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UNIT-1
Introduction to Satellite Communications
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Satellite Communication combines the
missileand microwave technologies
The space era started in 1957 with the
launching of the first artificial satellite
(sputnik)
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Satellite Communications
• Satellite-based antenna(e) in stable orbit above earth.
• Two or more (earth) stations communicate via one or
more satellites serving as relay(s) in space.• Uplink: earth->satellite.
• Downlink: satellite->earth.
• Transponder: satellite electronics converting uplink
signal to downlink.
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Satellite Communications
• Satellite Transponder is a microwave deviceconsisting of receiver, repeater and regenerator inorbit
• Satellite transmission involves sending signals tosatellite that receive, amplify, and transmit back to
earth
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Capabilities of Satellite Transmission
• Point-to-point transmission – To transfer large volume of data
– Voice, data, etc communication
– Video conference• Point-to-multipoint transmission
– Data communication
– Internet – Video conference
• Broadcast services such as television
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Satellite Network Configurations
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Satellite Services
• FSS Fixed Satellite Services (VSATnetworks,..)
• MSS Mobile Satellite Services (Inmarsat
systems,...)• BSS Broadcasting Satellite Services ( TV,
DVB..)
• RDSS Radiodetermination Satellite Services
(GPS)
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Satellite Orbits
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Satellite OrbitsGEO
advantages:
- the satellite appears to be fixed (immovable) when viewed from the Earth, notracking required for earth station antennas
- about. 40% of the earth`s surface is in view from the satellite
disadvantages:
- high attenuation level (power loss) (200dB) on the path
- large signal delay (238-284ms)
- polar regions (latitudes > 81 deg.) are not covered
LEO
advantages:
- much smaller attenuation compare GEO satellites
- low signal delay
disadvantages:
- short period satellite visibility (through earth station), many times during theday
- Doppler effect
- many satellites are required for establishing continuous transmission
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• Four different types of satellite orbits can be identifieddepending on the shape and diameter of the orbit:
• GEO: geostationary orbit,36000 km above earth
surface• LEO (Low Earth Orbit): 500 - 1500 km
• MEO (Medium Earth Orbit) or ICO (Intermediate
Circular Orbit): ca. 6000 - 20000 km
• HEO (Highly Elliptical Orbit) elliptical orbits
Orbits I
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Orbits II
earth
km
35768
10000
1000
LEO
(Globalstar,
Irdium)
HEO
inner and outer Van
Allen belts
MEO (ICO)
GEO (Inmarsat)
Van-Allen-Belts:
ionized particels
2000 - 6000 km and
15000 - 30000 km
above earth surface
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Geostationary satellites
• Orbit 35.786 km distance to earth surface, orbit in equatorial plane(inclination 0°)
complete rotation exactly one day, satellite is synchronous to earth rotation
• fix antenna positions, no adjusting necessary
• satellites typically have a large footprint (up to 34% of earth surface!),
therefore difficult to reuse frequencies
• bad elevations in areas with latitude above 60° due to fixed position above
the equator
• high transmit power needed
• high latency due to long distance (ca. 275 ms)
not useful for global coverage for small mobile phones and data
transmission, typically used for radio and TV transmission
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LEO systems
• Orbit ca. 500 - 1500 km above earth surface
• visibility of a satellite ca. 10 - 40 minutes
• global radio coverage possible
• latency comparable with terrestrial long distance connections, ca. 5 - 10
ms
• smaller footprints, better frequency reuse• but now handover necessary from one satellite to another
• many satellites necessary for global coverage
• more complex systems due to moving satellites
• Examples:
• Iridium (start 1998, 66 satellites)
• Globalstar (start 1999, 48 satellites)
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MEO systems
• Orbit ca. 5000 - 12000 km above earth surface
• comparison with LEO systems:
• slower moving satellites
• less satellites needed
• simpler system design
• for many connections no hand-over needed
• higher latency, ca. 70 - 80 ms
• higher sending power needed
• special antennas for small footprints needed
• Example:
• ICO (Intermediate Circular Orbit, Inmarsat) start ca. 2000
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GEO vs LEO
GEO
• advantages:- the satellite appears to be fixed (immovable) when viewed from the Earth, no tracking
required for earth station antennas
- about. 40% of the earth`s surface is in view from the satellite• disadvantages:
- high attenuation level (power loss) (200dB) on the path
- large signal delay (238-284ms)- polar regions (latitudes > 81 deg.) are not covered
LEO• advantages:
- much smaller attenuation compare GEO satellites
- low signal delay• disadvantages:
- short period satellite visibility (through earth station),
- many times during the day
- Doppler effect
- many satellites are required for establishing continuous transmission
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Communication Satellites
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Overview of LEO/MEO systems
Iridium Globalstar ICO Teledesic
# satellites 66 + 6 48 + 4 10 + 2 288
altitude(km) 780 1414 10390 ca. 700
coverage global ±70° latitude global globalmin.elevation
8° 20° 20° 40°
frequencies[GHz
(circa)]
1.6 MS29.2 ↑
19.5 ↓23.3 ISL
1.6 MS ↑
2.5 MS ↓
5.1 ↑6.9 ↓
2 MS ↑
2.2 MS ↓
5.2 ↑7 ↓
19 ↓
28.8 ↑
62 ISL
accessmethod
FDMA/TDMA CDMA FDMA/TDMA FDMA/TDMA
ISL yes no no yesbit rate 2.4 kbit/s 9.6 kbit/s 4.8 kbit/s 64 Mbit/s ↓
2/64 Mbit/s ↑
# channels 4000 2700 4500 2500Lifetime[years]
5-8 7.5 12 10
costestimation
4.4 B$ 2.9 B$ 4.5 B$ 9 B$
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Spectrum Allocation
Frequency Spectrum concepts:
• Frequency: Rate at which an electromagnetic wave reverts its polarity(oscillates) in cycles per second or Hertz (Hz).
• Wavelength: distance between wavefronts in space. Given in meters as:
λ = c/f
Where: c = speed of light (3x108 m/s in vacuum) f = frequency in Hertz
• Frequency band: range of frequencies.
• Bandwidth: Size or “width” (in Hertz) or a frequency band.
• Electromagnetic Spectrum: full extent of all frequencies from zero to infinity.
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Radio Frequencies (RF)
• RF Frequencies: Part of the electromagnetic spectrumranging between 300 MHz and 300 GHz. Interesting properties:
– Efficient generation of signal power
– Radiates into free space
– Efficient reception at a different point.
Differences depending on the RF frequency used:
- Signal Bandwidth
- Propagation effects (diffraction, noise, fading)- Antenna Sizes
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Microwave Frequencies
• Sub-range of the RF frequencies approximately from
1GHz to 30GHz. Main properties:
- Line of sight propagation (space and atmosphere).
- Blockage by dense media (hills, buildings, rain)
- Wide bandwidths compared to lower frequency bands.
- Compact antennas, directionality possible.
- Reduced efficiency of power amplification as frequency grows:
Radio Frequency Power OUT
Direct Current Power IN
R di F S t
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Radio Frequency Spectrum
Commonly Used Bands
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Frequency Bands
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Frequency allocation
B a n d D o w n l in k B a n d s , G H z U p l in k B a n d s , G H z
U h f M i l i t a r y0 . 25 - 0 . 2 7 (A p p ro x im a t e ly )0 . 29 2 - 0 . 31 2 (A p p r o xi m a t e ly )
S B a n d
L Ba n dC B a n d - C o m m e rc ia l 3 . 7 - 4 .2 5 . 92 5 - 6 . 42 5
X B a n d - M i lit a ry 7 . 25 - 7 . 7 5 7 . 9 - 8 .4
K u B a n d - C o m m e rc ia l 11 .7 - 1 2 . 2 14 .0 - 1 4 . 5
K a B a n d - C o m m e rc ia l 17 .7 - 2 1 . 2 27 .5 - 3 0 . 0
K a B a n d - M il it a ry 20 .2 - 2 1 . 2 43 .5 - 4 5 . 5Q / V G e o s t a t i o n a r y 37 .5 - 4 0 . 5 47 .2 - 5 0 . 2
Q / V N o n - g e o s t a t io n a ry 37 .5 - 3 8 . 5 48 .2 - 4 9 . 2
W B a n d 66 .0 - 6 7 . 0 71 .0 - 7 2 . 0
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C-Band Ku-band
Antennas
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Satellite Link Performance Factors
• Distance between earth station antenna and satellite
antenna
• For downlink, terrestrial distance between earth station
antenna and “aim point” of satellite – Displayed as a satellite footprint
• Atmospheric attenuation
– Affected by oxygen, water, angle of elevation, and higher
frequencies
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Applications
• Traditionally
– weather satellites
– radio and TV broadcast satellites
– military satellites
– satellites for navigation and localization (e.g., GPS)
• Telecommunication
– global telephone connections
– backbone for global networks – connections for communication in remote places or
underdeveloped areas
– global mobile communication
replaced by fiber optics
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Current GEO Satellite Applications:
• Broadcasting - mainly TV at present
– DirecTV, PrimeStar, etc.
• Point to Multi-point communications
– VSAT, Video distribution for Cable TV
• Mobile Services
– Motient (former American Mobile Satellite),
INMARSAT, etc.
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System Design Considerations
Basic Principles
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Signals
• Signals:
– Carried by wires as voltage or current
– Transmitted through space as electromagnetic waves.
– Analog: Voltage or Current proportional to signal. E.g.Telephone.
– Digital: Generated by computers.
Ex. Binary = 1 or 0 corresponding to +1V or –1V.
Current Trends in Satellite
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Current Trends in Satellite
Communications
• Bigger, heavier, GEO satellites with multiple roles
• More direct broadcast TV and Radio satellites
• Expansion into Ka, Q, V bands (20/30, 40/50 GHz)
• Massive growth in data services fueled by Internet
• Mobile services:
– May be broadcast services rather than point to point
– Make mobile services a successful business?
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Orbital Mechanics
Part 1
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Kinematics & Newton’s Law
• s = ut + (1/2)at 2
• v2 = u2 + 2at
• v = u + at
• F = ma
s = Distance traveled in time, t
u = Initial Velocity at t = 0
v = Final Velocity at time = t
a = Acceleration
F = Force acting on the object
Newton’s
Second Law
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FORCE ON A SATELLITE
• Force = Mass × Acceleration
• Unit of Force is a Newton
• A Newton is the force required to accelerate
1 kg by 1 m/s2
• Underlying units of a Newton are therefore
(kg) × (m/s2)
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ACCELERATION FORMULA
• a = acceleration due to gravity = µ / r 2 k m/s2
• r = radius from center of earth
∀ µ = universal gravitational constant G multipliedby the mass of the earth M E
∀ µ is Kepler’s constant and= 3.9861352 × 105 km3/s2
• G = 6.672 × 10-11 Nm2/kg2 or 6.672 × 10-20 km3/kg s2 in the older units
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FORCE ON A SATELLITE : 2
Inward (i.e. centripetal force)
Since Force = Mass × Acceleration
If the Force inwards due to gravity = F IN then
F IN = m × ( µ / r 2
)
= m × (GM E / r 2 )
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F 1(Gravitational
Force)
v (velocity)
Why do satellites stay moving and in
orbit?
F 2(Inertial-Centrifugal
Force)
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Orbital Velocities and Periods
Satellite Orbital Orbital Orbital
System Height (km) Velocity (km/s) Period
h min s
INTELSAT 35,786.43 3.0747 23 56 4.091
ICO-Global 10,255 4.8954 5 55 48.4
Skybridge 1,469 7.1272 1 55 17.8
Iridium 780 7.4624 1 40 27.0
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FORCE ON A SATELLITE
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Forces acting on a satellite in a
stable orbit around the earth.
Gravitational force is inversely
proportional to the square of
the distance between the
centers of gravity of the
satellite and the planet the
satellite is orbiting, in this case
the earth. The gravitational
force inward (F IN, the
centripetal force) is directed
toward the center of gravity of the earth. The kinetic energy of
the satellite (F OUT, the
centrifugal force) is directed
diametrically opposite to the
gravitational force. Kinetic
energy is proportional to the
square of the velocity of the
satellite. When these inward
and outward forces are
balanced, the satellite moves
around the earth in a “free fall”
trajectory: the satellite’s orbit.
If F OUT = F IN
the object is in
FREE FALL
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FREE FALL???
ORBIT LIMITS
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ORBIT LIMITS
Geographical Coordinates
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Geographical CoordinatesEarth Centric Coordinate System
The earth is at the center
of the coordinate system
Reference planes coincide
with the equator and the
polar axis
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Orbital Plane Coordinates
The earth is at the
center of the coordinate
system but ………
Reference is the plane
of the satellite’s orbit
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Balancing the Forces
Inward Force
r
mGM E
F 3
r
−=
−
Equation (2.7)
F −
G = Gravitational constant = 6.672 × 10-11 Nm2/kg2
M E = Mass of the earth (and GM E = µ = Kepler’s constant)
m = mass of satellite
r = satellite orbit radius from center of earth−
r= unit vector in the r direction (positive r is away from earth)
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Balancing the Forces
Outward Force F −
2
2
dt
d
m F
−
−
=
r
Equation (2.8)
Equating inward and outward forces we find
2
2
3 dt
d
r
−−
=−
rr
µ
Equation (2.9), or we can write
032
2
=+
−−
µ r dt
d rr Equation (2.10)
Second order differential
equation with six unknowns:
the orbital elements
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• We have a second order differential equation
• See text for a way to find a solution• If we re-define our co-ordinate system into polar
coordinates (see Fig.) we can re-write equation
as two second order differential equations in
terms of r 0 and ϕ 0
THE ORBIT
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Polar Coordinates
In the plane of theorbit
Polar coordinate system in the plane of the satellite’s orbit. The plane of the orbit
coincides with the plane of the paper. The axis z 0 is straight out of the paper from the
center of the earth, and is normal to the plane of the satellite’s orbit. The satellite’s
position is described in terms of the radius from the center of the earth r 0 and the angle
this radius makes with the x 0 axis, Φo.
O
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THE ORBIT
• We have a second order differential equation
• If we re-define our coordinate system into polar coordinates (see
Fig. 2.3) we can re-write equation (2.5) as two second order
differential equations in terms of r 0 and φ 0 .
and
THE ORBIT
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THE ORBIT
• Solving the two differential equations leads to six constants
(the orbital constants) which define the orbit, and three laws
of orbits (Kepler’s Laws of Planetary Motion)
• Johaness Kepler (1571 - 1630) a German Astronomer and
Scientist
KEPLER’S THREE LAWS
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KEPLER’S THREE LAWS
• Orbit is an ellipse with the larger body (earth) atone focus
• The satellite sweeps out equal arcs in equal time
(NOTE : for an ellipse, this means that the orbitalvelocity varies around the orbit)
• The square of the period of revolution equals a
CONSTANT ´ the THIRD POWER of SEMIMAJOR
AXIS of the ellipse
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Review: Ellipse analysis
• Points (-c,0) and (c,0) are the foci.
•Points (-a,0) and (a,0) are the vertices.
• Line between vertices is the major axis.
• a is the length of the semimajor axis.
• Line between (0,b) and (0,-b) is the minor axis.
• b is the length of the semiminor axis.
12
2
2
2
=+b
y
a
x
222cba +=
Standard
Equation:
y
V(-a,0)
P(x,y)
F(c,0)F(-c,0) V(a,0)
(0,b)
x
(0,-b)
ab A π =
Area of ellipse:
KEPLER 1: Elliptical Orbits
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The orbit as it appears in the orbital plane, The point O is the center of the earth and the
point C is the center of the ellipse. The two centers do not coincide unless the
eccentricity, e, of the ellipse is zero (i.e., the ellipse becomes a circle and a = b). The
dimensions a and b are the semimajor and semiminor axes of the orbital ellipse,respectively.
KEPLER 1: Elliptical Orbits
e = ellipse’s eccentricity
O = center of the earth (one
focus of the ellipse)
C = center of the ellipse
a = (Apogee + Perigee)/2
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KEPLER 1: Elliptical Orbits (cont.)
Equation 2.17 in text:
(describes a conic section,
which is an ellipse if e < 1)
)cos(*11
1φ e
pr +=
e = eccentricity
e<1 ⇒ ellipse
e = 0 ⇒ circle
r 0 = distance of a point in the orbit to the
center of the earth
p = geometrical constant (width of the conic
section at the focus)
p=a(1-e2 )
ϕ 0 = angle between r 0 and the perigee
p
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KEPLER 2: Equal Arc-Sweeps
Figure 2.5
Law 2
If t2 - t1 = t4 - t3
then A12 = A34
Velocity of satellite is
SLOWEST at APOGEE; FASTEST at PERIGEE
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Kepler’s Laws – 1 & 2
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KEPLER 3: Orbital Period
Orbital period and the Ellipse are related by
T 2 = (4 π 2 a3 ) / µ (Equation 2.21)
That is the square of the period of revolution is equal to a
constant × the cube of the semi-major axis.
IMPORTANT: Period of revolution is referenced to inertial space, i.e., to
the galactic background, NOT to an observer on the surface of one of the
bodies (earth).
µ = Kepler’s Constant = GM E
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Kepler’s 3rd Law: T² = (4π²a³)/μ
μ μ == 3.986004418 × 103.986004418 × 1055 km/s²km/s²
Numerical Example 1
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Numerical Example 1
The Geostationary Orbit:
Sidereal Day = 23 hrs 56 min 4.1 sec
Calculate radius and height of GEO orbit:• T2 = (4 π2 a3) / µ (eq. 2.21)
• Rearrange to a3 = T2µ /(4 π2)
• T = 86,164.1 sec
• a3 = (86,164.1) 2 x 3.986004418 x 105 /(4 π2)
• a = 42,164.172 km = orbit radius
•h = orbit radius – earth radius = 42,164.172 – 6378.14 = 35,786.03 km
S l Sid l D
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Solar vs. Sidereal Day
• A sidereal day is the time between consecutive crossings of any
particular longitude on the earth by any star other than the sun.
• A solar say is the time between consecutive crossings of any
particular longitude of the earth by the sun-earth axis.
– Solar day = EXACTLY 24 hrs – Sidereal day = 23 h 56 min. 4.091 s
• Why the difference?
– By the time the Earth completes a full rotation with respect to an
external point (not the sun), it has already moved its center
position with respect to the sun. The extra time it takes to cross
the sun-earth axis, averaged over 4 full years (because every 4
years one has 366 deays) is of about 3.93 minutes per day.
LOCATING THE SATELLITE IN
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LOCATING THE SATELLITE IN
ORBIT: 1
Start with Fig. 2.6 in Text ϕ o is the True
Anomaly
See eq. (2.22)
C is the
center of the
orbit ellipse
O is thecenter of the
earth
NOTE: Perigee and Apogee are on opposite sides of the orbit
LOCATING THE SATELLITE
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LOCATING THE SATELLITE
IN ORBIT
• Need to develop a procedure that will allow the
average angular velocity to be used
• If the orbit is not circular, the procedure is to use aCircumscribed Circle
• A circumscribed circle is a circle that has a radius equalto the semi-major axis length of the ellipse and also
has the same center
L S lli i O bi
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Locate Satellite in Orbit
η = Average angular velocity
E = Eccentric Anomaly
M = Mean Anomaly
M = arc length (in radians) that the
satellite would have traversed sinceperigee passage if it were moving
around the circumscribed circle
with a mean angular velocity η
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ORBIT CHARACTERISTICS
Semi-Axis Lengths of the Orbit
2
1 e
pa
−
=where
µ
2h p =
and h is the magnitude of
the angular momentum
See eq. (2.18)
and (2.16)
( ) 2/121 eab −= where µ
C he
2
= See eqn.(2.19)
and e is the eccentricity of the orbit
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ORBIT ECCENTRICITY
• If a = semi-major axis,
b = semi-minor axis, and
e = eccentricity of the orbit ellipse,
then
ba
ba
e +
−=
NOTE: For a circular orbit, a = b and e = 0
Time reference
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Time reference
• tp Time of Perigee = Time of closest
approach to the earth, at the same time, time
the satellite is crossing the x0
axis, according to
the reference used.
• t- tp = time elapsed since satellite last passedthe perigee.
ORBIT DETERMINATION 1
8/3/2019 Satellite Unit1
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ORBIT DETERMINATION 1:
Procedure:
Given the time of perigee t p, the eccentricity e
and the length of the semimajor axis a:
∀ η Average Angular Velocity (eqn. 2.25)
• M Mean Anomaly (eqn. 2.30)• E Eccentric Anomaly (solve eqn. 2.30)
• r o Radius from orbit center (eqn. 2.27)
∀ ϕ o True Anomaly (solve eq. 2.22)• x0 and y0 (using eqn. 2.23 and 2.24)