Rocket Propulsion

Rocket PropulsionPhysics

Montwood High School

R. Casao

• Momentum is particularly useful for analyzing a system in which the masses of parts of the system change with time. Rocket propulsion is a typical example of this kind of analysis.

• Newton’s 2nd law F = m·a cannot be applied directly because mass m changes.

• A rocket is propelled forward by rearward ejection of burned fuel that initially was in the rocket.

• The forward force on the rocket is the reaction to the backward force on the ejected material.

• The total mass of the system is constant, but the mass of the rocket itself decreases as material is ejected.

• Consider a rocket fired in outer space with no air resistance and no gravitational force acting on it.

• Consider the x-axis to be along the rocket’s direction of motion.

• At time t, the mass of the rocket is m and the x-component of the velocity is v.

• The x-component of the total momentum at this instant is P1 = m·v.

• In a short time interval dt the mass of the rocket changes by an amount dm.

– dm is negative because the rocket’s mass m decreases with time.

– During dt, a positive mass –dm of burned fuel is ejected from the rocket,

• Let vex be the exhaust speed of the ejected material relative to the rocket.

• The burned fuel is ejected opposite the direction of motion, so the x-component of velocity relative to the rocket is –vex.

• The x-component of the velocity vfuel of the burned fuel: vfuel = v + -vex = v – vex

• The x-component of momentum of the ejected mass (dm) is:

-dm·vfuel = -dm∙(v – vex)

• At the end of time interval dt, the x-component of velocity of the rocket and unburned fuel has increased to v + dv, and its mass has decreased to m + dm (remember that dm is negative).

• The rocket’s momentum is: (m + dm)·(v + dv)

• The total x-component of momentum P2 of the rocket plus the ejected fuel at time t + dt:

P2 = (m + dm)·(v + dv) + (-dm)∙(v – vex)

• Considering the rocket and the fuel as an isolated system, the total x-momentum of the system is conserved and P1 = P2:

m·v = (m + dm)·(v + dv) + (-dm)∙(v – vex)

dvdmdmvdvm

dvdmdmvdvm0

dvdmdmvdmvdmvdvmvmvm

dm)vdm(-vdv)dmdmvdvmv(mvm

ex

ex

ex

ex

• The dm∙dv term can be dropped because it is the product of two small quantities and is smaller than the other terms.

• Divide both sides by dt:

• The acceleration of the rocket is dv/dt; the left side of the equation, m∙a, is equal to the net force or thrust on the rocket.

dtdm

vdtdv

m ex

dtdm

vF ex

• Thrust is proportional to the speed of the ejected fuel vex and to the mass of fuel ejected per unit time, -dm/dt (remember that dm/dt is negative because it is the rate of change of the rocket’s mass).

• The x-component of the rocket’s acceleration is:

• The rocket’s mass decreases continuously while the fuel is being burned.

• If vex and dm/dt are constant, the acceleration increases until all the fuel is gone.

dtdm

mv-

dtdv

a ex

• An effective rocket burns fuel at a rapid rate (large dm/dt) and ejects the burned fuel at a high speed (large vex).

• Rockets work best with no air resistance. The rocket is not pushing against the ground to get into the air; the ejection of the burned fuel pushes the rocket forward.

• If the exhaust speed vex is constant, integrate to find the relationship between the velocity and remaining mass of fuel.

mdm

vdv ex

• Integrate each side of the resulting differential equation:

• The ratio mo/m is the original mass divided by the mass after the buel has been exhausted.

• This ratio is made as large as possible to maximize the speed gain, which means that the initial mass of the rocket is almost all fuel.

mm

lnvvv

mm

lnvvv

mlnvvv

dmm1

vv

mdm

vdv

oexo

oexo

m

mexo

m

mex

v

v

m

m ex

v

v

o

oo

oo

Vertical Motion• For a rocket fired straight up with velocity v

relative to Earth, initially to lift the rocket F th = Fw.

• Fuel is burned at a constant rate R.

• Rocket’s mass at time t is: M = Mo - R·t, where Mo = initial mass of rocket.

• Exhaust gas leaves rocket with velocity vex relative to the rocket.

• Rate at which fuel is burned equal to rate at which mass M decreases.

• Consider the rocket and unspent fuel within it as the system. Neglecting air drag, only external force on the system is that of gravity.

• Rocket equation:

• R∙vex is the force exerted on the rocket by the exhausting fuel, also called thrust, Fth.

RdtdM

andgMFnet

dtdv

MvRgM ex

exexth vdtdM

vRF

• Take up as the +y direction.

• Vertical component of rocket equation:

• Divide through by M and rearrange:

• Derivation for rocket starting at v = 0 m/s and t = 0 s and assuming g to be contant:

dt

dvMvRgM y

ex

gtRM

vRgv

MR

dt

dv

o

exex

y

t

0

t

0o

exy

t

0

t

0o

exv

0y

v

0

t

0 oex

y

exy

exy

dtgdttRM

1vR0v

dtgdttRM

vRv

tRMMdtgMvR

dv

dtgMvR

dvgMvR

dt

dv

y

y

tgMtR

Mlnvv

tgMtRlnMlnvv

tgMlnMtRlnvv

tgM0RlnMtRlnvv

0tgMtRlnvv

tgR

MtRlnvRv

o

oexy

ooexy

ooexy

ooexy

t

0oexy

t

0

t

0

oexy

• If the integration is from vyi to vyf and from mi to mf (no gravity):

• If the integration is from vyi to vyf and from mi to mf (with gravity):

f

iexyiyf

m

m

v

v y

mm

lnvvv

mdm

dvf

i

yf

yi

tgmm

lnvvvf

iexyiyf

Burnout Velocity• Burnout velocity is the final velocity of a

rocket when all the fuel is burned away.– If Mo is the total initial mass of the rocket and

mfuel is the mass of the fuel, the burnout velocity is:

tgmM

Mlnvv

fuelo

oex

Systems with Varying Mass: A Rocket

• So far we have always assumed that the mass of the system stayed constant

• There is an important class of problems where this is not so however – Rockets

• Most of the mass of a rocket is fuel; a rocket provides its thrust by burning its fuel and ejecting the fuel mass at a high velocity


• As you can imagine, it would be difficult to deal with a system where the mass was changing as a function of time

• But that’s really not necessary – we already know how to solve this problem…

• We do it by picking a system where the mass is not changing.


• So the system we pick is one which includes the exhaust gas (mass) as well as the mass of the rocket (which is rapidly becoming emptied its fuel


• Let’s assume that our rocket is isolated out in space to make things a little easier

• We begin by looking at the rocket at some time t and assume it has some initial velocity v


• Now we look at the rocket at some time dt later; the velocity of the rocket is now v + dv and the mass of the rocket is now M + dM where the value dM is a negative quantity


• We need to account for the exhaust however as it is part of our isolated system; it has a mass of -dM and a velocity U


• Since our system is isolated and closed, we can say that, according to the law of conservation of linear momentum, we know that: Pi = Pf

• This can be rewritten as:

dvvdMMUdMMv

Exhaust Rocket


• This equation can be simplified a bit if we consider the relative velocity of the exhaust to the rocket itself:

or

frame torelativeexhaust ofvelocity

exhaust torelativerocket ofvelocity

frame torelativerocket ofvelocity

rel

rel

vdvvU

Uvdvv


• Substituting that back and turning the crank a little we get:

• Dividing each side by dt we then get:

MdvdMv rel

dt

dvMv

dt

dM rel


• If we then replace dM/dt by -R (the rate at which the rocket loses mass) and note that dv/dt is just the acceleration of the rocket, we get the following (the 1st rocket equation):

MaRv rel


• The terms on the left hand side of the equation all depend solely on the design of the rocket engine

• Note that the left side of the equation must be a force as it is equal to mass times acceleration

• The term Rvrel is called the thrust of the rocket engine and we often represent it by T

MaRv rel


• To find the velocity of the rocket at any given time we have to integrate equation9-40

• Rearranging it a bit we get:

M

dMvdv

MdvdMv

rel

rel


• Now we need to integrate:

where ln means the natural logarithm, where vi and vf are the initial and final velocities respectively, and Mi and Mf are the initial and final mass of the rocket respectively

f

i

f

i

M

M

v

v M

dMvdv rel


• The result is (the 2nd rocket equation):

• Note an important point – the change in velocity is directly proportional to the relative velocity (the exhaust velocity) – the higher that is, the higher the change in velocity of the rocket itself

f

iif M

Mvvv lnrel


• Now let’s look at the 2nd term on the right:

• Let’s suppose that 90% of the rocket’s mass is ejected as exhaust – this means that the ratio Mi/Mf is 10:1

f

iif M

Mvvv lnrel


• The natural logarithm of 10 is about 2.3• This means that (given our parameters) the

rocket can never go faster than 2.3 times the exhaust velocity – again a reason to design an engine with the highest possible high exhaust velocity

• It also encourages the design of multi-stage rockets, so that the final mass is as close to that of the payload as possible

Sample Problem: A Shuttle Launch

• The first couple minutes of a shuttle launch can be described very roughly as follows:– The initial mass is about 2 x 106 kg– The final mass (after 2 minutes) is about

1 x 106 kg– The average exhaust speed is about 3000

m/s– The initial velocity is, of course, zero


• If all this were taking place in outer space, with negligible gravity, what would be the shuttle’s speed at the end of this stage?

• What is the thrust during the same period and how does it compare with the initial total weight of the shuttle (on earth)?

• Mi = 2 x 106 kg, Mf = 1 x 106 kg,vrel = 3000 m/s, vi = 0, Δt = 2 minutes


• Plugging the specified values into the 2nd rocket equation we get:

m/s 207910x1

10x2ln30000

ln

6

6

rel

f

f

f

iif

v

v

M

Mvvv


• To get the thrust we use the equation:

• We know that 1 x 106 kg of fuel is used in the first 2 minutes of launch, therefore:

relRvT

kg/s 8333s 120

kg10 x 1

min 2

kg10 x 1 66

R


• To get the thrust we plug the specified values into the equation:

N 25,000,000

m/s 3000kg/s 8333rel

T

T

RvT


• The mass of the shuttle before launch was2 x 106 kg so the shuttle’s weight was:

N 19,600,000

m/s 9.8kg2x10 26

W

W

maW


• The last problem was to determine the thrust-to-weight ratio at launch

• We now have both quantities so we plug them in and get:

28.1N 19,600,000

N 25,000,000

Weight

Thrust

Rocket Propulsion

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