APPENDIX

TEMPERATURE MEASUREMENT SCALES

ROCKET EQUATION

THERMOMETRIC SCALES

Sometimes in rocketry, temperature is measured in degrees Rankine, mysterious unit for non-professional people. The Rankine scale is an absolute Fahrenheit scale, that is to say it starts from the absolute zero. Exactly like Kelvin scale, which is an absolute Celsius scale.
I considered useful to publish the convertion formulas from Rankine degrees into Fahrenheit, Celsius o Kelvin degrees, and vice versa:

ROCKET MOTION EQUATION

The whole rocketry is based upon Tsiolkovsky's rocket equation, published in 1903:

which, integrated in the variable m, gives:

How is this equation obtained? A simplified demonstration is the following:
For Newton's 2^nd law, the force acting on the rocket (that is: thrust) equals its mass multiplied by acceleration:

where m is the (variable) mass of the rocket.
For Newton's 3^rd law (action-reaction principle) F is also the force which increased the momentum of exhausted gas; this force equals the derivative in respect of time of the momentum of exhausted gas; in an infinitesimal time interval dt, a mass -dm with speed v_e adds to the already ejected gas, so the momentum increase is -v_e·dm; thrust is therefore:

Note that the mass added to ejected gas is -dm because dm is just the rocket mass decrease.
By putting equal the 3) and 4):

from which, simplifying dt and dividing by m, 1) is obtained.

This demonstration is correct and elegant, but might seem not wholly convincing.
The rigorous demonstration involves use of all the terms present in the problem, considered as an isolated system made of rocket+ejected gas, observed from an inertial reference axis tern.
The thrust force, which rocket and ejected gas exchange each other, is an internal force and doesn't modify the total momentum of the system; so Newton's 2^nd law in its general form is applied:

with p = momentum,
considering the whole expression of the derivative:

Let's suppose that the rocket is initially located in a space region where there are no external forces, for example in the interplanetary space, and that it has a certain initial momentum. At a certain time the motor is ignited and the combustion gases are ejected out of the rocket, at constant rate, at speed v_e.
Since F = 0, indicating by capital letters the quantities relevant to rocket and by small letters those relevant to gas, the 7) may be written

The first term, bearing in mind that rocket mass decreases with the same rate the gas is expelled (-dm/dt), becomes

The second term needs to be considered with greater detail; the gas speed v is related to rocket velocity V and to ejection speed v_e (relative to rocket) by equation v = V - v_e, and so

The derivative dm/dt is simply the constant quantity of propellant burned in the time unit; the second term is zero because the gas, once expelled from the rocket, doesn't change its speed any more, so that

10)

Substituting the 9) and 10) in the 8) we obtain

from which

11)

The former defines rocket thrust, which is equal to the propellant mass burned in the time unit multiplied by ejection speed.
In order to obtain rocket velocity it's sufficient to bear in mind that, in the 11), the ejected gas quantity is equal to the rocket mass decrease, dm = -dM, and so

which is equal to 1).

From now on let's use the following symbology: we'll call c the exhaust speed v_e, M the mass at a generic time, M₀ the initial mass, and λ the propellant combustion rate, so M = M₀ - λ·t .
The 2) can be re-written therefore:

12)

If initial velocity is 0, the 12) becomes:

13)

which can be made explicit in respect of time:

14)

By integrating this last in respect of time, we can obtain the relation between time and space:

15)

which can also be expressed in respect of mass:

16)

or of velocity:

17)

It can be interesting to a non-professional to see how to integrate velocity in respect of time to obtain space:

Let's examine again the 14), in which v = v(t)

and let's integrate in respect of time in order to obtain space:

Now it's necessary to operate the following variable substitution:

which involves:

Replacing the new variable in the integral, we get:

Now we apply the well-known rule of integration by parts:

which gives:

which is equivalent to 15) and 16).

As last topic of this Appendix, I'd like to show that the two formulas quoted by Brinley for rocket velocity calculation at combustion end:

		formula 2) quoted at p.7-4, with corrections
		formula 8) quoted at p.8-1, with corrections

are actually equivalent, in case of adapted nozzle.
The meaning of symbols that appear in this two formulas is as follows:
ln = lg = natural logarithm (base e);

M_P =	propellant mass;
t_b =	combustion time;
γ =	launch angle from vertical;
I_sp =	specific impulse;

F =	thrust;
Q₀ =	initial weight of the rocket;
n =	mass ratio = M₀/M_f = M₀/(M₀ - M_P);

M_f =

rocket mass at combustion end.

If from 8), through proper transformations and substitutions, it's possible to arrive at 2), the equivalence is proved.
We get, through simple passages:

From the general formula 8) we have derived, assuming an adapted nozzle, the formula 2).
So, the formula 8), which uses the specific impulse, has a general validity, while the 2), which uses the gas ejection speed c, is valid for adapted nozzle only.