Multistage and singlestage rockets?

Question

I need to create a mathematical model where are two rockets: one with two stages and the other with one stage. At the beginning, each rocket has got the same amount of fuel.
I have to compute how many times bigger speed can reach a two-stage rocket than one-stage rocket?

Has anybody some ideas about it? Maybe there are some internet links where I could find this model?

Thanks in advance.

Answer 1

You need to create a equation to model the rocket flight, then integrate it. If the rocket changes suddenly (discontinuously) in mid-flight, you have to integrate the flight in two pieces, starting the second integration with the properties at the end of the first stage.

Let's assume a 1-stage rocket is floating in deep space, so there's no gravity involved. We define some variables:
m = rocket's mass
f = volume of fuel remaining (volume)
rho = density of the fuel (mass per volume)

The total mass of the system is:
M = m + rho*f

Now let's assume the fuel is used at a constant rate:
f0 = starting amount of fuel (volume)
r = rate at which fuel is used (volume per time)

So, the amount of fuel remaining, as a function of time, is:
f = f0 - r*t

Plugging f into our mass equation, we see that the mass changes in time as follows:
M = m + rho*(f0 - r*t)

That's the complete mass equation as a function of time. Now we need to know the force on the rocket. Let's assume the rocket engine provides a constant thrust until it runs out:
F = thrust force generated while burning fuel

By Isaac Newton's second law (F = M*a), the acceleration of an object is the force divided by the mass:
a = F / M

Plugging in our expression for the total mass of the system, we get:
a = F / (m + rho*(f0 - r*t))

The instantaneous change in velocity of an object is the acceleration multiplied by the time. However, our acceleration is changing constantly, so we need the equation in integral form:
v = integral(a,t,0,tfinal)

This means the integral of the acceleration with respect to time from t=0 to t=tfinal (the time when the fuel runs out). However, since we know the rate at which fuel is being used up, we know when tfinal occurs:
tfinal = f0/r

Integrating, we obtain the answer:
v = F/(r*rho)*log(m/(m + rho*f0))
The constant of integration here is v0, but it's zero since the rocket started from rest.

That's how fast a rocket of mass m carrying a volume f0 of fuel with density rho and producing constant thrust F will be going when it runs out of fuel.

I'm not gonna do a second stage for you, but it just involves doing a few things in-between integrations again. First, you lower the mass of the rocket instantaneously (the first stage falling off). Second, you tabulate the properties of the rocket after the first stage falls off. You now have a new starting mass, a new fuel load (for the second stage), and a starting velocity that isn't zero.

You should notice that the only difference between a 2-stage rocket and a single stage rocket is that you get to discard some mass in-between stages. You should end up with the result that throwing away mass allows you to reach a higher final velocity.

Take a look at the Saturn V rocket. The first stage is huge, since it has to lift the entire mass of the rocket. The second stage is smaller, since much of the mass has been discarded. The final stage is smaller still, since even more mass has been discarded.