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
¶
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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.
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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.
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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
.
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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.
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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.
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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.
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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)
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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).
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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.
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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.
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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).
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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.
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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.
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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