Rocket Propulsion

1. SATELLITE
COMMUNICATION
LECTURE-5
ROCKET PROPULSION
ASST. PROF. SANDIP DAS

2. ROCKET PROPULSION
• Isaac Newton stated in his third law of motion that "for every
action there is an equal and opposite reaction." It is upon this
principle that a rocket operates.
• Propellants are combined in a combustion chamber where they
chemically react to form hot gases which are then accelerated and
ejected at high velocity through a nozzle, thereby imparting
momentum to the engine.

3. THRUST
• Thrust is the force that propels a rocket or spacecraft and is measured
in pounds, kilograms or Newtons. Physically speaking, it is the result of
pressure which is exerted on the wall of the combustion chamber.
• The pressure distribution within the chamber is asymmetric; that is,
inside the chamber the pressure varies little, but near the nozzle it
decreases somewhat.
• The force due to gas pressure on the bottom of the chamber is not
compensated for from the outside. The resultant force F due to the
internal and external pressure difference, the thrust, is opposite to the
direction of the gas jet. It pushes the chamber upwards.

4. THRUST
• To create high speed exhaust gases, the necessary high temperatures
and pressures of combustion are obtained by using a very energetic fuel
and by having the molecular weight of the exhaust gases as low as
possible. It is also necessary to reduce the pressure of the gas as much as
possible inside the nozzle by creating a large section ratio. The section
ratio, or expansion ratio, is defined as the area of the exit Ae divided by
the area of the throat At.
• The thrust F is the resultant of the forces due to the pressures exerted on
the inner and outer walls by the combustion gases and the surrounding
atmosphere, taking the boundary between the inner and outer surfaces
as the cross section of the exit of the nozzle. As we shall see in the next
section, applying the principle of the conservation of momentum gives:

5. PRINCIPLE OF ROCKET PROPULSION
• It is based on the principle of conservation of momentum. In isolated system, the total linear
momentum remains conserved.
• Before a rocket is fired, the total momentum of rocket plus fuel is zero. Since, the system is an isolated
system, its momentum remains the same.
• When rocket is fired, fuel is burnt and hot gases are expelled from the back of the rocket.
• Since the momentum acquired by the gases is directed towards the rear, so the rocket must acquire an
equal amount of momentum in the forward direction to conserve momentum.

6. THEORY OF PROPULSION
• Consider a rocket is moving vertically upward in free space with no external forces. Thus the
momentum of the system remains the same.
• Suppose that t=0, the mass of the rocket is 𝑚0 and is moving with a velocity of 𝑉0 with respect to earth.
• Let at any time t, the mass of the rocket be m and its velocity be V as fuel is burnt, the velocity of the
rocket increases and its mass decreases. Therefore, m< 𝑚0 and V> 𝑉0.

7. THEORY OF PROPULSION (CONTINUED)
• In time ∆𝑡, the rocket ejects a mass of ∆𝑚 of the fuel in the form of gases with a velocity –v with
respect to the rocket. However, the velocity of the gases with respect to earth is (V-v).
• At the end of the time interval ∆𝑡, the mass of the rocket has decreased to (m- ∆𝑚) and its velocity has
increased to (V+ ∆𝑣)
• By the law of conservation of momentum, the initial momentum at the start of time interval ∆𝑡 must be
equal to the final momentum at the end of time interval ∆𝑡.

8. THEORY OF PROPULSION (CONTINUED)
According to law of conservation, Initial momentum=final momentum
mV= 𝑚 − ∆𝑚 𝑉 + ∆𝑉 +∆𝑚(𝑉 − 𝑣)
or, mV=mV-V∆𝑚 − ∆𝑚∆𝑉 + 𝑚∆𝑉 + 𝑉∆𝑚 − 𝑣∆𝑚
0= 𝑚∆𝑉 − 𝑣∆𝑚 −∆𝑚∆𝑉
The term ∆𝑚∆𝑉 is the product of two small quantities and therefore, can be neglected. After
simplification of the above equations-
0= 𝑚∆𝑉 − 𝑣∆𝑚
The quantity of fuel ejected ∆𝑚 is equal to the loss of mass of the rocket (−∆𝑚). The negative sign
indicates decrease in 𝑚. Therefore, momentum equation can be modified as:
∆𝑉= -
∆𝑚
𝑚
𝑣

9. THEORY OF PROPULSION (CONTINUED)
When ∆𝑡 tends to zero, then ∆𝑉 𝑎𝑛𝑑∆𝑚 can be replaced by dV and dm. Hence,
dV = - 𝑣
𝑑𝑚
𝑚
Integrating the above equation-
𝑣0
𝑉
𝑑𝑉 = - 𝑚0
𝑚
𝑣
𝑑𝑚
𝑚
Generally, the velocity of exhaust gases with respect to rocket is assumed to be constant.
𝑉 = 𝑉0 + 𝑣 log
𝑚0
𝑚
which gives velocity of the rocket having mass, m, at time t.
So, if the initial velocity of the rocket is zero (𝑉0=0), then the velocity of rocket becomes
𝑉 = 𝑣 log
𝑚0
𝑚

10. Thus, the velocity of the rocket at any instant is directly proportional to
• Exhaust speed of the ejecting gases with respect to rocket.
• Natural logarithm of the ratio of the initial mass to rocket to its mass at that instant.