The forces on a rocket change dramatically
during a typical flight.
During powered flight the propellants of the
propulsion system are constantly being
exhausted from the
nozzle. As a result, the
weight and mass of the rocket is constantly
changing.
Because of the changing mass, we cannot use the standard form of Newton's
second law of motion to determine the acceleration
and velocity of the rocket.
This figure shows a derivation of the change in velocity during
powered flight while accounting for the changing mass of the rocket.
In this derivation, we are going to neglect the effects of
aerodynamic lift and drag. We can add these
effects to the final answer.
Let us begin by considering the rocket drawing on the left of the figure.
M is the instantaneous mass of the rocket, u is the velocity of the
rocket, v is the velocity of the exhaust from the rocket, A is the area
of the exhaust nozzle, p is the exhaust pressure and p0 is the atmospheric
pressure. During a small amount of time dt a small amount of mass dm is
exhausted from the rocket. This changes the mass of the rocket and the velocity of the rocket
and we can evaluate the change in momentum of the rocket as
change in rocket momentum = M (u + du)  M u = M du
We can also determine the change in momentum of the small mass dm that is exhausted at
velocity v as
change in exhaust momentum = dm (u  v)  dm u =  dm v
So the total change in momentum of the system (rocket + exhaust) is
change in system momentum = M du  dm v
as shown on the figure. Now consider the forces acting on the system, neglecting the drag on
rocket. The weight of the rocket is M g (gravitational constant) acting at an angle a
to the flight path. The pressure force is given by (p  p0) A acting in the positive
u direction. Then the total force on the system is
force on the system = (p  p0) A  M g cos (a)
The change in momentum of the system is equal to the
impulse
on the system which is equal to the force on the system times
the change in time dt. So we can combine the previous two equations:
M du  dm v = [(p  p0) A  M g cos (a)] dt
If we ignore the weight force, and perform a little algebra, this becomes
M du = [(p  p0) A ] dt + dm v
Now the exhaust mass dm is equal to the
mass flow rate
mdot times the increment
of time dt. So we can write the last equation as
M du = [(p  p0) A + mdot v ] dt
On the web page describing the
specific impulse, we introduce the equivalent exit velocity
Veq which is defined as
Veq = v + (p  p0) * A / mdot
If we substitute the value of Veq into the
momentum equation we have
M du = Veq mdot dt
mdot dt is the amount of change of the instantaneous mass of the rocket.
The sign of this term is negative because the rocket is losing mass as the propellants
are exhausted.
mdot dt =  dM
Substituting into the momentum equation:
M du =  Veq d M
du =  Veq dM / M
We can now integrate this equation:
delta u =  Veq ln (M)
where delta u represents the change in velocity, and ln is
the symbol for the natural logarithmic
function.
The limits of integration are from the initial mass of the rocket to
the final mass of the rocket.
The instantaneous mass of the rocket M,
the mass is composed of two main
parts,
the empty mass me and the propellant mass mp.
The empty mass does not change with
time, but the mass of propellants on board the rocket does change with time:
M(t) = me + mp (t)
Initially, the full mass of the rocket mf contains the empty mass and
all of the propellant at lift off.
At the end of the burn, the mass of the rocket contains only the
empty mass:
M initial = mf = me + mp
M final = me
Substituting for these values we obtain:
delta u = Veq ln (mf / me)
This equation is called the ideal rocket equation. There are
several additional forms of this equation which we list here:
Using the definition of the
propellant mass ratio MR
MR = mf / me
delta u = Veq * ln (MR)
Veq is related to the specific impulse Isp:
Veq = Isp * g0
where g0 is the gravitational constant. So the change in velocity can be
written in terms of the specific impulse of the engine:
delta u = Isp * g0 * ln (MR)
If we have a desired delta u for a maneuver, we can invert this
equation to determine the amount of propellant required:
MR = exp (delta u / (Isp * g0) )
where exp is the exponential
function.
What good is all this math? Let's check the results of the equation
with a few numbers.
From our discussion of
Isp
and as shown in the
thrust simulator,
a reasonable value for Isp for a liquid hydrogen/liquid oxygen rocket
(like the Space Shuttle main engine)
would be about 350 seconds.
The delta u needed to get to a 200 mile high orbit is given by the
circular orbit simulator
as about 17,000 mph (~25,000 feet/sec).
The gravitational constant for Earth is 32.2 feet/second squared.
Putting these values into the ideal rocket equation, the resulting
mass flow ratio MR is equal to 10. From the ideal rocket equation,
90% of the weight of a rocket going to orbit is propellant weight.
The remaining 10% of the weight includes structure, engines, and payload.
So given the current stateoftheart, the payload accounts for
only about 1% of the weight of an ideal rocket at launch. Rockets are
terribly inefficient and expensive.
If you include the effects of gravity in the rocket equation, the equation becomes:
delta u = Veq ln (MR)  g0 * tb
where tb is the time for the burn.
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