4262: Rockets and Mission Analysis
Homework #2
Assigned: September 9, 2014
Due: September 23, 2014
Purpose of this homework assignment:
1. Review the use of the rocket equation
2. Consider the impact of initial acceleration on launch vehicle performance
3. Understand the benefits and drawbacks of rocket staging, using a slightly different
approach than that examined in lecture.
Question 1:
Our discussion of staging in class suggested it as a solution if a single-stage rocket was not
capable of reaching the desired velocity. Assume a rocket of total mass 100 tons, carrying a
spacecraft payload of 1 ton. The engines develop a constant exhaust velocity of 3,000 m/s. The
structural mass is assumed to be 10% of the fuel mass.
1) Determine the velocity of this configuration as a single stage rocket
2) If the rocket is divided into two smaller stages, each with half the fuel, and the
structural mass also shared equally, and the payload being the same, determine the
total velocity increment for the two stage configuration.
3) Repeat part (2), assuming 3 stages. What do you notice about the total velocity
increment as you add more and more stages? As an engineer, how would you
determine how many stages to use?
Question 2:
Our discussion of staging in class suggested it as a solution if a single-stage rocket was not
capable of reaching the desired velocity. It may also be desired to use a staged rocket even when
the desired final velocity can be reached in a single stage. In fact, one of the most contentious
issues about launch systems is whether single-stage-to-orbit is the way of the future. To help
your thinking on this important matter, it is suggested that you compare launch systems with 1, 2
and 3 stages, using as a Figure of Merit the ratio of Payload Mass / Initial Mass.
To make the comparison simple, assume the following:
Final velocity: 8,000 m/s
Exhaust velocity, constant for all stages: 4,000 m/s
Structural weight fraction (Structure Mass / Initial Mass) for all stages, including the
engine mass: 0.07
Initial acceleration fixed for all stages at ao/g=2.
1
Question 3:
The purpose of this question is to address the issue of whether or not there is an optimum
acceleration for an earth launched rocket (without drag).
Recall that the rocket equation can be written as:
 mt  
  gt
V  Vo  Ve ln 
 mo 
1
In this equation we have neglected the effects of drag and have assumed that the rocket is
traveling in vertical flight. Vo is the initial velocity, which is zero for vehicles being launched
from the surface of the earth, and Ve is the exit velocity of the propellant relative to the rocket.
In the case where Pe≠Pa, we can replace Ve in equation 1 with Veq, which is defined as:
 P  Pa 
Veq  Ve   e
 Ae
 m 
2
 is constant (which is a pretty good assumption), then we
If we make the assumption that that m
can write:
mo  mt   m t 
M a
T
t o ot
Ve
Ve
3
In equation 3, ao is the initial acceleration, and we can use this result to eliminate t in equation 1:
 mt   g
V  Vo
 
  ln 
Ve
 mo  ao
 mt 
1 

mo 

4
The last term is the gravity loss, since it is the reduction in velocity increment that the vehicle
m t 
suffers due to the acceleration of gravity. This is an implicit equation for
in terms of
mo
V  Vo
, with the initial acceleration ao as a parameter. As ao increases, the gravity loss decreases,
Ve
since the time over which gravity acts is reduced. So why not use a very large a o to get the
V  Vo
m t 
largest possible
for a given
? To answer this question, we can divide the mass of the
Ve
mo
rocket into parts:
m0  mstructure  mengine  m payload  m propellant
When all the propellant has been expended at time tb, we have:
2
5
mt b   mstructure  mengine  m payload
6
We can write:
m propellant
mt b  mo  m propellant

 1
mo
mo
mo
Examining equation 4, we can see that the smaller
the
7
m
m t 
or the closer propellant is to 1, the larger
mo
mo
V  Vo
can be achieved. But the limit is set by:
Ve
m propellant mstructure mengine m payload
mt b 
 1



mo
mo
mo
mo
mo
8
mengine
mstructure
and
are as small as
mo
mo
possible, given the loads and available technology. Some estimates are shown in the table below:
We want to design components of the rocket so that
Component or Component Ratio Relative to the Initial Mass Estimates
mstructure
 0.1
mo
1 T
mengine 
50 g e
mengine
1 T
1 ao


mo
50 mo g e 50 g e
9.a
9.b
9.c
Substitute these estimates into equation 8 and then into the rocket equation at time tb.
V tb   Vo a o m payload
V tb   Vo
Doing so, you should have 3 design parameters:
,
,
. Ordinarily
Ve
Ve
ge
mo
is set by the mission requirement, for example to get to LEO, and clearly we would like to have
m payload
a
as large as possible. The key question is: What is the best o ? It is a bit hard to see the
ge
mo
m payload
answer by looking at the rocket equation and trying to solve for
. To find an optimum is
mo
V tb   Vo
a
analytically messy. The best way to do this is to plot
as a function of o for various
Ve
ge
m payload
m payload
values of
. Plot these results for
=0, 0.01, 0.05, 0.1, 0.2, and 0.5.
mo
mo
3
Now, say we specify
best
V tb   Vo
at 1.0, which is the maximum efficiency condition. Is there a
Ve
ao
?. How would you describe the optimum?
ge
The optimization has considered only the effects of the gravity loss and the engine weight on the
a
choice of o . If we include other effects, we will get a different answer. This is shown in the
ge
next section.
The above example determines the initial acceleration that minimizes the initial mass, accounting
for the gravity loss and the engine mass. In doing so, we assumed that the structural mass
fraction was independent of the acceleration. In fact, it is not, because the structure must
withstand larger forces if the vehicle is accelerated more rapidly. We can repeat this analysis,
with the assumption that the structural mass fraction is proportional to the initial acceleration.
mstructure
a
The constant of proportionality is such that
 0.04 when o =2. From the rocket
ge
mo
equation we showed:
 mf
V t b   Vo
  ln 
Ve
 mo
 g
 
 ao
 mf
1 
 mo



4
mt b 
is given by equation 7. Modify the rocket
mo
mo
equation for this new assumption criteria and re-plot the results. How has the locus of
optima shifted? How would you explain this result? How would you expect these results to
differ if the effects of drag were included?
Again the quantity
mf
which is equal to
4