The above optimization problem is solved using calculus (see “Viva Rocketry! Part 2“). However, there is an alternative that requires only high school mathematics with the help of a Computer Algebra System (CAS). This non-calculus approach places more emphasis on problem solving through mathematical thinking, as all symbolic calculations are carried out by the CAS (e.g., see Fig. 2). It also makes a range of interesting problems readily tackled with minimum mathematical prerequisites.

The fact that

is a monotonic increasing function

where

or

(1) can be written as

where

,

.

Since means

.

That is

.

Solve

for gives if .

Hence, (1) is a quadratic equation. For it to have solution, its discriminant must be nonnegative, i.e.,

Consider

If , (3) is a quadratic equation.

Solving (3) yields two solutions

,

.

Since ,

(4) implies

and, the solution to (2) is

or

i.e.,

or

We prove that (4) is true by showing (5) is false:

Consider :

where

.

It can be written as

where

,

,

.

Since (see Exercise 1) and,

solve (7) for yields

.

It follows that for .

Consequently, is a negative quantity. i.e.,

which tells that (5) is false.

Hence, when , the global maximum is .

Solving for :

,

we have

.

Therefore,

attains maximum at .

In fact, attains maxima at even when , as shown below:

Solving for , we have

or .

Only is valid (see Exercise-2),

When ,

where

Solve quadratic equation for yields

.

The coefficient of in is , a negative quantity (see Exercise-3).

The implication is that is a negative quantity when .

A while back, we deemed that it is utterly impractical to calculate the value of using the partial sum of Leibniz’s series due to its slow convergence (see “Pumpkin Pi“)

Fig. 1

As illustrated in Fig. 1, in order to determine each additional correct digit of , the number of terms in the summation must increase by a factor of 10.

What we need is a fast converging series whose partial sum yields given number of correct digits with far fewer terms.

Looking back, we see that the origin of Leibniz’s series is the definite integral

To find the needed new series, we consider a variation of (1), namely,

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.

Let the velocity of rocket at the end of stage of flight.

Since , (2) becomes

i.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 approximately

Fig. 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 approximately

Fig. 7

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?

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. 9

Setting , we reach the goal:

Fig. 10

Fig. 11

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 .

A rocket is programmed to burn and ejects its propellant at the variable rate , where and are positive constants. The rocket is launched vertically from rest. Neglecting all external forces except gravity, show that the final speed given to the payload, of mass , when all the fuel has been burnt is

.

Here is the speed of the propellant relative to the rocket, the initial rocket mass, excluding the payload. The initial fuel mass is .

Exercise 2. Before firing, a single stage rocket has total mass , which comprises the casing, instruments etc, with mass , and the fuel. The fuel is programmed to burn and to be ejected at a variable rate such that the total mass of the rocket at any time , during which the fuel is being burnt, is given by

where is a constant.

The rocket is launched vertically from rest. Neglect all external forces except gravity, show that the height attained at the instant the fuel is fully consumed is

A single rocket expels its propellant at a constant rate .

Assuming constant gravity is the only external force, show that the equation of motion is

where is the rocket’s speed, the speed of the propellant relative to the rocket, the payload mass, and the initial rocket mass.

If the rocket burn is continuous, show that the burn time is and deduce that the final speed given to the payload is

where is the structural factor of the rocket.

Estimate the percentage reduction in the predicted final speed due to the inclusion of the gravity term if

, and .

Find an expression for the height reached by the rocket during the burn and estimate its value using the data above.

Let’s recall the governing equation of rocket’s flight derived in “Viva Rocketry! Part 1“, namely,

.

In the present context, . It implies that

and,

.

With , we have

,

i.e.,

or

.

The structural factor indicates the amount of fuel is . Since the fuel is burnt at a constant rate , it must be true that at burnt out time ,

.

Therefore,

.

The solution to initial-value problem

tells the speed of the rocket during its flight while fuel is burnt (see Fig. 1):

Fig. 1

Evaluate (1) at burnt out time gives the final speed of the payload:

Notice the first term of (2) is the burnt out velocity without gravity (see “Viva Rocketry! Part 1“)

It follows that the percentage reduction in the predicted final speed due to the inclusion of gravity is

Using the given values which are typical, the estimated value of (3) (see Fig. 2) is

.

Fig. 2

This shows the results obtained without taking gravity into consideration can be regarded as a reasonable approximation and the characteristics of rocket flight indicated in “Viva Rocketry! Part 1” are valid.

Since , (1) can be written as

To find the distance travelled while the fuel is burnt, we solve yet another initial-value problem:

Fig. 3

The solution (see Fig. 3) is

.

Hence, the height reached at the burnt out time is

.

Using the given values, we estimate that (see Fig. 4)

Fig. 4

Exercise 1: Find the distance the rocket travelled while the fuel is burnt by solving the following initial-value problem: