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PC4262 Remote Sensing
Satellite Orbits
Dr. S. C. Liew, Feb 2003
[email protected]
I. Orbital Motion In An Inverse-Square-Law Central Force Field
Equation of Motion
The motion of a satellite around the earth is governed by the Newton’s second law of motion,
i.e. F  mr , where F is the earth’s gravitational force acting on the satellite, m is the mass of
the satellite and r is the position vector of the satellite. We will first consider an idealized case
where the earth is approximated by a homogeneous sphere, then the gravitational force acting
on the satellite has a magnitude
(1)
GMm
F
r2
where G  6.672  10 11 N m 2 kg -2 is the universal gravitational constant, M is the mass of
the earth, and r is the distance of satellite from the centre of the earth. The force is directed
towards the centre of the earth, and hence the satellite is moving under the influence of a
central force field.
Let the center of the earth be the origin of the coordinate system. At time t = t0, the satellite is
at a location r(t0) = r0 having a velocity v(t0) = v0. The acceleration due to gravity is
(2)
K
a(t )   rˆ
r2
where K = GM, r̂ is the unit vector along the direction of r. At any time t, the velocity vector
is given by
t
(3)
v(t )  v 0   a(t ' ) dt '
0
and the position vector is
t
r(t )  r0   v(t ' ) dt '
(4)
0
So, in principal, given the initial position and velocity vectors, the position and velocity of the
satellite at any instant t can be evaluated. In practice, evaluating the integrals in (3) and (4) is
not straightforward. Note that the acceleration vector is expressed in terms of the
instantaneous position vector. To carry out the integration over time would require the
position vector to be known as a function of the time (which, by the way, is the solution we
are seeking). In later sections, methods for solving the equation of motion to obtain r(t) will
be derived.
The satellite always stays on the plane formed by the initial vectors r0 and v0 (why?). For
simplicity, let the x - y plane to be the orbital plane and the z axis be perpendicular to this
plane, and let the earth’s center be located at the origin of the coordinate system. It is
© S. C. Liew, Feb 2003.
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convenient to work with the polar coordinates (r, ) on the orbital plane. The three orthogonal
unit vectors in this coordinate system are rˆ , ˆ and zˆ , such that
(5)
zˆ  rˆ  ˆ , rˆ  ˆ  zˆ , ˆ  zˆ  rˆ
The time derivative of these unit vectors are

zˆ  0, rˆ   ˆ , ˆ   rˆ
(6)
In this system, the position vector is
r  r rˆ
(7)
v  r  rrˆ  rrˆ  rrˆ  r ˆ
(8)
v r  rr̂, v  r
(9)
The velocity vector is
with the components
The acceleration vector is

a  r  rrˆ  rrˆ  r ˆ  rˆ  r ˆ  (r  r 2 )rˆ  (2r  r)ˆ
(10)
with the components
(11)
a r  r  r 2 , a  2r  r
The equation of motion is thus,
r  
K
r2
(12)
rˆ
In components form,
K
r  r 2  
r2
(13)
2r  r  0
(14)
and
Constants of Motion
The angular momentum of the satellite is
L  m(r  v)  mr 2  zˆ
and the total orbital energy of the satellite is
1
Km
E  m(r 2  r 2  2 ) 
2
r
From Eq. (13) and (14), it can be shown that L and E are constants of motion, i.e.
  0, and E  0
L
(15)
(16)
(17)
We will show that another constant of motion is the Laplace-Runge-Lenz vector defined as
(18)
1
1
A
(r  L)  r
Km
r
Note that both the vectors r  L and r are on the plane of the orbit. Therefore A always lies
in the orbital plane.
From the equation of motion (12),
© S. C. Liew, Feb 2003.
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r  L  
Km
r2
(19)
[rˆ  (r  r )]
Using Eq. (8) for r , we get
Km
r  L  
[rˆ  (rrˆ  (rrˆ  r ˆ ))]   Km[rˆ  (rˆ  ˆ )]  Km ˆ
2
r
Now taking the time derivative of A,
  1 (r  L
  r  L)  rˆ  1 (r  L)  rˆ
A
Km
Km
Using relations in Eq. (20) and (6), we find that A is indeed a constant vector, i.e.
 0
A
(20)
(21)
(22)
Trajectory
We will now find the relation between r and , in order to obtain the equation for the
trajectory of the satellite. The derivative with respect to time can be converted to the
derivative with respect to ,
(23)
d d d
L d


dt dt d mr 2 d
Hence,
(24)
d2r
L d  L dr 


r 



dt 2 mr 2 d  mr 2 d 
Making a change of variable to u = 1/r, Eq. (13) and (24) are combined to get
(25)
d 2u
Km2
u 
d2
L2
For a closed orbit, the solution has the form
(26)
1 Km2

u 
1  e cos(  ) 
r
L2
The requirement that r is always positive constraints the value of the constant e to the range
between -1 and +1, and Eq. (26) describes an elliptical trajectory with the earth’s center at one
of the foci of the ellipse. If only positive e is considered, the minimum separation r occurs
when    , and the maximum separation occurs when      , with
(27)
1
Km 2


1  e
rmin
L2
1
rmax

Km 2
L2
(28)
1  e
The point on the ellipse where the satellite is closest to the center of the earth is called the
Perigee, while the point on the ellipse where the satellite is furthest away from the center of
the earth is called the Apogee.
Equating the sum of rmin and rmax to twice the semi-major axis of the ellipse:
rmin  rmax  2a
we get the relation
(29)
© S. C. Liew, Feb 2003.
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(30)
L2
a(1  e 2 ) 
Km 2
Hence, the elliptical orbit can be described by the equation,
r
(31)
a (1  e 2 )
1  e cos(  )
with
rmin  a(1  e)
(32)
rmax  a(1  e)
(33)
The angle  is determined by the orientation of the ellipse. It is the angle between the semimajor axis and the x-axis of the orbital plane coordinate system. The constant e is known as
the eccentricity of the ellipse. The center of the ellipse is located at a distance aefrom the
earth’s center.
y
y’
Perigee
S
x’
r


x
O
b
ae
O’
Apogee
a
y’
O = Center of the Earth
S = Satellite
O’ = Center of the ellipse
v
S
a
b
r

O’
a
ae
x’
O a(1 - e)
Transforming to the coordinate system where the center of the ellipse is the origin, with the
x’-axis lying along the semi-major axis, the equation of the ellipse is given by
© S. C. Liew, Feb 2003.
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x'  r cos(  )  ae 
a(1  e 2 ) cos(  )
a[e  cos(  )]
 ae 
1  e cos(  )
1  e cos(  )
y '  r sin(   ) 
a(1  e 2 ) sin(   )
1  e cos(  )
(34)
(35)
We will now find the length of the semi-minor axis b of the ellipse. Substituting x' 0 into
Eq. (34), we get cos(  )  e , r  a (from Eq. 31) and hence b can be obtained by
Pythagoras theorem,
(36)
b  a 1  e2
From Eq. (34), (35) and (36), it can be shown that the equation of the ellipse in this Cartesian
coordinate system is
2
2
(37)
 x'   y ' 


1
   
a b
In the (r ' , ' ) polar coordinate system with the origin at the orbital center
( x'  r ' cos ' ; y '  r ' sin ' ), the equation of the ellipse is
(38)
1  e2
r'  a
1  e 2 cos 2 '
Note that the ellipse is specified by any two of the following constants: the semi-major axis a,
the semi-minor axis b or the eccentricity e. The three constants (e, a, b) are related by the
equation
(39)
a2  b2
2
e 
a2
A more elegant way of deriving the equation of the orbital trajectory makes use of the
Laplace-Runge-Lenz vector defined in (18), which is a constant of motion.
Note that the vector A is perpendicular to the angular momentum vector L, i.e. A lies on the
orbital plane.
Taking the dot product of A with r,
1
1
1
Ar 
(r  L)  r  (r  r ) 
(r  L)  r  r
Km
r
Km
L2
(r  L)  r  [( rrˆ  r ˆ )  Lzˆ ]  rrˆ  Lr (rˆ  r rˆ )  rˆ  Lr 2  
m
(40)
(41)
Thus,
Ar 
L2
Km 2
r
(42)
i.e.
© S. C. Liew, Feb 2003.
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L2
Ar cos  
(43)
r
Km 2
where  is the angle between A and r. Rearranging, we get the equation for the orbital
trajectory,
r
L2
(44)
1  A cos  1
2
Km
Comparing with the previous result, we get
a(1  e 2 ) 
(45)
L2
Km 2
e A
(46)
  
(47)
Thus, the vector A is pointing along the semi-major axis from the earth’s center to the
perigee, and having a magnitude equal to the eccentricity e of the ellipse. So A can be
interpreted to be the eccentricity vector. In the jargons of satellite orbit calculation, the angle
 is called the True Anomaly of the satellite.
Total Energy and Angular Momentum
The total energy can be expressed as
1 
L2  Km
E  m r 2 

2 
r
m 2 r 2 
(48)
The total energy is a constant of motion. Thus, it can be evaluated at any part of the orbit. At
the point of closest approach, r is the minimum and hence r  0 . So E can be evaluated at this
point as
(48)
1 L2
Km
E

2 mrmin 2 rmin
Substituting the expression for rmin from Eq. (27), we get
E
K 2 m3
(49)
(1  e 2 )
2 L2
Hence, the eccentricity is related to the total energy and angular momentum by
2
e  1
2 EL
(50)
2
K 2 m3
This relation, together with Eq. (30) enable us to solve for the semi-major axis in terms of the
total energy:
(51)
Km
a
2E
Conversely, given a and e the total energy and angular momentum can be determined:
© S. C. Liew, Feb 2003.
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E
(52)
Km
2a
(53)
L  m Ka(1  e 2 )
Velocity, Angular Velocity and Orbital Period
The satellite velocity at a distance r from the earth’s center can be evaluated from the total
energy equation,
(54)
mv 2 Km
E

2
r
Equating this to the expression in (52) and rearranging, we get
(55)
2 1
v2  K  
r a
In terms of the angular position , this expression becomes
(56)
K [1  2e cos(  )  e 2 ]
v2 
a
(1  e 2 )
Note that the velocity is maximum when    , i.e. when the satellite is closest to the earth’s
center ( r  rmin ). The velocity is minimum when      when the satellite is furthest
away (i.e. at r  rmax ).
(57)
 2
1
K (1  e)
vmax  K 
  
a(1  e)
 rmin a 
 2
1
vmin  K 
  
 rmax a 
(58)
K (1  e)
a(1  e)
The angular velocity is obtained from the expression for the angular momentum (15),
   
L
mr 2
and in terms of the angular position,

Ka(1  e 2 )
(59)
r2
K 1 / 2 [1  e cos(  )] 2
 
a 3 / 2 (1  e 2 ) 3 / 2
(60)
Substituting the expression (59) for angular velocity into the expression for the total energy E
(16), we obtain the expression for the radial velocity r when the satellite is at a radial
distance r,
(61)
K 2 K Ka(1  e 2 )
r 2   

a
r
r2
Given the initial position of the satellite (r (t 0 ), (t 0 )) at time t  t 0 , the position of the
satellite (r (t ), (t )) at any time t can be found by integration of (60) or (61).
© S. C. Liew, Feb 2003.
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The angular position (t ) can be obtained by integrating (60):
(62)
a 3 / 2 (1  e 2 ) 3 / 2 (t )
d'

2
K1/ 2
 0 (1  e cos('))
where  0  (t 0 ) is the angular position of the satellite at time t  t 0 . The integral in (62)
does not have a closed form solution except when e = 0 (circular orbit). The function (t )
cannot be written explicitly. Numerical methods are required to solve for (t ) given a value
of t. The algorithm to obtain (t ) will be derived in the next section.
t  t0 
The orbital period is the time taken for the satellite to complete one revolution of the orbit.
The period can be computed from (62):
(63)
a 3 / 2 (1  e 2 ) 3 / 2 2
d
T

2
K1/ 2
0 (1  e cos )
The integral in (63) can be evaluated
2
(64)
d
2


2
(1  e 2 ) 3 / 2
0 (1  e cos )
So the orbital period is
(65)
a3
T  2
K
The orbital period depends only on the semi-major axis, and is independent of the eccentricity
of the ellipse.
Computing the Satellite Position at Any Time
y’
S’
S
b
a

O’
r
O

x’
 = True Anomaly
 = Eccentric Anomaly
We will now proceed to compute the satellite position (r(t), (t)) at any time t. For simplicity,
the x-axis is chosen such that it is parallel to the semi-major axis of the ellipse, i.e.  = 0, so
the angle  is the true anomaly at any time t. We will see later that evaluation of (t ) can be
simpler done via another parameter known as the eccentric anomaly, which is the angle  in
© S. C. Liew, Feb 2003.
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the above figure. In this figure, S is the location of the satellite at time t. A point S’ is
constructed such that it lies on a circle with radius a centered at O’, and the line S’S is
perpendicular to the semi-major axis. The eccentric anomaly is the angle OO'S' .
From the above figure, we have
a cos   ae  r cos 
a
r sin 
a sin   r sin  
b
1  e2
(66)
(67)
Using the expression (31) for the trajectory r () , we get the relations between  and :
1  e 2 sin 
sin  
1  e cos 
e  cos 
cos  
1  e cos 
(68)
1  e 2 sin 
sin  
1  e cos 
cos   e
cos  
1  e cos 
(70)
(69)
and
(71)
The time derivatives  can be obtained by differentiating any one of the four equations
above. For example, by differentiating (70), we get
(72)
 e sin  sin    (1  e cos ) cos   
 1  e 2 cos 

Substituting the expression for the time derivative  in (60), and after some algebraic
manipulations, we get a simple expression for 
(73)
2
 
T (1  e cos )
where T is the orbital period (Eq. 65). This equation can be easily integrated as,
(t )
(74)
2 t
d
t
'

(
1

e
cos

'
)
d

'


T t
0
0
i.e.
(75)
2
(t  t 0 )  (t )  e sin (t )
T
Here, t 0 is the time of the satellite’s closest approach, i.e. when  = = 0. The angle
2(t  t 0 ) / T is known as the mean anomaly of the satellite at time t. This equation relating
the eccentric anomaly to the mean anomaly is the Kepler’s Equation for the satellite orbital
motion. This equation cannot be inverted analytically, so an explicit expression for (t )
cannot be written down. Given the time t, the mean anomaly  can be calculated, and the
eccentric anomaly (t ) can be found by numerically solving the equation
(76)
    e sin 
One common method is to solve it iteratively. An initial guess for  is substituted into the
right hand side of the equation
© S. C. Liew, Feb 2003.
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(77)
    e sin 
to calculate a new value (and a better estimate) of . A good first guess is    if the
eccentricity e is close to zero. The procedure is iterated until the desired degree of precision
is reached. Once the value of  has been obtained, the true anomaly  can be calculated using
(70) and (71), or equivalently,
(78)

1  e2

tan 
tan


1  e2
The radial distance from the earth’s center is then obtained using (31).
II. Orbital Perturbation due to Nonsphericity of the Earth
The actual earth is not a perfect sphere. So the gravitational field in the space outside the earth
is no longer a central force field. Fortunately, the deviation from the ideal inverse-square-law
central force field model is slight, such that the effects of non-sphericity can be treated as a
perturbation to the elliptical orbit.
Earth’s Gravitational Potential
r
r - r
r
Consider a mass density distribution occupying a finite region in space. For a small element
of mass dm = dV located at r ' , the gravitational potential at a measurement point r due to
this mass element is
(79)
G(r' )
dV '
r  r'
The potential at the point r is then found by integrating over all space occupied by the mass
distribution:
(80)
G(r ' )
U (r )   
dV '
all space r  r '
dU (r)  
Note that
1
1
 r' 
 r' 
 (r 2  r '2 2rr ' cos  ) 1 / 2  1  2  cos    
r  r'
r
r
r

2  1 / 2



(1)
The Legendre’s Polynomials are defined as
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
(1)
n
2 1 / 2
 t Pn ( x)  (1  2 xt  t )
n0
Thus,
(1)
n
1
1   r' 
    Pn (cos  )
r  r'
r n 0  r 
The gravitational potential function can be expressed as
U (r )  
n

(1)
n
 r ' (r ' ) Pn (cos  ) dV '
GM
 R  all space
  
r n0  r 
MR n
The Legendre’s polynomials for the first few n are
P0 ( x)  1
(1)
P1 ( x)  x
1
(3 x 2  1)
2
1
P3 ( x)  (5 x 3  3x)
2
1
P4 ( x)  (35 x 4  30 x 2  3)
8
P2 ( x) 
The potential function can thus be expressed in power series of (1/r),
K
K
K

GM 
1  1  2  3  ...
U (r )  
2
3
r 
r
r
r

Kn 
1
n
 r ' (r ' ) Pn (cos  ) dV '
n
(1)
(1)
(n  0, 1, 2, ...)
MR all space
This expansion of the potential function is the multipole expansion.
The n = 0 is the monopole term,
1
 (r ' ) dV '  1
M all space
(1)
1
 r ' cos  (r ' ) dV '
MR all space
(1)
K0 
For the dipole term (n = 1),
K1 
Let r be located at the z axis, then
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1   R
K1 
   r ' cos ' (r ' , ' , ' ) r ' sin ' dr ' d' d'
MR  0 0
This term is zero if the mass density is symmetric about the x - y plane (the equatorial plane).
It can be shown that the gravitational potential field U (r ) satisfies the Laplace’s Equation:
© S. C. Liew, Feb 2003.
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(81)
 2U (r )  0
The general solution to the Laplace’s equation can be expressed as

n
U (r , , )    [ An r n  Bn r  ( n 1) ]Pnm (cos )[C nm cos m  Dnm sin m]
(82)
n0 m0
where Pnm (x) are the associated Legendre’s polynomial and An , Bn , Cnm , Dnm are constant
coefficients determined by the mass density distribution. Far away from the earth ( r  ) ,
the potential must drop to zero, so
(83)
An  0
In the limiting case of a perfect sphere, the potential reduces to U  GM / r . Thus, it is
convenient to rearranged (82) into the form,
(84)


n  R n
 K 
m

U (r , , )    1      J nm Pn (cos ) cos m(   nm )

 r  n  2 m  0  r 

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© S. C. Liew, Feb 2003.
[email protected]
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© S. C. Liew, Feb 2003.
[email protected]