航天动力学与控制

Lecture 2

2007 年 2 月

General Rigid Body Motion – The concept of Rigid Body

• A rigid body can be defined as a system of particles whose relative distances are fixed with time. The internal potential energy of a rigid body is constant

– Orbital Mechanics: translational motion of spacecraft under th

e influence of gravitational and other forces becomes orbital mechanics – Attitude Mechanics: Rotation about the center of mass under t

he influence of applied torques becomes attitude mechanics.

Basic Concept and Dynamical Equation

n-body problem – n-body problem has only 10 known integrals of motion: 3 velocity

components, 3 position components, 3 angular momentum components, and kinetic energy.

– only the two-body problem has an unrestricted solution. Special cases of the

three-body problem have been treated in closed form

Two-body and central force motion

•Assumption•Masses could be two bodies whose minimum distance apart is large compared to their largest dimensions or could have spherically symmetric mass distributions and never touch each other. This restriction allows these masses to be treated as particles in the following analysis.

Two-body and central force motion (basic differential equation)

Center of mass: c

1 1 2 2( ) ( ) 0c cm m r r r r

Geometry dictates that

2 1 r r r

Which permits expressions to become

21

1 2c

m

m m

r r r 1

21 2

c

m

m m

r r r

then

1 21 1 1 1

1 2c

m mm m

m m

F r r r 1 2

2 2 2 21 2

c

m mm m

m m

F r r r

Two-body and central force motion (basic differential equation)Mutual attraction requires that

1 2F F

1 2c cm mr rthen 0c rApplying this results leads immediately to

1 21 2

1 2

m m

m m

F F r

In terms of the gravitational attraction yields the basic differential

equation of motion for the two-body system. 2

2 30

d

dt r

rr

1 2( )G m m

Two-body and central force motion (Solution of the differential equation)

Using a straightforward vector approach. 2

2 30

d

dt r

rr r r

Defining angular momentum per unit mass as d

dt

rh r

and2

2( ) 0

d d d d d

dt dt dt dt dt

r r r rr r

Leads to the conclusion that / 0d dt h

constanth

r

Thus, angular momentum is conserved and three integrals of motion are

.

and this plane must be inertially fixed

(1)

Two-body and central force motion (Solution of the differential equation)Cross equation (1) with h

2

2 3 3( )

d d

dt r r dt

r rh r h r r

Applying the standard identity for triple vector products and noting that

d drr

dt dt

rr

Leads to 2

2( )

d d

dt dt r

r rh

(2)

Since h is constant, equation (2.) may be integrated directly

( )d

rdt r

rh r e

e is a constant of integration and is called the eccentricity vector. The orientation of e in this plane is taken as a reference direction

(3)

Two-body and central force motion (Solution of the differential equation)

Dot expression (3) with r and using the triple scalar product identity to obtain

2 /

1 cos

hr

e

cosre

r e

Vector e is parallel to the direction of minimum r, and angular is measured from this point to the position in the orbit. This angle is known as the true anomaly

Kinetic Energy 1 1 1 2 2 2

21 2 1 2

1 21 2 1 2

22 1 1 2

2 21 2 1 2

2 1 2 1

2 1

1 1

2 2

21

2 ( ) ( )

21

2 ( ) ( )

1( ) ( )

2 2

c c c

c c c

c c

T m m

m m m mm

m m m m

m m m mm

m m m m

m m m m

m m

r r r r

r r r r r r

r r r r r r

r r r r

The first term on the right side is just translational kinetic energy of two-body system. The remaining term is rotational kinetic

energy about the center of mass.

(4)

Central Force Motion

If 1 2m m then the motion of m2 about m1 is essentially the motion of a particle in an inertially fixed field. This type of motion

is known as central force motion.

1c r r 2 1 r r r 2

1

2T m r r

Notice that the two-body problem may be treated as central force motion of a particle of mass 2 1 2 1/( )m m m m

UF

The force which is a function of r can be written as the gradient of gravitational potential

Kinetic energy per unit mass is

d d

dt dt

r r

r r

d dr r

dt dt

ri i

Central Force MotionWhich permits writing kinetic energy in a more convenient form

2 2 2( )r r

The energy of a unit mass in a central fields is given

2 2 2( ) ( )r r U r

introducing a new variable as r

2

2

1d dr rr

d r d h

22

2 2

2 2 ( )d U r

h d h

thus

Central Force Motion

22

2( )

dd

U rh

Consider the inverse-square force law characterized by

( )U r

0

0

21

0 22

2 2 2

sin2 2( )

d h

U rh h h

Define a parameter p as 2h

p

0

2

1 1

sin(1 2

r p

p p

2

1 1

cos(1 2

r p

p p

Rearranging gives

0 Is replaced by

2

1 cos(

pr

e

2

1p

e

(5)

Eccentricity

(4) is identical to form (5)

0 Plus sign in (5)

minus sign in (5)

0 / 2 0 p

e=0, Circle

0<e<1 ellipse

e=1 parabola

e>1 Hyperbola

Kepler’s Time EquationThe relationship between time and position in orbit is considered

sin( , )d

h rvdt

r

r

Noting that sin( , )d d

v rdt dt

rr

2 dh r

dt

yields

3 (1 cos

ddt

p e

A new variable is introduced

coscos

ae r

a

Since 2(1 )p a e

cosa r

ae

Kepler’s Time EquationTaking the time derivative of this gives radial speed

cosr ae

Evaluation of energy at periapsis 2 2( )

2 2

r r

r a

Rearranging this leaves 2 2

2 2( )ar r

ae a r

ra

Then

3sinpt e

a

0 when 0pt

3n

a

Define

sinpnt e

This is known as Kepler’s equation for

relating time position in orbit

Kepler’s Time EquationWhen eccentricity is less than 1.0, the orbit is closed and periodic. The associated period can be determined by considering the rate of area swept out by the radius vector which is the areal velocity.

21 1

2 2

dAr h

dt

Integrating over the entire orbit gives the period as 2A

h

For an ellipse A ab

2 /

1p

hr

e

(1 )ph r e

From geometry, (1 )pr a e 21b a e

3 22

a

n

Kepler’s third law.

Kepler’s second law.

•相对运动 (加速度 )

( )b

d d d

dt dt dt

rr ω r ( )

b b

d d d dx y z x y z

dt dt dt dt

r ri j k i j k ω

The second term of the expression becomes

( ) ( )b

d

dt ω r ω r ω r ω r ω r ω ω r

0 2 ( )b b R R r ω r ω r ω ω r

0R is the acceleration of the origin of the moving frame br is the apparent acceleration of particle in the moving frame

第一章 基本的物理定律

2 bω r is the Coriolis acceleration due to the motion of p in x, y, z

ω r is the acceleration of p due to the changes of the angular velocity

( ) ω ω r is the centrifugal acceleration

•Motion of the Earth’s Surface

第一章 基本的物理定律

B

mg

m R

F RR

( ) 2 ( )B E b b R Ω Ω R a Ω v Ω Ω r

B

mg

m R

F RR

( ) 2 ( )b E bgR

R

a Ω Ω R Ω v Ω Ω r

40.728 10 rad/s Ωthe two centrifugal acceleration terms are negligible compared to the Coriolis term for low altitude orbits.

2b bgR

R

a Ω v

•哥氏力的影响

第一章 基本的物理定律

To illustrate the effect of this Coriolis acceleration, consider a satellite traveling due east over the ground station,

,b v mg v j F k ( cos ) ( sin ) Ω i k

( 2 cos ) (2 sin )b g v v a k i

This indicates that an observer would notice the satellite turing to the south or to the right of its initial flight path.