ECE 5233 Satellite Communications
Prepared by:
Dr. Ivica Kostanic
Lecture 2: Orbital Mechanics
(Section 2.1)
Spring 2014
Outline
Kepler’s laws of planetary / satellite motion
Equation of satellite orbits
Describing the orbit of a satellite
Locating the satellite in the orbit
Examples
Important note: Slides present summary of the results. Detailed
derivations are given in notes.
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Kepler’s laws of planetary motion
 Johannes Kepler published laws of planetary motion in solar system in early 17th century
 Laws explained extensive astronomical planetary measurements performed by Tycho Brahe
 Kepler’s laws were proved by Newton’s theory of gravity in mid 18th century
 Kepler’s laws approximate motion of satellites around Earth
 Kepler’s laws (as applicable to satellite motion)
1. The orbit of a satellite is an ellipse with the Earth at one of the two foci
2. A line joining a satellite and the Earth’s center sweeps out equal areas during equal intervals of time
3. The square of the orbital period of a satellite is directly proportional to the cube of the semi-major axis of its orbit.
Illustration of Kepler’s law
1.
r
p
1  e cos 
2.
r
dr
 const
dt
3.
T2
 const
3
a
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Derivation of satellite orbit (1)
 Based on Newton’s theory of gravity and laws of motion
 Satellite moves in a plane that contains Earth’s origin
 Acting force is gravity
 Mass of Earth is much larger than the mass of a satellite
Gravitational force on the satellite
F 
GM E mr
r3
Newton’s 2nd law
Constants
d 2r
F  ma  m 2
dt
G  6.672 10 11 Nm2 /kg 2
Combining the two
  3.983 105 km 3 /s 2
M E  5.98 10 24 kg
d 2r r
 3  0
2
dt r
Satellite in Earth’s orbit
Differential equation that determines the orbit
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Derivation of satellite orbit (2)
Solution of the motion differential equation gives trajectory
in the form of an ellipse
r0 
e=0.9
e=0.5
e=0.2
e=0
p
1  e cos 0
e  eccentrici ty
p
h2

h  angular moment
 Coordinate system – rotated so that the satellite plane is the
same as (X0,Y0) plane
p = 1;
e = 0.2
 Not all values for eccentricity give stable orbits
fi = 0:0.01:2*pi;
r = p./(1+cos(fi));
 Eccentricity in interval (0,1) gives stable elliptical orbit
polar(fi,r)
 Eccentricity of 0 gives circular orbit
 Eccentricity = 1, parabolic orbit, the satellite escapes the
gravitational pull of the Earth
 Eccentricity > 1, hyperbolic orbit, the satellite escapes
gravitational pull of the Earth
Note: Detailed derivations of the satellite
trajectory are given in the notes
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Describing the orbit of a satellite (1)
 E and F are focal points of the ellipse
 Earth is one of the focal points (say E)
 a – major semi axis
 b – minor semi axis
 Perigee – point when the satellite is closest to
Earth
 Apogee – point when the satellite is furthest
from Earth
 The parameters of the orbit are related
 Five important results:
r0 
p
1  e cos 0
ES  FS  2a
Elliptic trajectory –
cylindrical coordinates
Basic relationship of
ellipse
1.
Relationship between a and p
2.
Relationship between b and p
3.
Relationship between eccentricity,
perigee and apogee distances
4.
2nd Kepler’s law
5.
3rd Kepler’s law
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Describing the orbit of a satellite (2)
r0 
p
1  e cos 0
2. Relationship between b and p
Consider point P:
FP+EP=2a
Since FP=EP , EP=a
From triangle CEP

b  a  e a  1 e
2
b
1. Relationship between a and p
2a  r0 0  0  r0 0   

a
p
p
2p


1  e 1  e 1  e2
p
h /

1  e2 1  e2
2
2
2
p
1  e2
2

h2 / 
1  e2
2

p2
1  e 
2 2
p2

1  e2
; b  a 1  e2
3. Relationship between eccentricity, perigee and apogee distances
EB 
p
p
 rp ; EA 
 ra
1 e
1 e
ra  rp
ra  rp
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e
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Describing the orbit of a satellite (3)
4. 2nd Kepler’s law
The area swept by radius vector
dA 
1
1
r0 ds sin r0 , ds   r0 v dt sin r0 , ds 
2
2
dr
1
1
1
r0  v dt  r0  0 dt  hdt
2
2
dt
2
dA 1
 h  const
dt 2
5. 3nd Kepler’s law
dA 
1
h dt
2
Integrating both sides
T
1
1
ab   hdt  hT
20
2


 4 2  3
4 2 a 2 a 2 1  e 2 4 2 a 3
T 

p   2 a
2
2
h
h
h / p
2
 4 2  3
T 
a
  
T 2 ~ a3
2
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Locating the satellite in the orbit (1)
 Known: time at the perigee tp
 Determine: location of the satellite at arbitrary time t>tp
Definitions:
S – satellite
O – center of the Earth
C – center of the ellipse and corresponding circle
r0
0
E
A circle is drawn so that it
encompasses the satellite’s
elliptical trajectory
- distance between satellite and center of the Earth
- “true anomaly”
- “eccentric anomaly”
2  1/ 2

 3 / 2 - average angular velocity
T
a
M   t  t p  - mean anomaly
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Locating the satellite in the orbit (2)
Algorithm summary:
1. Calculate average angular velocity:
   1/ 2 / a 3 / 2
2. Calculate mean anomaly:
M   t  t p 
3. Solver for eccentric anomaly:
M  E  e sin E 


2

a
1

e
 r0 
1
4. Find polar coordinates: r0  a1  e cosE ;0  cos 

er
0


x0  r0 cos0 ; y0  r0 sin 0 
5. Find rectangular coordinates
Notes:
Detailed derivations provided in the notes
In 3, solution is determined numerically
In 4, equation for true anomaly gives two values. One of them needs to be eliminated
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Examples
Example 2.1.1.
Geostationary orbit radius
Example 2.1.2
Low earth orbit
Example 2.1.3
Elliptical orbit
Example C1.
Location of satellite in the orbit
Note: Examples are worked out in notes
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