Transcript central force motion

航天动力学与控制
Lecture 2
2007年2月
Basic Concept and Dynamical
Equation
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
the influence of gravitational and other forces becomes orbital mechanics
– Attitude Mechanics: Rotation about the center of mass under
the influence of applied torques becomes attitude mechanics.
Two-body and central force
motion
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
•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
m1 (r1  rc )  m2 (r2  rc )  0
Geometry dictates that
r2  r1  r
Which permits expressions to become
r1  rc 
m2
r
m1  m2
r2  rc 
 m1
r
m1  m2
then
F1  m1r1  m1rc 
m1m2
r
m1  m2
F2  m2r2  m2rc 
m1m2
r
m1  m2
Two-body and central force
motion (basic differential equation)
Mutual attraction requires that
F1  F2
m1rc  m2rc
then r  0
c
Applying this results leads immediately to
F1  F2 
m1m2
r
m1  m2
In terms of the gravitational attraction yields the basic differential
equation of motion for the two-body system.
d 2r 
 3r0
2
dt
r
  G(m1  m2 )
Two-body and central force
motion (Solution of the differential equation)
Using a straightforward vector approach.
d 2r 
r  2  3 .r  r  0
dt
r
(1)
Defining angular
momentum per unit mass as h  r 
r
and
dr
dt
d
dr
dr dr
d 2r
(r  )    r  2  0
dt
dt
dt dt
dt
Leads to the conclusion that
dh / dt  0
Thus, angular momentum is conserved and three integrals of
motion are h  constantand this plane must be inertially fixed
Two-body and central force
motion (Solution of the differential equation)
Cross equation (1) with
h
d 2r


dr

h


r

h


r

(
r

)
2
3
3
dt
r
r
dt
Applying the standard identity for triple vector products and noting that
r
Leads to
dr
dr
r
dt
dt
d 2r
d r
h   ( )
2
dt
dt r
(2)
Since h is constant, equation (2.) may be integrated directly
dr

dt
h 
r
(r  re)
(3)
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
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
h2 / 
r
1  e cos 
cos  
r e
re
(4)
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
 m1r1  r1    m2r2  r2 
2
2

m1m22
2m1m2
1
  m1rc  rc 
r

r

r

r
c
2
(m1  m2 ) 2
(m1  m2 )

T

m2 m12
2m1m2
1
  m2rc  rc 
r

r

r

r
c
2
(m1  m2 ) 2
(m1  m2 )

(
m2  m1
1 mm
)rc  rc  ( 2 1 )r  r
2
2 m2  m1
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.
Central Force Motion
If
m1
m2
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.
rc  r1 r2  r1  r T 
1
m2r  r
2
Notice that the two-body problem may be treated as central force motion of a particle of mass
m2 m1 /(m2  m1 )
The force which is a function of r can be written as the gradient of gravitational potential
F  U
Kinetic energy per unit mass is

 dr dr

 dt dt
dr
d
 ri r  r i r
dt
dt
Central Force Motion
Which permits writing kinetic energy in a more convenient form

 2
(r  r 2 2 )

The energy of a unit mass in a central fields is given

 2 2 2
(r  r  )  U (r )


r
d
1 dr
2
r
 2

r
d
r d

h

introducing a new variable as
thus
2  d  
2U (r )
2






h 2  d 
h2
2
Central Force Motion
d  
d
Consider the inverse-square force law characterized by
  0   

0


 
  2 
d
1 
h

  sin
  2 2 
2
2
 U (r )  
 2 

2
h2
h
h  0

2
Define a parameter p as
0
U (r )  
2
2

U
(
r
)




h2
Is replaced by  
cos(     

2
1 1

r p
1 2

p 2 p
p
h

sin(   0   
1 1

r p
1 2

2
p
p
Rearranging gives
r
p
1  e cos(   
(5)
e  1
2p

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 Equation
The relationship between time and position in orbit is considered
h  rv sin(r,
Noting that
v sin(r,

p
dr
)
dt
3
dt 
dr
d
)r
dt
dt
d
(1  e cos  
A new variable is introduced
ae  r cos 
a
p  a(1  e2 )
cos 
Since
cos 
ar
ae
h  r2
d
dt
yields
Kepler’s Time Equation
Taking the time derivative of this gives radial speed
r  ae cos
Evaluation of energy at periapsis
Rearranging this leaves
ar 2 r 2

Then
 ae2  (a  r )2
r 

a
 0
r 2  (r ) 2 


 
2
r
2a

a
t    e sin
3 p
when
Define
n

a3
nt p    e sin
This is known as Kepler’s equation for
tp  0
relating time position in orbit
Kepler’s Time Equation
When 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.
dA 1 2
1
 r h
dt 2
2
Kepler’s second law.
Integrating over the entire orbit gives the period as

A   ab
For an ellipse
h2 / 
rp 
1 e
From geometry,
2A
h
h  rp  (1  e)
rp  a (1  e)
  2
a3


2
n
b  a 1  e2
Kepler’s third law.
第一章 基本的物理定律
•相对运动(加速度)
r
d  dr 
d
 (ω  r)


dt  dt  b dt
d  dr 
d
 dr 
 ( xi  yj  zk )  xi  yj  zk  ω   


dt  dt  b dt
 dt  b
The second term of the expression becomes
d
(ω  r )  ω  r  ω  r  ω  r  ω  rb  ω  (ω  r )
dt
R  R 0  rb  2ω  rb  ω  r  ω  (ω  r )
R0
rb
2ω  rb
ωr
ω  (ω  r )
is the acceleration of the origin of the moving frame
is the apparent acceleration of particle in the moving frame
is the Coriolis acceleration due to the motion of p in x, y, z
is the acceleration of p due to the changes of the angular velocity
is the centrifugal acceleration
第一章 基本的物理定律
•Motion of the Earth’s Surface
RB 
F
mgR

m
R
R B  Ω  (Ω  R E )  ab  2Ω  v b  Ω  (Ω  r )
RB 
F
mgR

m
R
ab   g
R
 Ω  (Ω  R E )  2Ω  v b  Ω  (Ω  r )
R
Ω  0.728 104 rad/s
ab   g
R
 2Ω  v b
R
the two centrifugal acceleration terms are negligible compared to
the Coriolis term for low altitude orbits.
第一章 基本的物理定律
•哥氏力的影响
To illustrate the effect of this Coriolis acceleration,
consider a satellite traveling due east over the ground station,
vb  vj, F  mgk
Ω  ( cos  )i  ( sin  )k
ab  ( g  2v cos  )k  (2v sin  )i
This indicates that an observer would notice the satellite turing
to the south or to the right of its initial flight path.