chapter 2 satellite communications: satellite launching
TRANSCRIPT
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Chapter 2
Satellite Launching
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Launches and Launch
Vehicle
A satellite cannot be placed into a stable
orbit unless two parameters that are uniquely
coupled together => the velocity and the
orbital height simultaneously correct.
Example:
A Geostationary Satellite:
must be in orbit at a height of 35786.03 km above thesurface of the earth or 42164.17 km from center of the
earth.
Inclination zero degrees.
Velocity of 3074.7 m/s. tangential to the earth.
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The further out from the earth the orbit is, the greater theenergy required from the launch vehicle to reach the orbit.
In any earth satellite launch, the largest fraction of energyexpended by the rocket is used to accelerate the vehicle fromthe rest until it is about 20 miles (32 km) above the earth.
To make efficient use of the fuel, it is common to shed excessmass from the launcher as it moves upward on launch =>staging.
most launch vehicle have multiple stages and as each stageis completed, that portion of the launcher is expended untilthe final stage places the satellite into the desired trajectory.
Launch vehicles may be classified as expendable orreusablelaunch vehicles.
CONTINUED
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Expendable Launch Vehicles (ELV)- typical expendable launchers => US, Atlas-
Certaur and Delta Rockets and European SpaceAgency Ariane Rocket.
Reusable Launch Vehicles (RLV)- also referred as Space Transportation System
(STS).
- Partially reusable.
- The solid rocket booster are recovered and
refurbished for future missions and the shuttlevehicle itself is flown back to earth for refurbishedand reuse.
CONTINUED
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Placing Satellite Into
Geostationary Orbit
There are two way in placing the
satellite into geostationary orbit.
i. Geostationary transfer orbit and AKM
ii. Geostationary transfer orbit with sloworbit rising.
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Geostationary Transfer
Orbit and AKM the initial approach to launching geostationary orbit
satellite was to place the spacecraft with the final rocketstage still attached into LEO.
After been attached into LEO and the orbital elements
are measured, the final stage is reignited and thespacecraft is launched into a geostationary transferorbit (GTO).
The GTO perigee is the original LEO orbit altitude while
the GTO apogee is the GEO altitude. After a few orbitsin GTO while the orbital elements are measured, arocket motor (contained within the satellite itself) isignited at apogee and the GTO is raised until it is acircular, geostationary orbit.
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Since the rocket motor fires at apogee, it is commonlyreferred to as the apogee kick motor (AKM).
AKM => used to circularize the orbit at GEO and toremove any inclination error so that the final orbit of
the satellite is very close to geostationary.
CONTINUED
LEO
GTO
GEO
AKM
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Geostationary Transfer Orbit
With Slow Orbit Rising
in this procedures of launching, the spacecraftthrusters are used to raise the orbit from the GTO toGEO over a number of burns.
Many of the satellite elements are deployed while in
GTO (solar panel).
The satellite has two power levels of thrusters:
- for more powerful orbit raising maneuvers.
- for more on-board maneuvers.
since thruster take many time (hours) of operation toachieved the geostationary orbit, the perigee of theorbit is gradually raised over successive thrusterfirings.
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LEO
GTO
GEO
Successive
Orbit Rising
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Orbital Mechanic
How orbit is achieved,
Describe the motion of an object
orbiting another body,
How sat maneuver in space.
Most sat are placed atleast 250
miles (400km) above earthsurface.
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Newtons law
important to describe the motion of a sat.
The coordinates axes within which the orbit of
the sat can be set
Determine the various forces on the earthsatellite.
s = ut + at2
v2 = u2 + 2at
v = u + at
F = ma
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Two main forces that acting on the sat (in
stable orbit)
Me
Fin
Fout
Fout = (mv2)/ r
Fin = (GMem)/ r2
CONTINUED
i. Centrifugal Force (Fout)
ii. Centripetal Force (Fin)
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Centrifugal Force (Fout)
- Due energy for satellite.- Attempts to fling the satellite into a higher
orbit.
Centripetal Force (Fin)
- due to the gravitational attraction of theearth which the sat is orbiting.
- attempts to pull the sat to the earth.
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Keplers Law
First law
- states that the path followed by a satellitearound the primary will be an ellipse.
- eccentricity and semimajor axis are two of
the orbital parameters specified for thesat orbiting the earth.
- the eccentricity e is given by
e = [(a2 _b2)/a]
- when e = 0, the orbit become circular- when 0 < e < 1 , the orbit become ellipse
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Second Law- states that for equal time intervals, a sat
will sweep out equal areas in its orbitalplane.
- orbit is an ellipse.
- two shaded area of elliptical plane inwhich the orbit moves.
- from figure, sat moves in orbit betn:
t1 t2 and sweep area 1
t3 t4 and sweep area 2
- if t1 t2 = t3 t4 ,
then area 1 =area 2
CONTINUED
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Third Law
- states that the square of the
period time of orbit is proportional
to the cube of the mean distance
betn two bodies.
- the mean distance is equal to
the semimajor axis a.
T2 = (42 a3)/
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Geostationary Orbit
The most widely used orbit.
The sat appears stationary relative to theearth.
Lies in equatorial plane and inclination is
zero. The sat must orbit the earth in the samedirection as the earth spin and at the samespeed.
h
ae
Where:
h = 35786 km above the
equator.
= from earth surface to
geo orbit.
ae
= radius of earth
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Because of the orbit is circular:a = b = ae + h
The periodic time, T for the geo
orbit to the nearest minute is 23h56m in mean solar time. Less 4
min because the differences is
being accounted for the earths
movement around the sun.
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Exercise
1. The earth rotates once per sidereal day of23h 56m 4.09s. Find the radius of Geo inunit km.
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2. The space shuttle orbit at an altitude of
250 km above the earths surface. The
mean earth radius is approximately
6378 km. Using these figures,calculate the period of the shuttle orbit
the earth.
Find also the linear velocity of the
shuttle along the orbit.
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3. A sat is in an elliptical orbit with
perigee of 1000 km and apogee
of 4000 km. using a mean earth
radius of 6378.4 km, find theperiod of the orbit in hours, min
and sec.
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Past Paper Questions
Question 4 (Apr 2006)
Determine the apogee height of the ellipticalorbit satellite when the time period of the
satellite to rotate around the earth is 2 hours18 minutes and 45 seconds. Given theperigee height as 1000km with the earthradius of 6378 km and the Keplers constant is3.986005 105 km3/s2.
Compare mathematically the velocity of thesatellite while its rotating along perigee andapogee. By using the Keplers second law,comment on your answers.
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Question 5 (Nov 2005)
Using Keplers third law, find the period
of the orbit in hours, minutes and
seconds for an elliptical orbit satellite
with an apogee of 4500 km and perigeeof 1500km. Given the earth radius is
6378 km and the Keplers constant is
3.986005 105 km3/s2.
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Question 6
In stable orbit, name two main forces that acton the satellite due to kinetic energy andgravitational attraction.
Derive the equation to prove that the satelliteis stable in orbit when the orbit is situated31200km above the earth surface. Given theKeplers constant is 3.986005 105 km3/s2 .The earth radius of 6378 km and the satellitemass of 200 kg, find the velocity of the
satellite on the orbit.
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Question 7 (Apr 2007)
Find the height of the orbit and
the period of time in hour, minute
and second for the satellite
maneuvers on the orbit withvelocity 3.10 kms-1. Given the
Keplers constant is 3.9860045 x
105 km3s-2and the mean earthsradius is 6378 km.
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Antenna Look Angles
The user must be able to determine the azimuth andelevation angles of the ground stations antenna.
Figure indetermining thelook angles for
a geostationarysatellite.
d
rs
re
Local
horizontal
centre
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Look Angles Equations
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For Geostationary Satellite
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The azimuth look angle (Az) can be foundfrom:
Hemisphere
Station
Position of
Satellite With
Respect toStation
Relation
between Az and
Northern East Az = 180 -
Northern West Az = 180 +
Southern East Az =
Southern West Az = 360 -
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Azimuth Angle Diagrams
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Visibility Test
For a sat to be visible from an e/s, its El must be abovesome min value, which is atleast 0. A +ve elevationangle requires that:
This mean that the max central angular seperationbetn the e/s and the subsat point is limited by:
For a nominal geo orbit, the last equation reduces to: 83.1 for the sat to be visible.
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Exercise 1
1. An e/s situated in Docklands, London needsto calculate the look angle to a geo sat in theIndian Ocean operated by Intelsat. Thedetails of the e/s site and the sat are as
follows:
E/s latitude and longitude are 52.0 N and 0.
Sat longitude (subsat point) is 66.0 E
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Exercise 2(Apr 2007)
Consider an earth station situated atHoustan , Texas has a longitude of99.5 at West and latitude of 29.5 atNorth. The earth station received
signal from Westar V satellite whichsituated at a longitude of 119.5 atWest.
Find the elevation angle and theazimuth angle of the earth stationantenna. Determine whether thesatellite is visible or not.
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Exercise 3(Oct 2006)
An earth station located at 20.5 S and78.4 E is up linking a signal to a 90.5E geostationary satellite. Given theratio of the distances between the
satellite to the earth center and theradius of the earth as 6.6107345,calculate its azimuth, elevation anglesand the distance between the earth
station and the satellite. Verify thevisibility of the satellite from the earthstation.
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Exercise 4(Apr 2006)
An earth station located inHoustan, Texas with the latitudeof 29.5 N and longitude of 95.5
W transmit a signal to the SatcomI located at geosynchronous orbitwith longitude of 135 W.Calculate the elevation angle and
azimuth of the earth station.Determine whether the satellite isvisible from the earth station.
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Exercise 5(Nov 2005)
Consider an earth station located at a
longitude of 80 W and its latitude
located at a position somewhere in
North. The geostationary satellite islocated 120 West and is visible from
the earth station. If the intermediate
angle is 52.55, find the latitude
position of the earth station, theelevation angle and the azimuth angle
of the earth station antenna.