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Satellite Communications
Dr. Malgorzata Langer
The Institute of Electronics
Room 310; [email protected]
Why?
• Distance independent and fixed broadcast costs –
the same costs, regardless of the distance
between the transmitting and receiving earth
stations and of the number of ground terminals
receiving the transmission
• High capacity – high carrier frequencies
• Low error rates thanks to error correction
techniques
• Diverse user networks may be served
simultaneously, on large areas of the earth
When?
• The first launch:
October 1957; SPUTNIK I;
Soviet Union
• The first idea:
Arthur C. Clarke, 1945, US
Why? (How it works…)
• Satellite orbit determination is based on the
Laws of Motion,
published in 1609 (Astronomia Nova) and in
1619 (Harmonices Mundi), developed by
Johannes Kepler
and later refined by Isaack Newton, in 1665, in
Laws of Mechanics and Gravitation
Kepler’s Laws
• Kepler’s laws on planetary motion apply to
any two bodies in space that interact through
gravitation;
• In origin – they apply to the motions of the
planets around the Sun
Kepler’s First Law
F1 and F2 - localisation of the
two foci
The path followed
by the planet
around the Sun will
be an ellipse, with
the centre of mass
of the Sun as one of
the two foci of the
ellipse
Do you remember?
• When F1 and F2 are the foci places of the
ellipse – in any localization of the planet the
sum of the distances of the both foci is
constant and equals to the greater diameter
(doubled major axis)
The eccentricity
• It is a measure of the ‘circularity’ of the orbit. The higher
the eccentricity, the flatter the ellipse.
• The circular orbit means that the major axis equals to the
minor one and the eccentricity equals to zero
0<e<1
• Elliptical Orbit
• Circular Orbit
e=0
• The parabola (open orbit) e = 1; the hyperbola e > 1
e=
ra − rp
ra + rp
where e- the eccentricity of the
orbit; ra- the distance from the
center of the sun to the aphelion
point, and rp – to the perihelion
Some Numbers:
• In Solar System planet eccentricities are less
than 0.1
• The exception is Mercury (~ 0.2)
• The solar year (the tropical year) equals to
365 days 5 hours 48 minutes and 46 seconds
• The sidereal year (the determined star is the
constant point for the earth’s motion):
365 days 6 hours 9 minutes and 9 seconds
Can you see the problem?
• Auxiliary questions:
What time do we use and what we should?
What is the correct number for the day
period?
What is γ2000?
Why has the beginning of the spring been
changed lately?
How did ancient (and medieval) sailors find
their way across sees and oceans?
Geographic coordinates?
• Give the per cent difference between the
Solar day and the Sidereal one.
• Which coordinate (latitude or longitude, i.e.
parallels of latitude, meridians) shifts against
geographical places?
The Earth in The Solar System
• Medium distance from the Sun:
149,597,887 km (1.00000011 AU – astronomical
unit)
• Orbit’s perimeter: 0.940*109km
• Eccentricity : 0.01671022
• Perihelion: 147,098,074km
• Aphelion: 152,097,701km
• Sidereal year: 365.25696 days
• Mean orbital velocity: 29.783 km/s
(max: 30.287 km/s ; min. 29.291 km/s)
For a satellite
• The path followed by a satellite around the
Earth will be an ellipse, with the center of
mass of Earth as one of the two foci of the
ellipse.
• The size of the ellipse will depend on the
satellite mass and its angular velocity
Kepler’s Second Law
• For equal time intervals, the tracing radius of
the planet sweeps out equal areas in the
orbital plane
∆s
1 K
= const = 2
∆t
m
So, in perihelion (near the Sun)
the planet moves faster than in
aphelion
m – the mass of the planet; K – the angular momentum (moment of momentum) of the
planet
Likewise…..
• For equal time intervals the satellite sweeps
out equal areas in the orbital plane
• The area swept out by a satellite in one hour
period in the orbit’s apogee (the point farthest
from the earth) equals to the area swept out
during one hour in perigee.
• THE SATELLITE ORBITAL VELOCITY IS NOT
CONSTANT
Kepler’s Third Law
• The square of the periodic time of orbit is
proportional to the cube of the mean distance
between the two bodies
2
1
3
1
2
2
3
2
T
T
=
= const
a
a
In Details…
 4π  3
2
T =
a
 µ 
2
where T – orbital period in [s]; a –
distance between the two bodies in
[km]; µ - Kepler’s Constant:
3.986004x105km3/s2
 µ 
2/3
r = 2 T
 4π 
1/ 3
If the orbit is circular then a=r, and
ORBIT RADIUS = [Constant] x (Orbit Period)2/3
Newton’s Laws
• The gravitational force Fin and the angular
velocity force, Fout can be presented as:
µ
Fin = m 2 
r 
What is v for
Fout
 v2 
= m 
 r 
Fin = Fout
• m – satellite mass; v – satellite velocity in the
plane of orbit; r – distance from the centre of
the earth (orbit radius), µ - Kepler’s Constant
For The Satellite
3
a
T = 2π
G⋅m
•
•
•
•
Constant of gravitation G=6.67 * 10-11 Nm2/kg2
Earth’s mass m=5.973*1024 kg
Mean radius R=6.3*106m
‘a’ is R+h (the orbit’s altitude)
Results
• A specific orbit period is determined only by
the selection of the orbit radius
Orbit altitudes for specified orbital periods
Revolutions/day
Nominal period
(hours)
Nominal altitude
(km)
1
24
36000
2
12
20200
3
8
13900
4
6
10400
6
4
6400
8
3
4200
Orbital Parameters
• Apogee – the point farthest from the earth
• Perigee – the point of the closest approach to
the earth
• Line of Apsides – the line joining the perigee
and the apogee through the center of the
earth
• Ascending Node – the point where the orbit
crosses the equatorial plane going from south
to north
…
• Descending Node – the point where the orbit
crosses the equatorial plane going from north
to south
• Line of Nodes – the line joining the ascending
and descending nodes through the center of
the earth
• Argument of Perigee, ω – the angle from
ascending node to perigee, measured in the
orbital plane
Legend
Letters in the image denote:
A – Minor, orbiting body
B – Major body being orbited
by A
C – Reference plane
D – Orbital plane of A
E – Ascending node ☊
F – Reference direction (for
orbits in or near the ecliptic
plane).
Ω – Argument of the Ascending
Node
[wiki]
The Inclination Angle
• The angle between the
orbital plane and the
earth’s equatorial plane
• A satellite that is in orbit
with some inclination
angle is in an inclined
orbit
• A satellite that has an
inclination angle of 900 is
in polar orbit
• A satellite that has an
inclination angle of 00 is
in equatorial orbit
Wiki
Prograde and Retrograde Orbits
• An orbit in which the • An orbit in which the
satellite moves in
satellite moves in a
the same direction
direction counter to
as the earth’s
(opposite) to the
rotation is called a
earth’s rotation is
prograde orbit (the
called a retrograde
inclination angle is
orbit (the inclination
between 00 and 900
angle is between 900
and 1800
Answer the questions:
Most satellites are launched in a prograde orbit.
Why?
Calculate the orbit radius and the height (altitude
above the earth’s surface) for the satellite with the
orbit in the inclination angle =0, and when the
satellite appears to be motionless above the earth.