rocket physcs-engineering

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23 10 2012 Rocket Physics

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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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www.real-world-physics-problems.com/rocket-physics.html

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/


8/18



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.
http://www.real-world-physics-problems.com/drag-force.html


9/18



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)


10/18



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


11/18



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.


12/18



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


13/18



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:
http://www.grc.nasa.gov/WWW/K-12/airplane/rktthsum.htmlhttp://en.wikipedia.org/wiki/Rocket_engine


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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,


15/18



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.


16/18

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:


17/18



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


18/18


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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