2. Orbital Aspects
2.1 Orbits
Satellites in geostationary orbit:
Advantages:
• Satellite remains stationary, no tracking equipment for earth
station necessary
• Satellite is visible 24h per day
• Large coverage area: large number of earth stations can
communicate
• Almost no Doppler shift – keeps electronics simple
Disadvantages:
• Latitudes north of 81.5 deg are not covered
• Great distance – received signal is weak
• Launch cost higher than for low altitude orbits
• Only one geostationary orbit is possible
Satellites in inclined orbits:
• Molniya series i=65 deg, P=12h - provides communication
services to the northern regions of Russia
• Military satellites
• GPS satellites, i=55 deg, P=12h, 6 orbital planes oriented at 60
deg angles w.r.t. each other, 6 x 4 satellites, at least 9 satellites
are visible from any point on earth at any given time.
Satellites in polar orbits:
• Tiros – N series (1960 – 1966)
• NOAA, P=102 min, altitude about 800 km. Also provides
SARSAT service
• POES program – polar orbiting environmental satellites
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SARSAT service
• Based on utilization of Doppler shift and known satellite orbit
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Orbital Mechanics
Kepler’s Laws
• Empirically found but can be derived from Newton’s theory of
gravitation and laws of motion.
1. The orbit of each planet (satellite) is an ellipse with the
Sun (earth’s center) at one focus.
2. The line joining the earth’s center to a satellite sweeps
out equal areas in equal times
3. The square of the period of revolutions is proportional
to the cube of the semi-major axis.
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Newton’s Universal law of Gravitation:
Newton’s 2nd Law of Motion
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 Fundamental differential equation used in the study of artificial
satellites. It is a 2nd order vector linear differential equation.
• The solution will involve 6 constants, 2 for each coordinate
• The constants are called orbital elements of Keplerian elements
or Keplerian orbital elements
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 Problem of motion in 3 dimensions reduces to a 2-dimensional
problem of motion (motion in a plane) and to the problem of orienting
the plane in space.
The position of a satellite in space at any time is given by the set of 6
Keplerian elements.
Keplerian elements
Shape of the ellipse
1.
a:
semimajor axis
2.
e:
eccentricity
Timetable with which the satellite orbits Earth
3.
M:
mean anomaly at epoch, i.e., average value of the angular
position of the satellite w.r.t. the perigee. M↔ν, ν:true
anomaly. For circular orbit => M=ν.
Orientation of the ellipse in the orbital plane
4.
ω:
argument of perigee, i.e., the geocentric angle measured
from the ascending node to the perigee in the orbital plane in the
direction of the satellite’s motion.
Orientation of the orbital plane in space
5. i: inclination of the orbital ellipse. It is the angle measured
from the equatorial plane to the orbital plane at the ascending
node going from east to north.
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6.
Ω:
right ascension (RA) of the ascending node, i.e., the
geocentric angle measured from the vernal equinox to the
ascending node in the equatorial plane eastward.
There are three anomalies:
1. ν : true anomaly: geocentric angle measured from perigee to
the satellite in the orbital plane in the direction of the satellite’s
motion.
2. E: eccentric anomaly: angle measured at the center of the
orbit from perigee to the satellite’s projection onto a circle of
radius a in the direction of the satellite’s motion.
3. M: mean anomaly: M = n(t – tp); n: mean motion, tp: time of
perigee crossing.
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The mean, true and eccentric anomalies are related by:
(Kepler’s equation)
(Gauss’ equation)
M = E − esin E
for << 1;E = M + esin M +
e2
sin2M + O(e 3 )
2
Graphical description of the ellipse:
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1/ 2
ν 1+ e 
E
tan = 
 tan
2  1− e 
2
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ν = M + 2esin M + e 2 sin2M + O(e 3 )
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Other parameters used are;
P: period of orbit (instead of semi-major axis, a)
From Kepler’s 3rd law, P2 ∝ a3
Proportionality constant:
2
-2
G = 6.6726 •10-11 N m kg
M ⊕ = 5.98 •10 24 kg
-2 -2
GM ⊕ = 398,600.5 km €s
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€
4π 2 3
P =
a
µ
2
0
4π 2
,
µ
µ = GM ⊕
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[s]
n: mean motion (instead of semi-major axis, a)
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Vernal equinox or First Point of Aries or γ
Position on the sky where the sun crosses the celestial equator on its
way to higher declinations.
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Celestial position (in geocentric equatorial coordinate system)
Given in celestial coordinates, RA, dec
RA: right ascension, measured eastward in hours, minutes, seconds
(24h = 360 deg)
dec: declination, measured from celestial equator north (positive)
and south (negative) in degrees, arc-minutes, arc-seconds
Celestial position of vernal equinox: RA = 0h , dec = 0 deg
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2.2 Orbit Perturbations
Keplerian orbit is ideal. It is assumed that…
a) earth is spherically symmetric body with a uniformly
distributed mass; and that the only forces present are:
b) the gravitational forces of the earth with 12 r dependence,
r r
and
c) the centrifugal force from the satellite motion, and that
€ zero cross-section.
d) the satellite is a point-like body with
If assumptions were correct, Keplerian elements would not change.
But there are several effects that cause perturbations of the ideal orbit.
1. Effects of non-spherical earth
a) Effects of the equatorial bulge (3 effects: on n, Ω, ω)
i) Mean motion (n):
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€
 K (1−1.5sin 2 i) 

n = n 0 1+ 1
3

a 2 (1− e 2 ) 2 
2π
µ
n0 =
=
P0
a3
K1 = 66,063.1704
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ii) Regression of nodes (effect on Ω)
[rad/s]
[rad/s]
km2 ,
a in km, P0 in s
Nodes slide along the equator in a direction opposite to the
direction of satellite motion
dΩ
= −K cosi
dt
nK1
K= 2
a (1− e 2 ) 2
(K has same units as n)
dΩ
< 0 , westward slide of nodes
dt
orbit: dΩ > 0 , eastward slide of nodes
dt
for prograde orbit:
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for retrograde
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iii) Change of argument of perigee, ω, or rotation of line
of apsides  orbital ellipse rotates in orbital plane around
focus.
dω
= K(2 − 2.5sin 2 i
dt
(has units of K which has units of n)
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b) Effects of equatorial ellipticity
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c) Effects of tides
Tides change the mass distribution of the earth  very
small effect and only on low-earth orbit satellites
2. Direct third body effects (affects mostly i)
Direct attractions of the moon and the sun are significant.
For satellites in geostationary orbit:
di
≈ 0.8 [deg/yr]
dt
€
i increases to a maximum of 15 deg in 27 years and then decreases to
i=0 again.
For a geosynchronous orbit (i≠0) the satellite appears to move along a
figure “8” relative to the ground station.
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3. Atmospheric drag
(affects mostly a, e)
Important for satellites with perigee height ≤ 1000 km.
Perturbation of orbit depends on:
• Atmospheric density
• Satellite cross section
• Satellite mass
• Satellite speed
 lifetime is reduced
Satellite
Spy satellite
Sputnik
Explorer I
Transit
Geost. satellite
Perigee height
(km)
118
225
361
1000
36,000
Apogee height
(km)
430
946
2550
1000
36,000
Lifetime
15 d
3m
12 yr
1000 yr
1 Mill yr
4. Solar radiation pressure
Acceleration of satellite depends on:
• Solar radiation at satellite
• Satellite mass
• Satellite surface area exposed to sun
2.3 Visibility
There are three different planes and reference systems
• Orbital plane -perifocal coordinate system
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• Equatorial plane -- geocentric equatorial coordinate system
• Plane tangential to surface of earth -- topocentric horizon
coordinate system
 coordinate transformations necessary (definitions of time
necessary)
Problem: Determination of azimuth and elevation (look angles) of a
satellite and the range to the satellite for any point on earth.
Given: Keplerian elements of the satellite.
Solution path:
1) describe locations of satellite and earth station in same nonrotating coordinate system that travels with earth through space
(geocentric equatorial coordinate system).
2) Then form range vector and express it in topocentric horizon
coordinate system.
Steps to solve the problem:
1. Locate satellite in perifocal coordinate system
 r cos ν  rP 
rpf = 
= 
 r sin ν  rQ 
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2. Locate satellite in geocentric-equatorial coordinates system
 rI 


r

 
P
rr.eq. = R˜   =  rJ 
rQ   
rK 
elements are functions of Ω, ω, i
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R11 = cosΩcos ω − sinΩsin ω cosi
R12 = cosΩsin ω − sinΩcosω cosi
R21 = sinΩcos ω + cosΩsin ω cosi
R22 = .......
R31 = .......
R32 = .......
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This is a non-rotating coordinate system which travels with earth
through space.
3. Locate earth station in geocentric-equatorial coordinate system
 R⊕ + H cos λE cos(LST)  RI 


  
Rg.−eq. =  R⊕ + H cos λE sin(LST)  =  RJ 

 RK 
R⊕ + H sin λE
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How is LST related to standard time?
LST
GST
UT
standard time
a) LST=GST+ΦE
Example: York: longitude= 790 35’ W
 ΦE = -790 35’ W
if GST = 1200
= 8h

LST = 400 25’
= 2h 41m 40s
Both LST and GST are measured relative to fixed stars
Unit: sidereal day which is < mean solar day.
Mean sidereal day and mean solar day
⊕
Porbit
= 3.66.2422 • Prot⊕
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€
⊕
Porbit
= 366.2422 mean sidereal days
= 365.2422 mean solar days
1 calendar year
= 365 calendar days
1 calendar leap year = 366 calendar days (29 February is extra day)
Leap year: if year is divisible by 4, except if it is a centennial year.
However if the centennial year is divisible by 400, then it is also a
leap year.
GST[deg] = 99.6909833+36000.7689 Tc + 0.00038708 Tc2 +UT[deg]
Tc = (JD-2415020)/36525 Julian centuries
= elapsed time in Julian centuries between Julian day JD and noon
UT on Jan 0, 1900 (Jan 0.5, 1900).
UT or UTC: Universal time coordinated (based on atomic time
given by Cesium clocks; the atomic time is broadcasted).
UT[deg] = 360 [1/24(h + min/60 + sec/60)] [deg]
Example: GST on 28 January 1994 at 12:00 UT?
0.0 Jan 1994: JD = 2449352.5
28.5 Jan 1994:
+
28.5
2449381.0
Tc =(2449381.0 – 2415020.0)/36525
= 0.9407529
UT=180 deg
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
GST = 34,147.51907 [deg]
=
307.51907 [deg]
subtracted)
(94 • 360 deg
=
3070 31’ 8.652”
=
20h 30m 4.5768s
4. Locate satellite in topocentric-horizon coordinate system
a) calculate range vector in geocentric equatorial coordinate
system.
 ρI 
    
ρ = r − R = ρJ 
ρ K 
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b) Make coordinate transformation to the topocentrichorizon coordinate system
ρS 
 ρI 

 
 
ρ topo =  ρ I  = D˜  ρ J 
ρ Z 
ρ K 
Antenna look angles: Az, El
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tan AZ = −
sin El =
ρE
ρS
ρZ
ρ
ρ = ρ S2 + ρ E2 + ρ Z2
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Special case:
Antenna look angles: Az, El, range |ρ| and visibility limits
for geostationary satellite.
Given:
Azimuth, Az
tan A =
−tan( φ E − φ S )
sin λE
with station in southern hemisphere (λE < 0) , φE-φS< 0 : Az = A
φE-φS> 0 : Az = 3600 –A
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with station in northern hemisphere (λE >0) , φE-φS< 0 : Az = 1800 +A
φE-φS> 0 : Az = 1800 –A
Range:
1/ 2
ρ = [ R⊕2 + (RE + h) 2 − 2R⊕ (RE + h)cosc ]
Elevation:
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cos El =
RE + h
sinc
ρ
Limits of visibility:
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*  
 

*
−1 R⊕ cos El
 sinEl + sin 
 
 
 RE + h  
*
−1
(φ E − φ S ) = ± cos 

cos λE






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For given elevation limit EL*, only geostationary satellites with
longitudes within ±|(φE-φS)*| of station longitude are visible.
Earth eclipse of satellite and sun transit outage
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2.4 Launches
Country
Expendable
Max. payload (kg) Comment
multistage rockets
USA
EC
Titan III (till 2005)
Atlas Centaur V
Delta II
Delta III
Delta IV
Ariane 4, 4L
5000
8500
2000
3800
14000
4200
Russia
Ariane 5
Proton
10500
6000
….
launches geostat.sat.
launches GPS satel.
launches GPS satel.
launches geost. sat.
2-satellite capability
with SYLDA
launches geostat. sat.
launches geostat. Sat.
HI
H II
GSLV MkII
Long March
1100
4000
2500
5500
launches geostat. sat.
launches geostat. sat.
7700
Last launch: 2011
Japan
India
China
Reusable launch
vehicle
USA
Shuttle
The shuttle had advantages: large payload, checkout in space possible
The shuttle had disadvantages: safety restrictions, non-flammable
materials, PAM (perigee assist motor) necessary.
Stages of launch
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Rocket launch with Ariane
1) Satellite is sent into transfer orbit directly.
2) Acquisition is obtained via omnidirectional satellite antenna.
3) Keplerian elements of transfer orbit are determined.
4) Apogee kick motor is fired when satellite is at apogee.
5) Satellite velocity and orientation are adjusted, system is checked
out.
Shuttle launch
1) Satellite goes with shuttle into 300 km altitude orbit.
2) PAM (payload assist module, or perigee motor) sends satellite
into transfer orbit.
3) As for rocket launch.
4) As for rocket launch.
5) As for rocket launch.
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