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Delft University of Technology
Faculty Aerospace Engineering
Space Engineering & Technology I
Space Missions & Systems
AE1-801
First semester, first part
In-class problems, example problems, test and exam
problems
Issue 4.1
NOTE
This collection of problems only covers the part Space Missions & Systems of
the course Space Engineering and Technology I.
Problems of the part Orbital Dynamics can be found on the Blackboard site of
the course.
Delft, September 2005
Lecturer:
Ir. R.J. Hamann
System Integration
Design & Analysis of Space Systems
e-mail: R.J.Hamann@lr.tudelft.nl
Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
LEARNING GOALS AE1-801/SMS
To be able
• to place Space Flight in its general context
• to place major space development steps in time and to understand their mutual relation
• to distinguish space markets
• to summarise the major classes of space applications and their main characteristics
• to understand the characteristics of the space environment and their impact on
spacecraft characteristics
• to identify the impact of major mission parameters on the space vehicle concept
• to understand the break-down of space mission and space vehicle and the main
functions of these components
READING MATERIAL
Subject SET I, SMS
Introduction
Space and Space Flight
Mission Break-down and
Spacecraft Break-down
Space Flight History
Launch Vehicles
The Moon Program
Space Stations
Space Markets and Missions
Spacecraft: Scientific Research
Spacecraft: Earth Observation
Spacecraft: Communication
Spacecraft: Navigation
The Space Mission
The Space Lecture Series
Spacecraft Systems
Engineering [Fortescue]
-1
-1
7.3, 7.3.1, 7.3.2, 7.4, 7.5,
7.6, 7.7
---8.8, 11.8, 18.1, 18.10.2,
18.10.3
18.10.4, 19.3.3, 19.3.4
12.1, 18.10.1, 19.3.2
18.10.5
---
Space Mission Analysis
and Design [Larson]
-8.1, 8.1.4
1.2 (pp. 7-12), 10 pp.
(301-304)
----1.2, 1.3, 9.1
7.4, 9, 10.2
7.4, 9.4
13, 13.1, 13.1.1
11.7, 11.7.2
---
EXAMINATION
The regular exam on the material of Space Engineering & Technology I (ae1-801) is
scheduled after the second semester (June) together with the material treated in the part
“Orbit Dynamics” of that course:
• The part Space Missions & Systems (Hamann) is weighted 25%, equivalent to 45
minutes of exam time,
• The part Orbit Dynamics (Ambrosius/Visser) is weighted 75%, equivalent to 2 hours
and 15 minutes exam time.
A second exam opportunity is offered in August.
The part “Space Missions & Systems” also examined after the first period of the first
-2-
Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
semester (October/November): a light test of two hours duration. Bonus points are awarded
for the regular exam in June only:
• Grade for the test 10: 20% bonus in June,
• Grade 8 and 9: 15 % bonus,
• Grade 6 and 7: 10% bonus,
• Grade 4 and 5: 5% bonus,
• Grade 3 and lower: no bonus.
The exam problems for “Space Missions & Systems” are based on in-class problems and
example problems.
-3-
Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
IN-CLASS AND EXAMPLE PROBLEMS
Space and Space Flight
In-class problems
Heat Balance
Radiation heat balance of a flat surface with area A, front side looking at the sun and cold
space, rear side perfectly isolated
Heat balance
αAS = εAσ (T 4 − Tsp 4 )
where
α = solar absorptance (fraction of solar radiation absorbed; metal 0.1, black paint 0.9,
white paint 0.4)
ε = (infrared) emissivity (fraction of black body radiation emitted; metal 0.04, black
or white paint 0.9, glass 0.9)
σ = Stefan-Boltzmann constant (56.7051x10-9 W/(m2K4)
T = surface temperature in degree Kelvin
Tsp = temperature of cold space
Resulting surface temperature:
metal
498 K
black paint
396 K
rear side metalized glass mirror (second surface mirror)
228K
(glass may be replaced by plastic for a modest degradation in performance, but a
large gain in cost)
If you use a number of layers of highly reflecting material the absorptance and emittance can
effectively be reduced by a factor of 50 to 100 >> thermal blankets.
Using the different material properties the inside of the spacecraft can be kept between -10
and +40 C, the normal operating range of electronic equipment
Robot arm motion in the absence of gravity
EXAM AUGUST 2005 (15 minutes)
A robot arm of 10 m length (l) has to move a payload of 6000 kg mass. The actuator at the
root of the arm exercises 300 Nm torque. How fast has it moved the payload over 180
degrees?
Force and acceleration at the end of the arm:
Torque 300
=
= 30 N
10
l
30
F
a=
=
= 0.005 m / s 2
payload mass 6000
F=
-4-
Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
A 90 degree arc with full acceleration/deceleration is covered in
t=√ (2 s / a) = √ (2 l (π /2) / a) =
√ (10 π /0.005) = 79.3 s
180 degrees is covered in less than 160 seconds.
Example Problems
1-1
A Space Mission can be defined, designed and developed as an isolated item.
Do you agree or disagree with this statement?
Motivate your answer and illustrate it with an example.
1-2
List four or five typical characteristics of space, that can be exploited by space
missions
1-3
Calculate the temperature of a flat surface with area A, with its front side looking to
the sun and cold space and its rear side looking to cold space, both sides with the
same optical properties. Do this for a metal surface, for a black painted surface and
for a mirror surface.
Mission Break-down and Spacecraft Break-down
Example Problems
1-4
List the spacecraft subsystems and their functions.
1-5
List the mission segments according to [Larson]. Is this breakdown always used?
Space Flight History
Example Problems
2-1
Which were the driving forces for the rapid development in space technology
• in the period between 1940 and 1955?
• in the period between 1955 and 1970?
Launch Vehicles
In-class problem
Why do rockets have stages?
How fast can a rocket go?
-5-
Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
Tsiolkowsky: Ve = c ln
M0
Me
(for derivation see ae1-019, part Introduction Space
Technology)
where
final velocity in zero-gravity
nozzle exhaust velocity
Start mass of rocket
End mass of rocket
This is neglecting gravity and atmospheric drag; which will be addressed in detail in the
course Space Engineering & Technology I, part Mission Analysis I)
Ve
c
M0
Me
Practical values
c
3000 m/s (will be addressed in lecture ae2-s02).
M0
Me
5 to 6, i.e. 80 to 85 % of rocket mass are propellants (will be addressed in
lecture ae2-s02).
So final velocity is limited to:
Ve = 3000 ln 6 = 5375 m/s
When the rocket lifts off, it is in a vertical position and gravity is pulling it down; when it
achieves its final (circular) orbit, the velocity vector is purely radial and the gravity force is
compensated by the centrifugal force. It can be derived, that in the presence of a gravity field
the equation of Tsiolkowsky is modified into:
γ
tb
M
Ve = c ln 0 − ∫ g (t ) ⋅ sin(γ (t )) ⋅ dt
Me 0
local horizontal
F
g
where g(t) is the acceleration of gravity at time t , γ(t) the angle relative to the local horizontal
and tb is the burning time of the rocket engine.
The gravity term is equal to g0 (the acceleration of gravity at 0 m altitude) at lift off and equal
to zero (sin(γ) = 0), when the circular orbit is achieved. In a fair approximation we may
assume:
[ g (t ) ⋅ sin(γ (t ))]total trajectory = 0.5 ⋅ g0
Assuming a burning time of 5 minutes, the velocity loss due to gravity is then:
∆Vg = −0.5 ⋅ g 0 ⋅ t b = −0.5 ⋅ 9.81⋅ 5 ⋅ 60 = 1472 m / s
In the presence of gravity a single stage rocket can hence obtain a maximum velocity of
typically 3900 m/s.
The last velocity loss is due to aerodynamic drag, which is in the order of 300 to 400 m/s,
yielding a maximum single stage rocket velocity of 3600 m/s.
How fast does a satellite have to go?
Mean earth radius:
6370 km
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Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
Length of circular orbit at zero altitude:
L = 2π R = 40020 km
Orbital period:
90 minutes
(Source: lectures ae1-801, part Mission Analysis I)
Orbital velocity:
40020 *1000
= 7410 m/s
90 * 60
This is considerably larger than the 3600 m/s calculated above.
Conclusion
Single stage rocket does not deliver enough velocity to reach low earth orbit, even when
gravity and atmospheric drag are neglected. Feasibility of future developments depends on
•
•
achieving higher nozzle exhaust velocities,
designing lighter launcher structures.
Example Problems
2-2
Every launcher project starts developing a completely new concept. Do you agree or
disagree with this statement? Motivate your answer and illustrate with an example.
2-3
Calculate the theoretical final velocity of the V2 with Tsiolkovski, assuming that the
exhaust velocity of the rocket engine is 2250 m/s (ethanol + 25% water, liquid
oxygen). Is this sufficient to bring a satellite in a low earth orbit with 90 minutes orbital
period? Motivate your answer.
2-4
You intend to launch 50 satellites of 3000 kg each into GTO. Candidate launchers are
Ariane 5 launcher at a cost of 125 M$ each and an expected reliability of 0.9 and a
Proton D1e at a cost of 60 M$ and a reliability of 0.85. Which launcher do you select?
Motivate your answer with numerical results.
2-5
Why does Arianespace serve 60% of the commercial launcher market and what
upgrades are planned for Ariane 5 to maintain this position?
The Moon Program
Example Problems
3-1
Describe the sequence of events for the first successful Apollo mission to the Moon
and their purpose.
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Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
Space Stations
Example Problems
3-2
List the differences and commonalities between the Spacelab concept and the MIR
concept. Address mission, development concept and science in your list.
Space Markets and Missions
Example Problems
4-1
List the steps in Space System development and their definition according to [Larson]
4-2
Which are the essential elements in the Mission Need Analysis for a new Space
System?
4-3
Give two distinct ways to segment the Space Market and list the segments for each of
them
4-4
Calculate from the spacecraft market turnover in 1996 and the number of launches in
that year the average cost of one satellite (assuming a dedicated launch for each
satellite
Spacecraft: Scientific Research
In-class problem
Where do disturbance forces come from?
Aerodynamic disturbance force
Even in space (in the vicinity of the Earth or other planets up to several thousands of
kilometers altitude) there is some atmosphere present, hence an aerodynamic force will be
exerted on a satellite. As in "normal" aerodynamics, that force is proportional to the surface
area of the satellite, the density of the atmosphere and the square of the velocity of the
satellite. It also is a function of the shape of the satellite, but to a far lesser extent than for
aircraft or launchers. This force is usually called aerodynamic drag, and works always in the
direction opposite to the velocity and always leads to a loss of velocity.
The drag can be written as:
D = 0.5 CD A ρ V2 cos i
where
CD is the drag coefficient (about equal to 2 for most spacecraft configurations)
A is the surface area
ρ is the air density
V is the flight velocity
i is the angle between velocity and surface normal
-8-
Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
Aerodynamic drag can also be used to "capture" a satellite in an orbit around a planet (aero
braking).
Solar radiation disturbance force
A surface (hence also a satellite) experiences a force due to the pressure of the solar
radiation. This solar radiation force is proportional to:
• the surface area A
• the solar flux S in W/m2
• (2-α), where α is the solar absorptance (α = unity minus the reflection)
• cos i, where i is the angle between the sun vector and the surface normal
and inversely proportional to the speed of light c
Expressed in a formula
Fs =
S
⋅ A ⋅ (2 − α ) ⋅ cos i
c
So except for the term related to the solar absorptance it is very similar to the aerodynamic
drag. The force is always acting in the same direction as the sun vector (vector from the sun
to the satellite).
The solar radiation can be used to propel the spacecraft (solar sailing). It is only effective
relatively close to the sun, as the solar flux S decreases with the square of the distance from
the sun.
Gravity force
Of course the main force exerted on a body in space is the force of gravity; it keeps the
satellite in its orbit around a celestial body. We may calculate the orbital velocity orbital
period of a satellite in a circular orbit around the Earth in a relatively simple way.
The centripetal force for a rotational motion is:
V2
(1)
⋅ MS / C
R
where V is the circular velocity, R the radius of the circle and MS/C the mass of the spacecraft.
Fn = an ⋅ M S / C =
The gravity force is given by:
Fg = g ⋅ M S / C
where g is the acceleration due to gravity.
For a circular orbit centripetal and gravity force are in equilibrium:
V2
V2
⋅ MS / C = g ⋅ MS / C ;
= g; V = R ⋅ g
R
R
Newton’s law of gravity states that
Fn = Fg ;
-9-
(2)
Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
G ⋅ M earth ⋅ M S / C
= g ⋅ M S / C ; G ⋅ M earth = g ⋅ R 2
2
R
where G is the universal gravity constant and Mearth the mass of the Earth. Or:
Fg =
2
R earth
R2
where g0 is the acceleration due to gravity at the Earth surface (R = Rearth).
g ⋅ R 2 = g 0 ⋅ R earth ; g = g 0
2
(3)
Substituting (3) in (2) we obtain:
Vorbit = g 0 ⋅ R orbit ⋅
R earth
2
R orbit
2
= R earth ⋅ g 0 ⋅
1
R orbit
With Rearth = 6370 km and g0 = 9.81 m/s2 this becomes:
Vorbit = 6.37 ⋅ 10 6 ⋅ 9.81 ⋅
1
R orbit
=
1.995 ⋅ 10 7
R orbit
Orbital periodcan be written as:
2π ⋅ R orbit
t orbit =
Vorbit
For several orbital altitudes this results in:
Orbital altitude (km)
0
200
500
1000
20000
35850
Rorbit (m)
6.37 x 106
6.57 x 106
6.87 x 106
7.37 x 106
2.637 x 107
4.222 x 107
Vorbit (m/s)
7905
7784
7612
7349
3888
3071
torbit (minutes)
84.4
88.4
94.5
105.0
710.3
1439.7
Sometimes gravity is used in a very specific way:
• To accelerate a spacecraft when passing a celestial body (swing by),
• To achieve very specific orbits
- a sun-synchronous orbit (using the fact that the Earth is not spherical)
- an "orbit" in a libration point (the point where the gravity forces of two celestial bodies
are in equilibrium). The satellite maintains then a fixed position relative to those two
bodies.
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Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
Where do disturbance torques come from?
Aerodynamic disturbance torque
An aerodynamic disturbance torque results, if the aerodynamic force vector does not go
through the center of mass of the satellite. An example is given below.
Each of the components
of the drag and its
distance to the center of
total drag
mass is individually
Center of
torque
calculated. The sum is
Mass
the disturbance torque.
velocity
drag bottom
The aerodynamic torque
surface
can be used to align one
of the satellite axes with the velocity vector (aerodynamic stabilization).
drag top
surface
Solar radiation disturbance torque
A solar radiation disturbance torque results, if the solar radiation force vector does not go
through the center of mass of the satellite. An example is given below.
Again, each of the
components of the
solar force and its
total solar force
Center of
distance to the
torque
Mass
center of mass is
sun vector
individually
solar force
calculated. The
bottom surface
sum is the
disturbance torque. The solar radiation torque can be used to align one of the satellite axes
with the sun vector. This is done by manipulating the position and solar absorptance of the
satellite surfaces.
solar force top
surface
Gravity gradient torque
Let us now consider two equal masses, m1 and m2, connected by a boom. Mass m1 is further
away from the Earth than mass m2 (see the figure below). From (3) we may derive that mass
m1 and m2 will experience a force due to the gravity field:
2
R earth
Fg1 = m1 ⋅ g = m1 ⋅ g 0
2
R1
and
Fg 2 = m 2 ⋅ g = m 2 ⋅ g 0
R earth
R2
2
2
As R1 > R2 it follows Fg1 < Fg2.
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Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
m1
smaller force
Center of Mass
torque
m2
gravity field
larger force
For a circular motion V equals ω R and we may rewrite the centripetal force Fn given in (1) as
ω2 R2
V2
Fn = an ⋅ m =
⋅m =
⋅m = ω2 R m
R
R
The rotational velocity ω is equal for both m1 and m2. As R1 > R2 it follows Fn1 > Fn2.
So for m1 the centripetal force, which is directed away from the Earth, is larger than that
acting on m2 and the gravitational force, which is directed towards the Earth, is smaller than
that acting on m2. Consequently m2 will experience a resulting force that pulls it towards the
Earth, while m1 is pulled away from the Earth. As a result the boom, connecting the two
masses, will tend to align itself with the gradient of the gravity field. This is a stable
equilibrium position. There is another equilibrium position, when the boom is horizontal; this
position is, however, instable.
Note that the reasoning also applies, if the masses m1 and m2 are not equal.
In a mathematical sense the moment of inertia of the configuration shown in the figure in the
plane of the drawing is far larger than that around the axis of symmetry. This causes the
gravity gradient torque.
The gravity gradient torque may be used to stabilise a satellite passively. Next to the mass
needed to produce the desired mass distribution, no other energy and hardware is required.
Magnetic torque
Every satellite has a small (residual) magnetic dipole moment; you cannot avoid using small
amounts of magnetic material. This dipole moment will interact with a magnetic field, as is
the Earth magnetic field, as shown in the figure
magnetic field
below. The spacecraft acts as the needle of a
compass.
Note, that there is no resulting force in addition
to the torque. You can use this effect to
actively (by using switchable magnetic coils or
torque
torquers) or passively (by building permanent
magnets into the spacecraft) to stabilize a
dipole moment of spacecraft
- 12 -
Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
satellite.
Some disturbance torques are cyclic
Although generally disturbance torques are directed in arbitrary directions, they often show
some systematic behavior (pending the orbit and orientation of the satellite). Sometimes they
are cyclic, which means that over a certain period of time their integrated value equals zero.
So if you manage to store their energy over that period, you do not have to eliminate them by
spending "expensive" fuel (mass). An example is given below.
velocity
sun vector
A symmetrical satellite is
in a low circular equatorial
orbit, the solar panel
normal always pointing
towards the sun. Assume
that the sun vector lies in
the orbital plane. The
center of mass of the
satellite will be located in
the spacecraft body (it is
much heavier than the
solar panels). The center
of (aerodynamic) pressure
will be located in the
middle of the plane
through the solar panels,
so there will be an off-set
between center of mass
and center of pressure.
velocity V
Center of Mass
Center of
Pressure
off-set
between
CoM and
CoP a
α
- 13 -
Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
At α = 0 the satellite velocity vector is normal to the sun vector. If we assume the satellite
body to have small dimensions compared to the solar panel area, the aerodynamic drag is
then zero.
The projected area of the solar panel on the plane normal to the velocity vector is A sin α.
The offset of the center of pressure (in the center of the solar panel) and the center of mass
is a cos α. So the aerodynamic disturbance torque is:
Taero = 0.5 ⋅ a ⋅ C D ⋅ ρ ⋅V 2 ⋅ A sin α ⋅ a cos α = 0.25 ⋅ a ⋅ C D ⋅ ρ ⋅V 2 sin(2α )
Or the aerodynamic torque varies with twice the orbital frequency. This is shown in the
picture at the below.
cyclic aerodynamic torque
1
0.5
0
0
0.2
0.4
0.6
0.8
1
-0.5
-1
fraction of orbit
The vertical axis shows the variable
Taero
0.25 ⋅ a ⋅ C D ρ ⋅ A ⋅ V 2
If we now have an actuator that is able to store the angular momentum generated over one
quarter of an orbit, we do not have to spend electrical or chemical energy to dump this
angular momentum.
Example problems
5-1
Which objects have been observed by ANS and IRAS? Which properties of these
objects have been studied? Why have they been studied from space?
5-2
What is the minimum time it takes IRAS to map the total sky, taking into account the
maximum allowed offset of the instrument axis towards the sun vector and neglecting
the yearly variation of the angle between sun vector and equatorial plane?
5-3
Where should the IRAS ground station be located to have maximum contact time with
the satellite?
5-4
Why does an astronomy satellite observing the full sky in a sun-synchronous orbit
need to have a telescope field of view of at least 0.99o and a lifetime of at least 6
months?
5-5
Calculate the acceleration due to air drag of the LAGEOS satellite assuming a drag
coefficient of 2.0, an air density of 5.4 x 10-18 kg/m3 and an orbital period of 228.4
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Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
minutes. What is the total velocity change in one year? What is the total error in the
distance traveled in that year (both in absolute and relative (%) terms)? Assume that
the orbital altitude does not change over time. Is this satellite a stable position
reference? Motivate your answer.
Spacecraft: Earth Observation
In class-problem
How does a satellite see the Earth?
Low Earth orbit; distance to surface
Mean earth radius
Geostationary orbit; distance to surface
hLEO = 500 km
RE = 6370 km
hGEO = 35850 km
From the figure you may deduce some information about resolution obtained for LEO and
GEO orbit in an Earth observation mission.
alphaLEO
alphaGEO
RE
RS
h=500 km
h=35850 km
GEO
LEO
Or in detail:
RE
RS
RE+h
The radius of the Earth disc viewed from the satellite can be calculated from:
RS
=
RE
( RE + h) − RE
RE + h
2
Pythagoras
2
Calculating RS for LEO and GEO
RS LEO = 2386 km and RS GEO = 6297 km
If the total visible area of the Earth disc is viewed from the satellite with a 1024 x 1024 pixel
detector (digital camera), the LEO pixel measures (2386*2)/1024 = 4.66 km, the GEO pixel
(6297*2)/1024 = 12.26 km. In practice scanning techniques (and better detectors) are used to
improve the resolution, but the fact stays, that Earth observation from GEO is at least a factor
3 worse than from LEO.
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Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
How does a ground station see the satellite?
h
h = orbit altitude = 500 km
RE = Earth radius = 6370 km
h
RE
α
Ground
station
Calculating the angle α
RE
6370
=
= 0.927
RE + h 6370 + 500
α = 21.99 deg
cosα =
Assuming the orbital period to be equal to 90 minutes, the maximum ground contact time
becomes
t contact =
2α
2 ⋅ 21.99
t orbit =
⋅ 90 = 10.99 minutes
2π
360
In practice a gound station cannot receive the signal of a satellite, that is less than 10
degrees above the horizon and generally the ground station is not located directly under the
flight path of the satellite, so practical contact times are of the order of 6 to 7 minutes for a
satellite in a 500 km orbit.
What does a geostationary orbit mean for data transfer?
Assume both a geostationary and a low earth orbit satellite observe the same area on the
earth surface around a ground station from a circular, equatorial orbit and make a "picture"
of it once a day.
The LEO satellite "sees" the area and the ground station 11 minutes per orbit, which has a
period of 90 minutes (source: lectures Mission Analysis I of ae1-801). The GEO satellite
"sees" the area and the ground station continuously.
Per 24 hours the ratio of the time available to send the data of the picture to the ground
station is:
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Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
part of LEO orbit
that ground station
is visible
t contactLEO
t contactGEO
=
number of passes in LEO
orbit over ground station on
the equator
11 24 ⋅ 60 11
⋅
=
= 0.12
24 ⋅ 60 90
90
Hence the data rate of the LEO satellite has to be 8.2 times larger to transfer the same
amount of data. Note, that in this example the factor 0.12 is also the ratio of the time
available to take the picture of the object on the earth surface; the LEO satellite has to "work"
8.2 times as hard as the GEO satellite.
The data rate is proportional to the transmission power (source: Space Engineering &
Technology II). However, as the geostationary satellite is further away from the ground
station, the GEO satellite needs more power to transmit the same amount of data (see
chapter 7 of the lecture notes).
So even with the much smaller data rate, the GEO satellite still needs
distance
from Earth
narrower
antenna beam
hGEO 2 t contactLEO
1
5141 ⋅ 0.12
⋅
⋅
=
= 10.07
2 t
61.24
hLEO
contactGEO 61.24
times more communication power than the LEO satellite.
A geo-stationary orbit is generally chosen, if the mission needs
•
•
to overview the full earth disc (e.g. meteorological satellite),
to have continuous contact with a ground station (e.g. communication satellite).
The calculations show that you pay for these "services" in power required for the
communication link.
Example problems
6-1
List at least two reasons why you would perform Earth Observation from space
6-2
Calculate for the Meteosat satellite (see page 32) how many pixels the visual image
of the Earth disc contains (use the VIS resolution), assuming the pixels have equal
size in North-South and East-West direction. If we assume that each pixel uses 8 bits,
how many bits contains one Earth image? What is the corresponding bit rate for
sending that image to a ground station?
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Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
Spacecraft: Communication
In-class problem
What does a geostationary orbit mean for communication?
Low Earth orbit; distance to surface
Geostationary orbit; distance to surface
hLEO = 500 km
hGEO = 35850 km
For a given antenna and transmitter power, the energy of the signal expands in cone:
hLEO
hGEO
The area ratio of the "antenna" pattern at LEO and at GEO is:
ALEO hLEO 2
5002
=
=
= 0.000195
AGEO hGEO 2 358502
This is also the energy (power) loss the communication link experiences. Somewhere in the
communication chain transmitter power - transmitter antenna size - receiver antenna size receiver power this loss has to be compensated.
Suppose, your hand-held receiver has 0.2 Watt receiver power sufficient to receive the LEO
signal, it would need
1
⋅ 0.2 = 5141 ⋅ 0.2 = 1028 Watt
0.000195
power to receive the GEO signal.
Luckily there is some other factor to consider. We have assumed, that the antenna on the
LEO satellite and the GEO satellite is identical, that is it radiates its transmission power in a
beam of the same dimensions. That is not necessary, as shown in the following figure.
alphaLEO
RE
h=500 km
h=35850 km
LEO
GEO
alphaGEO
The angle alphaLEO is much larger than the angle alphaGEO, and any signal, that misses the
Earth is wasted. Hence we can make the antenna angle of the satellite smaller, and
concentrate the transmission power in a narrower beam.
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Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
The cone half angle alpha can be calculated as
alpha = arcsin
RE
RE + h
or alpha LEO = 68.01o and alphaGEO = 8.68o . The transmission energy content of the beam
reaching the Earth's surface is inversely proportional to the square of this angle, keeping the
transmission power constant. So the strenghth of the signal from the smaller beam is
(68.01/8.68)2 = 61.42 times larger than that of the wider beam. Note, that you see a larger part
of the earth from GEO than from LEO.
The receiver power of your hand-held receiver may hence at least be reduced by the factor
61.42; the required receiver power is then:
1028/61.42 = 16.74 Watt
This is still a factor 83.7 higher than required for the LEO satellite case.
Example problems
7-1
7-2
List the disadvantages and advantages of each of the following personal
communication systems:
• a system using geostationary satellites
• a system like Iridium
• a fully land-based system (GSM)
Address both space and ground segment
Assuming that
• a mobile telephone user is willing to pay a world-wide charge of 0.1 US$ per
minute for a 99% availability of the service
• the user makes 100 such one minute calls per month
• a MSS constellation with 6 years lifetime costs 4 G$
• operations of the system cost 200 M$ per year
• billing services cost 100 $ per user per year
• investors want to recover their money with 20% annual yield
How many customers does the constellation operator need? Do you think that is
economically feasible? Motivate your answer.
Assuming that
• the constellation has 66 satellites
• all calls take place during day time (8 hours)
• the users are uniformly distributed over the Earth’s surface
• all mobile calls are to a fixed telephone number
• a peak load of two times the average load (think about the 99% availability)
How many telephone lines should be available per satellite? Do you think this is
technically feasible with an Iridium-like constellation?
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Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
Spacecraft: Navigation
In-class problem
How expensive is eclipse?
h = orbit altitude = 20000 km
RE
h
RE = Earth radius = 6370 km
RE
α
sun
Calculating the angle α
RE
6370
=
= 0.242
RE + h 6370 + 20000
α = 14 deg
sin α =
Assuming the orbital period to be equal to 720 minutes (12 hours), the time the satellite is in
the Earth's shadow (eclipse) becomes
t eclipse =
2α
2 ⋅ 14
t orbit =
⋅ 720 = 56 minutes
2π
360
The sunlit period of the orbit is then 664 minutes.
The schematic of the electrical power provision to the satellite can be drawn as:
Bus Regulator;
efficiency 0.8
Solar Array;
efficiency 0.08
Satellite power
load
1000 W
1250 W
1000 W
Solar Array
charge
section;
efficiency 0.08
1708 W
Charge
Regulator;
efficiency 0.85
1452 W
Battery;
charge
efficiency 0.9
1307 W
Battery;
discharge
efficiency 0.9
1167 W
Discharge
Regulator;
efficiency 0.85
The charge regulator is used if the solar array voltage is not equal to the satellite bus voltage.
The battery charge regulator ensures that the battery charge voltage and current have the
correct value to achieve optimum battery charging. The battery discharge regulator ensures
that the bus voltage keeps the correct value during eclipse. Battery charge and discharge
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Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
efficiency are related to the internal resistance of the battery.
The schematic represents the worst case situation (low efficiency units and Si solar cells).
Using GaAs solar cells and higher efficiency units gives a better picture (more power or a
smaller solar array).
Once we know the power needed, we can calculate the solar array area (assuming the sun
vector is normal to the solar array) with
Psolar array = S sun Asolar arrayη solar array
where
S = solar flux = 1353 m/s2
η = efficiency
A = area
P = electrical power output
Now we can calculate the total energy needed to provide 1000 W to the satellite both in
sunlight and in eclipse
energy needed in eclipse
battery charge power during sunlight
satellite power during sunlight
total solar array delivered power
fraction of solar array power needed for eclipse
total solar array area
worst case
56 minutes 1708 W = 95648 Wmin
95648 Wmin/664 minutes = 144.0 W
1250 W
1394 W
10.3 %
12.88 m2
Note that for a low Earth orbit the fraction of solar array power needed for eclipse may
increase to some 45 to 50 %.
Example problems
8-1
Calculate the required solar array power for a satellite with the following
characteristics:
• Equatorial orbit 500 km, orbital period 90 minutes, satellite bus power 200 W
• Efficiencies:
• solar array 0.15 (GaAs solar cells)
• bus regulator 0.85
• battery charge and discharge regulator 0.85
• battery charge and discharge 0.9
• Solar flux 1353 W/m2; solar array normal always pointing to the sun
Draw a diagram of the power subsystem. Calculate the eclipse duration. Determine
the total solar array area required and the proportion of that area needed to generate
the eclipse power.
8-2
What is the principle used for position determination in the GPS system? How many
satellite signals are necessary to determine the position?
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Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
The Space Mission
Example problems
9-1
We want to observe the complete radiation spectrum of celestial bodies for long
periods of time from space. It seems attractive to go far from the Earth, as
• the Earth itself is a source of radiation (heat)
• close to Earth the time during which a celestial body (a point in inertial space) can
be observed uninterrupted is limited (the Earth passes through the satellite object line)
• we have long contact periods with the ground station, so we need less on-board
storage of data, less sophisticated autonomy and see “cosmic” events in real time
But there are disadvantages …...
List the disadvantages for this concept.
ANSWERS TO EXAMPLE PROBLEMS
Below the (Numerical) answers to some of the example problems are compiled.
Problem 1-3
solar absorptance
emissivity
Stefan-Boltzmann constant
Solar flux in space
Temperature of cold space
metal
0.1
0.04
black paint
0.9
0.9
5.67E-08
1400
4
mirror
0.1
0.9
419
333
192
Surface Temperature
[-]
[-]
[W/(m2K4)]
[W/m2]
[K]
[K]
Problem 2-3
Vmax = 2546 − 387 = 2159ms −1
vorbit = 7410ms −1
Problem 2-4
28 Arianes are selected at a total cost of 28*(125+800) = 25,900 M$. Corresponding figure
for the Proton are: 59 Protons at a total cost of 59*(60+400) = 27,140 M$.
Problem 5-2
To map the total sky requires a 180-60 = 120 degrees rotation around the Sun. This takes
IRAS 120/360 * 365 = 122 days.
Problem 5-5
a drag = 1.18 ⋅ 10 −13 m / s 2
Total velocity change in one year:
∆V = 3.72 ⋅ 10 −6 m / s
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Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
Position error:
∆s = 58.8 m
Problem 5-5
Total number of “pixels” is 5038*5038 = 25.4 E6 pixels
One Earth disc image: 25.4 E6 x 8 = 2.03 E8 bits
Data transfer rate: 2.03 E8/(30*60)=112.8 kbps.
Problem 7-2
Number of users
N = 2.08 x 107
Number of calls per satellite including reserve for peak load (factor 2): 2162 x 2 = 4324
Number of two-way channels available per Iridium satellite 2800
Problem 8-1
The solar array should be able to deliver 235W to the satellite during sunlight and 342W
during ecipse.
t eclipse = 34 min
Total solar array area: 2.18m2
Fraction of area needed for eclipse: 46.9%
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Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
TEST AND EXAM PROBLEMS
TEST OCTOBER 2001
Problem 1 (45 minutes)
Mission and System Break-down
A mission may be broken down in several segments.
1a
List the seven mission segments according to the convention used in “Space Mission
Analysis & Design” [Larson].
1b
List the function of each of the mission segments. Split the Space Segment in two
parts (Payload and Bus).
1c
Give the definition of each of the mission segments.
1d
Is this breakdown always used? Illustrate and motivate your answer.
Problem 2 (65 minutes)
Technical and Economic Feasibility of a Space-based GSM System
2a
Assuming that
• a mobile telephone user is willing to pay a world-wide charge of 0.15 US$ per
minute for a 99% availability of the service
• the user makes 100 such one minute calls per month
• a MSS constellation with 6 years lifetime costs 4 G$
• operations of the system cost 200 M$ per year
• billing services cost 100 $ per user per year (expenses of the mobile telephone
operator)
• investors want to recover their money with 20% annual yield
Calculate income, expenses and investor's profit as a function of the number of
customers N. How many customers does the constellation operator need?
2b Do you think that is economically feasible? Motivate your answer.
2c Assuming that
•
the constellation has 66 satellites
•
all calls take place during day time (8 hours)
•
the users are uniformly distributed over the Earth’s surface
•
all mobile calls are to a fixed telephone number
•
a peak load of two times the average load (think about the 99% availability)
How many telephone lines should be available per satellite? Hint: calculate first the
total average number of telephone calls per day of 8 hours.
2d The Iridium constellation has the following characteristics:
• World-wide system for mobile communication
• 2800 two-way channels for speech per satellite
• 66 satellites distributed over 6 orbit planes (constellation)
• Cost (all system): 4 G$
• Circular sun-synchronous orbit: 785 km
• Lifetime: 5-6 year
• Inter-satellite links
Do you think an Iridium-like constellation is a technically feasible solution for your
problem? If not, indicate why. Could you think of measures to improve the situation?
Motivate your answer.
Answers
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Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
2a:
2c:
Total income: 1080 N $; Total expenses: 600 N + 4 x 109+ 1.2 x 109 $; Total profit: 4.8
x 109 $; Number of users: 2.08 x109
Number of calls (lines) per satellite including reserve for peak load: 4324
Problem 3 (70 minutes)
Launching a Satellite by means of a Single Stage Launcher
A small satellite will be launched in a low, circular orbit with an altitude hE of 500 km. The
assignment is to determine which characteristics a single stage launcher shall have to enable
this. The Earth radius RE equals 6370 km.
3a Calculate the orbital velocity V500 in that low earth orbit, taking into account that the
orbital period equals 94.62 minutes.
3b During launch the rocket "loses" velocity due to the gravity forces. This velocity loss is:
∆Vg =
t =t b
∫ − g (t ) ⋅ sin(γ (t )) ⋅ dt
t =o
where g(t) is the acceleration due to gravity at time t, γ(t) is the angle between the
velocity vector of the launcher and the horizontal and tb is the burning time of the rocket
engine. At the start of the launcher γ equals 90o, when the engine stops (at the injection
in the circular orbit) γ equals 0o. An acceptable assumption is that during the full flight
g(t).sin(γ(t)) has an average value equal to 0.5 g0.
Can you list another loss occurring during the launcher ascent?
3c
3d
3e
3f
3g
3h
3i
3j
3l
Calculate the velocity loss (gravity loss) ∆Vg during the launch, assuming a rocket
engine burning time tb of 5 minutes.
The loss due to air drag during the flight equals 350 m/s. What is the total velocity
increment Vtotal that the launcher has to give to the satellite?
The launcher uses as propellants liquid hydrogen and oxygen, one of the most
energetic chemical propellant combinations. The combustion products have an exhaust
velocity we of 4200 m/s.
Calculate with Tsiolkowski the mass ratio M0/Me of the launcher, required to give the
velocity increment calculated in point 3d to the satellite (M0 is the mass of the launcher
at the start, Me the mass when all propellant has been used). Is that a realistic
(achievable) value? Calculate also Mp/M0 (the fraction of the total mass, that is used for
propellants, where the propellant mass Mp = M0 - Me).
Can you describe a method to decrease the required velocity increment?
The satellite is launched from the equator in exactly westward direction. Has this a
positive or negative impact on the required velocity increment to achieve the orbit?
Calculate the velocity increment ∆Veq of the launcher due to the rotation of the earth,
when the satellite is launched in the most favorable direction on the equator.
What is the total velocity increment Vtotal that the launcher has to impart to the satellite
in this case?
Calculate with Tsiolkowski the mass ratio M0/Me of the launcher required to give the
velocity increment to the satellite as calculated in point i. Calculate also Mp/M0 . Is the
difference with the results from point 3e large?
Which conclusion do you draw from the results of the calculations?
Answers
3a
V500 = 7603 m/s
3c
∆Vg = 1471.5 m/s
3d
Vtotal = 9425 m/s
3e
M0/Me 9.43; Mp/M0 = 0.894
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Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
3h
3i
3j
∆Veq = 463 m/s
Vtotal = 8963 m/s
M0/Me = 8.44; Mp/M0 = 0.882
EXAM JUNE 2002; JUNE 2005
Problem 1 (50 minutes)
Image Size and Bit Rate of a Weather Satellite
Meteosat is a geo-stationary satellite at an altitude hGEO of 35850 km, spinning at 100
revolutions per minute; its spin axis normal to the orbit plane. The sensor “scans” the Earth
once each revolution (one “line”) and builds one Earth image every 30 minutes in visible and
IR bands. After each scan (revolution) the sensor mirrors are adjusted, such that the next line
is scanned during the next revolution. The satellite sends the images and other data
collected from Earth located platforms to Earth. Its mass distribution is:
• 362 kg dry mass,
• 39 kg hydrazine propellant for attitude and orbit control,
• Apogee Boost Motor (ABM) propellant mass of 360 kg.
The resolution of the earth image is 2.5 km in the visual channel (VIS) and 5 km in the infrared channel (IR)
1a
1b
1c
1d
Draw a picture of the Earth and Meteosat in its orbit, indicating the Earth radius
REARTH , the orbit altitude hGEO and the half cone angle αGEO the Earth is included in,
when seen from the satellite.
Calculate the diameter of the Earth disc Ds as seen by Meteosat, assuming the Earth
radius REARTH equals 6370 km and the geostationary orbit altitude hGEO equals 35850
km. Calculate also αGEO.
What is the most Northern or Southern latitude of the Earth, that is visible from
Meteosat?
How many geostationary satellites would be needed to cover the whole Earth
between these latitudes? Motivate your answer.
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Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
1e
1f
1g
1h
1i
1j
1k
Calculate how many pixels NE-W the line of the image on the equator of the Earth disc
contains in the visual channel (use the VIS resolution).
What is the time needed to build up one line of the image?
Assuming the pixels have equal size in North-South and East-West direction and that
the image of the earth is a square, calculate how many pixels the image of the Earth
disc contains in the visual channel (use the VIS resolution).
If we assume that each pixel uses 8 bits, how many bits contains one Earth image?
What is the minimum bit rate for sending one full Earth image to a ground station?
Each revolution of the satellite one line of the Earth image is scanned. Deduce from
the answer on 1g how many lines the Earth image has.
In how much time is one full image of the Earth disc build up (using the results of 1j)?
Is this compatible with the time available to build up one image?
1l
Assume now that 2 pixels in North-South direction are scanned per scan and
that the mirrors are stepping 2 North-South pixels per revolution. How much
time is takes now to produce one full Earth image?
1m
How much time is left after completion of the scan according to 1l? For what purpose
this time will be used in your opinion?
What is the bit rate corresponding to the infra-red channel of Meteosat (use the infrared channel resolution)?
Assuming Meteosat has three visual and one infra-red channel: what is the total
minimum bit rate of Meteosat?
1n
1o
Answers
1b
Rs = 6297 km; Ds = 12594 km
1c
Latitude: 81.3 degrees
1d
Number of satellites: 2.21 ≈ 3; Overlap: 21.3 degrees
1e
Number of pixels on equator: 5038
1f
One line build up in 0.6 seconds
1g
Size of one visual image: 25.4 x 106 pixels
1h
Size of one visual image: 103.2 x 106 bits
1i
Bit rate min = 112.9 kbps
1j
Number of lines per image: 5038
1k
Scan time 5038 lines: 3022.8 seconds
1l
Scan time full Earth image 1511.4 seconds
1m
Time left after image scan: 288.6 seconds
1n
Bit rate IR channel: 28.2 kbps
1o
Total bit rate: 366.9 kbps
EXAM AUGUST 2002
Problem 1 (50 minutes)
Power Subsystem Design
A satellite is in a circular equatorial orbit of 800 km altitude with a period torbit of 100.87
minutes. Its payload observes the Earth by means of a powerful Synthetic Aperture Radar
(SAR) over the full orbit. Payload and satellite require 2500 W electrical power (PPL)
continuously. The power is provided by a solar array, the surface of which is pointed towards
the sun in the sunlit part of the orbit.
1a
The electrical power PPL of 2500 W is generated by the solar array. However, the
voltage output by the solar array is different from the one required by the satellite,
hence a bus voltage regulator is used to adapt the voltage. The efficiency ηBR of the
bus regulator is 90%.
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Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
1b
The solar array is covered by solar cells. They convert sunlight into electrical power
with an efficiency ηSA of 8%. (8% of the radiative energy of the Sun is converted into
electrical energy). All data are End of Life values.
Draw a block diagram showing the solar array, the bus regulator and
payload/satellite. Note the data given in the text above in the diagram.
Calculate the electrical power PBRin at the input of the bus regulator, required to
deliver 2500 W to the satellite and payload.
1c
The minimum solar flux SSUN equals 1350 W/m2. How much solar array area
ASAnominal is needed to provide the input power of the bus regulator, you have
calculated in 1b?
1d
Calculate the duration of the eclipse teclipse (period in the shadow of the Earth) in the
800 km orbit (draw a picture). The radius of the Earth RE is 6371 km and the Sun
delivers a perfectly parallel beam of light and that the Sun is in the orbit plane. How
long is the sunlit phase of the orbit tsunlit?
Compute the energy (in Ws or Joules) required during the sunlit period Esunlit.
The energy during eclipse is provided by a battery. The battery has an internal
efficiency ηbat of 90%, that is 10% of the energy or power is converted into heat, when
the battery is charged or discharged. Also, the voltage at the output of the battery is
not equal to the voltage needed by the satellite and the payload; a discharge
regulator is used to adapt the voltage. The discharge regulator has an efficiency ηDR
of 85%.
Draw the block diagram showing the battery, the discharge regulator and the satellite
and payload. Note in the diagram all data given above.
What is the power Pbatout required at the battery output to deliver the 2500 W to the
payload/satellite?
And which power losses Pintdischarge occur in the battery during discharge?
What is the energy Ebatdischarge the battery should contain prior to discharge to provide
the satellite with power during eclipse, assuming that the battery can be fully
discharged?
The battery has to be charged during the sunlit phase of the orbit. A special charge
section of the solar array is used for it, which converts the solar radiation with an
efficiency ηSA of 8% into electrical power. Current and voltage at the output of the
solar array have to be adjusted to values fit to charge the battery. A charge regulator
is used for that purpose, which has an efficiency ηCR of 80%. During charging the
battery converts 10% of the incoming power or energy into heat (power loss).
Draw a block diagram showing the charge section of the solar array, the charge
regulator and the battery. Note down the data known in it.
Assume, that the battery needs to contain 7x106 Ws, when it is fully charged. How
much energy Ebatin needs to be fed into the battery?
How much power Pbatin (Watt) needs to be fed into the battery during the sunlit phase
of the orbit?
How much power PCRin should the charge section of the solar array deliver to the
charge regulator in order to feed the required power into the battery?
Assuming the solar flux SSUN is 1350 W/m2 and the solar cell efficiency ηSA of 8%,
what is the area of the charge section ASAcharge of the solar array?
What is the total power PSA the solar array can deliver? What part (in percentage) of it
becomes available for satellite and payload?
1e
1f
1g
1h
1i
1j
1k
1l
1m
1n
1o
Answers
1b
Regulator input power: 2778 W
1c
Solar array area: 25.7 m2
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Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
1d
1e
1g
1h
1i
1k
1l
1m
1n
1o
Eclipse duration: 35.14 minutes; Sunlit period: 65.73 minutes
Energy required in sunlit phase: 10.96 x 106 Ws
Battery output power: 2941 W
Battery power prior to internal loss: 3268 W; Battery internal power losses: 327 W
Energy stored in battery: 6.89 x 106 Ws
Energy fed into battery: 7.8 x 106 Ws
Power fed into battery during sunlit phase: 1975 W
Charge section power: 2464 W
Charge section area: 22.8 m2
Total solar array power : 5669 W; Percentage power available for stallite/payload:
44.1%
TEST OCTOBER 2002
Problem 1 (55 minutes)
Orbit maintenance
An earth observation satellite is in a circular low earth orbit with an altitude of 500 km (h) with
an orbital period torbit of 94.62 minutes. The mission requires, that this altitude is accurately
maintained over a mission duration of 5 years. The satellite has a dry mass (without
propellants) Me of 1000 kg. The satellite carries large antennae to perform its observations
and has a large solar array to provide electrical power to the payload; the average surface
area of the satellite ASC equals 40 m2.
1a Which three external disturbance forces are exerted on the satellite in its orbit? Which
of these disturbance forces can directly and to a large extent influence the altitude of
the orbit? Motivate your answer.
1b Write down the expression for the force Datm exerted by the atmosphere on the satellite.
Which parameters determine the magnitude of the drag?
1c
Calculate the velocity VSC of the satellite in its orbit. The Earth radius RE equals 6371
km.
1d Calculate the drag force Datm exerted by the atmosphere on the satellite, assuming that
the drag CD equals 2.0 and that the maximum density of the atmosfeer on an altitude of
500 km ρ500 equals 2.82 x 10-12 kg/m3.
1e Compute the acceleration aSC of the satellite without propellants caused by this force.
What is the velocity loss ∆Vatm of the satellite due to drag during the life time tmission of 5
years? What will be the consequence of this velocity loss?
1f
To keep the satellite on an altitude of 500 km the velocity loss ∆Vatm must be
compensated by the orbit correction system. This rocket system uses as a propellant
hydrazine, which produces an exhaust velocity c of 1800 m/s. Compute with
Tsiolkowsky what must be the mass ratio M0/Me of the rocket system to compensate
the velocity loss. Compute the required value of the wet mass M0 of the satellite and
how much hydrazine Mprop is required.
1g Do you think that the acceleration aSC ,that you have computed in 1e, is correct? Is the
actual acceleration larger or smaller than this value? What is the consequence for the
required quantity of propellant?
Answers
1c
DSC = 6.52 x 10-3 N
1d
aSC =6.52 x 10-6 m/s2; ∆Vatm = 1028.8 m/s
1f
M0 = 1771; Mprop = 771 kg
1g
aSC = 3.68 x 10-6 m/s2
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Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
Problem 2 (30 minutes)
Panel surface finish and temperatures
A metal panel with an area A of 1m2 is attached to a spacecraft. It is perfectly isolated from
heat sources at its back side, while its front side is looking to the Sun and to cold space. Cold
space has a temperature Tsp of 2K. The solar absorptance α of the metal is 0.1 (10% of the
energy of the sunlight is absorbed), the infrared emissivity ε of the metal is 0.04 (4% of the
infrared energy of the panel is emitted). The panel must be kept between 250 and 400 K.
2a Calculate the solar radiation (in Watts) absorbed Qabs by the metal panel, assuming a
Solar flux S of 1420 W/m2.
2b Write down the formula for the heat balance for the metal panel. Does the area of the
metal panel influence the temperature the panel will reach?
2c
Calculate the temperature Tpa of the metal panel. The value of the Stefan-Boltzman
constant σ is 56.7051x10-9 W/(m2K4). Is this temperature in the range allowed?
2d To correct the temperature, you have two different surface finishes at your disposition:
black paint and rear side metalized glass mirror (Second Surface Mirror). Optical
properties of these materials are:
black paint
α = 0.9, ε = 0.9
rear side metalized glass mirror:
α = 0.1, ε = 0.9
Which surface finish will you use to keep the metal panel within its required
temperature range? Which temperature does the panel reach? Why do you not use the
other surface finish (motivate with numbers)?
Answers
2a
Qabs = 142 W
2c
Tpa = 500.2 K
2d
Black paint: Tpa = 398; Second surface mirror: Tpa = 230 K
Problem 3 (35 minutes)
Launcher selection
You are planning to launch 50 satellites of 3000 kg each in Geo-stationary Transfer Orbit
(GTO). Your satellites cost 100 M$ per piece. Candidate launch vehicles are the Ariane 5
launcher, costing 125 M$ per piece and having an expected reliability of 0.9, a Long March
2E of 47 M$ per piece and a reliability of 0.75 and a Proton D1e launcher of 60M$ per piece
and a reliability of 0.85. Each of these launchers may launch 1 or more satellites at the same
time up to the satellite mass in GTO. Performance of the launchers is:
Launcher
Satellite mass in GTO (kg)
Ariane 5
6800
Proton D1e
5500
Long March 2E
3700
3.1
3.2
3.3
3.4
Compute for each of the three launchers the cost per kg payload in GTO. Which
launcher is the cheapest according to this calculation?
How many launchers of each kind do you need to launch the 50 satellites, without
taking into account the reliability?
Calculate for each of the three launchers the required number launch vehicles, taking
into account the reliability (so also the loss of satellites).
What is for each of the three launchers the total launch cost, taking into account extra
launch vehicles as a consequence of failures to be expected (and also the cost of extra
satellites)? Which launcher do you select? Motivate your answer with numerical results.
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Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
3.5
Can you think of a way to exploit the unused launch capacity of the launchers you have
not selected?
Answers
3.1
Proton
3.2
Proton/Long March: 50 launchers; Ariane 5: 25 launchers
3.3
Ariane 5: 28 launchers plus 3 extra satellites; Proton: 59 launchers plus 9 extra
satellites; Long March: 67 launchers plus 17 extra satellites
3.4
Ariane 5: 4100 M$; Proton 4440 M$; Long March 4849 M$
EXAM JUNE 2003
Problem 1 (30 minutes)
Orbit choice for an astronomy mission
We want to observe the complete radiation spectrum of celestial bodies for long periods of
time from space. It seems attractive to go far from the Earth, as
• The Earth itself is a source of radiation (heat and reflected sun light)
• Close to Earth the time, during which a celestial body (a point in inertial space) can be
observed uninterruptedly, is limited (the Earth passes through the satellite – object line)
• We have long contact periods with the ground station, so we need less on-board storage
of data, less sophisticated autonomy and see “cosmic” events in real-time.
There are however also disadvantages.
Questions
List the disadvantages of this concept from the following points of view:
1a
Space environment
1b
Launch capacity
1c
Communication
1d
Pointing accuracy
1e
Electrical power
1f
Orbit and altitude control
Be specific in your answers; give a rationale for each disadvantage you can identify.
Problem 2 (15 minutes)
Robot arm performance
A robot arm of 10 m length has to move a payload of 6000 kg mass. The actuator at the root
of the arm exercises 300 Nm torque.
2a
2b
How fast has the robot arm moved the payload over 180 degrees?
To save electrical power the actuator is a small motor with a reduction gearbox (gear
ratio 1:400). What is the motor torque?
Answers
2a
F = 30 N; a = 0.005 m/s2; 180 degrees covered in 159.6 seconds
2b
Required motor torque: 0.75 Nm
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Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
EXAM AUGUST 2003
Problem 1 (25minutes)
Apollo mission sequence
During the first successful Apollo mission, leading to a landing on the Moon, a number of
events can be distinguished as shown in the picture below.
Questions
1a
Describe the sequence of events of the Apollo mission and their purpose or function.
[Example:
3.
accelerate:
Get in a transfer orbit to the Moon; approach the Moon]
1b
How many (rocket) stages can you distinguish in the Saturn V- Apollo Service Module
– Lunar Module combination?
1c
Assume that the Saturn-Apollo combination has six stages and that the typical
velocity increment of a rocket stage is 3300 m/s. What is the total delta V that the
Apollo mission can achieve? Is this a reasonable figure? Motivate your answer. Think
of the escape velocity.
Answers
1b
Six stages
1c
Total delta V: 19800 m/s used for escape velocity (11200 m/s), atmospheric drag
losses (250 m/s); gravity losses during powered flight (1000 m/s), moon decent and
moon ascent
Problem 2 (20 minutes)
Comparison GEO and LEO satellite power subsystem
A geo-stationary satellite GEOSAT orbits the Earth at 35850 km altitude.
2a
Assuming that the Sun is in the equatorial plane and that the radius of the Earth is
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Space Engineering & Technology I
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2b
2c
2d
2e
6370 km, what is the duration of eclipse? Illustrate by means of a drawing.
What is the duration of the sunlit phase of the orbit?
Calculate the maximum duration of the eclipse for LEOSAT, a satellite in a low,
equatorial Earth orbit of 500 km altitude and 90 minutes duration.
What is the duration of the sunlit phase of this low Earth orbit?
If both satellites need the same amount of power (1000 Watt), compare qualitatively:
• The size of the battery
• The size of the solar array.
Motivate your answer.
Answers
2a
teclipse = 69.4 minutes
2b
Sunlit period: 1371 minutes
2c
teclupse = 34 minutes
2d
Sunlit period: 56 minutes
2e
Ratio (size LEOSAT battery/size GEOSAT battery) ≈ 0.5; ratio (solar array charge
section LEOSAT/solar array charge section GEOSAT) ≈ 12
TEST OCTOBER 2003
Problem 1 (7 minutes)
Space technology development drivers
Which were the driving forces for the rapid developments in space technology
1a
In the period between 1940-1955?
1b
In the period between 1955-1970?
Illustrate your answer with some simple examples.
Problem 2 (17 minutes)
Gravity gradient stabilization
A simple way to control the attitude of a spacecraft is to use the gradient in the gravity field of
a celestial body.
1a
Explain be means of a drawing, how gravity gradient stabilization works.
1b
How do you achieve that a spacecraft is gravity gradient stabilized (think of the
Moments of Inertia)?
1c
What is the major advantage of gravity gradient stabilization?
Problem 3 (12 minutes)
IRAS orbit and ground station
The IRAS (Infra-Red Astronomy Satellite) has been designed to produce a map of the infrared sources in the sky (infra-red sky survey). Therefore the satellite was launched in a sun
synchronous orbit exactly over the borderline between the lit and the dark half of the Earth.
3a
What is approximately the local time in the point on the Earth surface directly under
the satellite (nadir)?
3b
Where should the IRAS ground station be located to have maximum contact time with
the satellite? Give a rationale for your answer.
Problem 4 (42 minutes)
LAGEOS orbit maintenance
LAGEOS is a spherical satellite with a dry mass Me of 411 kg and a diameter D of 0,6 m. Its
surface is covered with mirrors. By projecting a pulsed laser on the satellite from the Earth
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Space Engineering & Technology I
Space Missions & Systems
In-class problems, example problems, test and exam problems
Issue 4.1, September 2005
and by measuring the running time of the signal, the distance to the satellite can be
determined accurately. The circular orbital altitude horbit of LAGEOS is 5900 km and the
orbital period torbit is 228.4 minutes. The Earth radius RE is 6370 km.
4a
What is the velocity Vorbit of the satellite?
4b
Calculate the acceleration due to air drag adrag of the LAGEOS satellite assuming a
drag coefficient CD of 2.0, an air density ρ of 5.4x10-18 kg/m3.
4c
What is the total velocity change ∆V in one year?
4d
What is the total error in the distance travelled ∆s in that year, both in absolute and
relative (%) terms? Assume that the orbital altitude does not change over time.
4e
Assuming that the force executed by solar radiation is negligible, which force can be
determined very accurately by means of LAGEOS? Motivate your answer.
4f
Assuming that you use pressurized nitrogen gas and thrusters with an exhaust
velocity we of 80 m/s, how much gas (Mp in kg) would you need to compensate the
velocity loss during that year (use Tsiolkowski)? Does it seem useful to incorporate
such an orbit control system in the satellite? Motivate your answer.
Answers
4a
Vorbit = 5625 m/s.
4b
adrag = 1.16 x 10-13 m/s2
4c
Total velocity change in one year due to air drag: 3.66 x 10-6 m/s
4d
The absolute total error in distance traveled in one year due to air drag: 57.8 m;
relative position error 3.26 x 10-8 %
4f
Mp 0.0000188 kg
Problem 5 (35 minutes)
Impact of orbit altitude on receiver power
This problem compares the receiver power required for a mobile telephone for a system
using satellites in Low Earth Orbit (LEO; orbital altitude hLEO 500 km) or geo-stationary
satellites (GEO; hGEO orbital altitude 35850 km). The Earth radius RE is 6370 km.
5a
Assuming the energy of a radio signal for a given antenna and a given transmitter
power expands in a cone, what is the ratio of the energy (power) loss PLEO/PGEO of the
communication link with the LEO satellite and of the GEO satellite?
5b
Assuming the receiver of your hand-held telephone needs 0.2 W receiver power for
the system with LEO satellites, which power Prec(GEO) would it need for that same
system of satellites on geostationary altitude? Is this a reasonable and feasible
value?
5c
What is the minimum required angle of the satellite antenna beam α when it should
cover the whole visible Earth? Calculate this for the LEO case and the GEO case.
Draw a picture.
5d
The antenna beam needed in case of the GEO satellite is narrower than that needed
for the LEO case. What is the ratio of the area of the LEO beam and the GEO beam?
5e
Assuming the beam width of the GEO satellite is adapted to the minimum desired
dimension, what becomes now the receiver power Prec(GEO) required for your handheld phone for the system with GEO satellites (combine the answer of 5b and 5d)?
Answers
5a
The of the power loss of the communication link at GEO and at LEO altitude is
0.000195
5b
Prec(GEO) = 1028 W
5c
alphaLEO = 68.01o, alphaGEO = 8.68o
5d
The strength of the signal from the smaller beam is 61.42 times larger than that of the
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Space Engineering & Technology I
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5e
wider beam.
Prec(GEO) = 16.74 W
EXAM JUNE 2004
Problem 1 (45 minutes)
Solar sailing to the boundary of the Solar System
You are requested to estimate major system sizing and mission parameters for a mission to
the boundaries of our solar system. The propulsive method that will be used for the mission
is solar sailing, that is using the pressure exerted by the Sun’s radiation. The force Fs exerted
by the radiation can be expressed as:
Fs =
S
⋅ A ⋅ (2 − α ) ⋅ cos i
c
where A is the surface area of the solar sail, S is the solar flux in W/m2 (equals 1400 W/m2 at
the Earth), α is the solar absorptance of the solar sail (α = unity minus the reflection), i is the
angle between the sun vector and the surface normal and c is the speed of light (equals 3 x
108 m/s).
During the trip the spacecraft will not always have “the Sun in the back”, so you may assume
that the propulsive force is only active during 50% of the travelling time. The remainder of the
time the sail will be turned in such a way that no force will be exerted on the spacecraft. You
may also assume that during the propulsive phase of the trip the angle i equals 0 degrees.
1a
For which value of the solar absorptance α is the propulsive force maximum? What
kind of material has such properties?
You are now going to estimate the time it takes to travel towards the edge of the solar
system. To this purpose you are going to estimate the propulsive force at several locations
along the road. The solar absorptance of the solar sail equals 0.1. Remember that the solar
flux S is decreasing with the square of the distance to the Sun. The intermediate points in the
trajectory are:
Location
Distance to the Sun (km)
Earth (starting point)
149.6 x 106
Mars
227.9 x 106
Jupiter
778.3 x 106
Uranus
2870 x 106
Pluto (end of trip)
5900 x 106
1b
Calculate for each of these locations the magnitude of the solar flux in W/m2.
Your spacecraft has a mass of 250 kg and the solar sail has an area of 100 m2.
1c
Calculate for each of the intermediate points the solar radiation force Fs and the
acceleration of the spacecraft aS/C. Do not take into account the 50% “efficiency” of
the propulsive phase.
You may assume that the average acceleration during each stage of the trip is the average
of the acceleration at the beginning of stage and at the end of the stage (e.g. the acceleration
during the stage Earth - Mars is 0.5 x (acceleration at the Earth + acceleration at Mars).
1d
Calculate for each of the four stages the average acceleration aav.
1e
Calculate the length (in meters) and duration (in seconds) of the stage Earth - Mars,
assuming the propulsive force is acting continuously giving the average acceleration
as you have calculated in 1d (so not taking into account, that it is available only 50%
of the time) and that you are traveling in a straight line along the planets. Calculate
also the velocity your spacecraft has at the end of the stage.
1f
Estimate the time it takes to travel from Mars to Jupiter, from Jupiter to Uranus and
from Uranus to Pluto, assuming no acceleration by the solar sail (so traveling with the
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Space Engineering & Technology I
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In-class problems, example problems, test and exam problems
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velocity as you have calculated in 1e, which you set equal to 625 m/s from now on,
and you travel in a straight line). Calculate also what the maximum velocity increment
(delta V) due to the solar sail would have been during these stages, using the
average accelerations you have calculated in 1d. How does this compare with the
velocity you had at Mars? Do you think it is worthwhile to use solar sailing in these
later stages?
1g
What is the total mission duration to travel to Mars including the 50% availability of
the propulsive power?
1h
Do you think that the assumption of 50% availability of propulsive power and the
accuracy of the result of the calculations in 1f and 1g is acceptable, that is that the
order of magnitude of your estimate is correct? Motivate your answer.
The maximum time for the mission to Mars should be comparable to the one achieved with
chemical propulsion: two years.
1i
How much solar sail area is needed to achieve this mission duration? Try to answer
this by using the same assumptions as above and by observing the formulae you
have used to calculate the traveling time in question 1d and 1e.
Answers
2b
Location
S (W/m2)
Earth
1400
Mars
603
Jupiter
52
Uranus
4
Pluto
0.9
2c
Location
Fs (N)
aS/C (m/s2)
Earth
8.87E-4
3.55E-6
Mars
3.82E-4
1.53E-6
Jupiter
3.28E-5
1.31E-7
Uranus
2.41E-6
9.64E-9
Pluto
5.70E-7
2.28E-9
2d
Stage
Average acceleration aav (m/s2)
Earth-Mars
2.54E-6
Mars-Jupiter
8.30E-7
Jupiter-Uranus
7.03E-8
Uranus-Pluto
5.96E-9
2e
Stage
Distance (m)
Travel time (s)
aav (m/s2)
VMars (m/s)
Earth-Mars
7.83E10
2.48E8
2.54E-6
630
2f
Stage
Distance (m)
Travel time (s)
aav (m/s2)
Delta V (m/s)
Mars-Jupiter
5.50E11
8.66E8
8.30E-7
718
Jupiter-Uranus
2.09E12
3.29E9
7.03E-8
231
Uranus-Pluto
3.03E12
4.77E9
5.96E-9
29
2g
Travel time to Mars 15.7 years.
2i
Solar sail area 6200 m2.
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Space Engineering & Technology I
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Issue 4.1, September 2005
EXAM AUGUST 2004
Problem 1 (15 minutes)
Space Market Segmentation
Several distinct segmentations of the Space Market are used. One of them is a segmentation
according to application (where the product or service is for).
1a
List the market segments according to this segmentation.
Another segmentation is the one according to the type of customer.
1b
Give the two types of customer, that are generally distinguished.
1c
Which are the major drivers for each of these types of customer in term of
performance and cost?
In a third segmentation the term “Infrastructure” is used.
1d
Describe what is contained in that segment.
1e
List and describe two of the other three segments that are used in this segmentation.
Problem 2 (30 minutes)
International Space Station orbit maintenance
The International Space Station (ISS) is traveling in a circular, low Earth orbit at an altitude
horbit of 407 km. It has an orbital period torbit of 93 minutes.
2a
Calculate the orbital velocity Vorbit of the ISS in m/s, assuming that the radius of the
Earth REarth equals 6378 km.
The dimensions of the ISS are 80 x 109 x 44 m and the density of the atmosphere ρa at 407
km altitude varies between 7.3E-13 (solar minimum) and 7.5E-12 kg/m3 (solar maximum).
You may assume that the effective area of the ISS subjected to aerodynamic drag is 10% of
the total “frontal” area and that the drag coefficient CD equals 2.0.
2b
Calculate the maximum and minimum area of the ISS, assuming the velocity vector is
always normal to one of the sides of the Space Station “box” and taking into account
the 10% effective area.
2c
Write down the formula for the aerodynamic drag FD of an object in space. Is this
expression the same as the one used for aircraft? If so, why do you think that is the
case?
2d
Calculate the minimum and the maximum value for the atmospheric drag force the
ISS is experiencing.
The mass of the International Space Station MISS is 455 tonnes.
2e
Calculate the maximum and minimum value of the deceleration of the ISS.
2f
What is the minimum velocity loss (negative delta V) of the ISS in one orbit? And how
large is the maximum velocity loss of the ISS in three months?
2g
You are going to correct the orbital altitude for this maximum velocity loss of 66 m/s.
Assuming you have propellant on board with an effective rocket exhaust velocity we of
3000 m/s, calculate with Tsiolkowsky the mass ratio of the ISS (that is the ratio
between mass before orbit correction M0 and the mass after completion of the orbit
correction Me).
2h
If M0 equals to 455 tonnes, how much propellant Mp is needed for this orbit
correction? If a Progress vehicle can lift 2500 kg propellant to the ISS, what is the
maximum frequency of Progress flights during a solar maximum?
Answers
2a
Vorbit = 7640 m/s
2b
Maximum effective area of ISS 872 m2; minimum 352 m2
2d
Minimum: FD = 0.015 N; maximum 0.382 N
2e
Minimum deceleration: 3.3E-8 m/s2; maximum 8.4E-6 m/s2
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Space Engineering & Technology I
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2f
2g
2h
Minimum velocity loss is per orbit 1.84E-4 m/s; maximum velocity loss in three
months: 66 m/s
M0/Me = 1.022
4 Progress flights per 3 months
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