A rocket with stages is a composition of single stage rocket (see Fig. 1) Each stage has its own casing, instruments and fuel. The th stage houses the payload.
The model is illustrated in Fig. 2, the stage having initial total mass and containing fuel . The exhaust speed of the stage is .
The flight of multi-stage rocket starts with the stage fires its engine and the rocket is lifted. When all the fuel in the stage has been burnt, the stage’s casing and instruments are detached. The remaining stages of the rocket continue the flight with stage’s engine ignited.
Generally, the rocket starts its stage of flight with final velocity achieved at the end of previous stage of flight. The entire rocket is propelled by the fuel in the casing of the rocket. When all the fuel for this stage has been burnt, the casing is separated from the rest of the stages. The flight of the rocket is completed if . Otherwise, it enters the next stage of flight.
When all external forces are omitted, the governing equation of rocket’s stage flight (see “Viva Rocketry! Part 1” or “An alternate derivation of ideal rocket’s flight equation (Viva Rocketry! Part 1.3)“) is
It can be written as
Integrate (1) from to ,
At when stage’s fuel has been burnt, we have
Let the velocity of rocket at the end of stage of flight.
Since , (2) becomes
For a single stage rocket (), (3) is
In my previous post “Viva Rocketry! Part 1“, it shows that given and , (4) yields , a value far below , the required speed of an earth orbiting satellite.
But is there a value of that will enable the single stage rocket to produce the speed a satellite needs?
Let’s find out.
Differentiate (4) with respect to gives
since are positive quantities and .
It means is a monotonically decreasing function of .
Given , (5) yields approximately
This upper limit implies that for the given and , no value of will produce a speed beyond (see Fig. 4)
Let’s now turn to a two stage rocket ()
From (3), we have
If and , then
When and ,
This is a considerable improvement over the single stage rocket (). Nevertheless, it is still short of producing the orbiting speed a satellite needs.
indicates that is a monotonically decreasing function of .
Therefore, there is an upper limit to the speed a two stage rocket can produce. When , the limit is approximately
In the value used above, we have taken equal stage masses, . i.e., the ratio of .
Is there a better choice for the ratio of such that a better can be obtained?
To answer this question, let , we have
Since , by (8),
Substituting (8), (9) into (6),
a function of . It can be written as
This is a composite function of and .
Since is a monotonic increasing function (see “Introducing Lady L“),
Here, denote the maximum of and respectively.
To find , we differentiate ,
Solving for gives
By (8), the valid solution is
It shows that attains an extreme value at .
Moreover, we observe from (11) that
i.e., attains maximum at .
It follows that
attains maximum at .
Therefore, to maximize the final speed given to the satellite, we must choose the ratio
With , the optimum ratio , showing that the first stage must be about ten times large than the second.
Using this ratio and keep as before, (10) now gives
a value very close to the required one.
Setting , we reach the goal:
At last, it is shown mathematically that provided the stage mass ratios ( and )are suitably chosen, a two stage rocket can indeed launch satellites into earth orbit.
Exercise 1. Show that and .
Exercise 2. Using the optimum and , solving (10) numerically for such that .