowledge of the (current) mass distribution.

The major advantage of the second approach lies in the fact that it is often possible toidentify a frame which is fixed with respect to the rigid structure of the spacecraft, whichallows to describe intuitively the orientation of the spacecraft itself and its equipment(antennae, sensors, and so on). If the first approach is taken, the pseudo–body framemoves with respect to the spacecraft structure because of changes in the mass distribution,so that the knowledge of its attitude does not imply the knowledge of spacecraft andsensor orientation. Moreover, the expressions of the attitude equations are usually rathercomplex.

Chapter 2

Dynamics of Spinning Spacecraft

Spin stabilisation in a simple, low cost and effective means of attitude stabilisation. Priorto, or just after deployment, the satellite in spun up about its axis of symmetry. Forthis reason, spin stabilised satellites are usually short cylinders. The angular momen-tum accumulated about the spin axis the provides “gyroscopic stability” against externaldisturbance torques.

Although simple and reliable, spin stabilised satellites are inefficient for power gener-ation. Since the satellite is continually spinning, the entire surface of the satellite mustbe covered with solar cells. In addition, payload efficiency is particularly low when onlyone direction is fixed in space and maneuvering of the spin axis complex.

Most of these drawbacks were solved by the introduction of a dual–spin satellites,where a part of the satellite (the so–called platform) is despun and brought at rest withrespect to the desired reference frame, while gyroscopic stability to the whole system isprovided by the spinning rotor. Payload efficiency is greatly enhanced as it is usuallypossible to pivot antennae or sensors with respect to the (fixed) platform, thus gretlyreducing the need for manoeuvres, unless it is necssary to aim them towards a target ina completely different direction.

2.1 Torque–free motion of axi–symmetric satellites

The principal moments of inertia of an axi–symmetric satellite will be given by

Jt = Jx = Jy

Js = Jz

where the subscripts t and s stand for transverse and spin, respectively, and we assumethat the symmetry axis coincides with the third (e3) axis of the body frame FB.

During the spin–up manoeuver, the satellite will accumulate angular momentum aboutthe spin axis, but because of various perturbations or imperfections, such as thrustermisalignment, the final condition at the end of the spin–up will hardly be a pure spinabout the spin axis e3. The imperfections will cause some (hopefully residual) nutation.

For torque–free motionsM1 = M2 = M3 = 0

of axial symmetric spacecraft, Euler’s equations take the following form:

Jtω1 + (Js − Jt)ω2ω3 = 0

Jtω2 + (Jt − Js)ω1ω3 = 0

Jsω3 = 0

27

28 G. Avanzini, Spaceflight Dynamics – 2. Dynamics of Spinning Spacecraft

The first two equations are coupled, while the third one is independent of the othertwo. This means that the latter one can be integrated on its own. The resulting (trivial)solution is given by

ω3 = Ω

where Ω is the (constant) spin rate about the spin axis.Letting

λ =Js − Jt

JtΩ

the first two equations can be rewritten as

ω1 + λω2 = 0

ω2 − λω1 = 0

Multiplying the first equation by ω1 and the second by ω2 and summnig up, one gets

ω1ω1 + ω2ω2 = 0

that isω2

1 + ω22 = ω2

12 = constant

This means that the component of the angular velocity vector ~ω that lies in the e1 − e2

plane, namely~ω12 = ω1e1 + ω2e2

has a constant magnitude. As also ω3 is constant, we get that

||~ω|| = ω21 + ω2

2 + ω23 = ω2

12 + Ω2 = constant

The first two equations of motion,

ω1 + λω2 = 0

ω2 − λω1 = 0

can be easily integrated. In fact, deriving the first one with respect to the time t, one gets

ω1 + λω2 = 0

that, taking into account the second equation, becomes

ω1 + λ2ω1 = 0

which is formally identical to the well known equation of the linear harmonic oscillator.The general solution

ω1(t) = A cos(λt) +B sin(λt)

for initial conditionsω1(t = 0) = ω1,0 ; ω1(t = 0) = ω1,0

becomes

ω1(t) = ω1(0) cos(λt) +ω1(0)

λsin(λt)

= ω12 sin[λ(t− t0)]

G. Avanzini, Spaceflight Dynamics – 2. Dynamics of Spinning Spacecraft 29

−0.1

−0.05

0

0.05

0.1

ω1 [r

ad s

−1 ]

−0.1

−0.05

0

0.05

0.1

ω2 [r

ad s

−1 ]

0 0.5 10

0.05

0.1

0.15

t [103 s]

ω3 [r

ad s

−1 ]

Figure 2.1: Time–history of angular velocity components for a free spin condition.

Deriving the solution for ω1 w.r.t. t and substituting into the first equation, it is

ω2 = − ω1

λ= ω1(0) sin(λt) − ω1(0)

λcos(λt)

= −ω12 cos[λ(t− t0)]

The evolution of ω1 and ω2 shows that ~ω12 whirls around e3 with angular velocity λ.Writing the angular velocity as

~ω = ~ω12 + Ωe3

during the evolution, ~ω describes a cone around the axis of symmetry e3 of the spinningbody, which is called the body cone.

The angular momentum vector can be written in the form

~h = J1ω1e1 + J2ω2e2 + J3ω3e3

= Jt(ω1e1 + ω2e2) + Jsω3e3

= Jt~ω12 + JsΩe3

It can be observed that ~h and ~ω are both a linear combination of the vectors ~ω12 ande3. Thus, during the motion of the spinning body, the vectors ~h, ~ω and e3 lie in the sameplane Π, that rotates around e3, if we look at the motion from FB.

In the most general case ~h and ~ω are not aligned. They are aligned only if ω12 = 0,that is, if we have a pure spinning motion about the symmetry axis. This is the desiredspin condition, where a sensor placed on the satellite on its symmetry axis points a fixeddirection in space. If ω12 6= 0, the motion of the spinning body is more complex (and thepointing less accurate). A geometric description of this motion will now be derived.

The assumption of torque–free motion, ~M = 0, implies that

d~h

dt= ~M = 0 =⇒ ~h = constant


e1 e2

e3

~ω~h

~ω12

~h12

Π

AAA

BB

PPq) γ

PPPPq

θ

Figure 2.2: Torque–free spinning of an axi–symmetric satellite as seen in FB.

in the inertial frame, and Π rotates around ~h, if we look at the motion from FI .

It is possible to define two angles, θ and γ, that remains constant in time,

tan θ =h12

h3

=Jtω12

Jsω3

tan γ =ω12

ω3

=⇒ tan θ =Jt

Jstan γ

If we choose an inertial frame such that E3 ‖ ~h, θ is the (constant) nutation angle andit defines the orientation of the symmetry axis e3 in the inertial space. The angle γ is thesemi–aperture of the body–cone. This means that the angle between ~h and ~ω, equal to|θ− γ|, is also constant, and ~ω describes a cone around ~h, fixed in the inertial frame, thespace cone.

The body cone is attached to the body axes, but it is not fixed in space. On theconverse, the space cone, attached to the vector ~h, that is constant in the inertial frame,is fixed in FI . The two cones remains tangent along ~ω, that is the axis of instantaneousrotation of the body, and the motion of the satellite can be represented by the body conerolling along the surface of the space cone.

As a final observation, it should be noted that, when Js > Jt (oblate body, that isa disk–shaped body), it is γ > θ, and the space cone is inside the body cone. On theconverse, when Js < Jt (prolate body, that is a rod–shaped body), it is γ < θ, and thespace cone is outside the body cone.

For what concern the attitude resulting from such a motion, substituting the expres-


E1 E2

E3 ‖ ~h~ω

e3

Figure 2.3: Torque–free spinning of an axi–symmetric satellite as seen in FI .

sions for the angular velocity components determined above in the following equation,

ω1 = Ψ sin Φ sin Θ + Θ cos Φ

ω2 = Ψ cos Φ sin Θ − Θ sin Φ

ω3 = Ψ cos Θ + Φ

and remembering that Θ ≡ θ = constant ⇒ Θ = 0, one gets

Ψ sin Φ sin Θ = ω12 sin[λ(t− t0)]

Ψ cos Φ sin Θ = −ω12 cos[λ(t− t0)]

Ψ cos Θ + Φ = Ω

By squaring and summing the first two equations, it is evident that

Ψ2 sin2 Θ = (ω12)2

so that Ψ is constant; Ψ is called precession rate or coning speed, and it is the angularvelocity of the line of the nodes on the horizontal plane.

Dividing the first equation by the second, the spin angle Φ is determined,

tan Φ = − tan[λ(t− t0)] ⇒ Φ = −λ(t− t0) ± π

Deriving w.r.t. time, the inertial spin rate is obtained

Φ = −λ

that is also constant. At this point, it is possible to evaluate the precession rate from thethird equation,

Ψ cos Θ + Φ = Ω ⇒ Ψ =Ω − Φ

cos Θ


But from the definition of λ,

λ =Js − Jt

JtΩ ⇒ Ω =

Jt

Js − Jtλ =

Jt

Jt − JsΦ

one gets

Ψ =Js

Jt − Js

Φ

cos Θ=Js

Jt

Ω

cos Θ

If Jt > Js, that is we have a prolate body, Ψ and Φ have the same sign and we havedirect precession, that is the precession rate is in the same direction of the spin rate. Onthe converse, if Js > Jt and an oblate body is dealt with, Ψ and Φ have different signsand we have retrograde precession, the precession rate being in the opposite direction withrespect to the spin rate.

An observation

It is important to note that the derivation presented in this paragraph are valid for any

rigid body which has two equal principal moment of inertia. This is a category much widerthan that of axi–symmetric bodies, including any prism with a basis made of a regularpolygon, but also any other body of irregular shape such that there exists a set of principalaxes of inertia such that Jx = Jy 6= Jz.

The pure–spin and the flat–spin condition

When the nutation angle is zero, the symmetry axis, the angular velocity and the angularmomentum vectors are all parallel. This situation, which is referred to as the pure–spin

condition, is an equilibrium for the system, as it can be easily demonstrated by checkingthat for ωB = (0, 0,Ω)T the derivatives of all the components of the angular velocityvector are zero. This means that in the pure–spin condition the symmetry axis is (andremains!) fixed in the inertial space, as far as no transverse angular velocity component ispresent, unless an external perturbation affects the satellite dynamics. As a consequence,this is the desired condition for the axi–symmetric spinning spacecraft. Any displacementof e3 from the direction of ~h would cause a precession rate, that is, the symmetry axisrotates about the (inertially fixed) direction of ~h and the sensors mounted along e3 willnot point towards a fixed direction in space.

The pure–spin condition is stable, in the sense of Lyapunov: A small displacementfrom the ideal case will induce precession, but the nutation angle remains constant. Thismeans that no divergence takes over and the pointing error remains finite and, possibly,small, for both prolate and oblate bodies. Such an ideal situation will be dramaticallyinfluenced by the effects of energy dissipation, as described in Section 2.3.

It can be seen the also ωB = (ω1, ω2, 0)T is an equilibrium. In this case the satellitespins about a transverse axis. As far as any minor perturbation about e3 will inducea constant (although slow) drifting motion about the symmetry axis, this condition isunstable in the sense of Lyapunov, inasmuch as ω1 and ω2 will vary as the sine andthe cosine of λt, thus leaving the neighbourhood of the equilibrium. Also in this case,dissipation will take its toll.


~h

~ω

O

N

d

σ

Figure 2.4: The invariable plane.

2.2 Torque–free motion of tri–inertial satellites

An analytical solution of the motion of tri–inertial rigid body can be derived in terms ofJacobi elliptic functions. Luckily there is also a geometric description of the same motion,due to Poinsot, which is much simpler, nonetheless extremely useful for the descriptionof the dynamics of an arbitrary rigid body.

Remembering that the kinetic energy

T =1

2~ω · ~h

and the angular momentum vector ~h are constant, it is possible to consider the (constant)dot product

~ω ·~h

h=

2Th

= d (2.1)

as the length d of a (constant) segment ON along the direction of ~h (Fig. 2.4). The

invariable plane σ, which is the plane perpendicular to the direction of ~h, placed at adistance d from the body center of mass O, is thus fixed in FI , and it represents the locusof all the possible ~ω that satisfy Eq. (2.1). Remembering that the Poinsot ellipsoid isthe locus of all the possible ~ω that satisfy the kinetic energy equation, the intersectionbetween the ellipsoid and the invariable plane must contain the angular velocity vector.

It is easy to demonstrate that such an intersection is a single point P , i.e. the Poinsotellipsoid and the invariable plane are tangent. Since the time derivative of the kineticenergy is zero,

dTdt

=1

2

d~ω

dt· ~h = 0

the increment d~ω and ~h are perpendicular, thus d~ω lie on σ. But since the vector ~ω +d~ωmust be also on the Poinsot ellipsoid, d~ω must be tangent to its surface. These twoconditions can be satisfied only if the Poinsot ellipsoid is always tangent to σ. Moreover,the Poinsot ellipsoid is fixed in the body frame FB, so that d~ω is the same in both FI

and FB. This means that the Poinsot ellipsoid rolls without slipping on σ, as depicted inFig. 2.5.


~h

d~ω~ω

O

N

d

σ+

Figure 2.5: The Poinsot ellipsoid rolls on the invariable plane.

The locus of all the possible ~ω’s on the Poinsot ellipsoid is given by the polhodecurve, which is the intersection between the Poinsot ellipsoid and the angular momentumellipsoid. Thus, during the rolling motion, the tangent point moves on the Poinsot ellipsoidalong the polhode. The corresponding curve on σ is the herpolhode. When the body isaxisymmetric, both the polhode and the herpolhode are circles and the situation can bedepicted in terms of space and body cones. In general the herpolhode is not a closedcurve, but the polhode must be a closed curve on the Poinsot ellipsoid, inasmuch as aftera revolution around the spin axis, the vector ~ω must attain again the same value, in orderto satisfy conservation of both kinetic energy and angular momentum.

2.2.1 Drawing the polhode curves

In order to draw the shape of the polhodes on the Poinsot ellipsoid, it is sufficient to recallthe equations of the Poinsot ellipsoid and the angular momentum ellipsoid, that are

ω21

(2T /Jx)+

ω22

(2T /Jy)+

ω23

(2T /Jz)= 1

ω21

(h/Jx)2+

ω22

(h/Jy)2+

ω23

(h/Jz)2= 1

Subtracting the first equation from the second and multiplying the result by h2, one gets

Jx

(

Jx −h2

2T

)

ω21 + Jy

(

Jy −h2

2T

)

ω22 + Jz

(

Jz −h2

2T

)

ω23 = 0

the polhode equation. In order to have real solutions for the above equation, the threecoefficients cannot have the same sign. For this reason the parameter J∗ = h2/(2T ) mustlie between the maximum and the minimum moment of inertia. Assuming, without lossof generality, that Jx > Jy > Jz, it is

Jz ≤ J∗ ≤ Jx


−2−1

01

2

−2−1

01

2

−2

−1

0

1

2

ω1

ω2

ω3

Figure 2.6: Polhode curves on the Poinsot Ellipsoid.

The easiest way to determine the shape of the polhodes is to consider their projectiononto the planes of the three–dimensional space ω1–ω2–ω3. Eliminating ω3 between theequations of the two ellipsoids brings the equation

Jx(Jx − Jz)ω21 + Jy(Jy − Jz)ω

22 = 2T (J∗ − Jz)

which represents an ellipse, since all the coefficients are positive. Similarly, eliminatingω1 one gets

Jy(Jy − Jx)ω22 + Jz(Jz − Jx)ω

23 = 2T (J∗ − Jx)

which is again the equation of an ellipse, inasmuch as all the coefficients are negative. Onthe converse, eliminating ω2, which is the angular velocity component with respect to theintermediate axis, brings

Jx(Jx − Jy)ω21 + Jz(Jz − Jy)ω

23 = 2T (J∗ − Jy)

which represent a hyperbola, the coefficients of the left–hand side being of different sign.It should be noted that, depending on the sign of J∗ − Jy, which can be either positiveor negative, the axis of the hyperbola will be vertical or horizontal, in the ω1–ω3 plane.When J∗ = Jy, the polhode equation degenerates into the form

Jx (Jx − J∗)ω21 + Jz (Jz − J∗)ω2

3 = 0

which represents the separatrices, the boundaries of motion about the axis of maximumand minimum inertia.


2.2.2 Stability of torque–free motion about principal axes

Spinning about any of the principal axis is an equilibrium condition for a rigid body ofarbitrary inertias. In this case, the angular momentum vector, the angular velocity vectorand the principal axis used as the rotation axis are all parallel and the time derivativesof all the components of ωB are zero. The three spin condition ωx = (Ω, 0, 0)T , ωy =(0,Ω, 0)T , and ωz = (0, 0,Ω)T are the pure–spin condition for the three–inertial body.

The shape of the polhodes already provide an information about the stability of theseequilibria. The axes of maximum and minimum inertia are centres surrounded by finitesize closed orbits, which means that a perturbation will induce a motion in the neighbour-hood of the original equilibrium. On the converse, the intermediate axis of inertia is asaddle point, that is a small perturbation will take the angular velocity vector “far” fromthe initial point in the neighborhood of the saddle, so that spinning around the interme-diate axis is an unstable condition for the satellite, and cannot be used for gyroscopicstabilisation.

These facts can be demonstrated analytically. Let us consider a spinning motion aboutthe z axis of the principal frame (this time without making assumptions on the order ofthe principal moment of inertia), such that ~ω0 = Ωe3. Assuming that ~ω = ~ω0 + ∆~ω,where the body frame components of ∆~ω, given by ∆ωB = ω1, ω2, ω3T , are smallperturbations with respect to Ω, Euler’s equations can be rewritten as follows

Jxω1 + (Jz − Jy)Ωω2 = 0

Jyω2 + (Jx − Jz)Ωω1 = 0

Jzω3 = 0

where second and higher order terms were neglected.The third equation is decoupled, thus stating that a perturbation on the spinning axis

does not affect the other two. The first two equations can be rewritten in matrix formω1

ω2

=

[0 (Jy − Jz)Ω/Jx

(Jz − Jx)Ω/Jy 0

]ω1

ω2

The characteristic equation is thus given by

λ2 + Ω2 (Jy − Jz)(Jx − Jz)

JxJy= 0

and its roots are pure imaginary,

λ = ±iΩ√

(Jy − Jz)(Jx − Jz)

JxJy

if either Jy < Jz and Jx < Jz, or Jy > Jz and Jx > Jz, that is if the spin axis is eitherthe axis of maximum or minimum moment of inertia. In such a case, the solution of theperturbation equation is confined in the neighborhood of the spin condition and divergencebecause of neglected nonlinear terms is ruled out on energy conservation grounds, so thatthese conditions are Ljapunov (although not asymptotically) stable. On the converse, ifthe spin axis is the intermediate one, the product (Jy −Jz)(Jx−Jz) is negative, inasmuchas one of the two factor is positive and the other is negative. In this case, the roots areboth real,

λ = ±Ω

√

−(Jy − Jz)(Jx − Jz)

JxJy


and one of the two eigenvalues is strictly positive. This means that spinning around theintermediate axis is an unstable equilibrium for the spinning rigid body.

2.3 Nutation damping

Nutation damping is a simple yet effective way to restore a state of pure spin, if a nutationangle different from zero should be induced by some external cause. We will now inves-tigate how and under which circumstances it is possible to exploit the effects of energydissipation in order to make a pure spin condition asymptotically stable.

2.3.1 Effects of energy dissipation on axi–symmetric satellites

As a matter of fact, the rigid body is an abstraction. Usually flexible appendages and/orfuel sloshing induce some energy dissipation the effects of which can be easily determinedat least in the case of axi–symmetric satellites by use of the energy sink model.

The kinetic energy of a rotating rigid body is given by

T =1

2(Jxω

21 + Jyω

22 + Jsω

23

The expression for T can be rearranged for the axi–symmetric case in the form

T =1

2Jt(ω

21 + ω2

2)︸︷︷︸

transverse

+1

2JsΩ

2

︸︷︷︸

spin

At the same time, the angular momentum is

~h = Jtω1e1 + Jtω2e2 + Jsω3e3 ⇒ h2 = J2t (ω2

1 + ω22) + J2

s Ω2

so that the quantity (2JsT − h2) is

(2JsT − h2) = JsJt(ω21 + ω2

2) + J2s Ω2 − J2

t (ω21 + ω2

2) − J2s Ω2

= Jt(ω21 + ω2

2)(Js − Jt)

By taking the time derivatives of both sides of the equation, and remembering that,for zero external torque the angular momentum vector remains constant, one gets

dTdt

=1

2

Jt

Js

(Js − Jt)d

dt(ω2

1 + ω22)

If there is no dissipation, T = 0 and ω21 + ω2

2 is constant (as already demonstrated inSection 2.1). If there is dissipation, T < 0, which means that, assuming an approximatelyconstant mass distribution, so that the moments of inertia are unaffected, ω2

1 + ω22 must

either grow or decrease. It should be noted that, if the body is oblate, than Js−Jt > 0 andd(ω2

1 + ω22)/dt < 0, which meand that the transverse component of the angular velocity

vector is decreasing. Dissipation will stop only when it is ω21 + ω2

2 = 0, that is, when apure spin condition is reached. In such a case, dissipation turns the pure spin conditioninto an asymptotically stable equilibrium: a displacement from such a condition will bedamped out at the expenses of a reduction of the system energy.

On the converse, if the body is prolate and Js − Jt < 0, than dissipation induces anincrease of the transverse angular velocity component, as in order to satisfy the equation,


−0.1

−0.05

0

0.05

0.1

ω1 [r

ad s

−1 ]

−0.1

−0.05

0

0.05

0.1

ω2 [r

ad s

−1 ]

0 0.5 10

0.05

0.1

0.15

t [103 s]

ω3 [r

ad s

−1 ]

Figure 2.7: Time–history of angular velocity components with nutation damping.

d(ω21 + ω2

2)/dt must be positive. This means that the nutation angle is growing, beingproportional to ω12/Ω. In such a case the pure spin condition about the symmetry axis,which was stable in the sense of Lyapunov for the perfect rigid case, becomes unstable.Although divergence is usually slow, the nutation angle will steadily increase with time.

The rate of variation of the nutation angle is slow enough to make its rate negligiblewith respect to Ψ. Under this assumption it is possible to state that

ω21 + ω2

2 ≈ Ψ2 sin2 Θ

and, as a consequence, it is possible to formulate the energy decay rate as

dTdt

=1

2

Jt

Js

(Js − Jt)d

dt(Ψ2 sin2 Θ)

Moreover, for small nutation angles the precession rate can also be considered approx-imately constant, and differentiation with respect to time bring the following expression:

dTdt

=Jt

JsΨ2(Js − Jt) sin Θ cosΘΘ

When energy is dissipated and T < 0, the quantity (Js − Jt) sin(2Θ)Θ must also benegative, and a nutation rate is induced by dissipation. As qualitatively shown previously,if the satellite is an oblate body and Js > Jt, Θ is negative for small nutation angles, andtends to decrease with time because of dissipation. This means that spinning around thesymmetry axis is asymptotically stable, as Θ → 0. For a prolate body, when Js < Jt, theresulting nutation rate is positive and the nutation angle increases with time: the satellitetends towards a flat spin condition, Θ → π/2.

Note that both the pure–spin and flat–spin conditions are always equilibria for thesatellite, regardless of its shape, inasmuch as T = 0 and Θ = 0 if either Θ = 0 or π/2,but the stability of these equilibria is affected by the mass distribution.


A

O~b

(md/m)ξ

md

ξ

~n

e1

e2

e3

+u

Figure 2.8: A satellite equipped with a “ball–in–tube” damper.

Also note that as ω1 and ω2 are damped out in the stable case, ω3 increases becauseof conservation of angular momentum. Letting Ωf be the final spin rate, it is

h2 = J2t (ω2

1 + ω22) + J2

s Ω2 = J2s Ω2

f

Therefore,

Ω2f = Ω2 +

J2t

J2s

(ω21 + ω2

2)

and the final spin rate results higher than the initial one, as expected (Fig. 2.7). In theunstable case, dissipation will make the spin angular velocity component drop towardszero as the angular momentum is transferred by dissipation in the transverse direction.

2.3.2 More accurate models

In a real system there are several causes that induce energy dissipation, such as fuelsloshing. Most satellites use thruster to actively remove nutation and maintain the desiredpure spin condition, when the nutation angles increases beyond a prescribed threshold.For oblate bodies a nutation damper provides a very simple and useful means for passivelystabilizing the pure spin motion. An example of damper is the ball–in–tube: a plastic tubeis filled with viscous fluid and a ball bearing is free to move in the fluid, thus dissipatingenergy. The only drawback of this technique is that it may take several minutes (if nothours) to remove nutation.

In order to derive the equations of motion of a satellite equipped with a spring–mass–dashpot damper, it is necessary to consider the generalized form of the Euler equation,

d~h

dt+ ~S × ~aA = ~M


written in a reference frame FA attached to the satellite platform P, with origin in thepoint A, which is the center of mass of the satellite when the damper mass md is in theundeformed position P (ξ = 0) for the spring (Fig. FigSatDamp). All the moments aretaken with respect to A, while moment of inertia and position of the centre of mass O areaffected by mass displacement along the tube.

The derivation of the complete set of equations of motion of the satellite equipped witha damper is out of the scope of the course, but the results obtained from the analysis of themodel are of paramount importance. For a three–inertial satellite one finds out that, inpresence of energy dissipation, the only stable pure–spin condition takes place about theaxis of maximum inertia. Dissipation makes such a spin condition asymptotically stable,that is, it transforms the closed curves around the reference spin condition into convergingspirals. On the converse, spin about the axis of minimum inertia is transformed into anunstable equilibrium, the closed curves surrounding it being transformed into divergingspirals. The instability of the spin condition about the intermediate axis is unaffectedby dissipation. The spacecraft will simply converge towards a spinning motion about theaxis of maximum inertia, unless some external action prevents it from doing so.

2.4 Attitude Maneuvers of a spinning satellite

As stated since the beginning of the chapter, one of the major drawbacks of spin stabilisa-tion lies in the complexity of the technique necessary for performing attitude manoeuvres,that is, changing the direction of the spin axis of the satellite in a controlled way. In orderto rotate the spin axis of a spinning spacecraft, it is possible to exploit the precessionrate induced by a properly sized nutation angle. This requires a variation of the angu-lar momentum vector that can be obtained by applying a proper control torque to thespacecraft.

If we consider a thruster generating a force ~F at position ~r in the body frame centeredin the center of mass O, the resulting moment acting on the satellite is

~M = ~r × ~F

Given a firing time of ∆ton seconds, starting at t0, the overall change of angular momentumis equal to

∆~h =

∫ t0+∆ton

t0

(~r × ~F )dt.

If the firing time is sufficiently small with respect to the period of rotation of the spinningspacecraft Tspin = 2π/Omega, it is possible to approximate ∆~h as

∆~h = (~r × ~F )∆ton

The control torque is usually delivered by use of coupled pairs of thrusters, in sucha way that a net torque is produced, with no force perturbing the orbit motion of thespacecraft.

The torque pulse changes almost instantaneously the direction of the spacecraft’sangular momentum vector, thus creating a nutation angle θ. At this point the spacecraftis no longer spinning around its symmetry axis, and a precession rate builds up, equal to

Ψ =Js

Jt − Js

Φ

cos Θ


h0

∆h1h1

Or

F

θ+

:

O

h0

∆h1

h1 ∆h2

h2

O r

F

θ+

:θXXz

Figure 2.9: Changing the spin axis direction with nutation angle and precession rate.

It must be remembered the the direction of the precession motion depends on whetheran oblate or a prolate body is considered, but in the latter case the instability of the spincondition around the axis of minimum inertia when dissipation is present prevents theuse of this technique for changing the spin axis direction in most real spacecraft, unlessthe dissipation rate is very small.

Since the precession rate is constant, the precession angle will vary by

∆Ψ =Js

Jt − Js

Φ

cos Θ∆t

After a time interval equal to

T = π(Jt − Js) cos Θ

JsΦ

the precession angle is varied by π rad, and the direction of the symmetry axis of thespinning satellite has changed of 2θ. If the firing time ∆ton is such that the resultingnutation angle is half of the desired variation of the spin axis direction and the thrusteractivation time t0 is chosen in such a way that the resulting angular momentum incrementlies in the manoeuvre plane (that is, the plane containing both the initial and final positionof the satellite spin axis), then after T seconds the spin axis will be in the desired position,alhough moving at the coning speed Ψ. At this point a second thruster firing is requiredto stop the precession and achieving again a pure spin condition about the new re-orientedaxis.

From simple trigonometric consideration, it is easy to show that

||∆h1|| = ||∆h2|| = h0 tan θ = JsΩ tan θ

where h0 is the magnitude of the initial angular momentum. But being also

||∆h|| = M∆t


the required firing time for each pulse is equal to

∆ton =JsΩ tan θ

M

If ℓ is the thruster moment arm (i.e. the distance between the thruster axis and thesatellite centre of mass; in Fig. 2.9 this distance is equal to the radius of the spinningbody) and only one thruster is employed, it is M = Fℓ. In general at least two thrustersare activated in order to obtain a balanced control torque, as stated before, in which caseit is M = 2Fℓ

In order to estimate the amount of propellant necessary to perform the reorientation,it must be remembered that the specific impulse of an engine is given by

Isp =F

∆mg/τ

that is the thrust delivered divided by the weight of propellant burned in a unit time.The specific impulse is one of the most important characteristics of rocket engines of anysize, and from its value it is possible to determine

∆m =Fτ

gIsp

The total propellant mass required to rotate the spin axis of an angle equal to 2θ is

∆mtot = 4JsΩ tan θ

gIspℓ

if two thrusters are employed for delivering the control torque M .For small reorientation angles it is tan θ ≈ θ, and the propellant consumption is

roughly proportional to the reorientation angle itself. But as θ increases, the fuel con-sumption gets larger, so that it is more convenient to perform the overall reorientationin several smaller steps. Moreover, for a given Isp, the pulse duration may become a notnegligible fraction of the spin period, thus reducing the change in angular momentum.The drawback of a large reorientation performed in N steps is that the manoeuvre timeis increased and the overall manoeuvre is usually less accurate.

Figure 2.10 shows the results for reorientation 2θ angles between 0 and 70 deg, for asatellite with Js = 20 kg m2, Jt = 15 kg m2, spinning at Ω = 2 rad s−1, that is controlledby pair of thrusters with a 1 m moment arm, which produce 10 N of thrust each, with aspecific impulse Isp = 120 s.

The minimum pulse duration considered is 10 ms, that puts a limit on the resolutionof the reorientation, the minimum manoeuvre angle being 0.573 deg. On the other side,a thruster firing time longer than one tenth of the spin period cannot be considered a“pulse”. This limits the maximum reorientation angle that can be achieved by a sin-gle nutation manoeuvre, that is 17.9 deg. A stricter requirement on the pulse durationwill reduce the maximum available reorientation angle with a single manoeuvre. Using 2steps, the propellant savings at low angles are negligible (less than 1%), but it is possibleto achieve spin axis reorientation up to 35 deg. With 4 steps the maximum reorienta-tion angle becomes as high as 71 deg. Of course, the manoeuvre duration will increaseaccordingly.

The correct timing of the thruster pulses is of very important in order to obtaine anaccurate reorientation. In this framework it should be noted that, unless a very particular


0 20 40 600

0.01

0.02

0.03

0.04

0.05

Pro

pella

nt m

ass

∆ m

[kg]

θ2 [deg]0 20 40 60

0

1

2

3

4

5

Man

oeuv

re ti

me

T [s

]

θ2 [deg]

0 20 40 600

0.5

1

1.5

2

2.5

3

θ2 [deg]

Tot

al fi

ring

time

[s]

0 20 40 600

100

200

300

400

θ2 [deg]si

ngle

pul

se ∆

t [m

s]

Nman

= 1

Nman

= 2

Nman

= 4

∆ m ∝ θ

∆ m ∝ tanθ

Figure 2.10: Spinning satellite manoeuvre: Fuel consumption (a), Total manoeuvre time(b), Total firing time (c), single pulse duration (d).

case is dealt with, where the precession and the spin rate are in a rational ratio, the pair ofthrusters used for starting the manoeuvre will not be in the correct position for deliveringthe required torque in the manoeuvre plane after half of the precession revolution. Thismeans that two pairs of thrusters need to be activated for the second pulse, delivering atotal net torque equal to the required one. As a consequence, the above description ofthe manoeuvre technique will provide an estimate of fuel consumption slightly optimisticand the overall complexity of the manoeuvre technique is, if possible, even higher in theactual implementation.

Chapter 3

Dual–spin satellites

The presence of only one single inertially fixed direction on board of gyroscopically stabi-lized satellites greatly limits their use, as far as it is possible to mount a sensor payload ora communication antenna only along that direction. Moreover, maneuvering the satellitein order to reorient the spin axis is both difficult and expensive in terms of fuel consump-tion, the required propellant being proportional to the amount of angular momentumstored in the spinning satellite.

In order to overcome this negative features, an alternative satellite configuration wasproposed and used in the past, where the payload and the communication hardware ismounted on a platform P which has a relative rotational degree of freedom with respectto the spinning part of the satellite, the rotor R. During orbit injection the two elementsare locked and the satellite behaves like a standard rigid body. After the operationalorbit is reached, the spin–up motor accelerates the rotor, with respect to the platform,in order to transfer on the rotor the entire vehicle angular momentum, thus despinningthe platform. The de–spin maneuver is ended when the platform is fixed in space, whilethe rotor provides gyroscopic stability to the entire vehicle. The main advantage of sucha solution is that, pivoting the payload mounted on the platform allows aiming it in anydirection in space, with great flexibility, without changing the direction of the rotor spinaxis.

From the configuration point of view, the first dual–spin satellites were made of alarge rotor, containing most of the satellite equipment, with solar cells mounted on thesurface of the rotor. The configuration resembled that of a standard (for that time)spinning satellite. A small platform was attached to it, as represented in Fig. 3.1. Later,the platform became larger (about one half of the satellite). The Intelsat IV was thefirst prolate dual spin satellite (3.2). Finally, the rotor was substituted by a momentumwheel, which is a small (and light) wheel spinning at a very high speed, placed inside theplatform. In this latter case, the satellite is called a momentum–bias satellite.

The mathematical model that will be derived in the next section is not affected bythe relative size of platform and rotor. In this framework, both configurations (the dual–spinner, with a large rotor attached outside of the platform, and the momentum biassatellite, with a small spin–wheel inside the spacecraft bus) can be referred to as gyrostats.The presence of a relative rotational degree of freedom between the platform and the rotor(or spin–wheel) of the gyrostat makes it necessary to derive a new formulation for theangular momentum balance, since the body, featuring a rotating element, is no longerrigid.

45

46 G. Avanzini, Spaceflight Dynamics – 3. Dual–spin satellites

Figure 3.1: Dual spin satellite: configuration with large rotor (e.g. GOES-7 satellite).

Figure 3.2: Dual spin satellite: configuration with equivalent rotor and platform (e.g.Intelsat IV satellite).

G. Avanzini, Spaceflight Dynamics – 3. Dual–spin satellites 47

O

ΩW

Pa

e1

e2

e3

Figure 3.3: Sketch of a satellite equipped with a momentum wheel.

3.1 Mathematical model of a gyrostat

Figure 3.3 shows the generic arrangement of a momentum wheel W spinning about thespin axis a at an angular rate Ω with respect to the satellite platform P. The mathematicalmodel is derived under the assumption that both the platform and the wheel are perfectlyrigid. Moreover, a perfectly balanced and axi–symmetric wheel is assumed, the centre ofmass of which lies on the spin axis. As a result, the presence of a relative degree offreedom between W and P does not affect the mass distribution of the spacecraft, whichis independent of the angular displacement of W about a. For this reason it is stillpossible to define a body frame FB, fixed with respect to the platform, the inertia tensorI of the gyrostat P + W being constant in FB as for a standard rigid body. This meansthat I represents the total moment of inertia of the gyrostat, including the wheel.

3.1.1 Angular momentum formulation

If the wheel is locked to the platform, the angular momentum is given (as usual) by~h = I~ω, but if the wheel is spinning, than the overall angular momentum features asecond term,

~h = I~ω + IsΩa

where, letting Is be the moment of inertia of W about a, the quantity hr = IsΩ is therelative angular momentum. The absolute angular momentum stored in the spinningrotor

ha = hr + Isa · ~ωis made up of two contributions, the relative angular momentum due to the spin motionwith respect to P, and the angular momentum related to the motion of the platform.


Writing the components in body frame, it is

hB = IωB + IsΩaB

ha = Is(Ω + aT

BωB

)

The equation of motion of the gyrostat can be written in terms of evolution of the angularmomentum vector as

hB = −ωB × hB + MB

ha = ga

where MB is the external torque acting on the satellite, while ga is the torque applied bythe de–spin engine to the rotor. It should be noted how ga, which is an internal torque,does not appear in the equation of the time evolution of the overall angular momentum.Nonetheless its effects on the overall dynamics are significant.

Also, the presence of the additional rotational degree of freedom increases the orderof the system of first order nonlinear differential equations with respect to the rigid–bodycase, from 3 to 4 (three components for the total angular momentum plus the absolutewheel angular momentum about the spin axis). In order to integrate this fourth–order sys-tem and to evaluate the Euler angle rates, it is necessary to determine the angular velocityωB. Integrating the Euler angle rate equation (or the quaternion evolution equation), onegets the time evolution of satellite attitude.

The kinematic problem for the attitude of the satellite is solved expressing ωB as afunction of the total and wheel angular momentum, that is

ωB = I−1 (hB − IsΩaB)

The value of ωB thus depends on both hB and Ω. The rotor angular speed can beexpressed as

Ω = ha/Is − aTBωB

where ωB appears, so that the two equations must be solved simultaneously, but this canbe done easily, since the two equations are linear in the angular velocity components. Ifone assumes a set of principal axes of inertia, such that I = diag(Jx, Jy, Jz), the resultinglinear system is given by

Jx 0 0 Isa1

0 Jy 0 Isa2

0 0 Jz Isa3

Isa1 Isa2 Isa3 Is

ω1

ω2

ω3

Ω

=

h1

h2

h3

ha

Note that the 4 × 4 matrix with the moments of inertia must be inverted only once.

3.1.2 Angular velocity formulation

An alternative formulation of the equations of motion is also available, where gyrostatdynamics is expressed directly in terms of the angular velocity components and rotor spinrate. In this case, the angular momentum vector is writte as

hB = IωB +(ha − Isa

TBωB

)aB

=(I − IsaBaT

B

)ωB + haaB


The time derivative of the angular momentum vector components are thus expressed as

hB =(I − IsaBaT

B

)ωB + gaaB

At the same time it isha = ga = Is

(

Ω + aTBωB

)

so that, upon substitution of hB and ha into the angular momentum formulation, theequations of motion of the gyrostat become

ωB = I−1

[MB − ωB × (IωB + IsΩaB) − gaaB]

Ω = ga/Is − aTBωB

where I =(I − IsaBaT

B

)is the pseudo–inertia tensor. These equations of motion can

be used for describing the gyrostat spin–up dynamics, during platform de–spin, that is,a spin axis torque ga is applied until the platform angular velocity drops to zero. Theknowledge of ωB allows for a direct evaluation of the Euler angle rates.

3.2 Simplified models

3.2.1 The Kelvin gyrostat

In many application, especially once the de–spin manoeuvre is completed, it is possibleto assume that the rotor relative spin rate remains constant, under the action of a controlaction that drives the motor torque in order to compensate for fluctuations.

If one assumes that the angular speed of the rotor remains exactly constant, thesystem looses one degree of freedom, as Ω(t) = const and the associated dynamics can beneglected. In this case the so–called Kelvin gyrostat is obtained. The variable hr = IsΩbecomes a system parameter and the platform equation of motion decouples from rotorspin dynamics. In such a case, it is

IωB = −ωB × (IωB + hraB) + MB

In order to keep Ω constant (that is, ˙Omega = 0), it is necessary to apply a certain spintorque to the rotor, which, from the equation of motion of rotor spin dynamics can beevaluated as

ga = IsaTBωB

In real applications it is not possible to keep the rotor speed exactly constant and asimple feed–back loop is employed for determining the required spin torque. A tachometeris mounted on the wheel shaft, which measure the actual rotor spin rate Ω. The errorwith respect to the nominal (desired) rotor speed Ωdes is the input for the rpm regulator.As an example, the spin torque can be set equal to

ga = K(Ωdes − Ω)

where K is the gain of the rotor control system. In this way the rotor is accelerated(ga > 0) whenever the angular speed is lower than the prescribed one, and deceleratedwhen Ω > Ωdes. Of course, if the spin wheel dynamics is modeled including this simplecontrol scheme, it is no longer possible to treat Ω as a constant in the platform angularvelocity equations, and the full system equations must be accounted for.


3.2.2 The apparent gyrostat

If the rotor spin–up torque is zero (ga = ha = 0), the rotor absolute angular momen-tum becomes constant and the apparent gyrostat model is dealt with. Also in this caseone state variable is lost, since it can be determined from the knowledge of a constantparameter, namely ha. The equation of motion can thus be rewritten as

IωB = −ωB ×(

IωB + haaB

)

+ MB

with the usual meaning for the symbols.Note that, provided that the spacecraft moment of inertia, I, and wheel relative

angular momentum, hr, are replaced by the pseudo–inertia matrix, I, and the absoluteangular momentum, ha, respectively, the two reduced order models for the Kelvin andthe apparent gyrostats share the same expression.

3.3 Stability of axial gyrostat

Although of little practical use, the apparent gyrostat model is interesting for the deriva-tion of analytical results on the stability of the de–spun condition. The simple axial caseis discussed in this paragraph.

Assuming that FB is a set of principal axes of inertia and that the spin axis is parallelto e3 (axial rotor), the satellite and rotor absolute angular momentum can be written,respectively, as

hB =

Jxω1

Jyω2

Jzω3 + IsΩ

; ha = Is(Ω + ω3)

The equation of motion of the axial gyrostat can be written as follows:

Jxω1 + (Jz − Jy)ω2ω3 + IsΩω2 = M1

Jyω2 + (Jx − Jz)ω1ω3 − IsΩω1 = M2

Jzω3 + (Jy − Jx)ω1ω2 + IsΩ = M3

Is(Ω + ω3) = ga

When Ω = 0, the dual–spin satellite behaves like a standard rigid body. This situationis called all-spun condition. After platform de–spin, the satellite angular velocity must bezero and all the satellite angular momentum must be stored in the rotor. The stabilityof the equilibrium for the de–spun condition of a Kelvin gyrostat with axial rotor can beanalyzed as usual assuming the components ωB = (ω1, ω2, ω3) as small and neglectingsecond order terms in the equation of motion, which can be written for the torque–freecase as

Jxω1 + IsΩω2 = 0

Jyω2 − IsΩω1 = 0

Jzω3 = 0

The third equation is decoupled from the others, stating that a perturbation about e3

will remain constant. The first and the second equations can be recast in matrix form asω1

ω2

=

[0 −(Is/Jx)Ω

(Is/Jy)Ω 0

]ω1

ω2


It is easy to show that the eigenvalues of the state matrix are a pair of conjugateimaginary numbers, provided that Ω 6= 0. No unstable eigenvalues are present for anyvalue of Ω, so that exponential divergence is ruled out. It is possible to show, on angularmomentum conservation grounds, that the variation of ω1 and ω2 remains confined in aneighborhood of the origin, so that the system is stable in the sense of Lyapunov. Thismeans that the rotor provides the required gyroscopic stability to the entire satellite.

Adding damping (e.g. under the action of a nutation damper) it is possible to trans-form the L-stable de–spun condition into an asymptotically stable equilibrium.

Chapter 4

Active Stabilisation and Control ofSpacecraft

For many applications where accurate payload pointing is required full three-axis controlis used. The satellite does not spin (except during slew manoeuvres) and is stabilised byreaction wheels (gyros) and/or thruster relative to a given desired attitude (e.g. an Earthfacing attitude). Since the satellite is not spinning, large Sun facing solar panels canbe used for efficient power generation. The satellite is also capable of fast reorientationmanoeuvres (slews) in any arbitrary direction, if sufficient control power is available fromthe actuators. However, due to the added complexity, 3-axis stabilised satellites arefar more expensive than simple spin–stabilised ones. The application and its accuracyrequirements must justify the increase in costs and the reduced life expectation of thevehicle.

In order to actively stabilize the satellite and allow for autonomous maneuvering, thesatellite must be equipped with a suitable set of sensors, which provide the on–boardcomputer with the necessary information on the current attitude. The precision of thecontrol action depends on the accuracy of the attitude measurement. Moreover, thecontrol law itself depends on the type of actuators available. As a consequence, a briefoutline on the more common control actuator is given, prior to the discussion of attitudeactive control.

4.1 Actuators

Spacecraft control actuators can be based on different physical principles. The choice inthe type of actuator installed on board of the satellite greatly affects its configuration, theresulting pointing accuracy, and even the implementation of the control laws themselves.

A control action can be delivered to the satellite by either a direct torque producedby a (set of) thruster(s), by exchanging angular momentum between parts in relativerotational motion or by application of a magnetic torque, obtained from the interactionof an electric coil and the Earth magnetic field.

• Reaction control system (RCS)

– Cold gas jets

– Chemical propulsion (Mono–propellant or Bi–propellant propulsion system;can be used also for orbit control)

53

54 G. Avanzini, Spaceflight Dynamics – 4. Active Stabilisation and Control of Spacecraft

• Momentum exchange devices

– Reaction wheels

– Bias–momentum wheel

– Double–gimbal bias–momentum wheel

– Control Moment Gyroscopes

• Magneto–torquers

4.2 Linear model of rigid satellite attitude motion

For a general three axis motion, Euler’s equation states that

hB + ωB × hB = MB (4.1)

If there are no internal moving parts (e.g. no spinning rotors) this equation can be writtenin the form

IωB + ωB × (IωB) = MB

Assuming that the rotation is slow, the angular velocity can be considered as a first orderperturbation of the steady state ωB = 0, so that ||ωB|| ≪ 1 and ωB × (IωB) is a secondorder (negligible) term in the equations of motion. Euler’s equation thus becomes

IωB = MB

When dealing with a stabilization problem, that is small perturbations with respect toa prescribed nominal attitude are considered, also the angular displacement is “small”. Insuch a case, it is possible to demonstrate that, letting the variation of the Bryan’s anglesbe equal respectively to

∆ψ = θ3 ; ∆θ = θ2 ; ∆φ = θ1

and defining θ = θ1, θ2, θ3T , it is

θ = ωB

Euler’s equation in linear form become simply

Iθ = MB (4.2)

Finally, assuming that a set of principal axes of inertia is chosen as the body frame, theinertia matrix is in diagonal form, I = diag(J1, J2, J3), and three independent secondorder linear ordinary differential equations are obtained, one for each axis,

Jiθi = Mi , i = 1, 2, 3

It should be noted that for single axis rotations, a scalar equation in the same form asEq. (4.2) is obtained, regardless of the angular speed and angular displacement, that is

Jθ = M

This equation will be the starting point for the analysis of attitude slew maneuvers.

G. Avanzini, Spaceflight Dynamics – 4. Active Stabilisation and Control of Spacecraft 55

4.3 Linear model of gyrostat attitude motion

If the satellite is equipped with a momentum wheel with a moment of inertia Is spinningabout the axis a with a relative spin rate Ω, the total satellite angular momentum is

hB = IωB + hraB

where hr = IsΩ is the relative angular momentum; Ω can be written as

Ω = Ω0 + ∆Ω

being ∆Ω ≪ Ω0, i.e. ∆Ω is a “small perturbation” of the reference wheel spin rate Ω0.Equation (4.1) thus becomes

IωB + Is∆ΩaB + ωB × [IωB + Is (Ω0 + ∆Ω) aB] = MB

Neglecting higher order term one gets

IωB + Is∆ΩaB + IsΩ0ωB × aB = MB

The evolution of the perturbations of Euler’s angles is again

θ = ωB

so that the equations of motion becomes can be expressed in scalar form as follows:

J1θ1 + Is∆Ωa1 + IsΩ0(θ2a3 − θ3a2) = M1

J2θ2 + Is∆Ωa2 + IsΩ0(θ3a1 − θ1a3) = M2

J3θ3 + Is∆Ωa3 + IsΩ0(θ1a2 − θ2a1) = M3

This set of equations is coupled with the dynamics of the spin–wheel, described by the(linear) equation

Ω ≡ ∆Ω =ga

Is− ωT

BaB

When the rotor spin axis is parallel to the i–th principal axis of inertia (a ≡ ei), thei–th equation of motion becomes decoupled from the others, because aj = ak = 0, forj, k 6= i. As an example, let us assume that a ≡ e2, so that a1 = a3 = 0 and a2 = 1. Theequations of motion become

J1θ1 − IsΩ0θ3 = M1

J2θ2 + Is∆Ω = M2

J3θ3 + IsΩ0θ1 = M3

The pitch dynamics is decoupled from the roll and yaw motions, but is coupled with thespin–wheel dynamics

∆Ω =ga

Is− ω2

Summing up, the torque–free pitch dynamics (M2 = 0) is described by the equations

(J2 − Is) θ = −ga

Ω = Ω0 −J2

Isθ

On the converse, roll and yaw motions are coupled by the gyroscopic term. Moreover,if the rotor spin speed is time–varying and ∆Ω is not negligible, the roll and yaw couplingterms depend on the pitch dynamics through Ω.


4.4 Use of thrusters for attitude control

4.4.1 Single axis slews (open loop)

If we consider a constant torque acting around the generis i–th axis, the angular acceler-ation is

αi = θi = Mi/Ji

Integrating from the initial condition θ(t0) = θ0, θ(t0) = θ0 (and dropping the subscripti), one has

θ(t) = θ0 + α(t− t0)

θ(t) = θ0 + θ0(t− t0) +1

2α(t− t0)

2

The same equation can be integrated in the phase plane θ vs θ. By application of thederivation chain rule it is

θ =dθ

dθ

dθ

dt

so that one gets

α = θdθ

dθ

Integrating the previous equation in θ, one gets

∫ θ

θ0

ϑdϑ =

∫ θ

θ0

αdϑ

that, for constant applied torque, becomes

1

2

(

θ2 − θ20

)

= α(θ − θ0)

or,

θ2 = θ20 + 2α(θ − θ0)

Starting from a rest position in the nominal attitude (θ = 0 and θ = 0), one gets

θ2 = 2αθ =⇒ θ(t) =√

2αθ

A single–axis rotation manoeuvre can thus be performed in three phases: an initialacceleration phase (i) during which the spacecraft achieves the desired slew rate θd for thereorientation; the acceleration is followed by a drifting phase (ii), during which a constantangular velocity remains constant with no action from the thrusters, until, at a propertime instant, the drifting motion is stopped during the deceleration phase (iii) by a secondimpulse from the thrusters. Representing the manoeuvre in the phase plane, the statevariables evolve as reported in Fig. 4.1.

The configuration of the thrusters used for spinning the satellite up to the drift angularvelocity can be different. As reported in Fig. 4.2, a single thruster delivering a force ~Fwith a moment arm ℓ with resepct to the spacecraft centre of mass produces a torque themagnitude of which is Fℓ. Unfortunately, also a net force F is produced which will affectorbit parameters, thus coupling the orbit and the attitude control proboems.


θ

θ

θ0

θd

θf

acceleration drift deceleration

.

.

Figure 4.1: Evolution of θ and θ for a rest–to–rest manoeuvre.

XX -

j

ℓ

~F

XX

XX

-

j

ℓ

~F

ℓ

~F

Figure 4.2: Unbalanced and balanced thruster configurations.

On the converse, if a balanced configuration is employed, the simultaneous firing oftwo thrusters in opposite directions will produce only a torqe M = 2Fℓ, with a zero netforce, so that the orbit motion of the spacecraft is unaffected by attitude control pulses.

An approximate evaluation of the total manoeuvre time and the propellant necessaryfor performing it can be derived by neglecting the duration of the thruster firings foraccelerating and decelerating the angular motion of the vehicle, taking into account onlythe drift phase. Assuming that the drift angular velocity is θd, the time required for amanoeuvre of amplitude equal to θf is simply given by T = θf/θd.

Assuming a sharp thrust profile, it is

θd =2Fℓ

J∆t

and the pulse width (thruster on–time) for the spin–up is given by

∆ton,1 =Jθd

2Fℓ


From the definition of the thruster specific impulse

Isp =F∆ton

g∆mf

(that is the ratio between the momentum gained and the weight of the propellant used)the mass of propellant necessary for the acceleration is

∆m1 = 2F∆ton,1

gIsp=

Jθd

gℓIsp

The drift angular velocity can be determine as a function of the manoeuvre time, thatis, θd = θf/T , so that the propellant mass necessary for the spin–up becomes equal to

∆m1 =Jθf

gℓIspT

At the end of the drift interval, a second pulse of equal duration in the opposite directionwill stop the motion of the satellite and the new attitude will be acquired. The totalpropellant mass is thus equal to

∆mtot = ∆m1 + ∆m2 = 2Jθf

gℓIspT

It can be observed that allowing larger manoeuvre times, the same angle variationcan be achieved with a significantly smaller amount of fuel. Moreover, low energy thrustpulses are less likely to excite vibrations in the structure and/or in flexible appendagesattached to the spacecraft (such as antennas or solar panels). For large values of T , smalleron–time for the thruster are required, and the initial assumption that the acceleration anddeceleration phases have a negligible duration with respect to the overall manoeuvre timeis confirmed.

On the other side, for agile spacecraft that achieve high angular velocities, it is pos-sible to improve the accuracy of the estimate of manoeuvre time taking into account theduration of the initial and terminal phases. The time for accelerating the satellite up toθd is again given by

∆t1 =Jθd

Fℓ=θd

αwhich is still equal to the time ∆t2 necessary for bringing it back to rest at the end of themaneuver, with the difference that for fast manoeuvres the total firing time ∆t1 + ∆t2 isa not negligible fraction of the total manoeuvre time T .

During the acceleration, the variation of θ is

θ1 =1

2α∆t21 =

1

2

θ2d

α

and the total variation during the manoeuvre is given by

θf = θdtd + θ1 + θ2

where the angle variation during the acceleration and deceleration phases, θ1 and θ2, areequal. The drift time is td = T − 2∆t1, where T is the total manoeuvre time, so that

θf = θd(T − 2∆t1) + α∆t21

= θd

(

T − 2θd

α

)

+θ2

d

α


The total manoevure time becomes thus

T =θf

θd

+θd

α

In the limit, the minimum–time bang–bang control law brings to zero the drift timetd. The reader can demonstrate that in this limit case it is

T = 2√

θf/α

In both cases, the faster manoeuvres are performed at the expenses of a significantincrease in propellant consumption.

4.4.2 Closed–loop control

The ideal case: thrust modulation

If the satellite is equipped with a set of sensors that provides the necessary informationon the attitude motion, it is possible to implement a control law which provides a torquecommand that drives the satellite towards the prescribed attitude.

If the torque command is simply proportional to the attitude error, e = θdes − θ, theclosed–loop equation of motion becomes

θ = M/JM = K(θdes − θ) = Ke

=⇒ θ =K

J(θdes − θ) =⇒ θ + p2θ = p2θdes

where p2 = K/J . The solution of this second order equation for an initial condition ofrest for θ = 0 is

θ(t) = θdes [1 − cos (pt)]

i.e. an unacceptable undamped oscillation of amplitude θdes about the desired condition.In order to damp the oscillation and asymptotically reach the desired condition, it isnecessary to add a damping term in the control law, proportional to the angular rate θ.The command torque is thus

M = K(θdes − θ) +Kdθ (4.3)

and the closed–loop equation of motion becomes

θ + cθ + p2θ = p2θdes

where the coefficient of the damping term c = 2ζp = −Kd/J is positive (damped oscilla-tions) if the gain associated to the angular rate is negative.

There are two possible resulting closed–loop behaviours. For 0 < ζ < 1, the time–history of the angular motion is

θ(t) = θdes [1 − exp(−ζpt) cos (pdt)]

where pd = p√

1 − ζ2, that is, damped oscillations with an exponentially decaying ampli-tude.

If on the converse ζ > 1, the characteristic equation has two real solutions λ1,2 =

p(−ζ ±√

ζ2 − 1), which are both negative. In this case the evolution of the rotationangle is

θ(t) = θdes

[

1 − λ2

λ2 − λ1

exp(λ1t) +λ1

λ2 − λ1

exp(λ2t)

]


1Output

Schmidt trigger

2

.5s+1

Modulatorfilter

1Input

Figure 4.3: Pulse–Width/Pulse–Frequency (PWPF) modulator.

Two important issues need to be considered, at this point. The first one is the choiceof the gains. The case with ζ < 0 is not considered at all, inasmuch as one of the poles ofthe closed loop system would be a positive real number, that is, the closed loop systemwould be unstable. When ζ < 1 (sub–critical damping), the time τ to damp out 99% ofthe initial oscillation amplitude increases as ζ gets smaller according to the equation

exp(−ζpτ) = 0.01 =⇒ τ = − log(0.01)/(ζp)

At the same time, if ζ > 1 (super–critical damping) the time constant of the slowest modebecomes larger, being approximately equal to ζ/p, for large values of ζ . This means thata largely super–critical damping coefficient will cause a slower convergence.

The best performance in terms of maneuver agility are obtained for the critical damp-ing ζ = 1, which guarantees the fastest convergence time to the desired position. Theover–damped case may be of interest in those cases when fuel consumption is more astringent concern than maneuver time. On the contrary, the under–damped case, whichexhibits some overshoot, is of no practical interest, since it is characterized by a longersettling time and an increased fuel consumption, because of the thrusting system firingalternatively in both directions, during the oscillations.

The second issue concerns the actual implementation of a control law in the form (4.3).Such a control law would be realistic only if the control torque provided by the thrustercould be modulated. Unfortunately direct thrust modulation would require a complexhardware and, at the same time, would be extremely inefficient in large portions of therequired thrust range. For these reason, thruster are always on/off devices. As a conse-quence, some form of pulse modulation is required to provide a feasible implementation ofthe control torque demand. Two techniques will be presented, the Pulse–Width/Pulse–Frequency (PWPF) modulator, which is an analogic device, and the Pulse–Width Mod-ulator (PWM), which lends itself to a discrete time implementation.

Pulse–Width/Pulse–Frequency modulation

Figure 4.3 shows the structure of a Pulse–Width/Pulse–Frequency (PWPF) modulator.The main element of the modulator is the Schmidt trigger, which consist of a double relaywith hysteresis, separated by a dead band. In order to provide a quasi–linear steady–stateresponse, a modulation filter is added, the input of which is the signal e = em − ym, i.e.the difference between the feed–back signal error and the modulator output. The outputof the filter v is the activation signal for the Schmidt trigger.


The modulator filter is represented by a linear first order system, the dynamics ofwhich is

v =1

τ(Kem − v)

The response of the modulator filter to a steady input e is

v(t) = v0 exp (−t/τ) +Kem [1 − exp (−t/τ)]

until y = 0.The output y of the modulator remains zero until the signal v remains below the

activation threshold Uon. This threshold is never crossed if the error variable em remainssmaller than Uon/K, that is, the asymptote of v(t) → Kem lies below Uon.

If on the converse Kem > Uon, there will be a time tom when v reaches Uon and thetrigger output switches from 0 to Um (Fig. 4.4), so that the thruster is activated by thetrigger. At this point, the modulator input changes to e = em −Um. Assuming that em isunaffected by y, integration of the filter linear dynamics starting from the initial conditionv(ton) = Uon for e = em−Um brings to the following expression for the activation function

v(t) = Uon exp

(

−t− tonτ

)

+K(em − Um)

[

1 − exp

(

−t− tonτ

)]

= K(em − Um) + [Uon −K(em − Um)] exp

(

−t− tonτ

)

The thruster is switched off when the activation function becomes equal to Uoff , that iswhen

exp

(

−toff − tonτ

)

=Uoff −K(em − Um)

Uon −K(em − Um)

= 1 − Uon − Uoff

Uon −K(em − Um)(4.4)

It should be noted that, if K(em −Um) > Uoff , i.e. em > Uoff/K+Um, the above equationhas no real solution, that is the input is so great that the modulator calls for a continuousthruster engagement. This level constitutes the saturation level of the modulator. Onthe converse, if em becomes less than Uoff/K + Um, the thrusters are switched off whent = toff . Calling ∆ton = toff − ton the on–time of the thruster, and assuming that it issufficiently small, it is possible to approximate exp (−∆ton/τ) = 1 − ∆ton/τ , so that

∆ton ≈ τUon − Uoff

Uon −K(em − Um)

After the thruster pulse, the modulator input becomes again e = em and the responseof the filter for an initial condition given by v(toff) = Uoff is

v(t) = Uoff exp

(

−t− toffτ

)

+Kem

[

1 − exp

(

−t− toffτ

)]

The thruster will be activated again when v = Uon, that is when

exp

(

−t− toffτ

)

=Uon −Kem

Uoff −Kem


0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

Filt

er v

aria

ble

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

2

4

6

8

10

Out

put t

orqu

e

time [sec]

initial condition V = 0

first activation when V = Uon

Uon

Uoff

deactivation when V = Uoff

pulseduration

∆ ton

time betweenpulses ∆ t

off

Figure 4.4: Activation/deactivation sequence of a PWPF modulator.

Assuming again that the off time is also sufficiently small, one gets

∆toff = τUon − Uoff

Kem − Uoff

Figure 4.5 shows the activation sequence of a PWPF modulator for increasingly highervalues of the error signal em. In the first case it is clear how the error signal is so smallthat the thruster are not activated. For an error sligthly above the activation threshold,a sequence of small pulses (25 ms) separated by wide deactivation intervals (∆toff ≈ 0.6s) are present. This correspond to a small average control torque. A higher error signalincrease the duration of the pulses, while decreasing the time between the pulses, so thatthe average torque increases almost linearly until the saturation level is reached. Beyondsaturation (bottom–right corner in Figure 4.5) the deactivation threshold is no longercrossed and the thruster are not switched off.

The average output of the thruster, given by

y = Um∆ton

∆toff + ∆ton

is represented in Fig. 4.6. Its variation is almost linear for Uon/K < em < Uoff/K+Um, itis zero below the activation threshold and equal to the thruster force beyond the saturationvalue. The proper sizing of the trigger and filter characteristics through tuning of filter andtrigger parameters allows for realizing a modulator which takes into account the physicalcharacteristics of the thrusters, such as the minimum pulse time and the minimum timebetween pulses.


0 0.5 1 1.5 20

0.2

0.4

0.6

0.8

1

Filt

er v

aria

ble

0 0.5 1 1.5 20

2

4

6

8

10

Out

put t

orqu

e

time [sec]

0 0.5 1 1.5 20

0.2

0.4

0.6

0.8

1

Filt

er v

aria

ble

0 0.5 1 1.5 20

2

4

6

8

10

Out

put t

orqu

e

time [sec]

0 0.5 1 1.5 20

0.2

0.4

0.6

0.8

1

Filt

er v

aria

ble

0 0.5 1 1.5 20

2

4

6

8

10

Out

put t

orqu

e

time [sec]

0 0.5 1 1.5 20

0.2

0.4

0.6

0.8

1

Filt

er v

aria

ble

0 0.5 1 1.5 20

2

4

6

8

10

Out

put t

orqu

e

time [sec]

Figure 4.5: Activation sequences of a PWPF modulator for different input levels.

DC

0 2 4 6 8 10

0.0

0.5

1.0

e

Figure 4.6: Duty cycle of the PWPF modulator.


0 20 40 60 80 100−1

0

1

Tor

que

[N m

]

0 20 40 60 80 100

0

0.5

1

Ang

. spe

ed [d

eg/s

]

0 20 40 60 80 1000

10

20

Ang

le [d

eg]

time [sec]

Figure 4.7: Closed–loop behaviour (short term).

Figure 4.7 shows the response of a satellite controlled by a set of thruster driven by aPWPF modulator (continuous blue line). The response is compared with that obtainedby thrust modulation (red dashed line). In the initial part of the reorientation manoeuvre(t < 100 s), the modulator reproduce almost exactly the behaviour of the ideal linear PDcontroller.

In the following part of the simulation, shown in Fig. 4.8, the angular velocity ofthe ideal case goes asymptotically towards 0, while the small residual angular velocity(-0.002 deg/s) visible in the more realistic case, where the modulator is present, causes aslow drift towards the negative side of the dead–band. Since the error signal is within thedead–band no control action is produced, until for t = 273.36 s a pulse is fired which revertthe motion. The minimum pulse provides the required reversal, but the resulting angularvelocity is significantly higher (0.025 deg/s) so that after few seconds, t = 286.43 s, a newpulse in the opposite direction is required to revert again the motion on the positive sideof the dead–band. This bouncing between the edges of the dead–band characterizes thesteady–state condition of a spacecraft controlled with a reaction control system.

Pulse–Width Modulation

The PWPF modulator is a heritage of the analogic era, although it is still widely used. Themain advantage of PWPF modulation remains the possibility of commanding a sequenceof pulses with a quasi–linear response, in terms of averaged output. Nonetheless, manyactuators are now commanded in the framework of a digital implementation of the controllogic. It is well known that a digital controller obtained from the discretization of ananalogic one can attain at most the performance of the original analogic counterpart. If adigital controller is designed directly in the discrete time domain, taking into account thesampling frequency and A/D and D/A conversions, better performance can be obtained.


100 150 200 250 300 350 400 450 500−1

0

1

Tor

que

[N m

]

100 150 200 250 300 350 400 450 500−0.1

0

0.1

Ang

. spe

ed [d

eg/s

]

100 150 200 250 300 350 400 450 50019

20

21

Ang

le [d

eg]

time [sec]

Figure 4.8: Closed–loop behaviour (long term).

Pulse–Width modulation provide a means for designing a thruster control law wherethe implementation of the switching logic is inherently digital. As an advantage withrespect to PWPF (analogic) modulation, it is much easier to satisfy restrictions on theon–time ∆ton and the time–between–pulses ∆toff .

Figure 4.9 shows a sequence of pulses. If ts is the sampling period of the digitalcontroller, a pulse can be commanded for each duty cycle, the duration of which is ∆tdc =Nts. The minimum impulse bit (MIB) Non,min is chosen so that Non,mints > ∆ton,min.At saturation, the on–time is extended to the entire duty cycle. Below saturation, themaximum pulse must satisfy the constraint on minimum–time–between–pulses (MTBP),that is ∆toff = (N −Non)ts >MTBP.The average output of a thruster controlled in PWM is

y = Um∆ton

∆toff + ∆ton= Um

Non

N

The response of the PWM to a constant input error signal e is depicted in Fig. 4.10,assuming that a quasi–linear average output y = ke is desired between a deadband edb

and the saturation value esat, when the on–time equals the duty cycle and y = Um. Itshould be noted that, by a proper software implementation of the activation signal, it ispossible to obtain arbitrary pulse–width modulation curves.

4.4.3 Fine pointing control

After slew to target, the control system switches to fine pointing control, in order tomaintain the payload aimed at the desired target with a prescribed tolerance, in presenceof external disturbances. This can be done using momentum exchange devices and/or gasjets. A simple way to achieve such a fine pointing is to use cold gas jets, that produce


-

6y

Um

0t-

ts -∆ton

-∆toff

-∆tdc

Figure 4.9: Pulse–Width Modulation.

yUm

0 0.2 0.4 0.6 0.8 1 1.20

0.2

0.4

0.6

0.8

1

eesat

Modulator data: ts = 10 ms; Ndc = 20; ∆ton,min = 12 ms ⇒ Non,min = 2;MTBP = 17 ms ⇒ Non,max = 18.

Figure 4.10: Response of a PW Modulator.

thrust in the range between 0.05 and 20 N, with short pulse times (of the order of 10−2 s)for fine control. As far as the satellite will be bouncing back and forth between the edgesof a limit cycle, it is important to estimate the amount of fuel necessary as a function ofthe prescribed pointing accuracy and thruster characteristics.

Torque–free control

Let us consider the torque–free case, that is no external disturbance acts on the spacecraft,where the payload must be pointed in a given direction, within some deadband, which inthe case of astronomy payloads can become as small as 1/10 arcsec = 0.000278 deg!

The satellite is allowed to drift slowly inside the deadband, with drift velocity θd.When it reaches the limit of the deadband, θdb, the thruster is fired so as to revert thedrift rate in the opposite direction. The satellite will drift towards the other side of thedeadband, where the thruster are fired again, thus capturing the satellite into a limit cycleof width approximatley equal to 2θdb. In what follows it will be assumed that the limitcycle is symmetric, that is, the absolute value of the angular velocity during the drifting


−0.01

0

0.01

drift

drift

deadband

thru

ster

firin

g thruster firing

θ [10−4 deg]

θ [d

eg s

−1 ]

−3 0 3

.

Figure 4.11: Limit cycle for gas–jet fine pointing control.

phases is the same in both directions.1 The thrusters generates a moment

M = 2Fℓ = Jθ

For a thruster pulse width ∆ton, the change of the attitude rate is

∆θ = 2Fℓ

J∆ton = 2θd

for a symmetric limit cycle, in order to exactly revert the satellite motion after the firing.Remembering that

Isp =F∆tong∆mf

one gets, for each thruster firing, a fuel consumption of

∆mf = 2F∆tongIsp

= 2Jθd

gℓIsp

Considering both sides of the deadband, the total fuel consumption per cycle is

∆mf,tot = 4Jθd

gℓIsp

Ignoring the (usually negligible) thruster pulse time, the duration of the limit cycle is

TLC ≈ 4θdb

θd

1As it was evident from the simulation in the previous paragraph, this is not what happens in reality,since there is no way of enforcing a perfectly symmetric limit cycle with a simple PD controller, unlessmore complex control strategies are implemented. Nonetheless, the following analysis is conservaive,inasmuch as the symmetric limit cycle is the fastes one (its period is the shortest), thus requiring themaximum number of pulses per unit time. This means that the estimate of the fuel consumption in thissituation is a worst–case–scenario.


and the average fuel consumption (that is the fuel burned per unit time) is

¯mf =∆mf,tot

TLC=

Jθ2d

gℓIspθdb

which can be rewritten as

¯mf =(Fℓ∆ton)

2

gℓIspJθdb

This latter equation shows clearly that in order to achieve a great accuracy (small valueof θdb) without paying too a great penalty in terms of fuel consumption, the characteristicsof the thrusters are of paramount importance, inasmuch as ¯mf remains small onfly if highspecific impulse thrusters are employed, capable of delivering the force F for a very shortpulse time ∆ton.

Bias torque compensation

In most cases, the payload must be aimed at a target (be it on Earth or in the openspace) in presence of external disturbances. If the resulting torque is roughly constant,such as that due to the gravity gradient, it is possible to use this torque for reversing themotion on one side of the deadband. In this case, the environmental torque will bring thespacecraft drifting towards the top of the tolerance interval. At this point the thruster arefired, to reverse the motion, that will be slowed down by the adverse disturbance torque.Rather than allowing the spacecraft to reach the other side of the deadband and fire asecond thruster pulse, one allows the bias torque to slow and reverse the drift inside thedeadband interval. The limit cycle will be made of two parabolic arcs, one relative to thethruster firing phase, similar to that of the previous case, that reverse the satellite motionon one side of the deadband, and the second arc, relative to the slow deceleration andmotion reversal due to the external torque.

If the disturbance torque generates an angular acceleration

θ = αD = MD/J

it is

θ(t) = θ0 + αD∆t

θ(t) = θ0 + θ0∆t+ αD∆t2/2

where the subscript 0 indicates the initial condition at time t0. From the first equation,it is ∆t = (θ(t) − θ0)/αD, that, substituted in the second one, after some simple algebragives

θ2 = θ20 + 2αD(θ(t) − θ0)

If the initial condition is on the deadband limit, it is θ0 = θdb. In order to keep the limitcycle inside the dead band, the turning point at which θ drops to zero and reverts its signmust be crossed on the other side of the deadband, when θ0 = −θdb. In this case, it is

θ2 = 0 = θ20 + 2αD(−2θdb)

that is

θdb =Jθ2

0

4MD


−0.01

0

0.01

−3 0 3

drift

thruster firing

deadband

θ [10−4 deg]

θ [d

eg s

−1 ]

.

Figure 4.12: Limit cycle for gas–jet fine pointing control with constant disturbance torque.

In order to close the limit cycle, the thruster pulse must be such that the total ∆θ istwice the drift angular velocity θ0 at deadband crossing,

∆θ = 2θ0 = 2Fℓ∆ton/J

Substituting the latter expression in the equation of θdb one gets

θdb =(Fℓ∆ton)

2

4MDJ

and again the requirement on the thruster is to deliver a small duration pulse for keepingthe deadband small enough to satisfy fine pointing requirements.

Exercise

After determining the duration of the limit cycle in presence of a bias torque, where asusual the firing time can be considered negligible, prove that the average fuel consumption¯mf = ∆mf,tot/TLC is independent of the θdb, if the motion reversal under the action ofthe disturbance torque exploits the whole deadband amplitude.

4.5 Momentum exchange devices for attitude control

Gas jets can deliver very high torque to manoeuvre a spacecraft and change its orientation,but they have a limited life, due to the finite propellant mass budget available. For thisreason, it is important to rely on some different kind of actuators for stabilising andcontrolling satellite attitude during normal operations, without using propellant.

Reaction wheels (RWs) and momentum–bias wheels (MBWs) can exchange angularmomentum with the spacecraft bus using only electrical power, that is obtained from the


solar panels for the whole lifetime of the satellite. Each wheel is attached to the satellitestructure trough an electric motor, that can be used to accelerate or decelerate the wheel,relatively to the satellite. If we assume that initially both the satellite and the wheelare still in the inertial frame (zero angular momentum initial condition), spinning up thewheel in one direction will cause the bus to rotate in the opposite direction, because ofconservation of overall angular momentum.

Let us consider a single axis rotation about the principal axis of inertia ei. Thespacecraft has a total principal moment of inertia Ji about ei, including the wheel. Thewheel spins around the axis a = ei, having a spin moment of inertia Is about a andangular velocity Ω, relative to the spacecraft bus. Its rotation does not change the actualmass distribution, so that Ji is independent of the motion of the wheel.

The overall angular momentum about ei is

hi = Jiωi + IsΩ

where the first term is the angular momentum of the satellite induced by the overallrotation rate ωi, while the second one is the relative angular momentum stored in thewheel.

Note that the absolute angular velocity of the wheel is Ω + ωi with respect to theinertial frame FI . If initially ωi = Ω = 0, so that hi = 0, and taking into account thatthe total angular momentum is conserved, inasmuch as the wheel is spun up or slowedthrough the exchange of internal torques applied by the electric motor, one gets

ωi = −IsJi

Ω (4.5)

Usually it is Is ≪ Ji, so that ωi ≪ Ω, that is the angular velocity achieved by thespacecraft is only a small fraction of the relative spin angular velocity of the wheel.

In order to manoeuvre the spacecraft about 3 axes, it is necessary to equip the satellitewith at least 3 RWs, one for each body axis. The cluster of RWs usually contains afourth spare wheel, the axis of which is oriented in such a way so as to exchange angularmomentum components to or from any of the three axes, if one of the main wheels fails.

One of the wheels can be used to de–spin the satellite after orbit acquisition. In sucha case, this wheel shall be oriented as the spin axis of the satellite and apogee kick rocketmotor stack and it is called a momentum wheel. Its angular momentum is unlikely todrop back to zero and the control torques about its axis will be obtained accelerating anddecelerating the wheel.

The MBW will also provide a certain degree of gyroscopic stability, which is an in-teresting feature. At the same time, the presence of the momentum bias will couple themotion about the other two axis because of the gyroscopic effect observed when derivingthe equations of motion of gyrostats. As a consequence, it is no loger possible to considerseparately the control problem about the principal axis of inertia. In spite of this fact,only single–axis rotations will be considered in what follows, for the sake of simplicity.

4.5.1 Open–loop control with RWs

Assuming for simplicity torque–free motion, we will size a reaction wheel with respect torequirements in single axis slew manoeuvre. Dropping the i subscript, the total angularmomentum about one of the principal axis is

h = hs + hw


-t

6T

t1θ = 0

t2

t3 t4θ = θf

hs

−hs

Figure 4.13: Torque profile for a single axis slew manoeuvre.

where hs = (J − Is)ω is the spacecraft angular momentum (without the wheel), whilehw = Jsw(Ω + ω) is the spin wheel absolute angular momentum. Since there are noexternal toruqes, h = const and

h = hs + hw = 0 =⇒ hs = −hw

where ga = hw is the motor torque applied to the spin wheel.Note that, assuming single axis rotations, it is ω = θ ⇒ ω = θ, and the equation of

motion can thus be written in the form

(J − Is)θ = −ga

which has exactly the same form derived for the RCS case, with the only difference of thesign in the control torque, which, in the case of RWs or MBWs is a reaction torque.

As a consequence, a similar open–loop control technique can be used, where one pulseis used for spinning up the reaction wheel and consequently the vehicle in the oppositedirection, achieving a certain drift angular velocity, according to Eq. (4.5), while a secondpulse will spin down both the spacecraft and the wheel, achieving at the end of themanoeuvre the desired attitude θf .

After the first pulse, the spacecraft gains an angular momentum equal to

hs(t2 − t1) = (J − Is)θ = −hw(t2 − t1) = −ga(t2 − t1)

But it is alsohs = (J − Is)θ = (J − Is)α

where α = θ = −ga/(J − Is) is the angular acceleration of the spacecraft. The evolutionof the angular displacement is thus given, as in the RCS case, by

θ(t2) =1

2α(t2 − t1)

2

= −1

2

ga

J − Is(t2 − t1)

2

Note that simpler expressions can usually be employed for preliminary sizing purposes, byremembering that usually it is J ≫ Is. On the other side, the value of the available torqueis usually much smaller than that obtained from thrusters, so that the pulse duration isnot negligible with respect to manoeuvre time.


As an advantage of reaction and momentum bias wheels, the fact that electrical power(and no fuel) is employed for accelerating and decelerating the wheel make it possibleto perform fast manoeuvers, without paying any penalty in terms of operational lifetimewith respect to slower ones.2

4.5.2 Sizing a reaction wheel for single axis slews

For a minimum–time rotation, a bang–bang control must be used, so that t2 ≡ t3 and thefinal rotation angle is

θf = 2θ(t2) =hs

J − Is(t2 − t1)

2

But assuming t1 = 0, it is also tf = 2(t2 − t1), for bang–bang control, and one gets

θf =1

4

hs

J − Ist2f

Because of conservation of angular momentum, we recall that hs = −hw, so that themaximum torque that the electric rotor must produce is

ga,max = |hw| = 4(J − Is)θf

t2f

But for t = t2, it is also

hs(t2) = (J − Is)θ(t2) = −hw(t2) = −hw(t2 − t1) = −hwtf/2

when the wheel achieves the maximum angular momentum. This means that the momen-tum capacity of the wheel must be

hmaxw ≥ 2(J − Is)

θ2tf

Numerical example

We require a spacecraft with J = 100 kg m2 to perform a 0.2 rad slew in 10 sec. This meansthat the electric motor must apply torques in both direction equal to (neglecting the unknownvalue of Is)

hw ≈ 4Jθf

t2f

4 × 100 × 0.2

102= 0.8 Nm

The maximum momentum stored in the RW is reached when half of the reorientation is com-pleted, so that the RW capacity must be at least

hmaxw ≈ 2J

θ2

tf=

2 × 100 × 0.2

10= 4 kg m2s−1

Assuming a wheel inertia Jsw = 0.03 kg m2, the maximum wheel angular rate with respect tothe satellite bus is

Ω = − J

Isω = − J

Is

hs

J − Is

≈ hw

Jsw= 133 rad s−1 = 21.2rpm

2On the converse, a faster drifting phase requires more acceleration and deceleration from the reactioncontrol system, which means more fuel is burned for achieving the same final attitude. This means thatthe satellite will run out of fuel sooner, if fast manoeuvres are performed insterad of slow ones.


4.5.3 Closed–loop control with RWs for single axis slews

With the usual meaning of the symbols, the equation of motion for single–axis rotation is

d

dt

(

Jθ + IsΩ)

= Mext

so that it isJθ + IsΩ = Mext (4.6)

The reaction wheel dynamics is described by the first order equation

Is

(

Ω + θ)

= ga ⇒ IsΩ = ga − Isθ

Substituting the latter expression in Eq. (4.6) and assuming for simplicity that theexternal torque and the momentum bias are both zero, it is possible to rearrange thedynamics equation as

(J − Is)θ = −ga

By choosing a simple PD control in the form

ga = KP (θdes − θ) +KDθ (4.7)

the closed–loop dynamics is described by the second order differential equation

J∗θ +KDθ −KPθ = −KP θdes

where J∗ = J − Is is the moment of inertia of the platform without the spin–wheel.3

By defining the closed–loop system bandwidth as p2 = −KP/J∗ (with Kp strictly

negative for static stability) and a damping parameter ζ such that 2ζp = KD/J∗ (with

KD strictly positive for asymptotic stability), the equation of motion can be rewritten inthe form

θ + 2ζpθ + p2θ = p2θdes

which can be solved analytically and clearly presents a nice way of choosing the gains KP

and KD.

4.5.4 Bias torque and reaction wheel saturation

Let us assume for simplicity that the satellite is subject to a constant disturbance torqueMD. It is easy to see that the angular displacement θ evolves according to the differentialequation

θ + 2ζpθ + p2θ = p2θdes +MD/J∗

and asymptotically converge to the steady state

θss = θdes +MD/KP 6= θdes

This means that the use of the control law of Eq. (4.7) may not guarantee a sufficientpointing accuracy. The steady state error can be made smaller by increasing the propor-tional gain KP , but it will never vanish. Moreover, increasing the gain will likely degrade

3Note that the feedback control law duplicates exactly the control law employed for reaction controlsystems, Eq. (4.3), but because ga is a reaction torque, the gains appears with their sign changed in theequation of the closed loop system dynamics. Their values must be chosen accordingly.


the performance of the real system, since the sensor noise will also be multiplied by a highgain.

It is possible to asymptotically cancel the steady state error by adding an integralterm to the control law, i.e.

ga = KP (θdes − θ) +KDθ +KIy (4.8)

where

y = θdes − θ ⇒ y =

∫

(θdes − θ)dt

If an integral term is introduced, a steady–state can be achieved only if y = 0, i.e. θ = θdes,with a value of y such that

KIy = MD ⇒ ga,ss = MD

This means that if the spacecraft is subject to a constantly not null external torque, theaverage value of which is not zero (aerodynamic torque, gravity gradient), the platformcan be maintained aimed at a fixed position only if an equivalent motor torque is deliveredto the reaction wheel. As a consequence, the wheel will absorb these disturbances by spin–up and fix the orientation, but its angular velocity will constantly increases. Should thewheel reach the maximum spinning speed, the attitude control system is saturated.

In order to avoid such a problem, before reaching saturation it is necessary to dumpthe angular momentum excess. This can be done simply braking the wheel and firing thethrusters in the opposite direction. In this way the momentum accumulated in the wheelis brought back to zero, because of the external torque produced by the thrusters. Duringwheel desaturation the pointing accuracy of the control system is usually reduced.

The use of thrusters for despinning the wheels shows that also when using RWs forattitude control it is necessary to carefully design the propellant mass budget. Thrusterfirings will be also in this case the main source of fuel consumption during the stationkeeping phase, and will be the main limit to the satellite operative life.

Chapter 5

Environmental torques and otherdisturbances

Although small in terms of absolute value, environmental torques acting on a spacecraftfor a very long period of time can produce a sizable displacemtne of the satellite from thedesired attitude. A 3–axis stabilised spacecraft featuring only a reaction control system iscontinuously bouncing back–and–forth between the edges of the allowed dead–band, underthe alternating action of a (small) disturbance torque and the quick reversal provided bya (large) control pulse at one of the edges of the dead–band.

If the spacecraft is either gyroscopically stabilised or is characterised by a considerablemomentum bias the process may take longer, since a sizable variation of a large value of theangular momentum requires more time. In any case, sooner or later, some compensationis required. Disturbances of different nature may affect spacecraft attitude, especiallywhem flying a Low Earth Orbit. At the same time, internal sources of disturbances arealso present, which either affect the dynamic behaviour of the satellite with respect tothe ideal one, or directly affect sensor pointing precision because of oscillations inducedby flexible appendages.

Among many others, gravity gradient torque has the interesting property of stabilisingan Earth pointing attitude, provided that a certain allignment condition between orbitand principal frame is satisfied. When the requirement on pointing accuracy is weak (ofthe order of some tenths of a degree) this passive stabilisation technique, that does notrequire fuel, may be an option.

5.1 Environmental torques

The major sources of external disturbances that may influence the dynamic behaviour ofa satellite are:

• gravity gradient: the variation of gravity pull acting on different mass elementsof the spacecraft provide a small yet significant torque; in Low Earth Orbit (LEO)this torque can stabilise an Earth pointing attitude, although the resulting motionis not naturally damped;

• atmospheric drag: in Low Earth Orbit (LEO) the air density is small but notnull; the centre of pressure is usually not aligned with the center of mass and thisfact will produce a torque acting on the spacecraft;

75

76 G. Avanzini, Spaceflight Dynamics – 5. Environmental torques and other disturbances

• magnetic field: the satellite, with its metallic parts and currents flowing in theelectric system produces a magnetic field that interacts with the magnetic field ofthe Earth, exspecially in LEO; again a torque on the satellite is produced by thisinteraction; this coupling between a magnetic field produced by the satellite andthe environmental magnetic field can be used beneficially to provide a simple andeffective means of attitude control; a magneto–torquer is made of a magnetic coil(or a permanent magnet) the orientation of which inside the satellite bus can bechanged so as to produce a control torque; as an alternative, to coils at an angle canbe activated with different currents so as to provide an arbitrary torque in a plane;no torque can be delivered in the direction of the magnetic field, so that magnetictorques alone are not sufficient for three–axis control;

• solar radiation: the solar radiation produce a light pressure force on the satelliteparts exposed to the flow of nuclear particles emitted by the sun; again the forceoffset will determine a small torque, that is insignificant in LEO, but is the maindisturbance at higher distances from the Earth, such as in GEostationary Orbit(GEO).

A simple mathematical formulation is available only for the gravity gradient torque,that will be discussed later in this chapter. Atmospheric and solar radiation torquesdepend strongly on the shape of the spacecraft, and in particular on how the solar panelsand the spacecraft bus face the flow of the rarified gas moleculae or the stream of highenergy particles from the Sun. The magnetic torque, on the converse, depend on theelectric system architecture. As a consequence no general model is available for thesedisturbances.

5.2 Internal disturbances

There are also some sources of internal disturbances, that can be due to the uncertaintyof some parameters or to some simplifying assumptions the effect of which is not null onthe actual spacecraft:

• uncertain center of gravity: there can be an uncertainty on the position of thecenter of gravity up to some centimeters (1 to 3 cm);

• thruster misalgnment: the direction of the thruster can be different from thedesign one up to 1 deg;

• run–run thrust variation: there can be sizable variation (up to 5%) in the thrustproduced by a thruster in different firings;

• fuel slosh: fuel sloshing in the propellant tanks causes both shift in the positionof the centre of mass and in the mass distribution (that is, affects the value of themoments of inertia); moreover, as propellant is burned, a long term variation inmean mass properties is also experienced by the satellite;

• rotating parts: rotating parts inside the satellite (gyroscopes, reaction wheels, butalso tape recorders) produce microvibrations which in turn cause loss in pointingaccuracy;

G. Avanzini, Spaceflight Dynamics – 5. Environmental torques and other disturbances 77

• elastic modes of flexible appendages: several flexible items, such as antennas,booms and solar arrays can be attached to the spacecraft bus, and their flexibilitycan interact with the rigid body motion of the main platform, resonating at thebending frequencies, if properly excited, with serious consequences for the pointingaccuracy.

5.3 Gravity–gradient stabilization

Spin stabilization allows one to provide the satellite with gyroscopic stability, which canbe made asymptotic, by a proper use of dissipative devices. At the same time, somesignificant disadvantages affects the operational use of spin stabilized satellites, that are

• the presence of only one inertially fixed direction along which to point the payload;

• the inefficient production of electric power, inasmuch as the entire surface of thesatellite must be covered with photovoltaic cells, only half of which faces the sunduring the spinning motion (with less than one fourth at an angle greater than 45deg with respect to the direction of the sun rays);

• the complexity, limited accuracy and cost in terms of fuel of attitude maneuvers forreorientation of the spin axis.

A very interesting possibility for stabilizing an Earth facing satellite in low Earthorbit is to exploit the gravitational torque that acts on any rigid body of finite size.The differential gravitational acceleration across the satellite provides a small, but nonnegligible torque. Simply put, the side of the satellite closest to the Earth experiences aslightly greater gravitational acceleration that the opposite side. The resulting torque willtend to align the satellite with the local vertical direction. For some mission applicationsthe gravity gradient torque provides a disturbance which must be countered. However, foran Earth facing satellite, the gravity gradient provides a simple, low cost means of attitudestabilization, albeit rather inaccurate. The great advantage is that no fuel is required tomaintain the desired attitude, if a proper stability condition is met. Energy dissipationis still necessary in order to provide asymptotic convergence onto the prescribed Earth–pointing attitude.

5.3.1 Origin of the gravity–gradient torque

The gravity acceleration acting on a mass element dm is given by

~g = −GM~R + ~r

||~R + ~r||3

where G is the universal gravity constant, M is the mass of the primary body (the Earth,

for an Earth orbiting satellite), while ~R and ~r are the position vectors of the center ofmass of the satellite, CM, w.r.t. the centre O of the primary and the position vector ofthe mass element dm w.r.t. CM, respectively (Fig. 5.1). The total moment w.r.t. CM isthus given by

~M =

∫

B

(~r × ~g) dm


dm

CM

~g ~r

~R

o1

o2

o3

e1

e2

e3

E1 E2

E3

Figure 5.1: Gravity on a finite–size rigid body.

~M = −GM∫

B

~r ×(

~R + ~r)

||~R + ~r||3

dm

Since ~r × ~r = 0, it is

~r ×(

~R + ~r)

= ~r × ~R

and the latter expression becomes

~M = −GM∫

B

(

~r × ~R

||~R + ~r||3

)

dm

5.3.2 Pitch component of the gravity–gradient torque

If one is interested only in pitch motion about the axis perpendicular to the orbit plane,it is possible to obtain a simple expression for the pitch component of the gravity gradienttorque.The denominator of the expression of ~M can be manipulated as follows:

||~R + ~r||−3 =[(

~R + ~r)

·(

~R + ~r)]−3/2

=[

R2 + 2~r · ~R + r2

]3/2

where r and R are the magnitude of ~r and ~R, respectively. The latter expression can berewritten as

||~R + ~r||−3 = R−3

(

1 +2

R2~r · ~R +

r2

R2

)

−3/2

Being r ≪ R, the third term inside the parentheses can be neglected as it is (r/R)2 ≪ 1,and the second one is a first order perturbation compared to 1. But we can expand aterm like (1 + x)n in the neighbourhood of x = 0 as

(1 + x)n = 1 + nx+ O(x2)


HHHHHHHHHHHHHHHH

O

HHYeCM

@@Rrdm ~r

~Re2 e1

θ3

.................

orbitdirection

@@I

Figure 5.2: Scheme for the evaluation of gravity–gradient torque.

In the present case, given that x = 2(~r · ~R)/R2 and n = 3/2, one gets

R−3

(

1 +2

R2~r · ~R +

r2

R2

)−3/2

≈ R−3

(

1 − 3~r · ~RR2

)

= R−3

(

1 − 3

R2(xX + yY )

)

Assuming a pure pitch motion, the first two body axes, e1 and e2, lie in the orbitplane (see Fig. 5.2), so that the resulting pitch motion consists of rotations about the

body axis e3 only. In terms of body frame components, the position vectors ~R and ~r aregiven by

~R = Xe1 + Y e2

~r = xe1 + ye2

so that~r × ~R = (xY − yX)e3

The torque about e3 is thus given by

M3 = −GMR3

∫

B

(xY − yX)

[

1 − 3

R2(xX + yY )

]

dm

Developing the integrand one gets

(xY − yX)

[

1 − 3

R2(xX + yY )

]

= xY − yX − 3

R2(xY − yX) (xX + yY )

= xY − yX − 3

R2

(x2XY + xyY 2 − yxX2 − y2XY

)

and

M3 = −GMR3

[

Y

∫

B

xdm−X

∫

B

ydm +

− 3

R2

(

XY

∫

B

x2dm+ Y 2

∫

B

xydm−X2

∫

B

xydm−XY

∫

B

y2dm

)]

But, for the definition of center of mass, it is∫

B

xdm =

∫

B

ydm = 0

and if a frame of principal axis of inertia is chosen as the body frame, also∫

B

xydm = 0


and the expression of M3 reduces to

M3 = 3GM

R5XY

[∫

B

x2dm−∫

B

y2dm

]

= 3GM

R5XY

[∫

B

(x2 + z2)dm−∫

B

(y2 + z2)dm

]

= 3GM

R5XY [Jy − Jx]

With our choice of axes (see Fig. 5.2), it is

~R = −R cos θ3e1 +R sin θ3e2 = Xe1 + Y e2

and M3 is thus given by

M3 = −3GM

R3cos θ3 sin θ3 (Jy − Jx)

5.3.3 Attitude motion

Since we are considering only pitching motion (ω1 = ω2 = 0), we need only the thirdEuler’s equation, that can be written in the simple form

Jzω3 = −3GM

R3cos θ3 sin θ3 (Jy − Jx)

but, for the same reason, it is also ω3 = θ3. Moreover, the quantity

Ω2 =GM

R3

is the orbital angular velocity, and the equation of the pitching motion under the effectof gravity–gradient becomes

θ3 + 3Ω2 cos θ3 sin θ3Jy − Jx

Jz= 0

There are two equilibrium solutions, for θ3 = 0 and θ3 = π/2, the stability of whichis determined by the sign of the term Jy − Jx. For small perturbations about the localvertical – local horizontal reference frame it is sin θ3 ≈ θ3 and cos θ3 ≈ 1, and we get thesecond order linear equation of motion

θ3 +

(

3Ω2Jy − Jx

Jz

)

θ3 = 0

If Jy > Jx the term 3Ω2(Jy − Jx)/Jz is positive and we can write

λ2 = 3Ω2Jy − Jx

Jz

The equation of motion becomesθ3 + λ2θ3 = 0

the solution of which is given by

θ3(t) = α1 cos(λt) + α2 sin(λt)


PPPPPPPPPPPPPPPP

O

PPie

@

@R@@@@

@@e2 e1

)e:CCCWC

CCC C

CCC

e1

e2

divergence

oscillations

6

BB?

Figure 5.3: Scheme for the evaluation of gravity–gradient torque.

that is, the satellite is stable and oscillates in the neighborhood of the condition θ3 = 0with a frequancy equal to

λ2 =

√

3Ω2Jy − Jx

Jz

If on the converse Jy < Jx, it is 3Ω2(Jy − Jx)/Jz < 0, letting

µ2 = 3Ω2Jx − Jy

Jz

the equation of motion can be rewritte as

θ3 − µ2θ3 = 0

and its solution becomes

θ3(t) = α1 exp(−µt) + α2 exp(µt)

which is clearly unstable, because of the divergent exponential term. Because of such aninstability, the satellite will rotate away from the neighborhood of the initial conditionθ3 ≈ 0, and will acquire a periodic motion around the other equilibrium solution, forθ3 = π/2. In this case. It is left to the reader as an exercise to demonstrate that, ifJy < Jx, the equilibrium solution for θ3 = π/2 is stable.

As an example of the resulting motion, Fig. 5.4 shows the behaviour of the SpaceShuttle (I = diag(1.29, 9.68, 10.01)106 kg m2) on a low orbit with a period of 1 hour and35 minutes. The first simulation is performed starting from an Earth pointing attitude(α = 10 deg), that is the nose of the Shuttle faces the Earth, with a perturbation from thevertical of 10 deg. The resulting motion is confined and the angular velocity is less tham0.015 deg s−1. Internal dissipation (modeled as a term proportional to θ, with a dampingconstant C = 1.5 10−6 s−1) damps very slowly the oscillation in the neighbourhood of thelocal vertical.

If on the converse, the motion is started near the second equilibrium condition, forα = 90 deg (that is the nose of the Shuttle points the direction of the orbital velocity),the resulting motion is unstable and large amplitude oscillation will build up. From thesimulation it can be seen how the vehicle will reverse its attitude, and will oscillate aroundthe equilibrium at α = −90 deg, with the nose pointing towards the external space. It


0 2 4 6

−10

−5

0

5

10

No. of orbits

θ [d

eg]

−10 −5 0 5 10

−0.015

−0.01

−0.005

0

0.005

0.01

0.015

θ [deg]

dθ/d

t [de

g s−

1 ]

0 2 4 6

−250

−200

−150

−100

−50

0

50

100

No. of orbits

θ [d

eg]

−200 −100 0 100

−0.1

−0.05

0

0.05

0.1

θ [deg]

dθ/d

t [de

g s−

1 ]

Figure 5.4: Stable and unstable motion of the Space Shuttle under gravity–gradienttorque.

should also be noted how the effect of the nonlinear term in sin θ cos θ in the gravity–gradient torque affects the oscillation frequency, when the amplitude is large. In the lasttwo plots, the stable motion of the first case is reported as a comparison.

Part II

Advanced Orbital Dynamics: OrbitControl

83

Chapter 6

Basic concepts

6.1 Introduction

Orbital dynamics plays a central role in space system engineering. Two key functions areprovided:

• Mission planning (pre–flight):Orbital dynamics is used to assess the feasibility of a desired mission objective and isused to plan the spacecraft trajectory to achieve mission goals. This may be a brunsequence for a simple geostationary transfer orbit or a more complex interplanetarymission with multiple gravity assists. The ∆V budget for the mission is obtainedthrough simulations which then determine the spacecraft propellant mass. This inturn links in with the engineering design of the spacecraft. Once a trajectory hasbeen planned (either a complex interplanetary trajectory or a relatively simple Earthcentered orbit) a time–line of events may be determined from the orbital dynamics,for example, the periods when the spacecraft is in view of a ground station. Orbitaldynamics is also used to determine sensor coverage patterns for remote sensing andcommunication satellites.

• Mission Operations (in–flight):Along with pre–flight mission planning, orbital dynamics is used extensively inspacecraft mission operations. Principally, orbit prediction is used to determinethe spacecraft location and dynamical state at any given time. This is clearly of im-portance for planning contacts during ground station passes. Also, orbit predictionis used to determine the schedule of propulsvie burs for orbit manoeuvres and smallorbit trims. Scheduling is also of importance for planning payload actitivities, suchas activating remote sensing cameras for Earth orbit or interplanetary spacecraft.

The orbit (and the sequence of manoeuvres necessary to achieve it) is determined as afunction of predefined mission objectives. The constraints may be several, such as groundcoverage, for a remote sensing satellite, or planet intercept for an interplanetary missionwith gravity assists.

The best orbit, compatible with mission requirements and constraints, is selected andthe sequence of events, from launch to final orbit injection and operational manoeuvres,is planned. The correct determination of the ∆V budget is of paramount importance(as an example, the fuel fraction for orbit control of a communication satellite). Alsothe requirements for rendez–vouz and docking must be considered during the missionplanning phase and will depend on the capabilities of the vehicle of performing accurate

85

86 G. Avanzini, Spaceflight Dynamics – 6. Basic concepts

BB

BB

BB

BB

@@I

@@I

@@I

@@I

critical review of

mission requirements

constraints on

orbital manoeuvres

constraints on mass

(and payload) fraction

constraints on

spacecraft design

Mission Objectives

Orbit (to satisfy mission)

∆V and transfer times

Fuel mass fraction

DESIGN

?

?

?

?

Figure 6.1: Mission analysis and design flow–chart.

orbit manoeuvres. Once the fuel budget is known, the sizing of the launcher is performedand the launch windows is determined.

But orbit dynamics is also of crucial importance during mission operations, inasmuchit is at the base of orbit determination algorithm and orbit control techniques. Moreover,it may be necessary to perform orbit manoeuvres not scheduled in the original missionplan (satellite re–location), and the impact on mission effectiveness and satellite life–timeneed to be determined. In this framework, there is a strong need of reducing operationalcosts by moving to autonomous operations, that means that orbit determination algorithmwill run on-board and will be coupled with high–authority control systems and missionplanning algorithms.

In what follows, a brief review of basic concepts of orbit motion is proposed, inasmuchas Keplerian orbits provide a reference solution to the simplest possible dynamic modelof Earth–orbiting satellites. In this respect, the presence of other bodies (Moon and Sunin particular) or other phenomena such as the solar wind or the atmospheric drag will beseen as a perturbations of this reference solution. The effects of these perturbation willbe the subject of the following chapters. First of all, the variational equations for orbitparameters will be derived, which provide a means for representing the effect of either

G. Avanzini, Spaceflight Dynamics – 6. Basic concepts 87

disturbances or control action on the shape of the orbit and its posizion in the inertialframe. In the last Chapter, some problems of practical interest will be analyzed, suchas the station–keeping problem of satellites in Low–Earth and Geo–stationary orbits. Inparticular, some examples of control techniques and determination of fuel–budget will beprovided.

6.2 Keplerian motion: a review

6.2.1 Birth of Astrodynamics: Kepler’s Laws

Since the time of Aristotle the motion of planets was thought as a combination of smallercircles moving on larger ones. The official birth of modern Astrodynamics was in 1609,when the German mathematician Johannes Kepler published his first two laws of planetarymotion. The third followed in 1619. From Astronomia Nova we read

... sequenti capite, ubi simul etiam demonstrabitur, nullam Planetae relinqui figuram

Orbitæ, praeterquam perfecte ellipticam; conspirantibus rationibus, a principiis Physicis,

derivatis, cum experientia observationum et hypotheseos vicariæ hoc capite allegata.

and

Quare ex superioribus, sicut se habet CDE area ad dimidium temporis restitutorii, quod

dicatur nobis 180 gradus: sic CAG, CAH areae ad morarum in CG et CH diuturnitatem.

Itaque CGA area fiet mensura temporis seu anomaliae mediae, quae arcui eccentrici CG

respondet, cum anomalia media tempus metiatur.

The third law was formulater in “Harmonice Mundi”

Sed res est certissima exactissimaque, quod proportio quae est inter binorum quorum-

cunque Planetarum tempora periodica, sit praecise sesquialtera proportionis mediarum

distantiarum, id est Orbium ipsorum

In modern terms, the three laws can be formulated as follows:

First Law: The orbit of each planet is an ellipse with the Sun at a focus.

Second Law: The line joining the planet to the Sun sweeps out equal areas in equaltimes.

Third Law: The square of the period of a planet is proportional to the cube of itsmean distance from the Sun.

Kepler derived a geometrical and mathematical description of the planets’ motionfrom the accurate observations of his professor, the Danish astronomer Tycho Brahe.Kepler first described the orbit of a planet as an ellipse; the Sun is at one focus andthere is no object at the other. In the second law he foresees the conservation of angularmomentum: since the distance of a planet from the Sun varies and the area being sweptremains constant, a planet has variable velocity, that is, the planet moves faster when itfalls towards the Sun and is accelerated by the solar gravity, whereas it will be deceleratedon the way back out. The semi-major axis of the ellipse can be regarded as the averagedistance between the planet and the Sun, even though it is not the time average, asmore time is spent near the apocenter than near pericenter. Not only the length of theorbit increases with distance, also the orbital speed decreases, so that the increase of thesidereal period is more than proportional.


Kepler’s first law can be extended to objects moving at greater than escape velocity(e.g., some comets); they have an open parabolic or hyperbolic orbit rather than a closedelliptical one. Yet the Sun lies on the focus of the trajectory, inside the “bend” drawn bythe celestial body. Thus, all of the conic sections are possible orbits. The second law isalso valid for open orbits (since angular momentum is still conserved), but the third lawis inapplicable because the motion is not periodic. Also, Kepler’s third law needs to bemodified when the mass of the orbiting body is not negligible compared to the mass of thecentral body. However, the correction is fairly small in the case of the planets orbiting theSun. A more serious limitation of Kepler’s laws is that they assume a two-body system.For instance, the Sun-Earth-Moon system is far more complex, and for calculations of theMoon’s orbit, Kepler’s laws are far less accurate than the empirical method invented byPtolemy hundreds of years before.

Kepler was able to provide only a description of the planetary motion, but paved theway to Newton, who first gave the correct explanation fifty years later.

6.2.2 Newton’s Laws of Motion

In Book I of his Principia (1687) Newton introduces the three Axiomata sive Leges Motus:

Lex I - Corpus omne perseverare in statu suo quiscendi vel movendi uniformiter in direc-

tum, nisi quatenus a viribus impressis cogitur statum illum mutare.

Lex II - Mutationem motus proportionalem esse vi motrici impressæ, et fieri secundum

lineam rectam qua vis illa imprimitur.

Lex III - Actioni contrariam semper et æ qualem esse reactionem: sive corporum duorum

actiones in se mutuo semper esse æquales et in partes contrarias dirigi.

The laws of motion can be translated into the following English version:

First Law: Every object continues in its state of rest or of uniform motion in a straightline unless it is compelled to change that state by forces impressed upon it.

Second Law: The rate of change of momentum is proportional to the force impressedand is in the same direction as that force.

Third Law: To every action there is always opposed an equal reaction..

The first law requires the identification of an inertial system with respect to which itis possible to define the absolute motion of the object. The second law can be expressedmathematically as

~F =d~p

dt

where ~F is the resultant of the forces acting on the object and ~p = m~v is its momentum.For a constant mass m, it is

~F = m~a

where ~a = d~v/dt is the acceleration of the mass measured in an inertial reference frame.A different equation applies to a variable-mass system (as for instance a rocket launcherwhere the variation of the mass per unit time, m, is sizable). In that case the equation ofmotion becomes

~F = m~a + m~v

The third law permits to deal with a dynamical problem by using an equilibriumequation. Its scope is however wider: for instance, the presence of the action ~F on mimplies an action −~F on another portion of an ampler system.


6.2.3 Newton’s Law of Universal Gravitation

In the same book Newton enunciated the law of Universal Gravitation by stating thattwo bodies, the masses of which areM and m, respectively, attract one another along theline joining them with a force proportional to the product of their masses and inverselyproportional to the square of the distance between them, that is,

F = GMm

r2

where G = 6.673 10−11 m3kg−1s−2 is the Universal Gravitational Constant. The law iseasily extended from point masses to bodies with a spherical symmetry, but it is fairly ac-curate also for body of arbitrary shape, when their distance is several orders of magnitudelarger than their own dimension.

6.2.4 Equation of Motion in the Two-Body Problem

Consider a system composed of two bodies of masses m1 and m2, m1 > m2; the bodies arespherically symmetric and no forces other than gravitation are present. Assume an inertialframe Fi; vectors ~R1 and ~R2 describe the positions of masses m1 and m2, respectively.The position of m2 relative to m1 is

~r = ~R2 − ~R1

so that their mutual distance is r = ||~r|| and the following equations hold for the relativevelocity and acceleration,

~r = ~R2 − ~R1 ; ~r = ~R2 − ~R1

where all the derivatives are taken in a (at least approximately) inertial frame.The law of Universal Gravitation is used to express the force acting on each mass, so

that the second law of dynamics can be used to describe separately the motion of eachmass,

m1~R1 = G

m1m2

r3~r ; m2

~R2 = −Gm1m2

r3~r

By subtracting the second equation from the first (after an obvious simplification con-cerning masses) one obtains

~r = −Gm1 +m2

r3~r

In most practical systems one of the masses is several order of magnitude larger thanthe other one. The best balanced 2–body system in the Solar system is the planet Plutowith its satellite Caron, which has a mass equal to 1/7 that of the planet. The Earth–Moon system is the second best balanced pair, where the mass of the Moon is only 0.0123that of the Earth.

As a consequence, it is usually possible to let m1 = M and m2 = m, with M ≫ m, sothat the Equation of Relative Motion can be rewritten in the following simple form:

~r = −µr

3

~r

where µ = GM is the primary body gravitational parameter. Some numerical values ofmu are reported in Tab. 6.1

It is worthwhile to remark that


• the relative position ~r is measured in a non-rotating frame which is not rigorouslyinertial;

• the relative motion is practically independent of the mass of the secondary body,when m≪ M ; a unit mass will be assumed in the following.

6.2.5 Potential Energy

The mechanical work done against the force of gravity to move the secondary body fromposition 1 to position 2

L =

∫ 2

1

µ

r3~r · d~s =

∫ 2

1

µ

r2dr = − µ

r2+µ

r1= U2 − U1

does not depend on the actual trajectory from 1 to 2, so that one deduces that thegravitational field is conservative. The work can be expressed as the difference betweenthe values of the potential energy at point 1 and 2, the dependence of which on the radius(i.e. the distance of m from the M) can be expressed in the form

U = −µr

+ C

The value of the arbitrary constant C is conventionally assumed to be zero in Astrody-namics, that is, the maximum value of the potential energy is zero, when the spacecraftis at infinite distance from the primary body, otherwise it is negative, and equal to

U = −µr

6.2.6 Constants of the Motion

By taking the scalar product of the equation of motion with the velocity vector ~v = ~r,

~r · ~r = − µ

r3

(

~r · ~r)

and writing everything on the left–hand side by using Eq. (6.6) in Appendix, one gets

~v · ~v +µ

r3

(

~r · ~r)

= vv +µ

r3rr

that is,d

dt

(v2

2− µ

r

)

= 0

During the motion the specific mechanical energy

E =v2

2− µ

r

Table 6.1: Values of the gravitational parameter µ.

Sun 1.327 1011 km3 s−2

Earth 3.986 105 km3 s−2

Moon 4.903 103 km3 s−2


that is, the sum of kinetic and potential energy, is constant, even though it can be con-tinuously transferred from the kinetic to the potential form, and vice versa. This resultis the obvious consequence of the conservative nature of the gravitational force, which isthe only action on the spacecraft in the two-body problem.

The vector product of ~r with the equation of motion gives

~r × ~r = 0 ⇒ d

dt

(

~r × ~r)

=d

dt(~r × ~v) = 0

where Eq. (6.3) was applied.Therefore, the angular momentum

~h = ~r × ~v

is a constant vector. Since ~h is normal to the orbital plane, the spacecraft motion remainsin the same plane. This result is not surprising: the unique action is radial, no torqueacts on the satellite, and the angular momentum is constant.

Finally, the cross product of the equation of motion with the (constant) angular mo-

mentum vector ~h gives

~r × ~h +µ

r3~r × ~h =

d

dt

(

~r × ~h)

+µ

r3~r ×

(

~r × ~r)

= 0

By using Eqs. (6.6) and (6.8), one gets

d

dt

(

~r × ~h)

+µ

r3

[(

~r · ~r)

~r − (~r · ~r) ~r]

=

d

dt

(

~r × ~h)

+ µ

(

r

r2~r − ~r

r

)

=

d

dt

(

~r × ~h)

− µd

dt

(~r

r

)

= 0 (6.1)

As a consequence, the vector

~e =~v × ~h

µ− ~r

r

is another constant of the spacecraft motion in the two-body problem.

6.2.7 Trajectory Equation

The dot product

~r · ~e =~r · ~v × ~h

µ− ~r · ~r

r=~r × ~v · ~h

µ− r =

h2)

µ− r

allows one to determine the shape of the orbit. Letting ν be the angle between the(constant) vector ~e and the relative position vector ~r, it is ~r · ~e = re cos ν, so that theshape of the orbit in polar coordinates centered in M is given by

r =h2/µ

1 + e cos ν


P r d νF

a

s

Figure 6.2: Building a conic section.

The radius attains its minimum value for ν = 0 and the constant vector ~e is thereforedirected from the central body to the periapsis.

A conic is the locus of points P such that the ratio of their distance r from a givenpoint F (focus) to their distance d from a given line a (directrix) is a positive constant e(eccentricity), r/d = e. Letting s be the distance between the focus and the directrix itis (see Fig. 6.2.7)

d = s− r cos ν ⇒ r = e (s− r cos ν)

By collecting r on the r.h.s., one gets

r =es

1 + e cos ν

which is the equation of a conic section, written in polar coordinates with the origin at thefocus. The parameter p = se is called the semi–latus rectum. It is the distance betweenthe focus and the point P of the conic section for ν = π/2.

By comparison with the trajectory equation, one deduces that

• in the two-body problem the spacecraft moves along a conic section that has theprimary body in its focus; Kepler’s first law is demonstrated and extended fromellipses to any type of conics;

• the semi-latus rectum p of the trajectory is related to the angular momentum of thespacecraft (p = h2/µ)

• the eccentricity of the conic section is the magnitude of ~e, which is named eccentricity

vector.

6.2.8 Relating Energy and Semi-major Axis

One of the constant of the motion, the angular momentum, has been proved to be simplyrelated to the semi-latus rectum. Also the specific energy E will now be related to ageometrical parameter of the conics.


The constant values of the angular momentum and total energy can be evaluated atany point of the trajectory, in particular at the periapsis, where the spacecraft velocity isorthogonal to the radius vector and

h = rPvP ; E =v2

P

2− µ

rP

=h2

2r2P

− µ

rP

If the orbit is a closed ellipse, the pericentre rP and the apocentre rA (i.e. the closestand farthest distance of m from the focus where M lies) are given, respectively, by rP =p/(1 + e) and rA = p/(1 − e), so that the semimajor axis a of the ellipse is

a =rP + rA

2=

p

1 − e2⇒ p = a(1 − e2) =

h2

µ

It is thus possible to write

h2 = µa(1 − e2) and rP = a(1 − e).

By substituting these expressions in that of the specific energy at periapsis, one obtains

E =µa(1 − e2)

2a2(1 − e)2− µ

a(1 − e)= µ

(1 + e

2a(1 − e)− 1

a(1 − e)

)

= − µ

2a

A very simple relationship exists between the specific mechanical energy and the semi-major axis.

The last relationship between geometrical parameter of the conics and the constant ofthe motion is obtained by extracting the eccentricity e from the above expression of h2,that is,

e =

√

1 − h2

µa=

√

1 − 2E h2

µ2.

It should be noted how degenerate conics have zero angular momentum and therefore uniteccentricity.

6.2.9 Elliptical Orbits

Geometry of an Elliptical Orbit

An ellipse can be defined as the locus of points such that the sum of their distances fromtwo fixed point, the focii, is constant:

r1 + r2 = 2a.

The maximum width of the ellipse is the major axis, and its length is 2a, while themaximum width at the centre in the direction perpendicular to the major axis is theminor axis, 2b. The distance between the focii is 2c.

From the expression of an ellipse’s equation in polar coordinates

r =p

1 + e cos ∋


and defining its geometrical characteristics as in Fig. 6.2.9, it is easy to derive the followingrelations:

rP = p/(1 + e)

rA = p/(1 − e)

2a = rP + rA = 2p/(1 − e2)

2c = rA − rP = 2pe/(1 − e2) = 2ae

b2 = a2 − c2 = a2(1 − e2) = ap

As a consequence, the area of the ellipse

A = πab

can be rewritten asA = πa2

√

(1 − e2) = πa√

(ap)

Figure 6.3: Geometry of an ellipse.

Period of an Elliptical Orbit

The differential element of area swept out by the radius vector as it moves through anangle dν is

dA =1

2r2dν ⇒ dA

dt=

1

2r2ν =

1

2rvθ =

h

2= const

thus demonstrating Keplers third law. The constant value of dA/dt is evaluated byassuming that the radius vector sweeps out the entire area of the ellipse,

dA

dt=πab

Twhich permits to obtain the period T of an elliptical orbit:

T =2πab

h= 2π

√

a2b2

µp= 2π

√

a3

µ.


Figure 6.4: Geocentric frame.

6.3 Three-Dimensional Analysis of Motion

6.3.1 Non–rotating Frames

In order to describe a space mission, it is necessary to provide, together with the time–frame for the sequence of events and evolution of the orbit, a suitable reference systemand a corrisponent set of coordinates, that suits the application. The first requirementfor the spatial description of an orbit is a suitable (and at least approximately inertial)reference frame. A frame centered in the primary body (i.e. the most massive bodyin the system), with axes pointing a fixed direction with respect to the far stars (thusnon–rotating) usually does well the job.

For Earth orbiting satellites, a geocentric frame FG = (O; g1, g2, g3) centered in thecenter of the Earth and based on the equatorial plane can be used (Fig. 6.3.1), theequatorial plane being perpendicular to the Earth’s spin axis. The third unit vector,namely g3, is parallel to the z-axes and is therefore defined; its direction is towards theNorth. The x-axis is parallel to the equinox line, that is, the line of intersection of theecliptic plane (the plane where the Earth orbit around the Sun lies) and the equatorialplane; its positive direction is from the Earth to the Sun on the first day of spring or thevernal equinox. The unit vector g1 points towards the constellation Aries (the ram).

The Earth’s axis of rotation actually exhibits a slow precession motion and the x–axisshifts westward with a period of 26000 years (a superimposed oscillation with a periodof 18.6 years derives from the nutation of the Earth’s axis, which is due to the variableinclination of the lunar orbit on the ecliptic plane). For computations, the definition ofthe geocentric system is based on the direction of the line–of–intersection at a specifieddate or epoch. In astronomy, an epoch is a moment in time for which celestial coordinatesor orbital elements are specified; the current standard epoch is J2000.0, which is January


1st, 2000 at 12:00. The remanining unit vectors, E2 and g2, are uniquely defined by therequirement for both frames to be orthogonal and right–handed.

The spacecraft position can be described by the Cartesian components of its positionvector with respect to the origin of the reference frame; the use of the distance ρ fromthe center and two angles is usually preferred. The declination δ is measured northwardfrom the x–y plane; the right ascension α is measured on the fundamental plane, eastwardfrom the vernal equinox direction.

6.3.2 Classical Orbital Elements

A Keplerian trajectory is uniquely defined by 6 parameters. The initial values for theintegration of the second-order vector equation of motion, i.e. the spacecraft positionand velocity at epoch, could be used, but a different set of parameters, which providesan immediate description of the trajectory, is preferable. The classical orbital elements

are widely used for this purpose. Only 4 elements are necessary in the two–dimensionalproblem: three parameters describe size, shape, and direction of the line of apsides; thefourth is required to pinpoint the spacecraft position along the orbit at a particular time.The remaining 2 parameters describe the orientation of the orbital plane.

The classical orbital elements are

1. eccentricity e (shape of the orbit);

2. semi–major axis a or, equivalently, semi–latus rectum p (size of the orbit);

3. inclination i, that is, the angle between the third unit vector g3 and the angular

momentum ~h (inclination of the orbit plane with respect to the base plane of thenon–rotating reference frame);

4. longitude of the ascending node, Ω, that is, the angle in the fundamental planemeasured eastward from the first coordinate axis of the non–rotating frame g1 tothe ascending node, i.e. the point where the spacecraft crosses the fundamentalplane while moving in the northerly direction (orientation of the orbit plane);

5. argument of periapsis ω, that is the angle in the orbit plane between the ascendingnode and the periapsis, measured in the direction of the spacecraft motion (peri-center direction in the orbit plane);

6. true anomaly ν0 at a particular time t0 or epoch (spacecraft position); it is sometimereplaced by time of periapsis passage T .

Some of the above parameters are not defined when either inclination or eccentricityare zero. Alternate parameters are

• longitude of periapsis Π = Ω + ω, which is defined when i = 0 and the ascendingnode is not defined;

• argument of latitude at epoch u0 = ω+ν0, which is defined when the orbit is circular,e = 0, and the periapsis is not defined;

• true longitude at epoch ℓ0 = Ω + ω + ν0 = Π + ν0 = Ω + u0, which remains definedwhen either i = 0 or e = 0.


6.3.3 Modified Equinoctial Orbital Elements

It should be noted that geostationary orbits are circular and equatorial, which means thatboth the eccentricity and the inclination are zero. One possible alternative choice is to usethe modified equinoctial orbital elements. This is a set of orbital elements that are alwaysunivocally defined, for circular, elliptic, and hyperbolic orbits with any inclination, asthe modified equinoctial equations exhibit no singularity for zero eccentricity and orbitalinclinations equal to 0 and 90 degrees.

However, two of the components are singular for an orbital inclination of 180 degrees,that is, for retrograde orbits, that is, equatorial orbits flown in the direction opposite toEarth’s spin motion. As these orbits are seldom used, this singularity does not representa serious problem.

Relationship between modified equinoctial and classical orbital elements is given bythe following set of equations:

1. p = a(1 − e2);

2. f = e cos(ω + Ω);

3. g = e sin(ω + Ω);

4. h = tan(i/2) cos Ω;

5. k = tan(i/2) sin Ω;

6. ℓ0 = Ω + ω + ν0.

The inverse relations that allow to recover the classical orbital elements from the modifiedones are as follows:

semimajor axis: a =p

1 − f 2 − g2

eccentricity: e =√

f 2 + g2

inclination: i = 2 tan−1(√

h2 + k2)

= tan−1(2√h2 + k2, 1 − h2 − k2

)

argument of periapsis: ω = tan−1 (g/f) − tan−1 (k/h)

= tan−1 (gh− fk, fh+ gk)

right ascension of the ascending node: Ω = tan−1 (k, h)

true anomaly at epoch: ν0 = ℓ0 − Ω − ω

argument of latitude at epoch: u0 = ω + ν0

= tan−1 (h sin ℓ0 − k cos ℓ0, h cos ℓ0 + k sin ℓ0)

In the above equations the expression tan−1(a, b) indicates the calculation of the fourquadrant inverse tangent.

6.3.4 Determining the orbital elements

The orbital elements are easily found starting from the knowledge of the Cartesian compo-nents of position and velocity vectors ~r and ~v at a particular time t0 in either non–rating


reference frame defined in Section 6.3.1. One preliminarily computes the components ofthe constant vectors

~h = ~r × v and ~e =~v × ~h

µ− ~r

r

and therefore the unit vectors

k =~h

h; i =

~r

r; n =

~g3 × ~k

||~g3 × ~k||; p1 =

~e

e

These vectors, centered in the center of mass of the primary body, define respectively thedirections normal to the orbit plane, and towards the spacecraft, the ascending node, andthe pericenter, respectively.

The orbital elements are thus given by

1. p = h2/µ

2. e = ||e||3. cos i = k · g3 = k3

4. cos Ω = n · g1 = n3 (Ω > π, if n2 < 0)

5. cosω = n · p1 (ω > π, if e3 < 0)

6. cos ν0 = i · p1 (ν0 > π, if ~r · ~v < 0)

In a similar way one evaluates the alternate parameters

• cosu0 = i · n (u0 > π, if i3 < 0)

• cos Π = p1 · g1 (Pi > π, if e2 < 0)

• cos ℓ0 = i · g1 (ℓ0 > π, if i2 < 0)

The last two equations hold only for zero orbit inclination (i = 0).

6.3.5 Orbit propagation: the ideal case

After the orbital elements have been obtained from the knowledge of ~r and ~v at a specifiedtime, the problem of updating the spacecraft position and velocity is solved in the perifocalreference frame using the closed–form solution of the equation of motion, if the effectsof perturbations is neglected. This means that, provided that the orbit elements remain(approximately/sufficiently) constant, the satellite will keep on orbiting the Earth along aknown orbit described by the set of 6 parameters chosen. The ~r and ~v components in thegeocentric-equatorial (or heliocentric-ecliptic) frame can be obtained using a coordinatetransformation.

Appendix: Vector Operations

The result of some vector operations is given in this appendix.

~a · ~a = a2 (6.2)

~a × ~a = 0 (6.3)

~a × ~b = ~b × ~a (6.4)

~a · ~b × ~c = ~a × ~b · ~c (6.5)


Equation (6.5) is intuitively proved by considering that either side represents the volume

of the parallelepiped the sides of which are given by the vectors ~a, ~b, and ~c. In a morerigorous way, the scalar triple product of three vectors

~a · ~b × ~c = det

a1 a2 a3

b1 b2 b3c1 c2 c3

is a pseudoscalar and would reverse sign under inversion of two rows. Therefore

~a · ~b × ~c = ~b · ~c × ~a = ~c · ~a × ~b

By equating the first and the last expression, and remembering that the dot productcommutes, one gets the result.

~a · ~a = aa ⇐ d

dt(~a · ~a) =

d

dt

(a2)

(6.6)

Equation (6.6) is very important in orbital mechanics and rocket propulsion: when a vectorelementary increment d~a is considered, its component parallel to ~a increases the vectormagnitude, whereas the perpendicular component just rotates ~a, keeping its magnitudeconstant.

As for the vector triple product, the following results hold:

(

~a × ~b)

× ~c = (~a · ~c)~b −(

~b · ~c)

~a (6.7)

~a ×(

~b × ~c)

=(

~b · ~c)

~a − (~a · ~c)~b (6.8)

Chapter 7

Orbit Trim Manoeuvres

All spacecraft are subject to perturbations which induce a drift in the spacecraft orbitalelements from the nominal mission orbit. As an example, air drag causes a decreasein semi-major axis, altitude errors cause an along–track drift in the spacecraft groundtrack and lunar perturbations cause a drift in the inclination of geostationary satellites.These perturbations must be compensated for by periodic trim manoeuvres to returnthe spacecraft to its nominal orbit. The effect of small propulsive burns on the spacecraftorbital elements can be determined through the use of variational equations. The requiredaction can then be implemented with impulse thruster firings or continuous acceleration,by electric propulsion. In particular

• cold gas jets are characterised by a low specific impulse Isp, but are very simpleand cheap;

• monopropellant thrusters (usually hydrazine reacting on a heated catalyst bed)provide a good Isp but are expensive;

• bi-propellant thrusters provides a higher Isp but are more complex;

• electric propulsion is characterised by a very high Isp in a very low force range(few mN), providing an extended useful life.

7.1 Variational Equations

Consider an orbit frame FO, defined by the unit vectors

• r along the outward radial direction (that is from the spacecraft centre of massCM towards the outer space, along the line that connects the Earth’s centre to thespacecraft CM);

• n normal to the orbit plane, in the same direction as the satellite angular momentum~r × (m~v);

• t in the transverse direction (that is perpendicular to r, in the orbit plane andapproximately in the same directions as the orbital velocity vector1).

1To be more precise, the orientation of t is chosen so that t · ~v > 0.

101

102 G. Avanzini, Spaceflight Dynamics – 7. Orbit Trim Manoeuvres

An external propulsive force can be represented in the orbit frame as

~F = FRr + FT t + FN n

Also, in this frame, it is

~r = rr

~v = rr + rνt

The work done (per unit mass) by the external force over an element ds along thepath is

dE =1

m~F · ~ds

so that

dEdt

=1

m~F · ~v

=1

m(rFR + rνFT )

But for the two body problem, the specific energy of the orbiting body is

E = − µ

2a

and therefore it is alsodEdt

=µ

a2

da

dt

Comparing the two expressions of the rate of change of the energy E one gets

da

dt=

2a2

µm(rFR + rνFT )

The values of r and ν can be obtained from the orbit equations for the two bodyproblem, in which case it is

r =a(1 − e2)

1 + e cos ν

Using the chain rule, it is

r =dr

dt=

dr

dν

dν

dt

so that

dr

dt= − a(1 − e2)

(1 + e cos ν)2(−e sin ν)

dν

dt

= re sin ν

(1 + e cos ν)

dν

dt

Also, from the two body problem, the angular momentum vector magnitude h is givenby

h = r2ν =√

µa(1 − e2)

and, as a consequence, it is

ν =

√

µa(1 − e2)

r2

G. Avanzini, Spaceflight Dynamics – 7. Orbit Trim Manoeuvres 103

Considering the mean motion

n =

√µ

a3

ν can be rewritten as

ν =na2

r2

√1 − e2

Substituting the values found for r ad ν in the expression of the rate of variation ofthe semi–major axis, one gets

da

dt=

2a2

µ

na2

r2

√1 − e2

(re sin ν

1 + e cos ν

FR

m+ r

FT

m

)

Using the orbit equation,

1 + e cos ν =a

r(1 − e2)

the last formulation of a can be further simpliied,

da

dt=

2e sin ν

n√

1 − e2FR

m+

2a√

1 − e2

nr

FT

m

It can be noted how the effect of a radial force component FR over the semi-major axisvariation depends on the anomaly ν, while a force component FT along the transversedirection has an effect which is independent of the satellite position along the orbit.

In a similar way it is possible to demonstrate that the variational equation for theeccentricity e, the orbit inclination i and the right ascension Ω are given by

de

dt=

√1 − e2 sin ν

na

FR

m+

√1 − e2

na2e

[a2(1 − e2)

r− r

]FT

m

di

dt=

r cos(ω + ν)

na2√

1 − e2FN

m

dΩ

dt=

r sin(ω + ν)

na2 sin i√

1 − e2FN

m

It should be noted how only an out–of–plane force component can vary the orbit plane(that is its inclination and/or the ascending node).

7.2 Perigee Raise Manoeuvre

A transverse impulse burn at apogee can be used to raise the perigee, in order to com-pensate for the effects of air drag. At apogee it is

ν = π ; rap = a(1 + e)


From the variational equation calculated at apogee, with FR = 0, one gets

da

dt=

2

n

√

1 − e

1 + e

FT

m

de

dt= − 2

na

√1 − e2

FT

m

Assuming that the thruster burn is a pulse of small duration ∆t, the velocity incrementis

∆VT =FT

m∆t

so that the change of the semi–major axis and orbit eccentricity due to a tangential burnat perigee is

∆a =2

n

√

1 − e

1 + e∆VT

∆e = − 2

na

√1 − e2∆VT

Remembering that the perigee radius is

rpg = a(1 − e)

the change in perigee radius is

∆rpg =∂rpg

∂a∆a +

∂rpg

∂e∆e

= (1 − e)2

n

√

1 − e

1 + e∆VT − a

−2

na

√1 − e2∆VT

so that

∆rpg =4

n

√

1 − e

1 + e∆VT

With a similar method it is possible to demonstrate that

∆rap = 0

The change in orbit period

T = 2π

/√µ

a3

due to the manoeuvre is

∆T =∂T

∂a∆a = 2π

1

2√

a3/µ(3a2)

2

n

√

1 − e

1 + e∆VT

that is

∆T = 6πa2

µ

√

1 − e

1 + e∆VT

G. Avanzini, Spaceflight Dynamics – 7. Orbit Trim Manoeuvres 105

7.3 Plane Changes

From the variational equation of orbit inclination

di

dt=r cos(ω + ν)

na2√

1 − e2FN

m

assuming again a pulse of short duration, it is

∆i =r cos(ω + ν)

na2√

1 − e2∆VN

From this relation, it is evident that, for a given velocity increment ∆VN the variation ∆iof the orbit inclination is maximised is ω + ν = 0, that is the burn has to be applied atthe ascending node, when ν = −ω.

7.4 Finite Burn Manoeuvres

When low–thrust propulsion is employed (such as electrical thrusters, with a very high Ispbut a thrust in the range of some mN) a burn of finite duration needs to be considered.In this case it is necessary to perform an explicit integration of the variational equationrather then simply assuming a finite increment proportional to the rate of variation of theconsidered orbit element. Considering a perigee raise manoeuvre as a first example, it ispossible to determine the effects on a and e of a finite burn, for an arclength ∆ν/2 oneach side of the apogee. Changing the independent variable from time to true anomaly,one gets

da

dt=

da

dν

dν

dt

so that, assuming a transverse force only, the following equation is obtained:

2a

nr

√1 − e2

FT

m=

da

dν

na2

r2

√1 − e2

Rearranging, it is

da

dν=

2r

n2a

FT

m

=2(1 − e2)

n2(1 − e cos ν)

FT

m

For a small trim burn it is possible to neglect the variation of the orbit elements duringthe burn, so the total variation of the semi–major axis a is given by

∆a =2

n2(1 − e2)

FT

m

∫ νoff

νon

1

1 + e cos νdν

where

νon = π − ∆ν

2; νoff = π +

∆ν

2

are the position along the orbit at which the thruster are switched “on” and “off”, re-spectively.


It is possible to explicitly solve the integral

∫ νoff

νon

1

1 + e cos νdν

using the goniometric transformation

cosα =1 − tan2 α

2

1 + tan2 α2

and performing the change of variable

x = tanν

2

such that (

1 + tan2 ν

2

)

dν = 2dx

The result of the integration is

∫ νoff

νon

1

1 + e cos νdν =

2√1 − e2

[

arctan

(1 − e√1 − e2

tanν

2

)]π+∆ν

2

π−∆ν

2

and

∆a =4

n2

√1 − e2

FT

m

[

arctan

(1 − e√1 − e2

tanν

2

)]π+∆ν

2

π−∆ν

2

With a similar procedure it is possible to obtain the expression for the increments ∆eand ∆rpg of the eccentricity and the perigee radius, respectively.

In a similar fashion, it is also possible to analyze the effects of low–thrust propulsionsystems on orbit plane position (represented by Ω and i). As an example, when thevariational equation for orbit inclination is dealt with, one can write

di

dt=r cos(ω + ν)

na2√

1 − e2FN

m=

di

dν

na2

r2

√1 − e2

As thruster action along n (that is, the direction normal to the orbit plane) does notchange the orbit shape (namely its semi–major axis and eccentricity) it is possible tointegrate the above equation with respect to the true anomaly and evaluate the variation∆i induced by a normal acceleration FN/m along an orbit arc from νon to νoff .

Chapter 8

Orbit Control

As stated in the previous chapter, it is possible to vary spacecraft orbit parameters bypropulsive burns. For this reason, the equation that rule orbit shape (a and e) and orbitplane (i and Ω) were derived, where orbit parameters increment were determined as afunction of thrust pulse in the orbit reference frame FO.

Such maneuvres may be necessary as a consequence of the effects of external perturba-tions, which otherwise would result in a spacecraft drifting away from its nominal missionorbit. It is therefore important that the orbit be controlled through periodic burns tocompensate for these perturbations.

Longitude drift of a geostatonary satellite will lead to the satellite moving out of thefield of view of fixed ground antennae. Similarly, orbit decay will lead to in–track positionerrors as the orbital period decrease and the satellite advances ahead of its nominalposition.

At present most orbit control algorithms require a human operator to plan and up–link command sequences to arm the propulsion system and execute the required burns.However, there is move towards on–board orbit control where the spacecraft uses GPSsatellite navigation to determine its orbital elements and plans its own manoeuvres usingon–board software. This leads directly to significant operational cost reductions, at theexpenses of a much higher cost of the control system, with an equipmenst - hardware andsoftware - harder to be certified.

8.1 Air Drag Control

For circular Low Earth Orbit (LEO) air drag causes a decrease in orbit radius. It ispossible to use a two–impulse Hohmann transfer to manoeuvre the spacecraft back to itsnominal orbit.

Example - An Earth observation mission: Orbiting Volcano Observatory (OVO)

Parameterorbit inclination 70 degaltitude 518 kmperiod (approx.) 94 minorbit per day 15orbit per week 105

The orbit is specified for mission payloadeffectiveness, and there is an exact groundtrack repeat requirement.It is necessary to evaluate the station–keeping fuel budget for maintaining the nom-inal orbit.

The air drag is quite significant and as orbit decays, the orbit period shortens and the

107

108 G. Avanzini, Spaceflight Dynamics – 8. Orbit Control

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6

518 000

517 860

6

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66

1 2 3 4 t [weeks]

a [m]

deadband

exact repeat cycle is lost. The daily altitude loss is approximately 20 m. The resultingchange in orbit period is

∆T =∂T

∂a∆a = 3π

√a

µ∆a

which, for a nominal semi–major axis a = 6371 + 518 km and a variation ∆a = 20 m,gives a variation of the period ∆T = 0.0248 s per day.

The Earth rotates with an angular velocity ωE of

ωE =2π

24h= 7.27 · 10−5rad s−1

An error of ∆T in orbit period result in a ground track drift of

∆s = REωE∆T = 11.5m per day

Over a period of 1 month the ground track drift will be approximately of 320 m, enoughto miss targets with high resolution cameras and narrow field of view. So there is anevident need to regularly raise the orbit to null the ground track drift.

A 7 day station–keeping cycle must be designed to raise the orbit of ∆a = 140 m,using a two–pulse Hohmann transfer. Remembering that, for a Hohmann transfer, it is

∆a =2

n

√

1 − e

1 + e∆V

the ∆V necessary for each weekly correction is

∆V =1

2

√µ

a3∆a

that, for the case under consideration, is equivalent to ∆V = 0.077 m/s.

Over one year, the ∆V budget is ∆V = 0.077 × 52 = 4.0 m/s and thte total ∆Vbudget for an expected mission life of 3 years is approximately 12 m/s.

At this point, it is necessary to design a spacecraft with enough propellant for a 3years mission, with a 20% contingency.

G. Avanzini, Spaceflight Dynamics – 8. Orbit Control 109

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rpg0− ∆rpg

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Example - Perigee control

It is possible to use apogee burn to raise perigee, after air drag decay. For small pertur-bations, it is possible to assume a linear decrease of the perigee radius with time, thatis

rpg = rpg0− αt

where the decay rate α can be obtained from flight data of previous satellites or fromsimulation. A dead–band ∆rpg must be defined, that is the tolerance within which it isnecessary to maintain the satellite perigee radius with respect to the nominal value. Ifrpg is within the dead–band, no action must be taken, but if rpg crosses the boundary, astation–keeping manoeuvre must be performed. In the present case, the manoeuvre is anapogee tangential burn at the subsequent apogee pass, in order to raise the perigee at thedesired value.

The time between burns is approximately

∆t =∆rpg

α

The fuel budget for the perigee raise manoeuvre is obtained from the relation

∆V =n

4

√

1 + e

1 − e∆rpg

and the station keeping budget (that is the required ∆V per unit time) is

S =∆V

∆t

so that

S = αn

4

√

1 + e

1 − e

It should be noted how the station–keeping budget is independent of the dead–bandamplitude ∆rpg, which is defined by mission objectives, such as payload viewing require-ments or similar. Of course, if the tolerance is more strict, more manoeuvres must bescheduled and, although this does not affect the fuel budget, it increase the mission costsbecause of higher workload for the ground staff.


8.2 Geostationary Orbit Control

Also for geostationary orbits (GEO) the satellite may require accurate station–keeping, inorder to remain in the field of view of fixed ground antennae for direct broadcasting. Afterthe launch and the early operation phase (LEOP), during which the satellite is deployedand injected into its final orbit, the routine operation phase begins. Again, a dead–bandmust be specified that satisfy accuracy requirements avoiding rapid thruster firings. Ofcourse, the smaller the dead–band, the more accurate is the control, at the expenses of agreater frequency of manoeuvring.

Also in this case, station–keeping is usually under ground control, although the trendis to move to autonomous operations, to reduce operational costs.

There are two types of station keeping modes:

• North–South station keeping, required to compensate the effects of lunar andsolar perturbations on orbit inclination, with a significant cost in terms of ∆V forcontrol;

• East-West station keeping, mainly required to compensate the effects of tesseralharmonics of the Earth’s gravitational field, induced by the ellipticity of the equator(which is not a circle).

The location of the satellite on the geostationary ring can be defined in terms of itslongitude, relative to the Greenwhich meridian (λ = 0). The rate of change of longitudeis

λ =

√µ

a3− ωE

where ωE = 2π/24 h. For a geostationary orbit, the condition λ = 0 must be satisfied, sothat it is a =42 164.5 km.

The equatorial cross–section of the Earth is not perfectly circular. Its ellipticity inducea transverse acceleration λ, which, for small longitude dead–band, can be consideredconstant. Under this assumption, the variation of satellite longitude on the geostationaryring is

λ(t) = λ(t0) + λ(t0) (t− t0) +λ

2(t− t0)

2

Given a longitude dead–band δλ about the nominal position λm, the station keepingcycle begins at the time t0 and ends at t0 +T . The initial longitude drift rate λ0 must besuch that the longitude drift is reversed at the bottom of the dead–band, for t = t0 +T/2,so as to fully exploit the dead–band amplitude.

Assuming a positive longitude drift acceleration λ > 0, it is necessary to start thestation–keeping cycle on the East boundary

λ(t0) = λm + δλ

with a negative initial drift rate λ(t0) < 0, requiring that the drifting motion is reversedat the West bound

λ(t0 + T/2) = λm − δλ

where

λ(t0 + T/2) = 0

G. Avanzini, Spaceflight Dynamics – 8. Orbit Control 111

From the variation of the longitude drift rate

λ(t) = λ(t0) + λ (t− t0)

one gets

λ(t+ T/2) = λ(t0) + λT

2= 0

so that

λ(t0) = −λT2

which is negative, as expected. Also, it is

λ(t+ T/2) = λ(t0) + λ(t0)T

2+λ

2

(T

2

)2

= λm − δλ

Substituting in this equation

λ(t0) = λm + δλ ; λ(t0) = −λT2

one gets

λm + δλ− λ

(T

2

)2

+λ

2

(T

2

)2

= λm − δλ

so thatλ

2

(T

2

)2

= 2δλ

and, finally, the free drift time between burns is given by

T = 4

√

δλ

λ

Substituting for T into the expression of the initial longitude drift rate, one gets

λ(t0) = −2√

λδλ

At the end of the cycle the satellite longitude will return to the top of the dead–band,with

λ(t0 + T ) = λm + δλ

andλ(t0 + T ) = −λ(t0) = 2

√

λδλ

At this point a manoeuvre is required to revert the drift rate of

∆λ = −4√

λδλ

From the expression of the longitude drift rate

λ =

√µ

a3− ωE

it is easy to obtain the variational equation

∆λ =∂λ

∂a∆a = −3

2

√µ

a5∆a


from which it is evident that a longitude drift rate can be induced by a change ∆a ofthe semi–major axis. From the variational equations, a change of a can be obtained by atransverse burn ∆VT , where

∆a = 2

√

a3

µ∆VT

Therefore

∆λ = −3

a∆VT

and, knowing the required drift change ∆λ for longitude control, one gets

∆VT =4

3a√

λδλ

In this way the ∆V required for each cycle of duration T for station–keeping within adead–band of amplitude 2δλ is determined. The station–keeping budget for mission lifeTM with N = TM/T cycles is ∆VM = N∆V . Multiplication by a safety factor and theuse of the rocket equation allows one to find the required fuel mass.

The case of negative drift acceleration λ < 0 can be treated in a similar way, providedthat the station–keeping limit–cycle is started on the West bound λ− δλ with a positivedrift rate, and drift reversal is achieved at the East boundary of the allowed tolerance,λ+ δλ.

Space Craft Dyanmics

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