CONTENTS 2
Contents
1 Introduction to Space Exploration 6
1.1 Recommended literature and useful resources: . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.2 Background and motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.3 Time and Energy considerations and space exploration . . . . . . . . . . . . . . . . . . . . . 8
1.4 Solar system probes: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.5 History of rocket propulsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.5.1 First man-made satellite . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.6 Voyager mission . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
1.7 Solar Probe Plus mission . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
1.7.1 Benefits of space exploration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2 Rocket (Tsiolkovsky) equation 18
2.1 Rocket thrust, exhaust velocity and rocket equation . . . . . . . . . . . . . . . . . . . . . . . . 18
2.2 Rocket thrust . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.3 Example of rocket equation application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
3 Multi-staging, vertical motion in Earth gravitational field 28
3.1 Principle of multi-staging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
3.2 Vertical motion in Earth gravitational field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
3.3 Rocket launching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
3.4 Vertical range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
Exploring Planetary Systems II
CONTENTS 3
4 Rocket Launch aspects 38
4.1 Efficiency of delivering energy to payload . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
4.2 Rocket Launch with pitch angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
4.3 Flight path angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
4.4 Angle of attack and aerodynamic forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
4.5 Dynamic pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
5 Evolution of the flight speed and path angle 48
5.1 Gravity turn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
5.2 Path angle evolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
5.3 Uniformly changing path angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
5.4 Orbital injection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
5.5 Actual launch trajectory: Ariane 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
5.6 Actual launch trajectory: Pegasus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
6 Orbits and spaceflight 60
6.1 Elliptical transfer orbit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
6.2 Energetic considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
6.3 Orbital transfer fuel requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
6.4 Orbital transfer to a moving target . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
7 Gravity assist or slingshot 74
7.1 Gravity acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
7.2 Gravity deceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
7.3 Example: Pioneer 10 encounter with Jupiter . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
Exploring Planetary Systems II
CONTENTS 4
7.4 Maximum boost . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
7.5 Oberth effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
7.6 Lagrange point parking . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
7.7 L-points and astrophysical missions: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
8 Thermal rocket engines 91
8.1 Thrust and the effect of the atmosphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
8.2 Optimising the exhaust nozzle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
8.3 Exhaust velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
8.4 Optimising exhaust velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
8.5 Mass flow rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
8.6 Vulcaine engine of Ariane 5 launcher . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
8.7 Saturn V Rocket engine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
8.8 Liquid fuel characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
9 Electric propulsion 107
9.1 Basics of electric propulsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
9.2 Vehicle velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
9.3 Electric thrusters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
9.4 Electrothermal thrusters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
9.5 Ionising thrusters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
10 Electric thrust 118
10.1 Space charge limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
10.2 Electric field and potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
Exploring Planetary Systems II
CONTENTS 5
10.3 Choice of the propellant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
10.4 How efficient are ion thrusters? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
11 Solar sails and space environment 131
11.1 Solar sails . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
11.2 Radiation pressure vs gravitational force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
11.3 IKAROS solar sail . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
11.4 Solar wind pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
11.5 Space environment and hazards . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
11.6 Radiation in space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
11.7 Solar energetic particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
11.8 Plasma environment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
11.9 Space debris, micro-meteroids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
Exploring Planetary Systems II
1 INTRODUCTION TO SPACE EXPLORATION 6
1 Introduction to Space Exploration
1.1 Recommended literature and useful resources:
These lecture notes are based on the material from the following books:
Martin Turner, Rocket and spacecraft propulsion: Principles, Practice
and New Developments, Springer Praxis Books, 2006, e.g. Amazon
George P. Sutton and Oscar Biblarz, Rocket Propulsion Elements,
Wiley, 2001 see, e.g. 8th edition from Wiley, 2010
Exploring Planetary Systems II
1 INTRODUCTION TO SPACE EXPLORATION 7
1.2 Background and motivation
Since much of the current solar system knowledge is from exploration
rather than observation,
EPS1 course concentrates on planetary systems, i.e. observations or
‘what’ is observed. The observations are, of course, is based on remote
sensing.
EPS2 course is devoted to exploration, i.e. spaceflight technique or
‘how’ of exploration
Why most of the extra-solar system science is ‘observational’? - Be-
cause of the energy budget and timescales.
Exploring Planetary Systems II
1 INTRODUCTION TO SPACE EXPLORATION 8
1.3 Time and Energy considerations and space exploration
Time constraint: Suppose we have 1-tonne (10
3
kg) probe sent c/3
to a star 15 lightyears away (for simplicity we also ignore relativity)
Travel time is 45 years
Communication time is 15 years
In total we need 60 years to retrieve information from the probe, i.e. a
long time !
Energy budget: Kinetic energy of the spacecraft is mv
2
/2,
i.e. ∼0.5 ×10
3
×10
16
=0.5 ×10
19
J.
In 2008, total worldwide energy consumption was 470 exajoules ∼
5 ×10
20
J, (or the UK energy production in 2012
1
was 5 exajoules ∼
5 ×10
18
J)
1
see
Exploring Planetary Systems II
1 INTRODUCTION TO SPACE EXPLORATION 9
Taking into account efficiency and man-
ufacturing costs increases the energy
requirement, indeed just getting the probe
we need ∼ 1% of the entire planet energy
production (which is approximately the
entire UK energy production for 1 year)!
Hence interstellar probes are not a likely
prospects at the moment.
Exploring Planetary Systems II
1 INTRODUCTION TO SPACE EXPLORATION 10
1.4 Solar system probes:
Contrast this with solar system probes:
Time to travel 1 AU is about 0.5 year.
We need probe speed 10 km/s (Earth escape speed)
Hence energy requirements 0.5 ×10
3
10
8
J, e.g. 10
−7
of interstellar
probe - feasible.
Exploring Planetary Systems II
1 INTRODUCTION TO SPACE EXPLORATION 11
1.5 History of rocket propulsion
Konstantin Tsiolkovsky
(1857-1935)
Rocket propulsion has surprisingly
long history. Theory was done about
140 years ago:
1883 - space travel concept (escape
velocity and weightlessness)
1895 - artificial satellites
1903 - rocket equation
Initially all works were theoretical...
Exploring Planetary Systems II
1 INTRODUCTION TO SPACE EXPLORATION 12
Herman Oberth (1894-
1992)
His work on a static firing of his first
liquid-fueled rocket motor. His stu-
dent Wernher von Braun will be later
a leader in both German and Ameri-
can rocket engineering....
Exploring Planetary Systems II
1 INTRODUCTION TO SPACE EXPLORATION 13
Robert Goddard (1882-
1945)
Practical verification of rocket princi-
ples and novel design of engines. He
is credited with creating and building
the first liquid-fueled rocket.
Exploring Planetary Systems II
1 INTRODUCTION TO SPACE EXPLORATION 14
1.5.1 First man-made satellite
Animation
from youtube:
Launch of
Sputnik 1 -
October 4,
1957
Exploring Planetary Systems II
1 INTRODUCTION TO SPACE EXPLORATION 17
1.7.1 Benefits of space exploration
Discovery lifts off. Credit: NASA
The benefits can be cate-
gorized into three funda-
mental areas:
• innovation;
• culture and inspira-
tion;
• new means to ad-
dress global chal-
lenges.
Benefits from Space Ex-
ploration /NASA website/
Exploring Planetary Systems II
2 ROCKET (TSIOLKOVSKY) EQUATION 18
2 Rocket (Tsiolkovsky) equation
2.1 Rocket thrust, exhaust velocity and rocket equation
Figure 1: Exhaust
speed and thrust
To change momentum, the rocket must eject
mass. Rocket equation (Tsiolkovsky Equation)
says that rocket acceleration produced by ex-
pelling of propellant and decreasing the mass of
the rocket.
Suppose we have rocket of mass M expelling
expelling part of its mass at rate
˙
M with constant
effective exhaust speed v
E
.
Interestingly v
E
=const is actually an excellent
approximation for most boost systems.
Exploring Planetary Systems II
2 ROCKET (TSIOLKOVSKY) EQUATION 19
Rocket thrust, F, is the mass flow rate,
2
˙
M, multiplied by exhaust ve-
locity, hence
F =
˙
Mv
E
Acceleration force is based on Newton III law: acceleration and reaction
are equal and opposite. Hence acceleration is
dv
dt
=
F
M
=−v
E
˙
M
M
(2.1)
where v(t) is the speed of the rocket. Note the sign change in equation
2.1 - speed goes up as M decreases. From Equation (2.1) and since v
E
is constant
dv
dt
=−v
E
˙
M
M
=−
d
dt
[
v
E
ln(M)
]
and after integration with initial speed v(t = 0) = 0, one finds simple and
2
The mass flow rate is the negative of dM/dt
Exploring Planetary Systems II
2 ROCKET (TSIOLKOVSKY) EQUATION 20
revealing formula Tsiolkovsky used:
v(t) = v
E
ln
µ
M
0
M(t)
¶
, (2.2)
where M
0
is the initial mass (rocket and fuel).
The ratio M
0
/M = R is mass ratio or sometimes called rocket param-
eter.
Equation 2.2 says that the ratio of initial to current mass determines
current velocity. There is no history of how the mass loss is achieved is
required (e.g. independent on
˙
M).
Assumptions: v
E
= const (which is reasonable for chemical rockets)
and no other forces acting on the rocket (no gravity or atmospheric drag).
Hence mass ratio can be exploited in order to achieve orbital or escape
speeds - multi-stage rockets.
Exploring Planetary Systems II
2 ROCKET (TSIOLKOVSKY) EQUATION 21
Note: Although the final speed achieved is independent of
˙
M, the
distance that the rocket has to travel in order to reach this speed is not:
Indeed, if expelling all propellent instantaneously (shell from a gun), the
same final velocity is reached but the travelled distance is almost zero.
Exploring Planetary Systems II
2 ROCKET (TSIOLKOVSKY) EQUATION 22
2.2 Rocket thrust
Let us make another simplifying assumption that the thrust is constant
Thrust = const, i.e.
˙
M =const.
hence M(t) = M
0
−
˙
Mt and time t becomes
t =
M
0
−M(t)
˙
M
=
M
0
˙
M
µ
1 −
M
M
0
¶
(2.3)
So distance s(t) travelled by rocket is
3
3
To take the integral
R
ln[1/(1 −ax)]dx, we note that ln[1/(1 −ax)] = −ln(1 −ax), then integrating by
parts,
−
Z
ln(1 −ax)dx =−x ln(1 −ax) −a
Z
xdx
(1 −ax)
,
and using that −
R
ax
1−ax
dx =
R
−ax+1−1
1−ax
dx = x +
1
a
ln(1 −ax), we find
Z
ln
·
1
1 −ax
¸
dx =
1
a
(1 −ax) ln(1 −ax) +x.
Exploring Planetary Systems II
2 ROCKET (TSIOLKOVSKY) EQUATION 23
s(t) =
Z
t
0
v(t
0
)dt
0
=v
E
Z
t
0
ln
µ
M
0
M
0
−
˙
Mt
0
¶
dt
0
=
s(t) = v
E
M
0
˙
M
·µ
1 −
˙
Mt
M
0
¶
ln
µ
1 −
˙
Mt
M
0
¶
+
˙
M
M
0
t
¸
,
but since M(t) = M = M
0
−
˙
Mt, 1 −
˙
Mt/M
0
= M(t)/M, hence
s(t) = v
E
M
0
˙
M
·
M
M
0
ln
µ
M
M
0
¶
+1 −
M
M
0
¸
=v
E
M
0
˙
M
·
1 −
M
M
0
µ
ln
µ
M
0
M
¶
+1
¶¸
(2.4)
From equation (2.4) we see:
s(t) depends on exhaust velocity v
E
and inversely proportional to mass
rate
˙
M. Hence higher mass rate ⇒ larger thrust ⇒ shorter range to ac-
celerate.
Exploring Planetary Systems II
2 ROCKET (TSIOLKOVSKY) EQUATION 24
Note that this applies only to the range under power: after burn-out
the range will continue to increase with constant velocity (assuming no
other forces).
If spacecraft already has some initial speed v
i
(say from previous
rocket stage) then we need to add v
i
t = v
i
(1 − M/M
0
)M
0
/
˙
M to s(t) in
equation (2.4).
To sum it up, we have a set of equations for speed, time and distance
in the absence of gravity:
speed: v(t) = v
E
ln
µ
M
0
M
¶
,
time: t =
M
0
˙
M
µ
1 −
M
M
0
¶
,
distance: s(t) = v
E
M
0
˙
M
·
1 −
M
M
0
µ
ln
µ
M
0
M
¶
+1
¶¸
+v
i
M
0
˙
M
µ
1 −
M
M
0
¶
Exploring Planetary Systems II
2 ROCKET (TSIOLKOVSKY) EQUATION 25
where M = M(t) is the current mass.
Note, that in engineering literature, the exhaust velocity is often quoted
in terms of the specific impulse, so that
I
sp
=
v
E
g
(2.5)
where g is the acceleration of gravity. The units of specific impulse are
seconds.
Specific impulse is a measure of the propellant efficiency and the
rocket thrust can be written F = I
sp
˙
M g.
Exploring Planetary Systems II
2 ROCKET (TSIOLKOVSKY) EQUATION 26
2.3 Example of rocket equation application
Consider a third stage rocket burn, with initial horizontal speed of
2 km/s containing 4 tones of propellent and fuel empty mass of 700 kg
(typical of launch of small satellite into 500 km orbit). Solid propellant
produces vacuum exhaust speed of 2.9 km/s at mass flow rate 100 kg/s.
What is the velocity increment, burn time and distance travelled over the
burn time?
Mass ratio is M
0
/M =(4.0 +0.7)/0.7 '6.7,
Velocity increment is ∆v =2.9ln(6.7) '5.5 km/s
Burn time is ∆t =4.7/0.1(1 −0.7/4.7) '40 seconds.
Range travelled for zero initial speed 2.9×(4.7/0.1)×[1−1./6.7(1+ln(6.7))] '
77 km
and taking into account 2. ∗4.7/0.1 ∗(1 −0.7/4.7) =80 km/s, the total dis-
tance travelled during the third stage is 77 +80 =157 km.
Exploring Planetary Systems II
2 ROCKET (TSIOLKOVSKY) EQUATION 27
Note that relatively short distance, over which curvature of orbit is neg-
ligible. If acceleration was slower, would need to guide the trust in order
to keep it ⊥ to
−→
g to ensure a circular orbit.
Exploring Planetary Systems II
3 MULTI-STAGING, VERTICAL MOTION IN EARTH GRAVITATIONAL FIELD 28
3 Multi-staging, vertical motion in Earth gravitational field
3.1 Principle of multi-staging
Figure 2: Mutistaging (Turner, 2006)
Recall that
v(t) = v
E
ln
µ
M
0
M
¶
+v
i
here M
0
is the initial rocket
mass (structural, fuel and pay-
load) and M is the remain-
ing mass after the total burn.
Breaking the rocket into stages
allows the structural mass to
be reduced at each stage, per-
mitting a better overall velocity
boost (See Fig 2).
Exploring Planetary Systems II
3 MULTI-STAGING, VERTICAL MOTION IN EARTH GRAVITATIONAL FIELD 29
Single stage rocket:
v(t) = v
E
ln
(
R
0
)
,
where
R
0
=
M
s
+M
f
+M
p
M
s
+M
p
,
where M
s
, M
f
, M
p
is structural, fuel and payload masses.
Symmetric two stage rocket:
v(t) = v
E
ln
(
R
1
)
+v
E
ln
(
R
2
)
,
where
R
1
=
M
s
+M
f
+M
p
M
s
+M
f
/2 +M
p
, R
2
=
M
s
/2 +M
f
/2 +M
p
M
s
/2 +M
p
.
The total speed after two stages is larger than in case of a single stage.
Exploring Planetary Systems II
3 MULTI-STAGING, VERTICAL MOTION IN EARTH GRAVITATIONAL FIELD 30
Figure 3: NASA Saturn V
Although in principle, single stage is sufficient
to reach low orbit, rocket launchers use 2 or 3
stages. The larger number of stages generally
improves final v (For example, Saturn V used
3-5 stages, see Fig 3). However, modern rock-
ets seldom use more than three stages because
of increase of complexity in the system (higher
cost and additional risk into the success).
Exploring Planetary Systems II
3 MULTI-STAGING, VERTICAL MOTION IN EARTH GRAVITATIONAL FIELD 31
3.2 Vertical motion in Earth gravitational field
Let us assume that we have thrust velocity and
gravitational field acting and all vectors are along one
line.
Now the force equation is (recall that the thrust is
F = v
E
˙
M and
˙
M is the mass loss rate)
dv
dt
=
F −M g
M
,
where M g is the instantaneous weight of the rocket.
Hence
dv
dt
=−v
E
˙
M
M
− g ,
or
dv =−v
E
dM
M
− gdt,
Exploring Planetary Systems II
3 MULTI-STAGING, VERTICAL MOTION IN EARTH GRAVITATIONAL FIELD 32
Integrating both parts from t =0 to t
v(t) =−v
E
Z
M
M
0
dM
0
M
0
−
Z
t
0
gdt
0
Further assume that over this part of the flight, g is constant, then
v(t) =v
E
ln
M
0
M(t)
− gt
= v
E
ln
M
0
M(t)
| {z }
as before Eq (2.2)
−g
M
0
˙
M
µ
1 −
M
M
0
¶
| {z }
gravity loss
(3.1)
The new term in Equation (3.1) is referred as ‘gravity loss’, the first term
is the ‘initial velocity’.
A very short acceleration, with high thrust at high mass flow rate min-
imises gravity loss term; Inversely, slow acceleration with low mass flow
Exploring Planetary Systems II
3 MULTI-STAGING, VERTICAL MOTION IN EARTH GRAVITATIONAL FIELD 33
rate and low thrust lead to high gravity loss. This is probably not surpris-
ing: if the mass flow rate is high, less of the propellant has to be carried
to high altitude and vice-versa.
Exploring Planetary Systems II
3 MULTI-STAGING, VERTICAL MOTION IN EARTH GRAVITATIONAL FIELD 34
3.3 Rocket launching
As we saw, if the mass flow rate is high, less of
the propellant has to be carried to high altitude
and vice-versa. This is a real issue with rocket
launching: how to deal with the propellant un-
der gravity - exhausting most of the propellant
early in launch is beneficial; the work done in
transporting unspent propellant to high altitude
is much greater than the work done on payload.
Discovery lifts off.
Example: On Arian 5 (or space shuttle), the single stage burn is ∼
2 min, so the gravity loss is just g ×120 '1.2 km/s, which is not small.
Recall the mass ratio in velocity equation.
v(t) = v
E
ln
M
0
M(t)
− g
M
0
˙
M
µ
1 −
M
M
0
¶
Exploring Planetary Systems II
3 MULTI-STAGING, VERTICAL MOTION IN EARTH GRAVITATIONAL FIELD 35
which can introduce ‘thrust to weight ratio’
ψ =
F
gM
0
=
v
E
˙
M
gM
0
so that we can write
v(t)
v
E
=ln
M
0
M(t)
−
1
ψ
µ
1 −
M
M
0
¶
Typically thrust to weight ratio ψ '3. However, the Equation says that we
can reduce gravity loss by increasing ψ.
So why not use a very large ψ to get the largest possible v?
The answer is that a rocket carries not only fuel but has payload and
has structural mass. The propellant mass fractions are typically around
0.8-0.9 and the final mass ratio R = M(t)/M
0
will be 0.2-0.1. The value
v/v
E
is set by the mission requirement, and typically around 1-1.5.
Exploring Planetary Systems II
3 MULTI-STAGING, VERTICAL MOTION IN EARTH GRAVITATIONAL FIELD 36
For example, for v/v
E
=1.3, from
ψ =
1 −R
−1.3 −ln R
we calculate that for R =0.2, thrust to weight ratio ψ '2.6.
Exploring Planetary Systems II
3 MULTI-STAGING, VERTICAL MOTION IN EARTH GRAVITATIONAL FIELD 37
3.4 Vertical range
To calculate the range under vertical launch from surface, we proceed
as before integrating speed given by Equation (3.1) over time
s =
Z
v(t)dt = v
E
Z
t
0
ln
M
0
M(t)
dt −
Z
t
0
gtdt
0
Assuming g is constant, one finds
s(t) =v
E
M
0
˙
M
·
1 −
M
M
0
µ
ln
M
0
M(t)
+1
¶¸
−
1
2
gt
2
=v
E
M
0
˙
M
·
1 −
M
M
0
µ
ln
M
0
M(t)
+1
¶¸
−
g
2
M
2
0
˙
M
2
µ
1 −
M
M
0
¶
2
(3.2)
the range, e.g. distance travelled under acceleration.
Exploring Planetary Systems II
4 ROCKET LAUNCH ASPECTS 38
4 Rocket Launch aspects
4.1 Efficiency of delivering energy to payload
Let us consider the efficiency of delivering energy to payload via a
rocket. For simplicity let us neglect gravity.
The payload (and structural mass) M(t) ends up with kinetic energy
1
2
M∆v
2
=
1
2
Mv
2
E
ln
2
µ
M
0
M
¶
=
1
2
Mv
2
E
ln
2
R
where R = M
0
/M is the rocket parameter.
If we normalise this to the kinetic energy of the propellant in rest frame
of the rocket
1
2
(
M
0
−M
)
v
2
E
Exploring Planetary Systems II
4 ROCKET LAUNCH ASPECTS 40
The optimal R for the energy transfer can be found differentiating ϕ(R)
with respect to R
dϕ
dR
=0 =−
ln
2
R
(R −1)
2
+
2ln R
R(R −1)
hence the optimal R
c
satisfy the equation
ln R
c
=2 −2/R
c
i.e. R
c
' 4.9. Adding gravity and drag means that R > R
c
in practical
terms.
Exploring Planetary Systems II
4 ROCKET LAUNCH ASPECTS 41
4.2 Rocket Launch with pitch angle
Figure 5: Pitch angle and
thrust
The main requirement for a rocket is to place
payload in orbit or escape, e.g. gain horizontal
speed, not vertical. The vertical launch does not
contribute usefully to this, so we need to con-
sider an inclined launch for realistic dynamics.
The simplest case is one where the pitch an-
gle is constant. We need two components of
motion
vertical
dv
z
dt
=
F sin θ −M g
M
horizontal
dv
x
dt
=
F cos θ
M
(4.1)
where θ is the pitch angle, i.e. the angle of the
trust vector measured from the horizon to the direction of the thrust (see
Figure 5).
Exploring Planetary Systems II
4 ROCKET LAUNCH ASPECTS 42
The equations (4.1) can be integrated to obtain velocity components
vertical v
z
(t) =v
E
sinθ ln
M
0
M
− gt
horizontal v
x
(t) =v
E
cosθ ln
M
0
M
(4.2)
where we assumed zero velocity at t =0.
The total speed of the rocket is
v(t) =
q
v
2
z
+v
2
x
=
"
v
2
E
µ
ln
M
0
M
¶
2
−2v
E
gtsin θ ln
M
0
M
+ g
2
t
2
#
1/2
(4.3)
In case θ =π/2, we have a vertical launch result, while θ =0 does not
recover an ideal (without gravity) result. why not?
The answer is that as the rocket launched horizontally, it also falls
under gravity accelerating as it does. Practical result needs polar coordi-
nates not rectangular.
Exploring Planetary Systems II
4 ROCKET LAUNCH ASPECTS 43
4.3 Flight path angle
Figure 6: tanγ = v
z
/v
x
We note that the pitch angle is the angle be-
tween the horizontal and the thrust and, as we
see from Equations 4.2, it is not the angle of the
flight of the vehicle.
The flight path angle γ is given by
tanγ =
v
z
v
x
=
v
E
sinθ ln
M
0
M
− gt
v
E
cosθ ln
M
0
M
i.e.
tanγ =tanθ −
gt
v
E
cosθ ln
M
0
M
, (4.4)
where 2nd term is always finite and negative.
Hence we have that the flight path angle is always different from pitch
angle.
Exploring Planetary Systems II
4 ROCKET LAUNCH ASPECTS 44
The flight path angle γ varies throughout the flight, being at its greatest
offset from vertical axis immediately after vehicle axis departs from verti-
cal.
Indeed, for early times we can write
4
tanγ 'tanθ −
gM
0
v
E
˙
M cos θ
. (4.5)
4
Using Eq 4.4, derive the equation below for tanγ
Exploring Planetary Systems II
4 ROCKET LAUNCH ASPECTS 45
4.4 Angle of attack and aerodynamic forces
The angle between thrust vector and flight path is angle of attack see
Figure 7.
Figure 7: Angle of attack
Note that the angle of attack tends to zero as the flight burn pro-
gresses. Opposite of what is wanted: large angle of attack happens at
early flight stage, just where aerodynamic forces are dominant.
Exploring Planetary Systems II
4 ROCKET LAUNCH ASPECTS 46
4.5 Dynamic pressure
The aerodynamics of the vehicle flight is too complex to address in
details here, but the diagram (7) shows the main forces of interest: drag,
F
D
and lift, F
L
.
Both forces are growing with the speed of the rocket and can be written
F
D
=C
D
A
ρv
2
2
, F
L
=C
L
A
ρv
2
2
,
where
ρv
2
2
= dynamic pressure
A is the cross-sectional area, C
D
is the drag (lift) coefficients (dimension-
less numbers).
Without going into detail there is a critical point in the vehicle trajectory
at which aerodynamic forces are at a maximum, hence risk to structural
integrity (Figure 8). Since lift and drag forces ∝ v
2
, one method would
be to reduce the speed - not desirable. Instead the practical technique is
Exploring Planetary Systems II
4 ROCKET LAUNCH ASPECTS 47
to maintain the angle of attack at zero during the early part of the launch
(where the atmosphere is the densest). This reduces drag (by dropping
the effective area ) and sets lift to zero.
Figure 8: Dynamic pressure, velocity and altitude as a function of mass ratio.
Exploring Planetary Systems II
5 EVOLUTION OF THE FLIGHT SPEED AND PATH ANGLE 48
5 Evolution of the flight speed and path angle
At the last lecture we noted that the trajectory that results in a curved
path in witch the nose is gradually dropping (as opposed to an upwardly
curved path arising from constant pitch-angle flight) called gravity turn.
Exploring Planetary Systems II
5 EVOLUTION OF THE FLIGHT SPEED AND PATH ANGLE 49
5.1 Gravity turn
To model gravity turn, let us consider the equations for constant pitch
angle (Equations 4.1)
vertical
dv
z
dt
=
F sin θ −M g
M
horizontal
dv
x
dt
=
F cos θ
M
Now instead of continuing as before, lets replace θ with the flight path
angle γ
tanγ =
v
z
v
x
then we have
dv
z
dt
=
F
M
sinγ − g =
F
M
v
z
v
− g
dv
x
dt
=
F
M
cosγ =
F
M
v
x
v
Exploring Planetary Systems II
5 EVOLUTION OF THE FLIGHT SPEED AND PATH ANGLE 50
Multiplying equation for dv
z
/dt by v
z
and summing with dv
x
/dt equa-
tion multiplied by v
x
, we have
v
z
dv
z
dt
+v
x
dv
x
dt
=v
dv
dt
=
Fv
M
− gv
z
so that the equation for speed v =
p
v
2
x
+v
2
z
becomes
dv
dt
=
F
M
− g
v
z
v
=
F
M
− g sinγ (5.1)
Similarly multiplying equation for dv
z
/dt by v
x
and subtracting dv
x
/dt
equation multiplied by v
z
, we have
v
x
dv
z
dt
−v
z
dv
x
dt
=−gv
x
since tanγ = v
z
/v
x
d
dt
tanγ =
˙
v
z
v
x
−
v
z
˙
v
x
v
2
x
=
v
x
˙
v
z
−v
z
˙
v
x
v
2
x
(5.2)
Exploring Planetary Systems II
5 EVOLUTION OF THE FLIGHT SPEED AND PATH ANGLE 51
Left hand side (LHS) of Equation (5.2) can be rewritten
d tanγ
dt
=
˙
γ(1 −tan
2
γ) =
˙
γ
Ã
1 +
v
2
z
v
2
x
!
=
˙
γ
v
2
v
2
x
combing together, one finds equation for γ
dγ
dt
=−
gv
x
v
2
=−
g
v
cosγ (5.3)
Equations (5.1, 5.3) give the evolution of the flight speed v and path an-
gle γ.
dv
dt
=
F
M
− g sinγ,
dγ
dt
=−
g
v
cosγ
Exploring Planetary Systems II
5 EVOLUTION OF THE FLIGHT SPEED AND PATH ANGLE 53
5.3 Uniformly changing path angle
The equations (5.1, 5.3) describing motion in terms of speed v and
path angle γ are non-linear, but have a special case
˙
γ = const = −C, i.e.
γ is changing uniformly in time . Then from (5.3), we find
C =
g
v
cosγ
and so
v =
g
C
cosγ
˙
v =−
g
C
sinγ
˙
γ = g sin γ
Let us insert this expression into the equation for speed (Eq 5.1):
˙
v =
F
M
−
˙
v =⇒ 2
˙
v =
F
M
So we have
˙
v =
F
2M
=
1
2
v
E
˙
M
M
Exploring Planetary Systems II
5 EVOLUTION OF THE FLIGHT SPEED AND PATH ANGLE 54
which can be solved
v(t) = v
0
+
1
2
v
E
ln
M
0
M
(5.4)
since cosγ = Cv/g, substituting Equation (5.4), we find solution for the
path angle γ
cosγ(t) =cos γ
0
+C
v
E
2g
ln
M
0
M
(5.5)
Solution (5.5) shows that γ decreases with time (see Figure 9)
Note that pitch angle θ is equal to the flight path angle here, so that
angle of attack is zero.
Exploring Planetary Systems II
5 EVOLUTION OF THE FLIGHT SPEED AND PATH ANGLE 55
Figure 9: Flight path angle and speed for gravity turn.
Pitch angle and flight path angle decrease with time as rocket speed
increases (Figure 9).
Vehicle gains speed rapidly, but altitude is not gained so quickly be-
cause the curvature of the flight path.
Exploring Planetary Systems II
5 EVOLUTION OF THE FLIGHT SPEED AND PATH ANGLE 56
This helps to manage the aerodynamic stresses on the vehicle, pre-
serving structural integrity.
Exploring Planetary Systems II
5 EVOLUTION OF THE FLIGHT SPEED AND PATH ANGLE 57
5.4 Orbital injection
As soon as the atmosphere is sufficiently tenuous, the trajectory is to
maximise the velocity and altitude (no danger of air forces.) The only limi-
tation is the acceleration - it cannot be too great for a payload/astronauts.
The trajectory could be with constant pitch angle at this stage.
One factor to consider is the perigee of the orbit (longitude of the in-
jection point). This might affect the required trajectory.
Once the orbital injection point reached, the final segment is horizontal
acceleration to orbital velocity. A variety of orbits can be reached depend-
ing on the velocity specified.
If the velocity is
p
(GM/R +h) then circular orbit reached.
The higher velocity will lead to elliptical orbits.
The velocity
p
(2GM/R +h) leads to parabolic orbit.
Exploring Planetary Systems II
5 EVOLUTION OF THE FLIGHT SPEED AND PATH ANGLE 58
5.5 Actual launch trajectory: Ariane 4
Figure 10: Ariane 4 dynamic parameters: 1st stage burns for 204 seconds, short (60
s) segment followed by rapid gravity turn; 2nd stage burns for 124s, the pitch angle de-
clines; 3rd stage designed to increase velocity to more than 10 km/s to place spacecraft
to Geostationary Transfer Orbit with apogee at 36000 km. Note slow acceleration of
1.7g. The objective to have sufficient velocity to enter GTO.
Exploring Planetary Systems II
5 EVOLUTION OF THE FLIGHT SPEED AND PATH ANGLE 59
5.6 Actual launch trajectory: Pegasus
Figure 11: Pegasus dynamic parameters: 1st stage from altitude 11.6 km and reaches
velocity 2.3 km/s only after 76 seconds; 2nd stage burns out at 166 seconds and the
rocket coasts (converting kinetic into potential energy, e.g. velocity drops) gaining al-
titude to insertion value of 740km; 3rd stage then increases velocity from 4.6km/s to
required 7.5km/s.
Exploring Planetary Systems II
6 ORBITS AND SPACEFLIGHT 60
6 Orbits and spaceflight
Figure 12: Gravitation orbit
Let us recall that the spacecraft follows
a gravitational orbit (see Figure 12) given
by
1
r
=
GM
h
2
(1 +²cos θ) (6.1)
where M = M
⊕
r, θ are the radial and an-
gular coordinates respectively, h = r
0
v
0
is
the angular momentum per unit mass of
rocket (specific angular momentum), ² is
the eccentricity of the orbit, given by
² =
r
0
v
2
0
GM
−1 (6.2)
depends on the ratio of the specific kinetic energy and specific gravita-
tional potential, e.g. per unit of spacecraft mass.
Exploring Planetary Systems II
6 ORBITS AND SPACEFLIGHT 61
When r = r
0
, v =v
0
we have ² =0
radius is independent of θ,
e.g. circular orbit, and
v
0
=
s
GM
r
0
² =1
1/r −→ 0 as θ −→ π, escape velocity
achieved
v
0
=
p
2
s
GM
r
0
which is just 1.414 times circular orbit speed
(low orbit speed)
parabolic orbit
² >1
hyperbolic orbit
For Earth M
⊕
= 6 ×10
24
kg, mean radius R
⊕
= 6.3 ×10
3
km, at h =
Exploring Planetary Systems II
6 ORBITS AND SPACEFLIGHT 62
500 km above Earth surface, we have
v
0
=
s
GM
⊕
R
⊕
+h
'7.6 km/s, for circular orbit
Note that if velocity gain to spacecraft is greater then minimum value,
so that ² is not zero, then cos θ 6= 0 and radius is function of position in
orbit, e.g. elliptical orbit.
Key points:
1. Given that escape velocity is ∼1.4 times low orbit
speed, once in orbit its relatively easy to escape
Earth
2. once in orbit, escape is achieved by initial boost
parallel to Earth surface, not perpendicular, so we
want large horizontal boost. The vertical veloc-
ity gains height, where atmosphere drag is min-
imised.
Exploring Planetary Systems II
6 ORBITS AND SPACEFLIGHT 63
6.1 Elliptical transfer orbit
Figure 13: Two circular orbits, and
transfer orbit.
Horizontal acceleration is used
to move from one orbit to an-
other.
Note that in circular orbit, speed is
∝ 1/
p
R, i.e. the lower the orbit, the
faster the orbital speed.
So the orbit transfer requires two in-
crements for transfer between two cir-
cular orbits (see Figure 13):
One at perigee (velocity of intercept-
ing elliptical orbit is > circular one at
perigee)
One at apogee (the new circular intersection needs bigger speed than
that of intercepting elliptical orbit)
Exploring Planetary Systems II
6 ORBITS AND SPACEFLIGHT 64
For elliptical transfer orbit, semi-major axis is
a =
1
2
(a
2
+a
1
)
where a
1
is the radius of lower circular orbit, and a
2
is the radius of the
higher circular orbit. The eccentricity of the transfer orbit is
² =
a
2
−a
1
a
2
+a
1
Let us introduce parameter α = a
2
/a
1
, then
² =
α −1
α +1
1 +² =
2α
α +1
1 −² =
2α
α +1
Exploring Planetary Systems II
6 ORBITS AND SPACEFLIGHT 65
For two circular orbits we have
v
1
=
s
GM
a
1
v
2
=
s
GM
a
2
First boost at perigee, e.g. r = a
1
and θ =0
o
in Equation (6.1) for elliptical
orbit:
∆v
1
=v
ell i ptical
1
−v
circular
1
=
s
GM
a
1
(1 +²)
1/2
−
s
GM
a
1
=
s
GM
a
1
£
(1 +²)
1/2
−1
¤
(6.3)
Second boost at apogee, e.g. r = a
2
and θ = 180
o
in Equation (6.1)
Exploring Planetary Systems II
6 ORBITS AND SPACEFLIGHT 66
for elliptical orbit:
∆v
2
=v
circular
2
−v
ell i ptical
2
=
s
GM
a
2
−
s
GM
a
2
(1 −²)
1/2
=
s
GM
a
2
£
1 −(1 −²)
1/2
¤
(6.4)
Hence the total boost is, expressed in terms of v
1
:
∆v
v
1
=
∆v
1
+∆v
2
v
1
=(1 +²)
1/2
−1 +
r
a
1
a
2
£
1 −(1 −²)
1/2
¤
=(1 +²)
1/2
−1 +
s
1 −²
1 +²
£
1 −(1 −²)
1/2
¤
=
s
2α
1 +α
−1 +
1
p
α
1 −
s
2
1 +α
Exploring Planetary Systems II
6 ORBITS AND SPACEFLIGHT 67
6.2 Energetic considerations
Figure 14: Velocity change and trans-
fer orbit.
We can write down the total energy
per spacecraft mass
C =
1
2
v
2
−
µ
r
where v
2
=µ(2/r −1/a), and µ =GM.
Consider change of C as space-
craft changes orbit from a
1
to a
2
. How
much energy is needed to effect this
change?
The energies of two orbits are
C
1
=
1
2
v
2
1
−
µ
a
1
=−
1
2
µ
a
1
C
2
=−
1
2
µ
a
2
Exploring Planetary Systems II
6 ORBITS AND SPACEFLIGHT 68
hence the change of energy required is
∆C =C
2
−C
1
=
µ
2
µ
1
a
1
−
1
a
2
¶
this is the minimum change is kinetic energy required to effect the transfer.
Let us now work out the details of the transfer.
Energy in transfer orbit C
T
=−
µ
2a
T
=−
µ
a
1
+a
2
So the energy increment at A to inject vehicle into elliptical orbit is
∆C
A
=C
T
−C
1
=−
µ
a
1
+a
2
+
µ
2a
1
=
µ
2a
1
a
2
−a
1
a
2
+a
1
Similarly at B:
∆C
B
=C
2
−C
T
=−
µ
2a
2
+
µ
a
1
+a
2
=
µ
2a
2
a
2
−a
1
a
2
+a
1
Exploring Planetary Systems II
6 ORBITS AND SPACEFLIGHT 69
These energy changes are kinetic (effected by firing engines), so we have
due to changes in speed at A and B
∆C
A
=
1
2
(v
A
+∆V
A
)
2
−
1
2
v
2
A
∆C
B
=
1
2
v
2
B
−
1
2
(v
B
−∆V
B
)
2
Equating expressions for C
A
:
µ
2a
1
a
2
−a
1
a
2
+a
1
=
1
2
(v
A
+∆V
A
)
2
−
1
2
v
2
A
one finds
∆v
A
=
·
v
2
A
+
µ
a
1
a
2
−a
1
a
2
+a
1
¸
1/2
−v
A
=v
A
"
1 +
µ
a
1
v
2
A
a
2
−a
1
a
2
+a
1
#
1/2
−1
=
r
µ
a
1
"
µ
2a
2
a
2
+a
1
¶
1/2
−1
#
| {z }
=(1 +²)
1/2
−1
(6.5)
Exploring Planetary Systems II
6 ORBITS AND SPACEFLIGHT 70
where we used that v
A
=
p
µ/a
1
. And similarly for ∆v
B
∆v
B
=
r
µ
a
2
"
1 −
µ
2a
1
a
2
+a
1
¶
1/2
#
| {z }
=1 −(1 −²)
1/2
(6.6)
using v
A
=
p
µ/a
1
.
The velocity boosts ∆v
A
and ∆v
B
derived in Equations (6.5,6.6) as-
suming the minimum energy change are equal to ∆v
1
, ∆v
2
given by Equa-
tions (6.3,6.4). Hence, the elliptical transfer orbit with transfers at perigee
and apogee is minimum energy transfer orbit or Hohmann transfer orbit.
Exploring Planetary Systems II
6 ORBITS AND SPACEFLIGHT 71
6.3 Orbital transfer fuel requirements
Since boosts are tangential to the orbit and in vacuum, there are no
gravity or drag terms, so we can use the simple expressions (Rocket
Equation 2.2) for speed boost in terms of mass:
boost at A: ∆v
A
=v
E
ln
µ
M
0
M
A
¶
boost at B: ∆v
B
=v
E
ln
µ
M
A
M
B
¶
where M
0
−M
A
is the fuel used at A, and M
A
−M
B
is the fuel used at B.
Hence the total boost needed (in terms of fuel mass) between two
orbits
∆v =∆v
A
+∆v
B
=v
E
ln
µ
M
0
M
A
¶
+v
E
ln
µ
M
A
M
B
¶
=v
E
ln
µ
M
0
M
B
¶
Exploring Planetary Systems II
6 ORBITS AND SPACEFLIGHT 72
6.4 Orbital transfer to a moving target
Figure 15: Earth-Mars transfer orbit
Consider an orbital trans-
fer between Earth and Mars.
If the orbital transfer start at
angle θ = 0, as shown in
Figure (15) and the space-
craft will reach Mars orbit at
θ = π and 1/2-period later,
then in that time, Mars will
move along its path. In
this half-period (using Ke-
pler law T
2
=4π
2
a
3
/GM)
τ
1/2
=
1
2
τ =
1
2
2π
·
a
3
GM
¸
1/2
Exploring Planetary Systems II
6 ORBITS AND SPACEFLIGHT 73
Mars will move an angle
2π
τ
1/2
T
M
=2ππ
·
a
3
GM
¸
1/2
1
2π
"
GM
a
3
2
#
1/2
=π
·
a
a
2
¸
3/2
so the maneuverer has to begin when the target (Mars) is at an angle
π −π
·
a
a
2
¸
3/2
=π −π
·
τ
T
2
¸
before the target reaches the angle θ =π.
Exploring Planetary Systems II
7 GRAVITY ASSIST OR SLINGSHOT 74
7 Gravity assist or slingshot
Figure 16: Cassini mission
To conserve fuel for particularly de-
manding missions, it is possible to ’steal’
some orbital energy from a planet and use
it to boost (or brake) the space vehicle.
Such spacecraft maneuver is called grav-
ity assist or slingshot maneuver.
As an example consider Cassini mis-
sion. The spacecraft launched to Saturn
(6.7 year mission).
Spacecraft mass 5700 kg was launched
with speed 4 km/s but need to get 10 km/s
to reach Saturn.
So Cassini acquired extra speed from 4 intermediate planetary en-
Exploring Planetary Systems II
7 GRAVITY ASSIST OR SLINGSHOT 75
counters: Venus (2), Earth and Jupiter.
In each case, Cassini boosted by slingshot effect, where a close fly of
a gravitating object supply boosts the velocity with no energy gain in the
planet’s rest frame, but a clear boost in the Sun’s frame.
Exploring Planetary Systems II
7 GRAVITY ASSIST OR SLINGSHOT 76
7.1 Gravity acceleration
We can consider this graphically (see Figure 17,18).
Figure 17: Gravity assist (acceleration), e.g. |
−→
v
i
|<|
−→
v
f
|.
Let
−→
U be the planet velocity,
−→
v
i
is asymptotic initial spacecraft velocity
(e.g. far from the planet so we can ignore planet’s gravity),
−→
v
f
is asymp-
totic final (departure) spacecraft velocity (e.g. far from the planet so we
Exploring Planetary Systems II
7 GRAVITY ASSIST OR SLINGSHOT 77
can ignore planet’s gravity)
In the planet’s rest frame:
−→
v
0
i
=
−→
v
i
−
−→
U ,
−→
v
0
f
=
−→
v
f
−
−→
U
Important point is that these two velocities are equal by absolute value
¯
¯
−→
v
0
i
¯
¯
=
−→
v
0
f
|
but directed differently.
From the figure (17) we see that after rotation the final speed v
f
is
larger than v
i
.
Exploring Planetary Systems II
7 GRAVITY ASSIST OR SLINGSHOT 78
7.2 Gravity deceleration
If the spacecraft passes in front of the planet (Figure 18), we have
braking maneuver. There is no energy gain in the planet’s frame, but
Figure 18: Gravity assist (braking), e.g. |
−→
v
i
|>|
−→
v
f
|
clear gain in Sun’s frame. The best analogy is the tennis ball: accelerated
by the racket, but there is no energy change in the frame of the racket.
Another important analogy is Fermi acceleration of particles.
Exploring Planetary Systems II
7 GRAVITY ASSIST OR SLINGSHOT 79
Note that the energy for the spacecraft comes from the planet, e.g.
the spacecraft borrows the energy of the planet. The change of planet’s
velocity is small due to tiny spacecraft/planet mass ratio.
Exploring Planetary Systems II
7 GRAVITY ASSIST OR SLINGSHOT 80
7.3 Example: Pioneer 10 encounter with Jupiter
Figure 19: The projection on the ecliptic plane of the heliocentric path of
Pioneer 10 before and after its close encounter with Jupiter on 4 Decem-
ber 1973.
Exploring Planetary Systems II
7 GRAVITY ASSIST OR SLINGSHOT 81
An important example of gravitational assist is the encounter of the
Pioneer 10 spacecraft with the plant Jupiter in 1973.
5
. As Pioneer 10
approached, it was about 33 Jovian radii so that the gravitational force of
the Sun was less than 1% of that of the planet.
Pre-encounter speed was v
i
= 9.8 km/s at an angle of 49
o
counter-
clockwise (as viewed from the north ecliptic pole) of the line from the Sun
to the planet. Jupiter’s orbital speed U =13.5 km/s.
The pre-encounter speed v
0
i
of the spacecraft in the planetocentric
coordinate system
v
0
i
=|
−→
v
i
−
−→
U|=
q
v
2
i
−2Uv
i
cos(β) +U
2
'8.9 km/s
where β is the angle between
−→
v
i
and
−→
U and we used that the angle
β =90
o
−49
o
=41
o
(see Figure 19).
5
For detailed discussion see James A. Van Allen, Am. J. Phys. 71, 448 (2003)
Exploring Planetary Systems II
7 GRAVITY ASSIST OR SLINGSHOT 82
As a result of the encounter, the post-encounter velocity
−→
v
f
was di-
rected at 83
o
counterclockwise of the Sun-planet line (see Figure 19), so
the magnitude v
f
can be found from
v
0
i
=v
0
f
=|
−→
v
f
−
−→
U|
so that v
f
'22 km/s.
Hence, using gravitational assist the heliocentric speed of the space-
craft had increased from 9.8 km/s to 22 km/s and its kinetic energy had
increased by a factor of ∼5 !
Pioneer 10 was at a radial distance of 2.84 planetary radii at the space-
craft’s closest approach to the planet (Fig 20).
Exploring Planetary Systems II
7 GRAVITY ASSIST OR SLINGSHOT 83
Figure 20: The ecliptic plane projection of the December 1973 hyperbolic encounter
trajectory of Pioneer 10 with Jupiter as viewed in the planetocentric coordinate system.
Exploring Planetary Systems II
7 GRAVITY ASSIST OR SLINGSHOT 84
7.4 Maximum boost
Maximum boost is when
−→
v
f
is aligned exactly with
−→
U. Then the gain
is v
f
−v
i
=2|
−→
U| for head-on collision with rotation through 180
o
(see Fig
21).
Figure 21: Left: Maximum loss v
f
−v
i
=−2U and maximum gain v
f
−v
i
=
+2U (right)
Head-on slingshot with large rotation is difficult to organise, since the
probe motion must be first retrograde and close encounter is needed, with
risk of surface impact.
Exploring Planetary Systems II
7 GRAVITY ASSIST OR SLINGSHOT 85
To increase the gain due to slingshot, the rocket engines are fired at
periapsis to provide additional boost to the spacecraft.
This is the application of the Oberth effect, where the energy gain
of the spacecraft is larger when the boost ∆v is applied when probe is
travelling at high speed (rather than at low speed).
Exploring Planetary Systems II
7 GRAVITY ASSIST OR SLINGSHOT 86
7.5 Oberth effect
Let us consider the rocket equation
M
dv
dt
=F ,
where F is the rocket thrust. The change of specific (per rocket mass)
kinetic energy e =v
2
/2 is
v
dv
dt
=
F
M
v, ⇒
de
dt
=
F
M
v,
hence we can see that the gain/loss of specific energy of is proportional
to rocket speed.
The Oberth effect says that the propellant has more usable energy
(when spacecraft moves with higher speed) due to its kinetic energy on
top of its chemical potential energy.
Exploring Planetary Systems II
7 GRAVITY ASSIST OR SLINGSHOT 87
7.6 Lagrange point parking
The Lagrange points are the five positions in an orbital configuration
where a small object affected only by gravity can be stationary relative to
two larger objects.
The Lagrange points are the positions where the combined gravita-
tional pull of the two large masses produce precisely the centripetal force
required to rotate with them.
L3,L4,L5 are slightly outside the m orbit, when M >> m.
For Sun-Earth system L4, L5 are about 60
o
ahead and behind Earth
as it orbits Sun (interplanetary dust can be found there). L4 and L5 are
stable and L1-3 are unstable.
Key points: only minimal amounts of orbital correction is needed
here, hence fuel reserves can be exploited very efficiently.
Exploring Planetary Systems II
7 GRAVITY ASSIST OR SLINGSHOT 89
ESA Herschel Space Observatory
(see ESA pages)
L2 spacecrafts: Herschel
Space Observatory, Plank
spacecraft, GAIA . Earth
shadow helps in astrophysi-
cal observations.
Sun-Earth L3 is unstable because of the other planet’s attractions.
L4, L5: Combined gravitational effects of M and m are such (given
equidistance to masses) that the resultant force acts through the barycen-
tre (centre of mass and centre of rotation), producing the requisite orbital
Exploring Planetary Systems II
8 THERMAL ROCKET ENGINES 91
8 Thermal rocket engines
The thermal rocket is the basis of all launchers and almost all space
propulsion (electric propulsion will be discussed in the next lecture).
Thermal rocket motor is a heat engine, it converts the heat generated
by burning propellant (fuel and oxidiser) in the combustion chamber into
kinetic energy of emerging exhaust gas.
Liquid-fuelled rocket engine consists of combustion chamber into which
fuel and oxidant are pumped and an expanding nozzle which converts the
high pressure hot gas into high velocity exhaust stream (Fig 22). The
expansion of the gas against the walls of the nozzle does the work and
accelerates the rocket.
Solid-fuelled rocket engine operates in the same way, but the fuel and
oxidant are pre-mixed in solid form and are contained within combustion
chamber (Fig 22). Normally the combustion takes place on the inner sur-
face of the propellant charge.
Exploring Planetary Systems II
8 THERMAL ROCKET ENGINES 92
Figure 22: Liquid-fuelled rocket engine (top), solid-fuelled rocket engine
(bottom).
Exploring Planetary Systems II
8 THERMAL ROCKET ENGINES 93
8.1 Thrust and the effect of the atmosphere
Figure 23: Gas flow in the nozzle
The force accelerating
the exhaust gas, the reac-
tion of the walls, is equal to
F =
I
pd A
where integral is taken over
the whole inner surface of
the chamber and nozzle.
There is also retarding force (see Fig 23). The pressure decreases in
the nozzle as the gas moving through the nozzle. The retarding force is
the pressure at the exit plane, p
E
, multiplied by the area at exit plane A
E
.
The accelerating force acting on gases can be written
F
G
=
I
pd A − p
E
A
E
=
˙
Mu
E
Exploring Planetary Systems II
8 THERMAL ROCKET ENGINES 94
where
˙
M is the mass flow rate and u
E
is the exhaust velocity.
Similarly, the accelerating force acting on the rocket (thrust) can be
written
F =
I
pd A − p
a
A
E
,
where p
a
is the atmospheric pressure.
Now we can eliminate the surface integral, so we find for the rocket
thrust
Thrust: F =
˙
Mu
E
+ p
E
A
E
− p
a
A
E
, (8.1)
this is thrust equation. The difference between Equation 8.1 and F =
˙
Mv
E
we used (see Equation 2.1) is that u
E
is the true exhaust velocity.
The expression for the effective exhaust velocity v
E
can be derived
v
E
= u
E
+ A
E
(p
E
− p
a
)
˙
M
(8.2)
Hence the thrust of the rocket depends on ambient conditions.
Exploring Planetary Systems II
8 THERMAL ROCKET ENGINES 95
8.2 Optimising the exhaust nozzle
The pressure drops along the expanding nozzle, and if nozzle is longer
the exhaust pressure decreases. For maximum exhaust velocity (Equa-
tion 8.2), the design of exhaust nozzle should be optimised so that the exit
pressure p
E
is equal to the ambient pressure p
a
.
If the nozzle is made longer, then extra area will either add to the thrust
or to the retarding force, depending on whether the internal pressure ex-
ceeds the atmospheric pressure. Thus adding to the nozzle length will
increase the thrust provided p
E
≥ p
a
.
This is one of the most important issues in rocket design. Launchers
mostly begin at sea level (high atmospheric pressure), but atmospheric
pressure decreases with hight as the rocket moves.
Exploring Planetary Systems II
8 THERMAL ROCKET ENGINES 96
8.3 Exhaust velocity
The change in internal energy for the gas in the nozzle is given
c
p
M(T
c
−T
E
)
where c
p
is the specific heat at constant pressure, T
c
and T
E
are the
temperatures of the gas in the combustion chamber(initial) and at the exit,
respectively.
Assuming no energy is transferred to the walls, the change in the in-
ternal energy is equal to the gain in kinetic energy of the exhaust gas
Mu
2
E
/2:
u
2
E
=2c
p
(T
c
−T
E
)
Using adiabatic gas expansion in the nozzle, we can write
pV
γ
=const, T p
γ/(γ−1)
=const
where γ is the ratio of the specific heats at constant pressure and at con-
Exploring Planetary Systems II
8 THERMAL ROCKET ENGINES 97
stant volume. The specific heat at constant pressure can be written:
c
p
=
γ
γ −1
R
µ
where R is the universal gas constant, µ is the molecular weight of the
exhaust gases.
Substituting T
E
and c
p
, we can find the velocity expressed in terms of
pressure
u
2
E
=
2γ
γ −1
RT
c
µ
"
1 −
µ
p
E
p
c
¶
(γ−1)/γ
#
(8.3)
where T
c
, p
c
are the temperature and pressure in the combustion cham-
ber.
From equation (8.3), we see that the exhaust velocity depends on noz-
zle design, which determines the pressure ratio p
E
/p
c
(Fig 24). The per-
fect nozzle has p
e
=0 and the exhaust velocity u
E
has a maximum value.
This demonstrates that the rocket engine is the most efficient in a vacuum.
Exploring Planetary Systems II
8 THERMAL ROCKET ENGINES 98
Figure 24: Gas velocity u
E
as the function of the pressure ratio
Exploring Planetary Systems II
8 THERMAL ROCKET ENGINES 99
8.4 Optimising exhaust velocity
For p
e
equal to zero, the exhaust velocity has a maximum value
u
2
E
=
2γ
γ −1
RT
c
µ
Expectedly, the velocity is dependent on temperature T
c
.
For small values of γ, the velocity is a strong function of γ.
As γ →1, u
E
tends to ∞. Typical values of γ are around 1.2
The velocity also depends inversely on the molecular weight of the
exhaust gas µ. Low molecular weight has advantage if combustion tem-
perature can be kept high. Often extra hydrogen is added, although it
plays no part in combustion.
Exploring Planetary Systems II
8 THERMAL ROCKET ENGINES 100
8.5 Mass flow rate
The mass flow rate (along the nozzle) can be expressed
˙
M =ρuA
where
˙
M is the (constant) mass flow rate, ρ is the density, u and A are
the gas speed and the cross-sectional area.
The expression for the exhaust velocity given by (8.3) can be used
to give velocity at any point along the nozzle. Hence the mass flow rate
becomes
˙
M =ρ A
Ã
2γ
γ −1
RT
c
µ
"
1 −
µ
p
p
c
¶
(γ−1)/γ
#!
1/2
where p is along the nozzle.
Using the gas laws
ideal gas: p =
ρ
µ
RT, adiabatic: p ∝ρ
γ
,
Exploring Planetary Systems II
8 THERMAL ROCKET ENGINES 101
we can present density via pressure. From adiabatic law we have p/p
c
=
(ρ/ρ
c
)
γ
, hence
ρ =ρ
c
µ
p
p
c
¶
1/γ
and using the ideal gas law ρ
c
=µp
c
/RT
c
, we find
ρ =
p
c
µ
RT
c
µ
p
p
c
¶
1/γ
and derive the mass flow rate per cross-sectional area
˙
M
A
= p
c
Ã
2γ
γ −1
µ
RT
c
µ
p
p
c
¶
2/γ
"
1 −
µ
p
p
c
¶
(γ−1)/γ
#!
1/2
(8.4)
The flow density given by (8.4) is presented in Figure 25.
Exploring Planetary Systems II
8 THERMAL ROCKET ENGINES 102
Figure 25: variation of flow density through the nozzle.
The flow density first grows (as p/p
c
drops) , then it starts to decrease.
Since the mass flow rate is constant, so this curve implies that for optimal
expansion the cross-sectional area of the stream should first decrease
then increase.
Exploring Planetary Systems II
8 THERMAL ROCKET ENGINES 103
From Equation (8.4), the ideal cross-sectional area of the nozzle for
any pressure is
A =
˙
M
p
c
Ã
2γ
γ −1
µ
RT
c
µ
p
p
c
¶
2/γ
"
1 −
µ
p
p
c
¶
(γ−1)/γ
#!
−1/2
(8.5)
While it shows the relation between local A and p, it cannot show the
shape of the nozzle.
The proper shape of the nozzle, de Laval nozzle, has a smoothly
curved converging part and a bell-shaped diverging cone. The flow lines
are axial over the whole cross-sectional area, so the thrust is developed
along the axis and is not lost.
Exploring Planetary Systems II
8 THERMAL ROCKET ENGINES 104
8.6 Vulcaine engine of Ariane 5 launcher
Basic design of a liquid propel-
lant engine comprises a combustion
chamber and nozzle, with propellant
tanks and a mechanism (injector) to
deliver the propellants to the com-
bustion chamber. The injector has
to fulfill 3 functions: deliver the fuel
and oxidiser in a fine spray; enable
rapid mixing; and deliver the mixture
at a high pressure to the combustion
chamber, to ensure a high flow rate.
Exploring Planetary Systems II
8 THERMAL ROCKET ENGINES 105
8.7 Saturn V Rocket engine
First stage F-1 engine on display at
the Alabama Space & Rocket Cen-
ter.
Exploring Planetary Systems II
8 THERMAL ROCKET ENGINES 106
8.8 Liquid fuel characteristics
The characteristic velocity c
∗
c
∗
=
p
c
A
∗
˙
M
measures the efficiency of conversion of thermal energy in the combustion
chamber into high-velocity exhaust gas. A
∗
is the throat area.
Exploring Planetary Systems II
9 ELECTRIC PROPULSION 107
9 Electric propulsion
Chemical rockets use energy stored in the propellant to create a hot
gas, which on expulsion, generate thrust. Power output is rigidly defined
by chemical properties and flow rate, so that the exhaust speed v
E
is
preset by thermodynamics/nozzle design.
Objective: For more ambitious space missions (e.g. interplanetary
missions), we want to breach v
E
'4.5 km/s limit.
Method: If we can separate the energy input to the engine from the
propellent flow then more energy can be given to a unit mass of propellant
than is available from only its chemical properties.
Advantages: Can get around the v
E
limiting problem, and can do this
electrically.
Exploring Planetary Systems II
9 ELECTRIC PROPULSION 108
9.1 Basics of electric propulsion
Let us revisit basic quantities, this time with possibility of v
E
being
adjustable: This means that we have to include the electrical power supply
mass M
E
and power output P
E
in our general description, so that the
mass ratio
R =
M
S
+M
P
+M
E
M
S
+M
E
, (9.1)
where M
S
is the structural mass, M
P
is the propellent mass, and M
E
is
the new addition of electric power supply mass.
Now, let us introduce thrust efficiency:
thrust efficiency =η =
˙
Mv
2
E
/2
P
E
(9.2)
where, for simplicity we assume
˙
M = M
p
/t is constant.
Exploring Planetary Systems II
9 ELECTRIC PROPULSION 109
Similarly, we can define power to mass ratio:
power to mass ratio =ξ =
P
E
M
E
(9.3)
We apply electrical energy to the propellant from an external power
supply (e.g. battery, solar panels, generator). It breaks the link between
the energy supply and the propellant, which is the case for a chemical
rocket.
Hence from Equations (9.2,9.3), we have:
v
E
=
µ
2ηP
E
˙
M
¶
1/2
=
µ
2ηξM
E
˙
M
¶
1/2
=
µ
2ηξM
E
t
M
P
¶
1/2
(9.4)
and the corresponding thrust:
F =
˙
Mv
E
=
¡
2ηξM
E
˙
M
¢
1/2
=
µ
2ηξM
E
M
P
t
¶
1/2
(9.5)
So the exhaust velocity is no longer a free parameter, it is determined by
power and mass flow rate.
Exploring Planetary Systems II
9 ELECTRIC PROPULSION 110
For a chemical rocket, v
E
is determined by propellant choice and en-
gine design, and the rocket speed is given by the mass ratio.
For electrical rockets, have to fold in the power supply mass, efficiency,
and power to mass ratio, e.g.
v =
µ
2ηξM
E
t
M
P
¶
1/2
ln
·
1 +
M
p
M
S
+M
E
¸
=v
E
ln
"
1 +
2ηξt
2ηξt(M
s
/M
p
) +v
2
E
#
(9.6)
where we used from (9.4) M
E
/M
P
=v
2
E
/(2ηξt).
Note the departure on the burn time (it wasn’t there before in chemical
rocket)
The burn time is important because it defines the rate at which propel-
lant is used, and hence the power that has to be applied. For the same
onboard mass of propellant, a short burn time requires higher power and
a heavier power supply.
Exploring Planetary Systems II
9 ELECTRIC PROPULSION 111
9.2 Vehicle velocity
The equation (9.6) for vehicle velocity shows that the mass ratio for a
given propellant mass decreases as the exhaust velocity increases, due
to the increased power supply mass.
Let us find rewrite equation (9.6)
v =v
E
ln
"
1 +
a
b +v
2
E
#
so, we find:
for v
2
E
<< b, we have v ∝v
E
,
for v
2
E
>> b, we have v ∝1/v
E
.
Figure (26) shows that the vehicle velocity does not increase mono-
tonically with exhaust velocity, and peaks for a certain value. It can also
be seen that increasing the burn time increases the peak value, both of
the vehicle velocity and the optimal exhaust velocity. The decrease in ve-
hicle velocity further is due to the increasing mass of the power supply
Exploring Planetary Systems II
9 ELECTRIC PROPULSION 112
and hence a reduction in mass ratio, as mentioned above.
Figure 26: Vehicle velocity vs exhaust velocity and burn time.
Exploring Planetary Systems II
9 ELECTRIC PROPULSION 113
9.3 Electric thrusters
There are two broad categories
(a) electrothermal - heat propellant electrically to produce high-speed
neutral exhaust (e.g. super-chemical rocket)
(b) ionising - convert propellant to ions and use electric fields to expel
ions at high speed
Exploring Planetary Systems II
9 ELECTRIC PROPULSION 114
9.4 Electrothermal thrusters
In case (a), hot wire heats propellant to produce high temperature
(. 2200 K) and pressure. As an example, consider thruster working with
hydrogen gas. Using H
2
, heated gas can be ejected as exhaust very
efficiently, with v
E
∼10 km/s, which is much higher than chemical rocket.
However, the mass flow rate is low.
Using equation (9.2), for 1 kW thruster, we have
˙
M =
2ηP
E
v
2
E
'
2000
(10
4
)
2
η ∼2 ×10
−5
η kg/s (9.7)
which is very small compared to chemical rocket.
The thrust F =
˙
Mv
E
is also very small
F ∼0.2η N
Low thrust and low mass rate, but burn time is long, so such thrusters
are good for small orbital corrections and long-life missions.
Exploring Planetary Systems II
9 ELECTRIC PROPULSION 115
9.5 Ionising thrusters
In case (b), fuel is ionised and ejected out of the engine using strong
electric fields.
Ions are ejected by strong electric field producing a fast charged ex-
haust (Figure 27). Free electrons are freed into the stream to neutralise
the ions (they have very low momentum, so do not affect the thrust.)
Exploring Planetary Systems II
9 ELECTRIC PROPULSION 116
Figure 27: Diagram of NSTAR ion thruster used on Deep Space 1, from Turner, 2006
Exploring Planetary Systems II
9 ELECTRIC PROPULSION 117
Deep Space 1 mission by
NASA (1998-2001) was to
test a number of technolo-
gies, e.g. NSTAR ion
thruster. The spacecraft en-
countered comet Borrelly and
returned the best images and
other science data ever from
a comet. See mission pages
here.
Note: there is no need for nozzle, since ion motion is already ordered.
Exploring Planetary Systems II
10 ELECTRIC THRUST 118
10 Electric thrust
10.1 Space charge limit
Figure 28: Accelerating grids of
ion thruster (see Figure 27)
Expulsion of the ions can cause
space charge build-up which can cancel
field at the grids. This places a limit on
the ion current, e.g. mass flow rate. Pois-
son equation says
d
2
V
dx
2
=−
N q
²
0
(10.1)
where V is the electric potential at x, N
is the number density of ions, q is the
ion charge, x is the axial coordinate, ²
0
is vacuum permittivity (see Figs 28 and
27).
Now the kinetic energy of gained by
Exploring Planetary Systems II
10 ELECTRIC THRUST 119
ions (see Eq 10.1) comes as the result of
potential difference:
1
2
mv
2
= q(V
0
−V )
where V
0
is the potential difference, m is the ion mass.
Hence the ion speed at point x
v =
·
2q(V
0
−V (x))
m
¸
1/2
(10.2)
With the ion number density N, charge q, speed v, we can construct
the ion current density
j = N qv (10.3)
Substituting (10.3, 10.2) into Poisson equation (10.1), we find
d
2
V
dx
2
=−
j
²
0
[
2q(V
0
−V (x))/m
]
1/2
Exploring Planetary Systems II
10 ELECTRIC THRUST 120
Now multiplying both parts by dV /dx, we have
d
2
V
dx
2
dV
dx
=−
r
m
2q
j
²
0
1
p
V
0
−V (x)
dV
dx
and assuming constant current, we integrate over x
µ
1
2
dV
dx
¶
2
=
r
m
2q
j
²
0
2
p
V
0
−V (x) +const
and finally
µ
dV
dx
¶
2
=4
r
m
2q
j
²
0
p
V
0
−V (x) +
µ
dV
dx
¶
2
¯
¯
¯
¯
¯
x=0
now we note that (dV /dx)
2
=
−→
E
2
, electric field strength squared.
The value of electric field at the first grid, dV /dx
|
x=0
= −E
0
= 0 and
we obtain for electric field
−E =
dV
dx
=−2
s
j
²
0
µ
m
2q
¶
1/4
[
V
0
−V (x)
]
1/4
(10.4)
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10 ELECTRIC THRUST 121
Making the change of variables α =2( j/²
0
)
1/2
(m/2q)
1/4
,
[
V
0
−V (x)
]
−1/4
dV
dx
=−α
we can integrate (10.4) again
4
3
[
V
0
−V (x)
]
3/4
=αx
so that the potential V (x) becomes
V (x) =V
0
−
·
3
4
αx
¸
4/3
=V
0
−
3
2
s
j
²
0
µ
m
2q
¶
1/4
x
4/3
(10.5)
Equation (10.5) determines the potential at any point between the grids
as a function of the current density.
Extracting the current density j from Equation (10.5), we find the cur-
rent density at the space charge limit
j =
4
9
²
0
µ
2q
m
¶
1/2
|∆V |
3/2
x
2
(10.6)
Exploring Planetary Systems II
10 ELECTRIC THRUST 122
where ∆V = V −V
0
. Equation (10.6) is known as Child-Langmuir law for
plasma sheath potential.
Noting that |E
0
| = |∆V |/x, where x is the gap between the grids/elec-
trods, we simplify
j =
4²
0
9
µ
2q
m
¶
1/2
µ
|E
0
|
3
x
¶
1/2
(10.7)
where |E
0
| is the field in the absence of ions.
Hence the thrust, ∝ j is strongly dependent on electric field and elec-
tric gap.
For same field |E
0
|, smaller gap produces higher current, and hence
higher thrust. For given potential drop ∆V , smaller gap increases field.
So small gaps are advantageous.
Exploring Planetary Systems II
10 ELECTRIC THRUST 123
10.2 Electric field and potential
What is the maximum field that can be expected ? Recall that the
plasma (ionised gas) is controlling the field.
The field strength varies with position as a result of the space charge
effect, but we can estimate the maximum value by substituting the limiting
current j from Equation (10.7) into the equation for electric field E (10.4).
This yields
E
max
=
4
3
E
0
(10.8)
which is the maximum electric field.
Exploring Planetary Systems II
10 ELECTRIC THRUST 124
Figure 29: Variation of potential V (x) and electric field E
When ions are present, the space charge cancels the electric field
Exploring Planetary Systems II
10 ELECTRIC THRUST 125
close to one of the electrodes and enhances it near the other. Electric
field is structured by plasma and differs from the case of empty space
(Figure 29). The max value given by Equation (10.8) is reached near the
right electrode.
Hence for a given potential drop and gap size, the current density is
proportional to the number of ions unit area of grid - makes a limiting value
due to sheath formation.
This limit is analogy to the mass-flow rate limit in chemical rockets.
The mass flow rate per unit area A is
˙
M
A
= j
m
q
(10.9)
where m is the ion mass.
Increase of in the mass flow rate requires higher electric field E
0
or
smaller gap x, but there are practical limits. Indeed, the thrust per unit
Exploring Planetary Systems II
10 ELECTRIC THRUST 126
area is
F
A
=
˙
M
A
v
E
= j
m
q
µ
2q∆V
m
¶
1/2
=
8
9
²
0
E
2
0
so high thrust needs high E
0
.
However, it is better to increase the diameter of the thruster itself,
maintaining maximum j, but increasing the total current.
Exploring Planetary Systems II
10 ELECTRIC THRUST 127
10.3 Choice of the propellant
The exhaust speed, which is the ion speed, depends on charge to
mass ratio (see Equation 10.2):
v
E
=
·
2q
m
∆V
¸
1/2
However for typical potential drops of ∼1 kV, v
E
is already rather large for
most elements (Fig 30), e.g. typically 10
4
−10
5
m/s.
The ion thrusters are naturally high speed/low thrust engines, and re-
call that rocket velocity v has optimum v
E
(see Figure 26), after which
engine efficiency is lower.
Indeed for Argon, Ar
+
q
m
∼5 ×10
6
, gives v
E
∼7 ×10
4
m/s
Even for Mercury, Hg
+
v
E
∼3 ×10
4
m/s
Exploring Planetary Systems II
10 ELECTRIC THRUST 128
In fact, Xenon is the best choice, sufficiently heavy to restrict v
E
usefully,
but non-toxic and can be stored as liquid gas.
Figure 30: Exhaust speeds for various ions.
Exploring Planetary Systems II
10 ELECTRIC THRUST 129
10.4 How efficient are ion thrusters?
Electrical power P
E
is just current multiplied by the voltage drop, so the
power per unit area ( with maximum current density from Equation 10.6)
becomes:
P
E
A
= j∆V =
4
3
µ
2q
m
¶
1/2
|∆V |
5/2
x
2
The thrust generated by this electrical power, e.g. the ratio is
P
E
/A
F/A
=
j∆V
(
˙
M/A)v
E
Using Equation (10.9), we find
P
E
/A
F/A
=
j∆V
jMv
E
/q
=
q∆V /m
v
e
=
v
2
E
/2
v
E
=
v
E
2
so we find that
P
E
F
=
v
E
2
Exploring Planetary Systems II
10 ELECTRIC THRUST 130
so the thrust F is small and the challenge of ion propulsion is to reduce
v
E
and increase the thrust.
Hence ion engines are best suited to very high velocity increments
missions in which the time taken to achieve ∆v is not critical, e.g. time
for station keeping and interplanetary missions, where low thrust is not a
problem.
Exploring Planetary Systems II
11 SOLAR SAILS AND SPACE ENVIRONMENT 131
11 Solar sails and space environment
11.1 Solar sails
Is it possible to avoid carrying fuel altogether ? A solar sail harnesses
radiation pressure from the Sun. Radiation pressure at distance r from
the Sun
P
rad
=
L
¯
4πr
2
c
where L
¯
is the solar Luminosity, c is the speed of light.
Incident photon has momentum p = hν/c, where ν is the frequency, h
is Planck constant.
If we have a perfect reflector, then the change of photon momentum is
2hν/c
before: −−→
hν
c
| after: ←−−
−
hν
c
|−−→
2
hν
c
Exploring Planetary Systems II
11 SOLAR SAILS AND SPACE ENVIRONMENT 132
11.2 Radiation pressure vs gravitational force
Suppose we have a perfect solar sail reflector at distance r from the
Sun, ratio of radiation pressure to gravitational force is
L
¯
A
4πr
2
c
µ
GM
¯
m
r
2
¶
−1
=
L
¯
4πGM
¯
c
A
m
'10
−3
A
m
where L
¯
is the solar Luminosity, M
¯
is the mass of the Sun, A is the sail
area, m is the sail mass.
Hence for balance we need mass per unit area ∼10
−3
kg/m
2
.
For comparison, A4 printer paper is ∼75 g/m
2
=7.5 ×10
−2
kg/m
2
.
Realistically, expect solar sail material to have mass per unit area of
1/10 of that of paper.
Also if carrying a 1-tonne payload, expect sail area to be 10
6
m
2
or
one square kilometer sail.
Exploring Planetary Systems II
11 SOLAR SAILS AND SPACE ENVIRONMENT 133
11.3 IKAROS solar sail
image from JAXA, Japan
IKAROS (Japanese mission)
was launched in 2010 and
made the world’s first demon-
stration of solar power sail.
The sail is 20 m on the di-
agonal and is made of a
7.5-micrometre thick sheet of
polyimide.
The polyimide sheet had a
mass of about 10 grams per
square metre, resulting in a
total sail mass of 2 kg.
Exploring Planetary Systems II
11 SOLAR SAILS AND SPACE ENVIRONMENT 134
11.4 Solar wind pressure
How does the radiation pressure compare with power from solar wind?
The solar wind pressure P
wind
is
P
wind
=
˙
M
¯
v
w
4πr
2
where
˙
M
¯
= 10
−14
M
¯
year
−1
is the solar mass loss rate in units of solar
mass per year.
Assume typical solar wind speed v
wind
∼300 km/s, then
P
wind
P
rad
=
v
w
c
˙
M
¯
L
¯
'10
−4
,
so the solar wind pressure is negligible.
Exploring Planetary Systems II
11 SOLAR SAILS AND SPACE ENVIRONMENT 137
Energetic charged particles surround the Earth (Van Allen radiation
belts).
The solar activity can create additional energetic particles which can
come in sudden bursts. High energy heavy charged particles also reach
the Earth vicinity from the Sun or from outside the solar system.
The energetic particles affect spacecraft internal charging, displace-
ment damage, and ionizing dose.
• Trapped Particles (in magnetosphere, e.g. radiation belts)
• Solar Particles (from solar flares, Coronal Mass Ejections)
• Cosmic Rays (galactic cosmic rays, anomalous cosmic rays)
Geomagnetic field which strongly influences plasma/particle motions and
locations (see Figure 31).
Exploring Planetary Systems II
11 SOLAR SAILS AND SPACE ENVIRONMENT 140
11.8 Plasma environment
Ionised gas (plasma) can be encountered around Earth and beyond
magnetosphere in the solar wind. Various plasma environments include:
• ionosphere, the layer of Earth upper atmosphere (∼ 60 −800 km),
where photoionization by solar X-rays and extreme ultraviolet rays
creates free electrons
• inner magnetosphere, includes the plasmasphere, region of approx-
imately dipolar Earth magnetic field
• plasma sheet, in the outer magnetosphere on the midnight side of
the Earth
• magnetosheath, interface that separates interplanetary space and
the magnetosphere
• solar wind, high-speed (400 −700 km/s) plasma flow
Exploring Planetary Systems II
11 SOLAR SAILS AND SPACE ENVIRONMENT 141
Figure 33: Typical Space Plasma Environments shown in Figure 31, data from Grand
et al, Space Science Reviews, 1983
Plasma causes various problems to spacecraft, one of most serious
of which is the electrostatic charging of satellite surfaces. Discharges can
severely disturb operations and even result in the loss of satellites.
The cold ionospheric plasma is also a problem for operating high power
systems because of its conductivity, and electromagnetic systems be-
cause of its dispersive and refractive properties.
Exploring Planetary Systems II
11 SOLAR SAILS AND SPACE ENVIRONMENT 142
11.9 Space debris, micro-meteroids
Due to high speed meteoroids and space debris have large kinetic
energy and hence can seriously damage satellites (Figure 34 shows the
distribution of numbers and sizes):
• space debris problem is due to growing space activities
• meteoroid environment is an ever-present feature
Around 2 ×10
4
tons of natural materials dust, meteoroids, asteroids,
and comets hit Earth every year (from US Federal Aviation Administration
www.faa.gov)
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11 SOLAR SAILS AND SPACE ENVIRONMENT 143
Figure 34: The distribution of numbers and sizes of debris particles in low Earth orbit.
Extract from ‘Technical Report on Space Debris’, UN Committee on Peaceful Uses of
Outer Space.
Exploring Planetary Systems II
