Boosting rocket flight performance without calculus (Viva Rocketry! Part 2.1)

Fig. 1

The stages of a two-stage rocket have initial masses m_1 and m_2 respectively and carry a payload of mass P. Both stages have equal structure factors e and equal relative exhaust speed c. The rocket mass, m_1+m_2 is fixed and \frac{P}{m_1+m_2}=b.

According to multi-stage rocket’s flight equation (see “Viva Rocketry! Part 2“), the final speed of a two-stage rocket is

v = -c \log(1-\frac{em_1}{m_1+m_2+P}) - c\log(1-\frac{em_2}{m_2+P})

Let a = \frac{m1}{m_2}, it becomes

v = -c\log(1-\frac{ea}{a+1+b(a+1)})-c \log(1-\frac{e}{1+b(a+1)})

where a>0, b>0, c>0, 0< e < 1. We will maximize v with an appropriate choice of a.

That is, given

v = c\log(\frac{(a+1)(b+1)((a+1)b+1)}{(1-e+(a+1)b)((1-e)a+(a+1)b+1})

where a>0, b>0, c>0, 0<e<1. Maximize v with an appropriate value of a.

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

\log is a monotonic increasing function \implies v_{max} = c\log(w_{max})


w = \frac{(a+1)(b+1)((a+1)b+1)}{(1-e+(a+1)b)((1-e)a+(a+1)b+1)}


w(1-e+(a+1)b)((1-e)a+(a+1)b+1) - (a+1)(b+1)((a+1)b+1)=0\quad\quad\quad(1)

(1) can be written as

A_1 a^2 + B_1 a +C_1= 0


A_1 = -bew+b^2w+bw-b^2-b

B_1 = e^2w-2bew-2ew+2b^2w+3bw+w-2b^2-3b-1,

C_1 = -bew-ew+b^2w+2bw+w-b^2-2b-1.

Since A_1 = 0 means

-b e w + b^2  w + b w - b^2 - b =0.

That is

w = -\frac{b+1}{e-b-1}.


\frac{(a+1)(b+1)((a+1)b+1)}{(1-e+(a+1)b)((1-e)a+(a+1)b+1)} = -\frac{b+1}{e-b-1}

for a gives a = 0 \implies A_1 \neq 0 if a > 0.

Hence, (1) is a quadratic equation. For it to have solution, its discriminant B_1^2-4A_1C_1 must be nonnegative, i.e.,

(e^2-2be-2e+b+1)^2 w^2-2(b+1)(2be^2+e^2-2be-2e+b+1) w +(b+1)^2 \geq 0\quad(2)


(e^2-2be-2e+b+1)^2 w^2-2(b+1)(2be^2+e^2-2be-2e+b+1)w+(b+1)^2 = 0\quad(3)

If e^2-2be-2e+b+1 \neq 0, (3) is a quadratic equation.

Solving (3) yields two solutions

w_1 = -\frac{(b+1)(2\sqrt{b(b+1)}e^2-2be^2-e^2-2\sqrt{b(b+1)}e+2be+2e-b-1)}{(e^2-2be-2e+b+1)^2},


Since 0 < e < 1,

w_1 - w_2 = -\frac{4(b+1)\sqrt{b(b+1)}(e-1)e}{(e^2-2be-2e+b+1)^2} > 0\quad(4)

(4) implies

w_1 > w_2

and, the solution to (2) is

w \leq w_2 or w \ge w_1


w \leq \frac{(b+1)(2\sqrt{b(b+1)}e^2+2be^2+e^2-2\sqrt{b(b+1)}e-2be-2e+b+1)}{(e^2-2be-2e+b+1)^2}\quad\quad\quad(4)


w \ge -\frac{(b+1)(2\sqrt{b(b+1)}e^2-2be^2-e^2-2\sqrt{b(b+1)}e+2be+2e-b-1)}{(e^2-2be-2e+b+1)^2}\quad\quad\quad(5)

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

Consider w - w_1= 0:

\frac{(b+1)(e-1) e \cdot f(a)}{(1-e+ab+b)(a(1-e)+ab+b+1)(e^2-2be-2e+b+1)^2} = 0\quad\quad\quad(6)


f(a) = 2a\sqrt{b(b+1)}e^2+a^2be^2+be^2+e^2-2a^2b\sqrt{b(b+1)}

-4ab\sqrt{b(b+1)}e - 2b\sqrt{b(b+1)}e - 4a\sqrt{b(b+1)}e


-2e+2a^2b^2\sqrt{b(b+1)}+4ab^2\sqrt{b(b+1)} + 2b^2\sqrt{b(b+1)} +2a^2b\sqrt{b(b+1)}

+6ab\sqrt{b(b+1)} + 4b\sqrt{b(b+1)} + 2a\sqrt{b(b+1)} + 2\sqrt{b(b+1)}

+2a^2b^3 + 4ab^3 + 2b^3 +3a^2b^2 + 8ab^2 +5b^2+a^2b+4ab+4b+1.

It can be written as

A_2a^2 + B_2a + C_2\quad\quad\quad(7)


A_2 = be^2-2b\sqrt{b(b+1)}e-2b^2e-2be+2b^2\sqrt{b(b+1)}+2b\sqrt{b(b+1)}


B_2 = 2\sqrt{b(b+1)}e^2-4b\sqrt{b(b+1)}e -4\sqrt{b(b+1)}e




Since A_2 > 0 (see Exercise 1) and,

solve (7) for a yields

a = -\sqrt{1+\frac{1}{b}}.

It follows that for a > 0, f(a) > 0.

Consequently, w-w_1 is a negative quantity. i.e.,

w-w1 <  0

which tells that (5) is false.

Hence, when e^2-2be-2e+b+1 \neq 0, the global maximum w_{max} is w_2.

Solving w = w_2 for a:

\frac{(a+1)(b+1)((a+1)b+1}{(1-e+(a+1)b)((1-e)a+(a+1)b+1)} = \frac{(b+1)(2\sqrt{b(b+1)}e^2+2be^2+e^2-2\sqrt{b(b+1)}e-2be-2e+b+1)}{(e^2-2be-2e+b+1)^2},

we have

a = \sqrt{1+ \frac{1}{b}}.


e^2-2be-2e+b+1 \neq 0 \implies w attains maximum at a = \sqrt{1+ \frac{1}{b}}.

In fact, w attains maxima at a = \sqrt{1+\frac{1}{b}} even when e^2-2be-2e+b+1 = 0, as shown below:

Solving e^2-2be-2e+b+1 = 0 for e, we have

e_1= -\sqrt{b(b+1)}+b+1 or e_2 = \sqrt{b(b+1)} + b + 1.

Only e_1 is valid (see Exercise-2),

When e = e_1,

w(\sqrt{1+\frac{1}{b}} )- w(a) = - \frac{(b+1)g(a)}{4 \sqrt{b(b+1)} (\sqrt{b(b+1)}+ab) (a\sqrt{b(b+1)}+b+1)}\quad(8)


g(a) = (2a^2b+4ab+2b+2a+2)\sqrt{b(b+1)}-2a^2b^2-4ab^2-2b^2-a^2b-4ab-3b-1

Solve quadratic equation g(a) = 0 for a yields

a = \sqrt{1+\frac{1}{b}}.

The coefficient of a^2 in g(a) is 2b\sqrt{b(b+1)}-2b^2-b, a negative quantity (see Exercise-3).

The implication is that g(a) is a negative quantity when a \neq \sqrt{1 + \frac{1}{b}}.

Hence, (8) is a positive quantity, i.e.,

e^2-2be-2e+b+1 = 0, a \neq \sqrt{1+\frac{1}{b}} \implies w(\sqrt{1+\frac{1}{b}})-w(a) > 0

We therefore conclude

\forall 0 < e < 1, b > 0, w attains its maximum at a = \sqrt{1+\frac{1}{b}}.

Fig. 2

Exercise-1 Prove:0<e<1, b>0 \implies

be^2-2b\sqrt{b(b+1)}e-2b^2e-2be+2b^2\sqrt{b(b+1)}+2b\sqrt{b(b+1)}+2b^3+3b^2+b > 0

Exercise-2 Prove: b > 0 \implies 0 <-\sqrt{b(b+1)} + b +1 <1

Exercise-3 Prove: b > 0 \implies 2b\sqrt{b(b+1)}-2b^2-b < 0

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