All rockets work on the same basic principle: Throw something out the back of the rocket and it will provide a thrust force. Really, it’s just like a collision between two objects. Let’s start with the simplest case: a rocket that just shoots one ball of fuel out the back.

Since there is an interaction between the rocket and the fuel, the force on the fuel has the same magnitude (but opposite direction) as the force on the rocket. Also, since the rocket pushes on the fuel for the same time that the fuel pushes on the rocket they will have opposite changes in momentum.

This is the essence of a rocket—throw stuff out the back. There are four basic types of rockets.

- Chemical Rocket. A chemical reaction shoots some of the propellant out the back of the rocket.
- Ion Engine. An ion engine uses an electric field to accelerate charged particles out the back of the rocket.
- Water rocket. Just like a chemical rocket, but instead of a chemical reaction high pressure gas (maybe air) pushes the propellant (maybe water) out of the rocket.
- Nuclear propulsion. Place a small nuclear bomb behind your spacecraft (hopefully with some type of shield). Boom. The debris flies away from the rocket and pushes the rocket forward.

Now for some theoretical rockets. Here are three starting assumptions.

- When fuel is expelled from a rocket, it leaves with a constant velocity that is relative to the rocket. I will call this the exhaust velocity and use the symbol
*u*. - Momentum is conserved when stuff is shot out the back of the rocket (I already said this).
- As fuel is consumed, the total mass of the rocket decreases. In fact, it might be easier to separate the mass into two parts: the mass of the fuel (
*m*_{f}) and the mass of the payload (*M*). Here, “payload” mass includes everything that is not fuel (rocket parts, humans, robots, computers, iPhones).

This leads to a conceptual question. Yes, I really want you to think about this answer. Yes, this will be on the test.

Suppose you have a rocket with a massMand total fuel of massm_{f}. The fuel willalwaysbe shot such that it leaves the rocket at a relative velocityu. Which would give the greatest change in velocity for the rocket, shooting out all the fuel at once (as with a nuclear propulsion engine) or letting it out slowly (like an ion engine)? Be sure to support your answer.

Just to separate the question from the answer below, here is a video of a rocket made from pouring butane on top of soda. Yes, it’s as awesome as it looks and yes, I will be writing a future post on this topic (but we need to collect more data first).

### The Rocket Equation

Before I give my answer to the question above, I am going to derive the classic rocket equation. This tells you the change in speed for a rocket that shoots out fuel. Yes, there will be some maths, so hold on. So, let’s start with a rocket. It has a mass

*m*and is moving with a velocity*v*. Next it shoots out some fuel of mass*dm*with a speed*u*relative to the rocket giving the rocket a new velocity of*v*+*dv*.Since momentum is conserved (and I am writing everything in just one dimension), the momentum of the initial rocket must be equal to the momentum of the new rocket (with a smaller mass) and the ejected exhaust. Note that the exhaust has a velocity of

*u*with respect to the rocket. This means that when it is ejected it has a velocity of*v*–*u*with respect to the background axis. Also notice that the velocity and mass of the rocket change. So, here is the conservation of momentum equation.Now I just need to multiply everything and some stuff will cancel (do this for yourself to check) and I get:

Most textbook derivations will take the following step. They will say “hey look, we have

*dm*multiplied by*dv*—both of those are tiny so the product will be super tiny.” So they make the approximation that as the fuel size goes to zero (for continuous fuel use), this term also goes to zero. Let’s go with that assumption for now.This gives a differential equation. All of the velocity terms are on one side and all the mass terms are on the other (remember that

*u*is constant). Integrating both sides:Remember that when I started I said everything was in one dimension. For this case, the exhaust is in the negative direction (so it has a negative value). If I just want

*u*to be the speed of the exhaust gas, I can write the rocket equation in the following (and more common) form:But does this expression answer my rocket question from above? According to this derivation, the change in velocity for a rocket depends only on the exhaust velocity and the fuel to rocket (payload) ratio. It doesn’t matter how fast you use the fuel, just the amount. So, the answer should be that releasing all the fuel at once is the same as releasing it slowly. This is incorrect though.

### Another Example

Let me show you a simple situation we can use to answer the rocket question from above. Suppose I have two rockets with a mass

*M*and fuel mass*m*. Rocket A shoots all the fuel at once (again, like a nuclear propulsion engine) with a fuel speed of*u*and rocket B shoots two blobs of fuel—first a shot of*m*/2 and another one of*m*/2. Both start from rest. Here’s a diagram showing the two rockets after shooting the fuel out.Maybe you can already see the solution. The key is that second blob of fuel shot from the second rocket. It leaves the rocket with a speed

*u*relative to the rocket. However, the rocket is already moving in the opposite direction. This means that the speed of the second blob is less than*u*. That makes the total momentum of the propellant coming out rocket B has a smaller momentum than the one ball of fuel coming out of rocket A. Since momentum is conserved, rocket B must also have a smaller momentum than rocket A. These rockets have the same mass which means that rocket B has a smaller velocity. It’s better to shoot all your fuel at once instead of spreading it out.### A Better Rocket Equation

You might protest and say that the “rocket equation” says that the rate of fuel use doesn’t matter. Well, that’s what the equation says—but go back to the derivation. Remember the part where we let

*dm*dv*be zero? We said that it was just really small? Well, if you shoot out fuel all at once,*dm*isn’t so small and neither is*dv*. Leaving that term back into the derivation, I have the following (same as before). Now I can solve this for*dv*: Note that this*dv*should really be Δv since we are doing discrete fuel instead of continuous—the same is true for the*dm*mass. But now that we have an expression for the change in velocity for each pellet shot, I can easily make a numerical model. With that expression and the change in mass, I can calculate the total change in velocity for the different number of fuel pellets. Notice also that the change in velocity only depends on the fuel speed (*u*), the fuel pellet size*dm*and the mass of the rocket. It does not depend on time or the rate at which fuel is used. The fewer fuel pellets shot, the greater the change in speed compared to continuous rockets. However, this increase in speed is only really noticeable when the fuel to payload ratio is high. If the mass of the fuel is small compared to the total mass of the rocket, it doesn’t really matter how you use the fuel.### But What Does It All Mean?

Which is better? A fusion powered nuclear explosion propulsion or a fusion powered ion engine? The nuclear explosion engine isn’t completely crazy. There’s even a Wikipedia page on it. Of course in actual use, a nuclear explosion would send “fuel” out in many directions, not in one big mass as in the simulator. But still, it similar to the theoretical case above. And here is the answer: If the mass of the rocket is large compared to the mass of all the fuel—it doesn’t matter. If your rocket consists of mostly fuel, the nuclear propulsion would be best (theoretically).