MAB120  MAN120  MAB125
semester one 2010
see Blackboard site for additional details, including submission instructions
Problem Solving Assignment
1.
Rocket launching:
In this problem we are looking
at some of the aspects that need to be considered
in designing and describing the launching a single
stage rocket with a payload. In what follows we will
be ignoring any eﬀects of crosswinds, rotation of
Earth or air resistance.
We begin by assuming that the rocket is launched
from Earth’s surface in such a way that it follows a
simple vertical trajectory. As a singlestage rocket,
immediately after launch there will be a period of
rapid acceleration as the fuel consumed provides
thrust for the upward motion. One of the complex
issues to be considered is that as the fuel is used, the
mass of the rocket (
i.e.
fuel + engine + payload)
decreases, and hence the acceleration will increase
for the same amount of thrust. The period of time
over which the rocket engine is ﬁring is known as the
‘boost phase’ of the trajectory, hence when the fuel
runs out, the ‘boost phase’ is at an end.
Representing the rocket’s total mass with the un
known function
m
(
t
), if we assume that one of the
design parameters for the rocket engine is to set a
constant fuel burn rate, then we will have the rela
tionship
dm/dt
=

b
where
b
is the fuel burn rate
constant measured in kg s

1
. The initial total mass
of the rocket can be represented by the constant m
0
,
and we can assume that the fuel is a signiﬁcant frac
tion of this mass. In this case, the quantity of fuel
can again be a design parameter, so we will take the
mass of fuel to be
X
m
0
, where 0
< X <
1.
If we further assume that the ‘boost phase’ is short
enough, and we assign the function
h
(
t
) to be the
rocket’s height above groundlevel, then we can de
scribe the acceleration experienced by the rocket
during this period as
d
2
h
dt
2
=
κb
m
0

bt

g ,
where
g
is the acceleration due to gravity, and
κb
is
the thrust provided by the rocket engine,
κ
being an
experimentally determined constant. If we launch
the rocket from a stationary position, we can jus
tiﬁably include the initial conditions
h
(0) = 0 and
dh/dt
= 0 when
t
= 0.
(a) Show that the expression for the height
h
(
t
) of
the rocket when it just runs out of fuel is given
by
κ
m
0
b
±
(1

X
) ln(1

X
) +
X
±
1

g
2
κb
X
m
0
²²
.
(b) What is the velocity of the rocket at the end of
the ‘boost phase’?
(c) Astronauts aboard the ‘vomit comet’ or its
current equivalent
1
, experience a rapid loss of
weight as the acceleration due to gravity is
counteracted by the acceleration provided by
the aircraft.
The
rate
of weightloss expe
rienced is equivalent to a rate of decrease in
the acceleration due to gravity of the order of

10
/
15 ms

2
/
s =

2
/
3 ms

3
.
For our rocket and its payload, the acceleration
due to gravity decreases as the rocket gets fur
ther from the launch site  still assuming a ver
tical trajectory. If we take the radius of Earth
to be R
0
= 6
.
38
×
10
6
m, then the acceleration
due to gravity goes like
a
g
=
M G
(
h
(
t
) +
R
0
)
2
,
where M and G are constants
2
and as before,
h
(
t
) is the rocket’s height above Earth’s surface
as a function of time. For an object of constant
mass, but changing height, the rate of weight
loss experienced would be given by
d
(
a
g
)
/dt
.
Using the information gained previously for the
height and velocity of the rocket in the ‘boost
phase’, ﬁnd an expression for the rate of weight
loss experienced by the rocket and it’s payload
as it continues to climb just at the end of the
boost phase. You should use the operating pa
rameters of the Shahab 3 missile employed by
Iran (see notes
3 4
), which has an initial mass
of 16000 kg of which 90% can be considered as
fuel. Furthermore the Shahab 3 has a boost
phase of 95 s by which time it reaches a height
of 75 km. How does this rate of weightloss
compare with that experienced in the ‘vomit
comet’?
1
Wikipedia article at
http://en.wikipedia.org/wiki/Vomit
Comet
2
The constants are the mass of Earth (M = 5
.
9742
×
10
24
kg) and the universal gravitational constant (G = 6
.
673
×
10

11
m
3
kg

1
s

2
)
3
http://www.princeton.edu/sgs/publications/sgs/pdf/12
1
2
wilkening.pdf
4
http://www.armscontrol.org/print/2617
version 01  updated April 15, 2010
1