Spacecraft Dynamics and Control

Matthew M. Peet
Arizona State University

Lecture 11: Intro to Rocketry

Introduction

In this Lecture, you will learn:

Introduction to Rocketry
• Mass Consumption
• Specific Impulse and Rocket Types
• ∆v limitations
• Staging

Numerical Problem: Suppose our mission requires a dry weight of 30kg. How
much propellant is required to achieve a circular orbit of altitude 200km?

M. Peet Lecture 11: Spacecraft Dynamics 2 / 29

Questions about Propulsion

We have talked a bit about ∆v.

• How is ∆v created?
• How expensive is it?
• Is it really instantaneous?

M. Peet Lecture 11: Spacecraft Dynamics 3 / 29

∆v budget
A Typical mission uses a lot of
∆v.
• How much propellant will
we need?
• What is the maximum ∆v
budget?

M. Peet Lecture 11: Spacecraft Dynamics 4 / 29

Some Definitions

In a staged Launch system, the mass

varies with time.
• Dry weight is the weight without
propellant.
I This is the final weight.
I Craft plus payload
• There are several variations of dry
weight.

M. Peet Lecture 11: Spacecraft Dynamics 5 / 29

How to create Thrust: Newton’s Second Law
Approximation: Consider the expulsion of a piece of propellant, ∆m.
Initial State:
• Propellant and Rocket move together.
• Total Momentum:
hi = (mr + ∆m)v
Final State:
• Propellant and Rocket move separately.
• Rocket has velocity v + ∆v
• Propellant has velocity v − c.
I c is the exhaust velocity

• Total Momentum:
hf = mr (v + ∆v) + ∆m(v − c)
Conservation of Momentum:
• Setting hi = hf , we obtain:
(mr + ∆m)v = mr (v + ∆v) + ∆m(v − c)
• Solving for ∆v, we obtain
∆m
∆v = c
mr
M. Peet Lecture 11: Spacecraft Dynamics 6 / 29
 How to create Thrust: Newton’s Second Law
Lecture 11 Approximation: Consider the expulsion of a piece of propellant, ∆m.
Initial State:
• Propellant and Rocket move together.

2018-05-03 Spacecraft Dynamics • Total Momentum:

Final State:
hi = (mr + ∆m)v

• Propellant and Rocket move separately.

• Rocket has velocity v + ∆v
• Propellant has velocity v − c.
I c is the exhaust velocity

How to create Thrust: Newton’s Second Law • Total Momentum:

hf = mr (v + ∆v) + ∆m(v − c)
Conservation of Momentum:
• Setting hi = hf , we obtain:
(mr + ∆m)v = mr (v + ∆v) + ∆m(v − c)
• Solving for ∆v, we obtain
∆m
∆v = c
mr

• In this slide, hi is the initial linear momentum of rocket and propellant mass
• v is the initial velocity of the spacecraft
• mr is the mass of the rocket
• ∆m is the mass of the propellant.
• hf is the final linear momentum of the rocket combined with the propellant. By
conservation of momentum, hi = hf
• ∆v is the change in velocity of the rocket.
• Note: This is for a single particle of propellant - ∆m and can not be used to
calculate ∆v directly. We will integrate this equation of many particles of
propellant to get the true ∆v.
Continuous Thrust
For a single particle of propellant, we have
∆m
∆v = c
mr
Dividing by ∆t and taking the limit as ∆t → 0,
we get ṁr (t)
v̇(t) = c
mr (t)
where we often assume constant mass flow rate
ṁr (t).

Returning to the differential form, we can directly integrate

c
dv = dmr
mr
to obtain 
m(t0 )

∆v = v(tf ) − v(t0 ) = c ln
m(tf )
∆m
Which is quite different from the approximation ∆v = mr c!
M. Peet Lecture 11: Spacecraft Dynamics 7 / 29
 Continuous Thrust
Lecture 11 For a single particle of propellant, we have
∆m
∆v = c

2018-05-03 Spacecraft Dynamics mr

Dividing by ∆t and taking the limit as ∆t → 0,

we get ṁr (t)
v̇(t) = c
mr (t)
where we often assume constant mass flow rate
ṁr (t).

Continuous Thrust Returning to the differential form, we can directly integrate

c
dv = dmr
mr
to obtain 
m(t0 )

∆v = v(tf ) − v(t0 ) = c ln
m(tf )
∆m
Which is quite different from the approximation ∆v = mr c!

∆m
• ṁr = lim∆t→0 ∆t
is the rate at which we are using up propellant.
• Although liquid and hybrid rockets can control this rate, in practice, we want to
make it as large as possible so that the ∆v happens quickly.
• m(t0 ) is the mass before the burn. m(tf ) is the mass after the burn.
• v(tf ) is the velocity after the burn. v(t0 ) is the velocity before the burn.
Sizing the Propellant
Now we have an expression for thrust as a function of weight.
 
m(t0 )
∆v = v(tf ) − v(t0 ) = c ln
m(tf )
where recall m0 = m(t0 ) is the mass before thrust and m(tf ) is the mass after.
• ∆v is a function of the ratio of wet weight to dry weight
• For a given maneuver, we can calculate the required propellant
∆m ∆v
= 1 − e− c
m0

M. Peet Lecture 11: Spacecraft Dynamics 8 / 29

 Sizing the Propellant
Lecture 11 Now we have an expression for thrust as a function of weight.

m(t0 )

∆v = v(tf ) − v(t0 ) = c ln

2018-05-03 Spacecraft Dynamics m(tf )

where recall m0 = m(t0 ) is the mass before thrust and m(tf ) is the mass after.
• ∆v is a function of the ratio of wet weight to dry weight
• For a given maneuver, we can calculate the required propellant
∆m ∆v
= 1 − e− c
m0

Sizing the Propellant

• m(t0 ) is the wet weight (with propellant).

• m(tf ) is the dry weight (after all propellant has been used up)
• First equation is called the rocket equation, derived by Konstantin E. Tsiolkovsky
(1857-1935)
Effective Exhaust Velocity
The efficiency of the rocket depends on the relative velocity of the propellant, c.
• However, there is also a force due to pressure, F = Ae (Pe − P∞ ).

The effective velocity, c, of propellant is determined by configuration of the

rocket:
Ae
c = Ve + [Pe − P∞ ]
ṁ
Note: Pe gives a boost to thrust, but at the cost of a lower Ve
• As Ve increases, Pe drops (particles accelerate out of high-P regions)
• It is always best to maximize Ve (we want Pe = P∞ .
• In space, this implies we want A At as large as possible.
e

• Propellant is usually rated by c and not Ve !

M. Peet Lecture 11: Spacecraft Dynamics 9 / 29
 Effective Exhaust Velocity
Lecture 11 The efficiency of the rocket depends on the relative velocity of the propellant, c.
• However, there is also a force due to pressure, F = Ae (Pe − P∞ ).

2018-05-03 Spacecraft Dynamics

The effective velocity, c, of propellant is determined by configuration of the
Effective Exhaust Velocity rocket:
c = Ve +
Ae
ṁ
[Pe − P∞ ]
Note: Pe gives a boost to thrust, but at the cost of a lower Ve
• As Ve increases, Pe drops (particles accelerate out of high-P regions)
• It is always best to maximize Ve (we want Pe = P∞ .
Ae
• In space, this implies we want A as large as possible.
t
• Propellant is usually rated by c and not Ve !

• In previous slides, we have been using c instead of Ve for exhaust velocity so as

not to confuse you later. However, these formulae should be used with effective
exhaust velocity, c.
• P∞ is the atmospheric pressure.
• Pe is the pressure at exit from the nozzle.
• At is the area of the throat.
• Ae is the area of the nozzle exit.
• The effective exhaust velocity of H2 -O2 propellant in space is 4,440 m/s
• On the ground, there is an optimal Ae corresponding to Pe = P∞ .
Pressure Changes affect Efficiency on Saturn V

M. Peet Lecture 11: Spacecraft Dynamics 10 / 29

Specific Impulse

Definition 1.
The Specific Impulse is the ratio of the momentum imparted to the weight (on
earth) of the propellant.
∆mc c
Isp = =
∆mg g
h i
m0
Since ∆v = c ln mf , specific impulse gives a measure of how efficient the
propellant is.

M. Peet Lecture 11: Spacecraft Dynamics 11 / 29

 Specific Impulse
Lecture 11 Definition 1.

2018-05-03 Spacecraft Dynamics The Specific Impulse is the ratio of the momentum imparted to the weight (on
earth) of the propellant.
Isp =
∆mc
=
c
∆mg g
h i
m0
Since ∆v = c ln mf , specific impulse gives a measure of how efficient the
propellant is.

Specific Impulse

• measured in second, Isp tells, for any amount of propellant mass, how many
seconds the rocket will provide thrust equal to the weight (g = 9.81) of the
propellant consumed.
• Because the effective velocity depends on atmospheric pressure, Isp is different on
the surface of the earth vs. in space.
• Typically, Isp assumes a perfectly expanded rocket.
Example
Problem: Suppose our mission requires a dry weight of mL = 30kg. Using an
Isp of 300s, how much propellant is required to achieve a circular orbit of
altitude 200km?
Solution: A Circular orbit at 200km requires a total velocity of
r r
µ 398600
v= = = 7.78km/s
r 6578
Add 1.72km/s to account for gravity and drag. This totals
9.5km/s = 9500m/s. The Isp = 300s, which means c = 3000m/s. Thus we
have
mp ∆v
= 1 − e− c = .9579
m0
 
.9579
Since m0 = mL + mp , mp = mL 1−.9579 = 682kg.
• Which is sort of a lot!
• What about structural mass and ∆v for orbital maneuvers?
• We’ll return to this problem later
M. Peet Lecture 11: Spacecraft Dynamics 12 / 29
 Example
Lecture 11 Problem: Suppose our mission requires a dry weight of mL = 30kg. Using an
Isp of 300s, how much propellant is required to achieve a circular orbit of
altitude 200km?

2018-05-03 Spacecraft Dynamics Solution: A Circular orbit at 200km requires a total velocity of
r
µ
r
398600
v= = = 7.78km/s
r 6578
Add 1.72km/s to account for gravity and drag. This totals
9.5km/s = 9500m/s. The Isp = 300s, which means c = 3000m/s. Thus we
have
Example mp
m0
∆v
= 1 − e− c = .9579
 
.9579
Since m0 = mL + mp , mp = mL 1−.9579 = 682kg.

• Which is sort of a lot!

• What about structural mass and ∆v for orbital maneuvers?
• We’ll return to this problem later

• m0 is wet mass, dry mass plus propellant.

Solid Rocket Motors

Advantages: Disadvantages:
• Simple • Limited Performance
• Reliable • Not Adjustable (Safety)
• Low Cost • Toxic Byproducts

M. Peet Lecture 11: Spacecraft Dynamics 13 / 29

Solid Rocket Motors

Figure: Thiokol (ATK Launch Systems) = STAR, LEASAT; United Technologies =

IUS

M. Peet Lecture 11: Spacecraft Dynamics 14 / 29

Liquid Monopropellants

Figure: Typical Hydrazine Monopropellant

Advantages: Disadvantages:
• Simple • Lower Performance than
• Reliable bipropellant
• Low Cost

M. Peet Lecture 11: Spacecraft Dynamics 15 / 29

Liquid Bipropellants

Figure: Centaur O2 -H2 upper stage.

Advantages: Disadvantages:
• High Performance • Complicated
• Adjustable • Dangerous
• Sometimes Toxic
M. Peet Lecture 11: Spacecraft Dynamics 16 / 29
Liquid Bipropellants

M. Peet Lecture 11: Spacecraft Dynamics 17 / 29

Hybrid Rockets

Advantages: Disadvantages:
• Throttled • Requires Oxidizer
• Non-Explosive • Bulky
Flown on SpaceShipOne (Developed by SpaceDev, N02, Isp = 250s, Max
Thrust 74kN)

M. Peet Lecture 11: Spacecraft Dynamics 18 / 29

Hybrid Rockets

Figure: American Rocket Company = SpaceDev

M. Peet Lecture 11: Spacecraft Dynamics 19 / 29

Electric Propulsion
Electrothermal: Electrostatic: Electromagnetic:
• Ohmic Heating • Repulsion/Attraction • Ions accelerated by
EM waves
neutralizing electron gun

magnets

electron
gun

negative grid
electron positive grid
neutral propellant atom

positive ion

Advantages: Disadvantages:
• Very High Performance • Low Thrust
• Requires Electricity
M. Peet Lecture 11: Spacecraft Dynamics 20 / 29
 Electric Propulsion
Lecture 11 Electrothermal: Electrostatic: Electromagnetic:
• Ohmic Heating • Repulsion/Attraction • Ions accelerated by

2018-05-03 Spacecraft Dynamics magnets

EM waves
neutralizing electron gun

electron
gun

Electric Propulsion electron

neutral propellant atom

positive ion
negative grid
positive grid

Advantages: Disadvantages:
• Very High Performance • Low Thrust
• Requires Electricity

• SRP average generation is 3.2MW.

Electric Propulsion

The choice of Electrothermal/Electrostatic/Electromagnetic depends on

available power.

M. Peet Lecture 11: Spacecraft Dynamics 21 / 29

Electric Propulsion

M. Peet Lecture 11: Spacecraft Dynamics 22 / 29

Staging
Previously, we assumed the rocket only consisted of payload and propellant:
m0 = mL + mp .
   
mp − ∆v m(t0 ) mL + mp
=1−e c ∆v = c ln = c ln
m0 m(tf ) mL
Which would mean the only way to increase ∆v is to decrease payload or
increase the size of the rocket.
However: Payload is not the only part of the rocket.
• Rocket engines and storage tanks are heavy.
• Typically, structure accounts for ∼
= 1/7 of the propellant weight
m0 = mL + ms + mp = (mL + 1/7mp ) + mp
• While 1/7 may not seem a lot, without staging, it limits the total ∆v to
mp
∆v = c ln( = 2c ∼
) = c ln 7 ∼ = 6km/s (assuming mL = 0)
mp /7
But ∆v = 8km/s is needed for low earth orbit (LEO) - not accounting for drag
or gravity losses (2km/s)!
OMG: Space flight is IMPOSSIBLE!
• It’s a conspiracy.
M. Peet Lecture 11: Spacecraft Dynamics 23 / 29
Structural Mass on Saturn V

M. Peet Lecture 11: Spacecraft Dynamics 24 / 29

Structural Coefficient
Definition 2.
The ratio of structure to total mass is called the structural coefficient, :
ms
=
ms + mp

In the ideal case, structural weight would be discarded as soon as it is no longer

required.
• Continuous Staging
In this ideal scenario, we would have
 
m0
∆v = (1 − )c ln
mL

• The structure simply decreases the efficiency of the fuel!

In Staging, we discard structure at discrete points in time.
h i
m0
• Staging can never be better than ∆v = (1 − )c ln m .
L

M. Peet Lecture 11: Spacecraft Dynamics 25 / 29

 Structural Coefficient
Lecture 11 Definition 2.
The ratio of structure to total mass is called the structural coefficient, :

2018-05-03 Spacecraft Dynamics =

ms
ms + mp

In the ideal case, structural weight would be discarded as soon as it is no longer

required.
• Continuous Staging

Structural Coefficient In this ideal scenario, we would have

∆v = (1 − )c ln

m0


mL

• The structure simply decreases the efficiency of the fuel!

In Staging, we discard structure at discrete points in time.
h i
m0
• Staging can never be better than ∆v = (1 − )c ln m .
L

• Prussing uses structural coefficient 

• Prussing uses the term “mass ratio” to refer to the full weight of a stage over the
empty weight.
mp + ms + mL
Z=
ms + mL
• Prussing uses the term “payload ratio” to indicate the ratio of payload to
structural mass plus propellant mass.
mL
λ=
mp + ms

• I use the term “structural mass fraction” to indicate the mass of structure needed
for every mass of fuel
ms 
=
mp 1−
3-stage Scenario
Suppose we divide the structural and propulsive weight into three components
1. First Stage: ms1 , mp1
2. Second Stage: ms2 , mp2
3. Third Stage: mL , mp3
Then ∆v is the combined ∆v of all three stages.
∆vT = ∆v1 + ∆v2 + ∆v3
 
mp1 + mp2 + mp3 + ms1 + ms2 + mL
= c ln
mp2 + mp3 + ms1 + ms2 + mL
   
mp2 + mp3 + ms2 + mL mp3 + mL
+ c ln + c ln
mp3 + ms2 + mL mL

Optimal choice of mp1 , mp2 and mp3 is difficult.

For a fixed total mass, m0 , we can maximize
• Payload weight
• Total ∆v
A good rule is mp1 = 3mp2 = 9mp3 .
M. Peet Lecture 11: Spacecraft Dynamics 26 / 29
1,2 and 3stage Scenarios (Fixed Total Mass)

M. Peet Lecture 11: Spacecraft Dynamics 27 / 29

Staging on Minuteman ICBM

M. Peet Lecture 11: Spacecraft Dynamics 28 / 29

Summary

This Lecture you have learned:

Introduction to Rocketry
• Mass Consumption
• Specific Impulse and Rocket Types
• ∆v limitations
• Staging

M. Peet Lecture 11: Spacecraft Dynamics 29 / 29