Transcript Rocket Propulsion

Rocket Propulsion
Physics
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 SF = 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)
m  v  (m  v  m  dv  v  dm  dm  dv)  (-v  dm  v ex  dm)
m  v  m  v  m  dv  v  dm  v  dm  v ex  dm  dm  dv
0  m  dv  v ex  dm  dm  dv
m  dv  v ex  dm  dm  dv
• 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: m  dv   v  dm
dt
ex
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.
F  v ex
dm

dt
• 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:
dv - v ex dm
a


dt
m dt
• 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.
• 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.
dv  v ex
dm

m
• 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.
dm
vodv  mo v ex  m
m 1
v
v v  v ex  
 dm
o
mo m
v
m
m
v  v o  v ex  ln m m
m
v  v o  v ex  ln
mo
mo
v  v o  v ex  ln
m
o
Vertical Motion
• For a rocket fired straight up with velocity v
relative to Earth, initially to lift the rocket Fth
= 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.
dM
Fnet  M  g and
 R
dt
• Rocket equation: M  g  R  v  M  dv
ex
dt
• R∙vex is the force exerted on the rocket by
the exhausting fuel, also called thrust, Fth.
Fth  R  v ex
dM

 v ex
dt
• Take up as the +y direction.
• Vertical component of rocket equation:
 M  g  R  v ex  M 
dv y
dt
• Divide through by M and rearrange:
dv y
R  v ex
R

 v ex  g 
g
dt
M
Mo  R  t
• Derivation for rocket starting at v = 0 m/s
and t = 0 s and assuming g to be contant:
dv y
R  v ex

g
dt
M

vy
0
 R  v ex

dv y  
 g   dt
 M

 R  v ex

dv y   
 g   dt
0
 M

t
M  Mo  R  t
t
 R  v ex 
  dt   g  dt
v y   
0
0 M Rt
0
 o

t
t
1
v y  0  R  v ex  
 dt  g   dt
0 M Rt
0
o
vy
t
v y  R  v ex 
 ln R  t  Mo
R
v y  v ex   ln R  t  Mo
t
t
g t0
0
t
0
 g  t  0 
v y  v ex   ln R  t  Mo  ln R  0  Mo   g  t
v y  v ex   ln R  t  Mo  ln  Mo   g  t
v y  v ex  ln Mo  ln R  t  Mo   g  t
Mo
v y  v ex  ln
g t
R  t  Mo
• If the integration is from vyi to vyf and from
mi to mf (no gravity): v yf
mf dm
v yi dv y  mi m
mi
v yf  v yi  v ex  ln
mf
• If the integration is from vyi to vyf and from
mi to mf (with gravity):
mi
v yf  v yi  v ex  ln
gt
mf
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:


Mo
  g  t
v  v ex  ln
 Mo  mfuel 
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
Systems with Varying Mass: A Rocket
• 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.
Systems with Varying Mass: A Rocket
• 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
Systems with Varying Mass: A Rocket
• 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
Systems with Varying Mass: A Rocket
• 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
Systems with Varying Mass: A Rocket
• 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
Systems with Varying Mass: A Rocket
• 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:
Mv  dM U   M  dM v  dv
Exhaust
Rocket
Systems with Varying Mass: A Rocket
• This equation can be simplified a bit if we
consider the relative velocity of the
exhaust to the rocket itself:



velocityof rocket  velocityof rocket  velocityof exhaust
relativeto frame
relativeto exhaust
relativeto frame
or
v  dv   vrel  U
U  v  dv  vrel

Systems with Varying Mass: A Rocket
• Substituting that back and turning the
crank a little we get:
 dMvrel  Mdv
• Dividing each side by dt we then get:
dM
dv

vrel  M
dt
dt
Systems with Varying Mass: A Rocket
• 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):
Rv rel  Ma
Systems with Varying Mass: A Rocket
• The terms on the left hand side of the equation
all depend solely on the design of the rocket
engine
Rv rel  Ma
• 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
Systems with Varying Mass: A Rocket
• To find the velocity of the rocket at any
given time we have to integrate equation
9-40
• Rearranging it a bit we get:
 dMvrel  Mdv
dM
dv  vrel
M
Systems with Varying Mass: A Rocket
• Now we need to integrate:
vf
Mf
 dv  v 
rel
vi
Mi
dM
M
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
Systems with Varying Mass: A Rocket
• The result is (the 2nd rocket equation):
Mi
v f  vi  vrel ln
Mf
• 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
Systems with Varying Mass: A Rocket
• Now let’s look at the 2nd term on the right:
Mi
v f  vi  vrel ln
Mf
• Let’s suppose that 90% of the rocket’s
mass is ejected as exhaust – this means
that the ratio Mi/Mf is 10:1
Systems with Varying Mass: A Rocket
• 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
Sample Problem: A Shuttle Launch
• 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
Sample Problem: A Shuttle Launch
• Plugging the specified values into the 2nd
rocket equation we get:
Mi
v f  vi  vrel ln
Mf
6
2x10
v f  0  3000 ln
6
1x10
v f  2079 m/s
Sample Problem: A Shuttle Launch
• To get the thrust we use the equation:
T  Rv rel
• We know that 1 x 106 kg of fuel is used in
the first 2 minutes of launch, therefore:
6
6
1 x 10 kg 1 x 10 kg
R

 8333 kg/s
2 min
120 s
Sample Problem: A Shuttle Launch
• To get the thrust we plug the specified
values into the equation:
T  Rv rel
T  8333 kg/s 3000 m/s 
T  25,000,000 N
Sample Problem: A Shuttle Launch
• The mass of the shuttle before launch was
2 x 106 kg so the shuttle’s weight was:
W  ma


W  2x10 kg 9.8 m/s
6
W  19,600,000 N
2

Sample Problem: A Shuttle Launch
• 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:
Thrust 25,000,000 N

 1.28
Weight 19,600,000 N