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Rocket Physics
Picture of Saturn V Launch for Apollo 15 Mission. Source: NASA
Rocket physics is basically the application ofNewton's Laws to a system
with variable mass. A rocket has variable mass because its mass
decreases over time, as a result of its fuel (propellant) burning off.
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A rocket obtains thrust by the principle of action and reaction (Newton's
third law). As the rocket propellant ignites, it experiences a very large
acceleration and exits the back of the rocket (as exhaust) at a very high
velocity. This backwards acceleration of the exhaust exerts a "push" force
on the rocket in the opposite direction, causing the rocket to accelerate
forward. This is the essential principle behind the physics of rockets, and
how rockets work.
The equations of motion of a rocket will be derived next.
Rocket Physics Equations Of Motion
To find the equations of motion, apply the principle of impulse and
momentum to the "system", consisting of rocket and exhaust. In this
analysis of the rocket physics we will use Calculus to set up the governing
equations. For simplicity, we will assume the rocket is moving in a vacuum,
with no gravity, and no air resistance (drag).
To properly analyze the rocket physics, consider the figure below which
shows a schematic of a rocket moving in the vertical direction. The two
stages, (1) and (2), show the "state" of the system at time t and time
t+dt, where dt is a very small (infinitesimal) time step. The system
(consisting of rocket and exhaust) is shown as inside the dashed line.
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Where:
m is the mass of the rocket (including propellant), at stage (1)
me is the total mass of the rocket exhaust (that has already exited the
rocket), at stage (1)
v is the velocity of the rocket, at stage (1)
Je is the linear momentum of the rocket exhaust (that has already exited
the rocket), at stage (1). This remains constant between (1) and (2)
dme is the mass of rocket propellant that has exited the rocket (in the
form of exhaust), between (1) and (2)
dv is the change in velocity of the rocket, between (1) and (2)
ve is the velocity of the exhaust exiting the rocket, at stage (2)
Note that all velocities are measured with respect to ground (an inertialreference frame).
The sign convention in the vertical direction is as follows: "up" is positive
and "down" is negative.
Between (1) and (2), the change in linear momentum in the vertical
direction of all the particles in the system, is due to the sum of the
external forces in the vertical direction acting on all the particles in the
system.
We can express this mathematically using Calculus and the principle ofimpulse and momentum:
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where Fy is the sum of the external forces in the vertical direction acting
on all the particles in the system (consisting of rocket and exhaust).
Expand the above expression. In the limit as dt0 we may neglect the
"second-order" term dmedv. Divide by dt and simplify. We get
Since the rocket is moving in a vacuum, with no gravity, and no air
resistance (drag), then Fy = 0 since no external forces are acting on the
system. As a result, the above equation becomes
The left side of this equation must represent the thrust acting on the
rocket, since a = dv/dt is the acceleration of the rocket, and F=ma
(Newton's second law).
Therefore, the thrust T acting on the rocket is equal to
The term v+ve is the velocity of the exhaust gases relative to the rocket.
This is approximately constant in rockets. The term dme/dt is the burn
rate of rocket propellant.
As the rocket loses mass due to the burning of propellant, its acceleration
increases (for a given thrust T). The maximum acceleration is therefore
just before all the propellant burns off.
From equation (2),
which becomes
The mass of the ejected rocket exhaust equals the negative of the masschange of the rocket. Thus,
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Therefore,
Again, the term v+ve is the velocity of the exhaust gases relative to the
rocket, which is approximately constant. For simplicity set u = v+ve.
Integrate the above equation using Calculus. We get
This is a very useful equation coming out of the rocket physics analysis,
shown above. The variables are defined as follows: vi is the initial rocket
velocity and mi is the initial rocket mass. The terms v and m are the
velocity of the rocket and its mass at any point in time thereafter
(respectively). Note that the change in velocity (delta-v) is always thesame no matter what the initial velocity vi is. This is a very useful result
coming out of the rocket physics analysis. The fact that delta-v is
constant is useful for those instances where powered gravity assist is
used to increase the speed of a rocket. By increasing the speed of the
rocket at the point when its speed reaches a maximum (during the
periapsis stage), the final kinetic energy of the rocket is maximized. This in
turn maximizes the final velocity of the rocket. To visualize how this
works, imagine dropping a ball onto a floor. We wish to increase the
velocity of the ball by a constant amount delta-v at some point during its
fall, such that it rebounds off the floor with the maximum possible velocity.
It turns out that the point at which to increase the velocity is just before
the ball strikes the floor. This maximizes the total energy of the ball(gravitational plus kinetic) which enables it to rebound off the floor with
the maximum velocity (and the maximum kinetic energy). This maximum
ball velocity is analogous to the maximum possible velocity reached by the
rocket at the completion of the gravity assist maneuver. This is known as
the Oberth Effec t (opens in new window). It is a very useful principle
related to rocket physics.
In the following discussion on rocket physics we will look at staging and
how it can also be used to obtain greater rocket velocity.
Rocket Physics Staging
Often times in a space mission, a single rocket cannot carry enough
propellant (along with the required tankage, structure, valves, engines and
so on) to achieve the necessary mass ratio to achieve the desired final
orbital velocity (v). This presents a unique challenge in rocket physics.
The only way to overcome this difficulty is by "shedding" unnecessary
mass once a certain amount of propellant is burned off. This is
accomplished by using different rocket stages and ejecting spent fuel
tanks plus associated rocket engines used in those stages, once those
stages are completed.
The figure below illustrates how staging works.
http://en.wikipedia.org/wiki/Oberth_effect -
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Source: NASA
The picture below shows the separation stage of the Saturn V rocket for
the Apollo 11 mission.
Source: NASA
The picture below shows the separation stage of the twin rocket boosters
of the Space Shuttle.
http://www.nasa.gov/http://www.nasa.gov/ -
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Source: NASA
In the following discussion on rocket physics we will look at rocket
efficiency and how it relates to modern (non-rocket powered) aircraft.
Rocket Physics Efficiency
Rockets can accelerate even when the exhaust relative velocity is moving
slower than the rocket (meaning ve is in the same direction as the rocket
velocity). This differs from propeller engines or air breathing jet engines,
which have a limiting speed equal to the speed at which the engine can
move the air (while in a stationary position). So when the relative exhaust
air velocity is equal to the engine/plane velocity, there is zero thrust. This
essentially means that the air is passing through the engine without
accelerating, therefore no push force (thrust) is possible. Thus, rocket
physics is fundamentally different from the physics in propeller engines and
jet engines.
Rockets can convert most of the chemical energy of the propellant into
mechanical energy (as much as 70%). This is the energy that is converted
into motion, of both the rocket and the propellant/exhaust. The rest ofthe chemical energy of the propellant is lost as waste heat. Rockets are
designed to lose as little waste heat as possible. This is accomplished by
having the exhaust leave the rocket nozzle at as low a temperature as
possible. This maximizes the Carnot efficiency, which maximizes the
mechanical energy derived from the propellant (and minimizes the thermal
energy lost as waste heat). Carnot efficiency applies to rockets because
rocket engines are a type of heat engine, converting some of the initial
heat energy of the propellant into mechanical work (and losing the
remainder as waste heat). Thus, rocket physics is related to heat engine
physics. For more information see Wikipedia's article on heat engines
(opens in new window).
In the following discussion on rocket physics we will derive the equation of
motion for rocket flight in the presence of air resistance (drag) and
gravity, such as for rockets flying near the earth's surface.
Rocket Physics Flight Near Earth's Surface
The rocket physics analysis in this section is similar to the previous one,
but we are now including the effect of air resistance (drag) and gravity,
which is a necessary inclusion for flight near the earth's surface.
For example, lets consider a rocket moving straight upward against an
atmospheric drag force FD
and against gravity g (equal to 9.8 m/s2 on
earth). The figure below illustrates this.
http://en.wikipedia.org/wiki/Heat_enginehttp://www.nasa.gov/ -
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where G is the center of mass of the rocket at the instant shown. The
weight of the rocket acts through this point.
From equations (1) and (3) and using the fact that acceleration a = dv/dt
we can write the following general equation:
The sum of the external forces acting on the rocket is the gravity force
plus the drag force. Thus, from the above equation,
As a result,
This is the general equation of motion resulting from the rocket physics
analysis accounting for the presence of air resistance (drag) and gravity.
Looking closely at this equation, you can see that it is an application of F
=ma (Newtons second law). Note that the mass m of the rocket changes
with time (due to propellant burning off), and T is the thrust given from
equation (3). An expression for the drag force FD can be found on the
page on drag force.
Due to the complexity of the drag term FD, the above equation must be
solved numerically to determine the motion of the rocket as a function oftime.
Another important analysis in the study of rocket physics is designing a
rocket that experiences minimal atmospheric drag. At high velocities, air
resistance is significant. So for purposes of energy efficiency, it is
necessary to minimize the atmospheric drag experienced by the rocket,
since energy used to overcome drag is energy that is wasted. To minimize
drag, rockets are made as aerodynamic as possible, such as with a pointed
nose to better cut through the air, as well as using stabilizing fins at the
rear of the rocket (near the exhaust), to help maintain steady orientation
during flight.
In the following discussion on rocket physics we will take a closer look atthrust.
Rocket Physics A Closer Look At Thrust
The thrust given in equation (3) is valid for an optimal nozzle expansion.
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This assumes that the exhaust gas flows in an ideal manner through the
rocket nozzle. But this is not necessarily true in real-life operation.
Therefore, in the following rocket physics analysis we will develop a thrust
equation for non-optimal flow of the exhaust gas through the rocket
nozzle.
To set up the analysis consider the basic schematic of a rocket engine,
shown below.
Where:
Pi is the internal pressure inside the rocket engine (which may vary with
location)
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Pa is the ambient pressure outside the rocket engine (assumed constant)
Pe is the pressure at the exit plane of the rocket engine nozzle (this is
taken as an average pressure along this plane)
Ae is the cross-sectional area of the opening at the nozzle exit plane
The arrows along the top and sides represent the pressure acting on the
wall of the rocket engine (inside and outside). The arrows along the
bottom represent the pressure acting on the exhaust gas, at the exit
plane.
Gravity and air resistance are ignored in this analysis (their effect can be
included separately, as shown in the previous section).
Next, isolate the propellant (plus exhaust) inside the rocket engine. It is
useful to do this because it allows us to fully account for the contac t
force between rocket engine wall and propellant (plus exhaust). This
contac t force can then be related to the thrust experienced by the
rocket, as will be shown. The schematic below shows the isolated
propellant (plus exhaust). The dashed blue line (along the top and sides)
represents the contact interface between the inside wall of the rocketengine and the propellant (plus exhaust). The dashed black line (along the
bottom) represents the exit plane of the exhaust gas, upon which the
pressure Pe acts.
where Fw is the resultant downward force exerted on the propellant (plus
exhaust) due to contact with the inside wall of the rocket engine
(represented by the dashed blue line). This force is calculated by: (1)
Multiplying the local pressure Pi at a point on the inside wall by a
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differential area on the wall, (2) Using Calculus, integrating over the entire
inside wall surface to find the resultant force, and (3) Determining the
vertical component of this force (Fw). The details of this calculation are
not shown here.
(Note that, due to geometric symmetry of the rocket engine, the resultant
force acts in the vertical direction, and there is no sideways component).
Now, sum all the forces acting on the propellant (plus exhaust) and then
apply Newton's second law:
Where:
mp is the mass of the propellant/exhaust inside the rocket engine
a is the acceleration of the rocket engine
u is the velocity of the exhaust gases relative to the rocket, which is
approximately constant
dme/dt is the burn rate of the propellant
The right side of the above equation is derived using the same method
that was used for deriving equation (1). This is not shown here.
(Note that we are defining "up" as positive and "down" as negative).
Next, isolate the rocket engine, as shown in the schematic below.
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where Frb is the force exerted on the rocket engine by the rocket body.
Now, sum all the forces acting on the rocket engine and then apply
Newton's second law:
where mre is the mass of the rocket engine. The term PaAe is the force
exerted on the rocket engine due to the ambient pressure acting on the
outside of the engine.
Now, by Newtons third law Frb acting on the rocket body is pointing up
(positive). Therefore, by Newtons second law we can write
where mrb is the mass of the rocket body (which excludes the mass of the
rocket engine and the mass of propellant/exhaust inside the rocket
engine).
Combine the above two equations and we get
Combine equations (5) and (6) and we get
where the term (mp+mre+mrb) is the total mass of the rocket at any
point in time. Therefore the left side of the equation must be the thrust
act ing on the rocket (since F=ma, by Newton's second law).
Thus, the thrust T is
This is the most general equation for thrust coming out of the rocket
physics analysis, shown above. The first term on the right is the
momentum thrust term, and the last term on the right is the pressure
thrust term due to the difference between the nozzle exit pressure and
the ambient pressure. In deriving equation (3) we assumed that Pe = Pa,
which means that the pressure thrust term is zero. This is true if there is
optimal nozzle expansion and therefore maximum thrust in the rocket
nozzle. However, the pressure thrust term is generally small relative to the
momentum thrust term.
A subtle point regarding Pe = Pa is that Je (the momentum of the rocket
exhaust, described in the first section) remains constant. This is because
the pressure force pushing on the exhaust at the rocket nozzle exit (due
to the ressure P exactl balances the ressure force ushin on the
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remainder of the exhaust due to the ambient pressure Pa. Again, we are
ignoring gravity and air resistance (drag) in the derivation of equation (3)
and (7). To include their effect we simply add their contribution to the
thrust force, as shown in the previous section on rocket physics.
The rocket physics analysis in this section is basically a force and
momentum analysis. But to do a complete thrust analysis we would have
to look at the thermal and fluid dynamics of the expansion process, as the
exhaust gas travels through the rocket nozzle. This analysis (not
discussed here) enables one to optimize the engine design plus nozzle
geometry such that optimal nozzle expansion is achieved during operation
(or as close to it as possible).
The flow of the exhaust gas through the nozzle falls under the category of
compressible supersonic flow and its treatment is somewhat complicated.
For more information on this see Wikipedia's page on rocket engines (opens
in new window). And here is a summary of the rocket thrust equations
provided by NASA (opens in new window).
In the following discussion on rocket physics we will look at the energy
consumption of a rocket moving through the air at constant velocity.
Rocket Physics Energy Consumption For Rocket Moving At
Constant Velocity
An interesting question related to rocket physics is, how much energy is
used to power a rocket during its flight? One way to answer that is to
consider the energy use of a rocket moving at constant velocity, such as
through the air. Now, in order for the rocket to move at constant velocity
the sum of the forces acting on it must equal zero. For purposes of
simplicity let's assume the rocket is traveling horizontally against the force
of air resistance (drag), and where gravity has no component in the
direction of motion. The figure below illustrates this schematically.
The thrust force T acting on the rocket is equal to the air drag FD, so that
T = FD.
Let's say the rocket is moving at a constant horizontal velocity v, and it
travels a horizontal distance d. We wish to find the amount of mechanical
energy it takes to move the rocket this distance. Note that this is not the
same as the total amount of chemical energy in the spent propellant, but
rather the amount of energy that was converted into motion. This amount
of energy is always less than the total chemical energy in the propellant,
due to naturally occurring losses, such as waste heat generated by the
rocket engine.
Thus, we must look at the energy used to push the rocket (in the forward
direction) and add it to the energy it takes to push the exhaust (in the
backward direction). To do this we can apply the principle of work and
energy.
For the exhaust gas:
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Where:
me is the total mass of the exhaust ejected from the rocket over the flight
distance d
u is the velocity of the exhaust gases relative to the rocket, which is
approximately constant
Ue is the work required to change the kinetic energy of the rocket
propellant/exhaust (ejected from the rocket) over the flight distance d
For the rocket:
Where:
T is the thrust acting on the rocket (this is constant if we assume the
burn rate is constant)
Ur is the work required to push the rocket a distance d
Set R = dme/dt (constant burn rate), and assuming optimal nozzle
expansion we have
Therefore,
Now, the time t it takes for the rocket to travel a horizontal distance d is
The total mass of exhaust me is
Substitute this into equation (8), then combine equations (8) and (9) to
find the total mechanical energy used to propel the rocket over a distance
d. Therefore,
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Since T = Ru,
But T = FD, which means we can substitute T with the drag force in the
above equation.
Now,
Where:
C is the drag coefficient, which can vary along with the speed of the
rocket. But typical values range from 0.4 to 1.0
is the density of the air
A is the projected cross-sectional area of the rocket perpendicular to the
flow direction (that is, perpendicular to v)
v is the speed of the rocket relative to the air
Substitute the above equation into the previous equation (for T) and we
get
As you can see, the higher the velocity v, the greater the energy required
to move the rocket over a distance d, even though the time it takes is
less. Furthermore, rockets typically have a large relative exhaust velocity
u which makes the energy expenditure large, as evident in the above
equation. (The relative exhaust velocity can be in the neighborhood of a
few kilometers per second).
This tells us that rockets are inefficient for earth-bound travel, due to theeffects of air resistance (drag) and the high relative exhaust velocity u. It
is only their high speed that makes them attractive for earth-bound travel,
because they can "get there" sooner, which is particularly important for
military and weapons applications.
For rockets that are launched into space (such as the Space Shuttle) the
density of the air decreases as the altitude of the rocket increases. This
decrease, combined with the increase in rocket velocity, means that the
drag force will reach a maximum at some altitude (typically several
kilometers above the surface of the earth). This maximum drag force must
be withstood by the rocket body and as such is an important part of
rocket physics analysis, and design.
In the following discussion on rocket physics we will look at the energy
consumption of a rocket moving through space.
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Space
A very useful piece of information in the study of rocket physics is how
much energy a rocket uses for a given increase in speed (delta-v), while
traveling in space. This analysis is conveniently simplified somewhat since
gravity force and air drag are non-existent. Once again, we are assuming
optimal nozzle expansion.
We will need to apply the principle of work and energy to the system
(consisting of rocket and propellant/exhaust), to determine the requiredenergy.
The initial kinetic energy of the system is:
Where:
mi is the initial rocket (plus propellant) mass
vi is the initial rocket (plus propellant) velocity
The final kinetic energy of the rocket is:
Where:
mf is the final rocket mass
vf is the final rocket velocity
The exhaust gases are assumed to continue traveling at the same velocity
as they did upon exiting the rocket. Therefore, the final kinetic energy of
the exhaust gases is:
Where:
dme is the infinitesimal mass of rocket propellant that has exited the
rocket (in the form of exhaust), over a very small time duration
ve is the velocity of the exhaust exiting the rocket, at stage (2), at a
given t ime
The last term on the right represents an integration, in which you have to
sum over all the exhaust particles for the whole burn time.
Therefore, the final kinetic energy of the rocket (plus exhaust) is:
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Now, apply the principle of work and energy to all the particles in the
system, consisting of rocket and propellant/exhaust:
Substituting the expressions for T1
and T2
into the above equation we get
Where:
U is the mechanical energy used by the rocket between initial and final
velocity (in other words, for a given delta-v). This amount of energy is
less than the total chemical energy in the propellant, due to naturally
occurring losses, such as waste heat generated by the rocket engine. U
must be solved for in the above equation.
and
and
From equation (4),
where u is the velocity of the exhaust gases relative to the rocket
(constant).
Hence,
Note that all velocities are measured with respect to an inertial reference
frame.
After a lot of algebra and messy integration we find that,
This answer is very nice and compact, and it does not depend on the
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initial velocity vi. This is perhaps a surprising result coming out of this
rocket physics analysis.
If we want to find the mechanical power Pgenerated by the rocket,
differentiate the above expression with respect to time. This gives us
where dme/dt is the burn rate of rocket propellant.
Note that, in the above energy calculations, the rocket does not have to
be flying in a straight line. The required energy U is the same regardless of
the path taken by the rocket between initial and final velocity. However, it
is assumed that any angular rotation of the rocket stays constant (and
can therefore be excluded from the energy equation), or it is small enough
to be negligible. Thus, the energy bookkeeping in this analysis of the
rocket physics only consists of translational rocket velocity.
This concludes the discussion on rocket physics.
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