In today’s class we continue with the theory
of rocket propulsion. In order to focus
ourselves and just make sure we are on the
right track; I will go through a few slides
which will illustrate what we did in the earlier
class. We will also find out that not only
rockets eject momentum to propel; but in nature
itself we have some creatures which
make use of the same principle.
Let us briefly go through some of the slides.
Rockets are used to launch satellites or may
be objects in space. We talked in terms of
circular orbits, we talked in terms of
geosynchronous orbits, polar orbits, sun synchronous
orbits and retrograde orbits,
different orbits.
We also talked in terms of elliptical orbits
and we said elliptical orbit is one wherein
you
have two foci or focal points. This is shown
here; the Earth is at the focal point and
this
spacecraft going round in an elliptic orbit.
We defined something known as eccentricity
which was the distance between the two foci
and the major axis which is 2a over here.
We also said for an elliptical orbit you could
have an inclination, even a circular orbit
could have an inclination and the inclination
is between the equatorial plane and the
plane of the orbit. I want to make it clear
because we defined one orbit known as a
Molniya orbit and we told that in a country
like Russia in the northern hemisphere
wherein the satellite has to stay in the orbit
above the northern hemisphere for a longer
time; we have highly elliptical orbits wherein
the apogee is at a distance of something
like almost like 60000 to 70000 kilometers
and the perigee is quite small of the order
of
6000 kilometers.
With the result that say I have the Earth
over here, the northern hemisphere is on the
upper portion, the satellite is in the northern
hemisphere for a much longer time.
Something like 11 hours it is over this northern
part and only one hour in the southern
region. The countries in the upper north can
view the satellite for a longer time. This
Molniya orbit is particularly important and
was developed by Russia. The apogee is of
the order of 70,000 kilometers. The inclination
of this orbit was something like 63.4
degrees.
We also talked in terms of the rocket principle
and in the example of the sled we
reviewed the velocities achieved when stones
were thrown out simultaneously and one
after the other.
We got the net velocity of the sled was something
like the mass of the stones thrown
divided by the total mass into the velocity
with which the stones were thrown. This was
when the stones were thrown simultaneously.
The important thing here is when we are
looking at the sled which is moving, we are
talking with the inertial frame in mind. I
look
at the sled. I am standing outside as shown
by the observer.
You know, I am observing it from the inertial
frame of reference and writing the
momentum conservation equation.
When the two stones are thrown not together
but one after the other; we got a higher
value of velocity because instead of 2m by
M when the two stones were thrown
simultaneously, we got m by M plus m over
M minus m when the two stones were
thrown one after the other. Since M minus
m (capital M minus small m)is less than small
M, we got an increased velocity for the sled
and so if I have a series of stones thrown
one
after the other; well I can get a higher velocity
than when the stones thrown together. The
question is why is this so?
Because when the net mass is decreasing I
accelerate a lesser mass and I get a higher
value of velocity. Therefore, we find that
when I throw one stone after the other, I
do not
have something like stones being thrown once
at the initial time in which case I have the
net velocity is equal to u plus a t where
a is the acceleration; but, since I am accelerating
gradually in view of the reduced mass, I get
much higher velocity.
Let us look at some examples. I chose the
above example from the national geographical
magazine. What is shown here is something
known as a squid. Squid is like a large fish
something like 10 meters long and is found
off the coast of Japan and off New Zealand.
It periodically visits these coastal areas
and it is an endangered specious and it propels
using the rocket principle. Let us look at
the parts of this particular squid. You have
something like a mantle or a funnel here through
which it sucks in water.
The above slide shows a squid which was caught
off the coast of new Zealand and you
see the length compared to this man is something
like 10 meters long.
Let us go to the principle of is motiont.
You know there is something like a funnel
here
at the rear of the squid in this slide. To
move itself it opens its funnel or mantle
and gulps
in water. As it gulps in water it gulps in
some sand may be eggs may be small fish or
whatever it is available around it in the
water. Then it closes the funnel i.e., it
closes its
gate to the funnel or mantle or the mouth
as it were and then it through its muscles
it
contracts the muscles such that it builds
some pressure of water which it has gulped.
And then when it wants to move, it just opens
the gate again and spouts out the water in
this direction and when it spouts out the
water in this direction it moves in the opposite
direction. It keeps on squirting the water
and it propels itself. Therefore, when it
squirts
out the water it moves. Whenever it wants
to move; it expels water, waste eggs and all
that and that is how it moves. The principle
is very similar to a rocket. It collects water,
pressurizes the water, releases the water
gradually and it moves.
And the type of velocity what it gets is quite
substantial. It pressurizes the water to
around 0.4 times the atmosphere and once when
it pushes it out, it is able to leap
something like a 50 meters and gets a velocity
of something like 2.5 kilometers per hour.
This is quite phenomenal when you consider
that this is in water is viscous liquid. It
is
able to go at that speed because of the gradual
release of pressurized water from its
mouth i.e., funnel or mantle
Now I show the funnel again. It opens its
mouth, water enters, it contracts itself and
then
it opens it, closes the gate, compresses it,
releases the pressure, then the water squirts
out
and whatever is available along with the water
is squited out and it moves.
And this is the principle of what we call
as the GIANT SQUID. In fact in US we have
had rocket projects named as Project Squid.
In what way do you think is the motion of
a
Squid different from a fish?
A fish has fins by which it displaces the
water. That means it slowly displaces the
water
like I go on a boat let us say. I am sitting
in the boat I have an ore. With the ore I
displace
water. I displace some water i.e., provide
velocity to the water but, the mass of water
displaced is large but, the velocity during
the displacement is small.
In the case of a squid: It takes in a small
amount of water, it pushes it out at high
velocity
and its able to do a much better job. This
same principle is used in a water rocket.
I show
a water rocket; if I want to make one, all
what I do is I take one of these bottles,
partially
fill in with water and then I invert it and
pressurize the water. I remove the cap and
when
I do so water squirts out as a jet and the
water rocket moves up.
The principle of any rocket is quite identical.
Only major difference is that you need
higher velocities. Therefore, you put more
enthalpy into the working medium of the
rocket which is then expanded out. The medium
which is expanded would have a higher
value if its enthalpy is higher. Therefore,
we get higher velocity with a gas heated in
chemical or nuclear rockets.
Having said about giant squid, I discuss another
example. There is a creature by name
bombardier beetle. You know what a beetle
is: in Tamil we call as “vandu” and in
Malayalam it is known as “nandu”. It is
a harmless creature. It cannot fly well as
it is
bulky and it goes round either by flight or
by walk. Whenever it comes into our house,
all what we do is push it over it a piece
of cardboard or thick paper and throw it out.
A particular form of beetle known as bombardier
beetle has a different means of
locomotion. Beetle we had said is a harmless
creature; ants sting it and nature has given
its some means to protect itself.
It has something like two stomachs, and after
the two stomach there is another receptacle
which is like a third stomach. In one of these
stomachs secretes hydrogen peroxide which
is an oxidizer. In the second stomach it secretes
hydroquinone which is a fuel. Hydrogen
peroxide being an oxidizer and hydroquinone
being a fuel can react to form hot gases.
When the bombardier beetle is chased by the
ants or when it is bitten by the ants it
immediately squirts the hydrogen peroxide
and hydroquinone into its third stomach
which is coated with enzymes. The fuel and
oxidizer react to generate hot gases in the
presence of the enzymes and these hot gases
are forced out in the form a jet which kills
the ants or else they are chased out as the
bombardier beetle moves forward.
Let us examine the processes again. In the
above slide, the two stomachs are shown
which form hydrogen peroxide and hydroquinone.
They are secreted here. In the third
stomach below is a lining of mucus (an enzyme)
which acts as a catalyst. The catalyst
promotes the reaction and whenever it is attacked
it just squirts out the hot gases. This is
similar to the processes of combustion and
expansion in a liquid propellant rocket.
What does a liquid propellant rocket consist
of? We are yet to study it. You have a fuel
tank, you have an oxidizer tank, you pump
the fuel and oxidizer into it, you ignite
it and
you push the gases out. So, this small insect
which is available in nature works on the
principle of a liquid propellant rocket.
I think we should look at nature to understand
many of the things what we are studying.
Everything is there may be we have to be more
observant. Having said that let us get into
some more details. Let us illustrate the rockets
used in satellites because we told that
whenever a satellite is there in space in
geostationary orbit and its life is nearing
completion, we have to push it out. A satellite
has something like 16 rockets which are
there at the edges of this satellite and these
are used for correcting the attitude. May
be
station keeping of the satellite and whenever
the life time of the rocket is near to being
over, we fire some of these rockets such that
we remove it from the geostationary orbit
and push it into deep space. That means we
make it escape to deep space.
Based on the above background how to develop
the theory of rocket propulsion which
leads to the rocket equation? The rocket equation
was developed not very early, only in
the year 1903 and that also by a Russian school
teacher by name TSIALKOWSKI. Now,
let us see how this is done and what is this
rocket equation also referred to as
TSIALKOWSKI equation.
Let us derive it first. We will follow the
same procedure what we adopted while finding
out the velocity gain by the sled, these two
boys standing on it, throwing one stone after
the other. We will make some simplifying assumptions.
Let us assume I have a rocket as shown in
this figure. We will assume rocket has a shape
something like this. It need not really be
the only shape and may be you will be come
out
with better configurations of rockets. Let
say at time t it is moving with a velocity
let us
say V. Let its mass at time t be capital M.
After the time t plus a small time let us
say
delta t or rather a small time period delta
tau seconds, it gains a small velocity delta
V.
Now how does it gain this velocity? It is
moving forward and then during this small
time,
a small mass delta m is being ejected out.
The rocket is ejecting matter and in a small
time delta tau a small mass delta m is ejected
out with a velocity VJ. I do not know the
direction in which the mass is ejected out.
We will presume it to be in the direction
of
motion of the rocket.
And therefore, the final mass of this rocket
is going to be M minus delta m over here.
Let
us presume that the velocity of the rocket
at time t plus delta tau seconds is V plus
delta
V. Since the 
rocket has lost a 
mass delta m in this small time, the final
mass of the rocket
is M minus delta m.
Now we talk in terms of inertial frame of
reference and therefore we watch the rocket
from the ground. I watch rocket go up with
a velocity V and then after a time delta tau
I
am looking at it going with a velocity V plus
delta V. Now, I do the momentum balance
in the inertial frame of reference, and what
is it I get?
The initial momentum of the rocket is equal
to MV. It is mass into velocity at time t.
Now what is the momentum of this rocket at
time t plus delta tau as a I am seeing from
the inertial frame of reference? I have M
minus delta m into V plus dv plus you know
you also find that delta m has been pushed
out of the rocket with velocity VJ.. The
velocity VJ is with respect to the rocket
and from the inertial frame of reference when
the
rocket is moving with velocity V plus delta
V, the mass delta m will appear to leave with
a velocity V plus delta V plus VJ. This is
what is shown in the slide above. That is
the
velocity of this parcel of mass delta m and
is the relative velocity VJ plus V plus delta
V.
This is how we wrote the equation for the
sled. We ignore the gravitational field and
the
resistance to motion of the rocket by the
air and we get the momentum to be conserved
exactly in the same way as in the sled problem.
In the case of the sled which was initially
stationary, the initial momentum was 0. In
this case, the initial momentum is MV. This
equals the momentum after t plus delta tau
seconds.
If we simplify, what is it that get. MV is
equal to MV minus delta m into V. Then I get
plus M into delta V minus delta m into delta
V. Now, from this set of equations I get
delta m into VJ plus delta m into delta V.
I should have written here small delta because
I want to reserve the capital delta for something
else.
And now I find that this delta m delta V and
delta m delta V cancels. So also MV and
MV cancels. Therefore, I am left with the
term M into delta V plus delta m into VJ is
equal to zero. This gives delta V as equal
to minus delta m into VJ divided by capital
M.
Therefore, we get delta V is equal to minus
VJ into delta M by M or let us say delta M
if
we use the capital delta M here.
We need to solve the above equation. What
is the value of delta m? Let us assume that
the mass which gets exhausted from the nozzle
is something you are constantly pushing
out. Mass at the rate let us say m dot. Therefore,
the value of delta m should be equal to
m dot into the small time what I said here,
delta tau. You know the rate at which mass
is
leaving the nozzle is m dot over a small time
delta tau it is equal to m dot delta tau and
what should be the value of the capital M?
Capital M must be equal to the initial mass
of
the rocket at time t is equal to its mass
at time zero minus m dot into t. This is the
mass
of the rocket at time t.
In other words at time t is equal to 0; the
rocket had a mass initial mass it continues
to
eject mass at the constant rate m dot and
therefore, at time t its value M is equal
to initial
mass M i minus m dot into t.
And therefore, now I can erase this part and
now I can write the value of delta V is equal
to minus VJ into delta m which is equal to
m dot into delta tau divided by M i minus
m
dot into t.
I want to integrate this equation and if I
have to integrate this equation from initial
time
of 0 to a final time tf at which I get the
total velocity increment of delta V. I know
the
value of m dot. The mass at time t is the
initial mass M i minus m dot t into time t.
The
value of delta m equals minus m dot into delta
tau. With the two negatives the minus
sign will not be there in the expression for
delta V. I show the derivation of delta m
as
being minus of mdot into delta tau in the
following slide.
The minus sign shows that the mass has left
the system and therefore, when I substitute
the value of delta m. I should have substituted
minus m dot delta tau that means this
should have been a negative sign and this
negative sign and this negative sign would
have given me a positive sign and the expression
should have been plus V square.
To summarize: We solved the momentum equation
in the inertial frame of reference and
balanced MV with M minus delta m into V plus
delta V plus delta m into VJ plus V plus
delta V and then we got an expression which
gave us delta V was equal to minus VJ into
delta M divided by m and we find delta m is
equal to minus of this particular expression
and therefore, we got it as VJ m dot del d
tau into M i minus m dot V.
We now integrate it to determine the net velocity
gained by the rocket.
And what is the expression I get now? Let
us assume that VJ which is the velocity at
which the gases leave the rocket is a constant.
The limits of integration are from t = 0 to
t= tf. Integral of mdot divided by Mi minus
mdot into t dt is natural log of Mi minus
mdot into t. We also get a negative sign from
the minus mdot t in the denominator and
this expression is between the limits t = 0
and t= tf.
The value of the expression Mi minus mdot
t at t= 0 is Mi viz the initial mass of the
rocket while Mi minus mdot t at t= tf is its
final mass Mf. And therefore, the value delta
V is equal to minus VJ into natural log Mf
by Mi. The negative sign can be removed by
inversing the term within the logarithm. The
velocity change or increment provided by
the rocket is therefore VJ into logarithm
of the initial mass to the final mass of the
rocket.
This velocity change is spoken of as incremental
velocity and the equation is known as
the rocket equation.
All what the rocket equation tells is when
I burn a quantity of fuel between the initial
value and the final value and I am exhausting
it out at a constant velocity VJ, the final
velocity of the rocket is equal to the velocity
with which I am ejecting matter out into
logarithm of the initial mass to the final
mass. This is what we call as the rocket
equation. The Russian school teacher Tsialkowski
derived it and postulated that a high
value of jet velocity VJ is required and a
large value of mass ratio (Mi toMf) is desirable
We can go into interplanetary missions because
all what we want is we want a jet
velocity and the mass should keep getting
depleted and this is what is the theory of
rocket propulsion states.
Can I repeat it to some extent because this
forms the basis and you should know the
limitations.
The jet velocity or the velocity which with
mass is being removed into natural logarithm
of the initial mass divided by the final mass
decides the velocity provided by a rocket.
The value of final mass of a rocket to the
initial mass of the rocket is also called
as the
mass ratio of a rocket. We did not consider
the gravitational forces nor the drag forces
in
the derivation of delta V.
Therefore, delta V is also spoken of as ideal
velocity increment. Now, in the earlier
classes we saw that the rocket has to supply
the necessary orbital velocity and the total
velocity for which expressions were derived.
If I can write it, VT is equal to under root
G ME by R into RE plus 2 h divided by RE plus
h so many meters per second is the velocity
what was a required.
Now, if I have a rocket and on top of this
I put whatever I want to launch; I give it
a
velocity I can put it in orbit and but, we
found that we need a velocity of the order
of
something like 10 to 12 kilometers per second.
Therefore, what it is required in order to
achieve high velocities? See, you have to
have high value of mass ratios which means
that the ratio of initial mass to the final
mass must be a large. That means that difference
between the initial and the final must be
large and also the VJ must be a large value.
That
means the jet velocity is a controlling parameter
and higher the jet velocity you can have
higher delta V..
Therefore, the figure of merit of a rocket
if somebody were to ask us we must say well
one is the jet velocity, the other is something
related to the masses. Therefore, let us go
back and look at this term because this is
fairly clear to us. Supposing, I were to exhaust
at some jet velocity and I can get as high
a value. Apparently you cannot get the speed
of
light, you cannot get more than some amount.
But, I have some limitations. I will come
back to these limitations. In addition I am
talking in terms of Mi by Mf should be a large
number.
Let us just examine this number before I can
make a rocket because as of today you
cannot get more than something like 3000 to
5000 meters per second as jet velocity. We
will find out where and what the limitation
are there but, let us first examine this.
Let us therefore write the ideal velocity
increment as equal to jet velocity into ln
of the
initial mass to the final mass of the rocket.
What does the initial mass of the rocket
consist of? It will anyway have the useful
part of a rocket? What is the useful part
of a
rocket? The object which is going round and
round that is the useful part of the rocket
which is we call as payload. Then we have
the structure of the rocket that means it
must
have some metal and other structural materials
including inert materials to protect it from
heat such as insulation. Thus we would have
a structural mass.
Plus it should also have some fuel and we
said fuel is used for propelling the rocket
and
is known as propellant. Therefore, we have
a mass corresponding to mass of propellant
Mp. That means that the initial mass of a
rocket comprises of the payload mass, the
mass
of the structure plus the mass of the propellant
and we say Mi is equal to Mu plus the
structural mass plus the fuel or the propellant.
This is the initial mass. When the rocket
has done its job, what must be the final mass
of the rocket? It has done its job that means
all the propellant has burned out. Therefore,
the final mass will be the mass of the useful
component viz., the payload plus the mass
of the structure. But, it is also possible
that the
structure could be removed the payload after
the rocket functions and thrown out. But,
otherwise the structure will remain as part
of the final mass of the rocket. This is
generally the case. Therefore, the initial
mass is Mu plus Ms plus Mp while the final
mass is Mu plus Ms.
Now, let us analyze the masses. The incremental
velocity is VJ ln of Mu plus Ms plus
Mp divided by Mu plus Ms. Let us express these
mass as non-dimensional terms or as
mass fractions. Let us say I call the useful
mass of a rocket Mu divided by the total initial
mass namely Mu plus Ms plus Mp as equal to
alpha (α). The value of α is proportional
to
the initial mass of the rocket and is the
non-dimensional payload mass. Similarly, I
make
structural mass divided by the initial mass
of the rocket as the structural mass fraction
and call it as β. The payload mass fraction
is the ratio of the payload mass or useful
mass divided by the initial mass of the rocket.
This is denoted by γ.
Dividing the mass terms within the logarithmic
term by initial mass of the rocket, we get
deltaV is equal to VJ ln of alpha +beta +
gama divided by alpha+ bet. However, the sum
of the masses is the total mass of the rocket
and therefore the sum of alpha, beta and
gama is unity. And therefore, I get this equation
for delta V as VJ into ln of one over
alpha plus beta.
Now, what is it what we want to do in a rocket?
We want to have as much payload as
possible. May be we would like the useful
mass to be high. Let us say what is the
fraction of the useful mass α.
I get therefore, from the above expression
one over alpha plus beta is equal to
exponential of delta V by VJ. This is done
by taking exponential on both the sides and
the
exponential of ln becomes unity. Taking the
inverse on both sides we get alpha plus beta
is equal to the exponential of the negative
of delta V divided by VJ. the payload mass
fraction alpha is therefore equal to exponential
of the negative of delta V by VJ minus the
structural mass fraction Beta. .
Now, let us examine under what conditions
will we get the value of the useful payload
mass fraction to be high. Let us plot it out
for different values of VJ at given values
of
velocity increments. The velocity increment
required could be between 8 to 12 km/s as
seen earlier and depends on the mission. There
might be some variations in the jet
velocity between 3000 and 5000 meters per
second. What are the values of payload
fraction what we get? Let us just plot it
out and see.
It is shown in the above wherein alpha is
plotted as a function of deltaV by VJ. As
the
incremental velocity increases for a given
value of VJ, the value of alpha decreases
for a
given specified value of Beta. As the structural
mass fraction Beta increases, the useful
payload fraction alpha decreases. If the efflux
velocity VJ is higher for a given
incremental velocity delta V, the payload
fraction increases. And generally this value
of
the payload fraction might be around 0.4 and
keeps falling and in the operable regions,
the fraction could be around 0.1 or lower
depending on the structural mass fraction
beta.
Further, as beta increases alpha decreases.
We have still not put any numbers here
because we do not know what it is the range
of beta. But, generally beta should be
around let us say 0.1. As mass of structure
increase beta increases to around 0.12 to
0.15.
That means the payload fraction will keep
decreasing as the structural mass increases.
If
we can have a high value of VJ then I can
get a higher value of the payload fraction.
Or if
I have a rocket or object which requires more
ideal velocity then I get a lower value of
payload. You know we are just looking at the
rocket equation and trying to draw some
conclusions from it. The conclusions that
we draw are if we want to put payload of
higher mass then I need a structure which
much be very light. I must have a large value
of jet velocity VJ or else if I can have rocket
which does not have to go very far away it
and the orbit is nearby then I can carry a
higher mass. Well this is all about the rocket
equation and the conclusions from it.
But then the problem is that we need an incremental
velocity of about 10 to 12
kilometers per second. We called it as VT.
This is what we said as the total velocity
to
climb up and orbit. But, then to get a reasonable
value for the payload mass fraction for
this incremental velocity with the existing
jet velocities VJ and structural mass fraction
Beta is very difficult if not impossible.
If I have a single rocket, top of which I
have a
payload, I may not be able to get a useful
mass fraction for the payload because I have
a
definite mass of the structure. I have a limitation
on VJ and therefore, to be able to launch
a payload into orbit using a single rocket
is difficult. I use the word difficult though
it
appears to be impossible at this time. So
far it has been impossible but the quest for
the
rocket engineer is to make a single rocket
do the job.
But the job can be done by using multiple
rockets. Let us examine this point. Let us
say
instead of having a single rocket, I make
two rockets. I put rocket over here and on
the
top of it I put another rocket. This is my
first rocket, this is my second rocket. Now,
the
first rocket gives a value of the ideal velocity
delta V1. The second rocket already has a
delta V1 when its starts functioning. It gives
me a value of velocity delta V2 and the total
velocity of this composite two stage rocket,
gives me a delta V equal to delta V1 plus
delta V.2. Therefore, by putting one rocket
on top of the other, we are able to achieve
higher incremental velocities. We call these
rockets as multi stage rockets and most of
the rockets used today are multi stage rockets.
That means you want a velocity increment
of something like ten kilometers may be the
first one could give you 1 km/s, the second
one could give you 3km/s and the third one
could be you still higher at 6 km/s and
therefore, you keep on adding stages of a
rocket and this is what we say as multi stage
rockets.
For a multistage rocket, we have delta V is
equal to VJ of the first rocket into logarithm
of the initial mass to the burn out mass of
the first rocket plus VJ of the second stage
multiplied by the logarithm of the initial
mass to final mass of the second rocket and
so
on. I get the final ideal velocity. Maybe
we should we should try to analyze this in
some
detail. This is about staging of rockets.
We have something as the base or core stage,
then on the top of it we have the first stage,
then the second stage and so on.
Let us taken an example: Let us examine the
construction of India’s GSLV rocket,
which is very much in the news. We know that
this rocket consist of a core rocket, it
consist of four rockets attached to the core
stage, then it consists of the second stage
and
the third stage on the top of the third stage
sits the satellite.
Therefore, you have the first stage, second
stage, third stage. This gives you delta V1
delta V2 delta V3 the sum of which gives you
the velocity to put it into orbit. Therefore,
we talk in terms of staging. Staging means
one after the other but, what about these
four
rockets attached to the core? Why should they
be required?
We have put one stage on the other to get
higher incremental velocity; however, in the
process we have increased the mass of the
total rocket. When you have increased the
mass and you want this to be lifted, the core
stage should develop sufficient force.
However, the single core stage may not be
able to generate that level of forces.
Therefore you need additional rockets so that
the force or the thrust is able to take off
from the ground and that is why we put rockets
together and this is known as clustering.
Why do we need clustering of rockets? To give
you sufficient force for propelling. Even
the upper stages may need clustering.
Let us just put things together and summarize
what we have learnt so far.
We found that a rocket gives you incremental
velocity delta V. How does it give you
delta V because of change of momentum produced
by the efflux of the jet. We wrote the
momentum change equation from an inertial
frame of reference and found out the value
of delta V. What is the change of momentum
known as? It is known as impulse. What is
the unit of impulse? Same as momentum kilogram
meter per second. But, kilogram
meter per second can also be written as kilogram
meter per second square into second
which is same as Newton second. Therefore,
I can write the impulse as equal to Newton
second.
Impulse in Newton second is what gives delta
V therefore, what is the force with which
the rocket is pushed up? Rate of change of
momentum that means d by dt of mv or d by
dt of I. That is, we get so much force which
is equal to d by dt of momentum; this is
equal to mass flow rate mdot of the exhaust
which is going out with velocity VJ. This
equals mdot into VJ and therefore, I can also
write the force is equal to m dot VJ
(Newton) or compared to momentum which is
equal to m V, I write force is equal to m
dot V and this is the force pushing the rocket.
There is a limit to the mass which can be
released.
And therefore, we allow more masses to be
going through by clustering and thus achieve
the desired force. That means that we need
a larger force to push and that is why we
require clustering. Sometimes, we have a booster
rocket to whose sides we attach two
rockets. These are like straps. I strap something
on to it. The side rockets are also known
as strap-on.
Let me take one or two examples. May be 
I will show it through slides when we meet
in
the next class. But, to be able to just conclude
at this point of time all what I would like
to say is we derive the 
rocket equation through the change of momentum.
We looked at
the inertial frame of reference, watched the
rocket go up and we found out what is the
ideal 
velocity increment.
We also discussed about some creatures in
universe which make use 
of the rocket
principle. Then 
we found out that 
the structural mass 
of 
the rocket plays an important
role just 
as much as the jet velocity plays a role.
Then to get a high value 
of 
delta V, we
found the need to operate in stages and to
be 
able to 
take off with the larger mass of 
the
stages we needed some additional side rockets,
which are known as clustering and strapon.
