Fig. 1

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.Fig. 2

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 , gives.

At when stage’s fuel has been burnt, we have where and,.

Let the velocity of rocket at the end of stage of flight. Since , (2) becomesi.e.,

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 . Moreover, Given , (5) yields approximatelyFig. 3

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

.

Consequently, When and ,Fig. 5

This is a considerable improvement over the single stage rocket (). Nevertheless, it is still short of producing the orbiting speed a satellite needs. In fact, indicates that is a monotonically decreasing function of . In addition,.

Therefore, there is an upper limit to the speed a two stage rocket can produce. When , the limit is approximatelyFig. 6

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 where 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 givesor .

Fig. 7

By (8), the valid solution is.

It shows that attains an extreme value at . Moreover, we observe from (11) that and.

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.Fig. 8

Setting , we reach the goal:Fig. 9

Fig. 10

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 .