How much fuel does it take to accelerate a rocket to a given velocity? Here we navigate a step-by-step derivation of the rocket equation, explain the underlying concepts, then discuss how it does (and does not) model actual rockets.

Rocket equation

Every rocket ever made works by taking advantage of Newton’s Third Law: a rocket engine pushes mass out the back, which means the mass is pushing the rocket toward the front, accelerating the rocket.

In theory, we can throw anything out the back to accelerate rockets. Heck, you could have an astronaut onboard with a rolling chair and a fire extinguisher, like so:

But studies show this is not an effective way to propel rockets. Many rockets rely instead on chemical propellants, where an oxidizer (such as liquid oxygen) and fuel (such as liquid hydrogen) react, producing gases that are expelled in the opposite direction. 🔥 🚀

⚠️ Unlike when an ice skater pushes off a wall to gain speed, a rocket is not pushing against Earth’s atmosphere to gain speed, as evidenced by the fact that rockets still work in space where there is little atmosphere. Whereas an ice skater’s mass doesn’t change when they push off to accelerate, a rocket’s mass decreases.

So how much fuel would you need to get a rocket to a certain speed? More precisely, if we know we need to accelerate our payload mass of $m_f$ to a certain speed, how much total mass $m_i$ (including fuel and payload) do we need to start with? It turns out we can use conservation of momentum and some inspired uses of single-variable calculus to find the answer.

Why not burn all your fuel at once

Let’s start simple and ignore gravity and atmospheric drag. Imagine a rocket somewhere out in space, perhaps on its way to Alpha Centauri, the closest start outside of the Solar System.

The most fuel-efficient way to accelerate the rocket would be to burn all the fuel right at the beginning. Otherwise, some of the energy is spent accelerating fuel only to burn it off later.

Conservation of momentum before and after all the fuel is burnt.

Conservation of momentum requires that the system’s momentum before the fuel is burnt and the momentum after the fuel is burnt is the same.

If this is confusing, check out this post on momentum.

Mathematically, we have

$0 = mv_{1} + Mv_2 \\ \Rightarrow v_2 = \frac{mv_1}{M}$

✏️ This is essentially an inelastic collision in reverse. Although we think of an inelastic collision as two objects colliding to form one, there’s no reason we can’t use the same math to explain one object splitting apart into two. Nothing in the math forbids us from reversing the order of events. In general, classical mechanics is time-invariant: the laws of classical mechanics work just as well forward in time as they do backward.

There, problem solved right? Plug in your values and solve for the unknown.

💣 Unfortunately, it’s not that simple. Instantly burning all of the fuel right at the beginning is essentially just exploding a bomb under the rocket. While fuel-efficient, practically speaking:

our rocket is not strong enough to redirect all the gases in the opposite direction of intended travel, which decreases efficiency (since energy is spent pushing the rocket in a direction that is not up) and likely destroys the entire rocket (the rocket would explode)
even if the rocket’s structural engineering overcame these concerns, neither our cargo nor our human occupants would be able to sustain the massive acceleration

Therefore, we’re left with burning the fuel over time. However, this complicates our setup. Instead of a constant rocket mass, we have a mass that changes over time, making it seemingly impossible to apply the same conversation of momentum principle.

Treating the rocket mass as constant for small time intervals

💡 But what if we only looked at conservation of momentum for a small enough time period that the mass is approximately constant? In other words, let’s apply the same logic, but restrict our setup to a sufficiently small time interval that we can treat the rocket mass as not changing.

Change in masses and velocities for a rocket firing over an extremely small time interval such that the mass is approximately constant.

Fig. 2 shows this setup. At the start, we treat the rocket as having mass $m$ moving at velocity $v$ . Then, the rocket ejects mass $\Delta m$ with velocity $v_e$ , causing the rocket to decrease in mass to $m - \Delta m$ and increase in velocity to $v + \Delta v$ .

Note that this is the exact same setup as Fig. 1, but we’ve modified some of the variable names in a way that will become clear as we derive the rest.

⚠️ Note that the exhaust velocity $v_{e}$ is relative to the rocket, not to the ground. So if we were an observer on Earth looking at a rocket launch, the exhaust velocity at the moment of launch (when the rocket is not moving) would be 4500 m/s backward. Once the rocket is moving at, say, 6000 m/s, the exhaust velocity would still be 4500 m/s backward, but as an observer we would see the exhaust moving 6000 m/s-4500 m/s = 1500 m/s forward!

Conservation of momentum requires the total system momentum before and after the mass is ejected to be the same:

$\begin{aligned} mv &= (m-\Delta m)(v+\Delta v) + (v-v_{e})\Delta m \\ &= mv - v\Delta m + m\Delta v - \Delta m\Delta v + v\Delta m - v_{e}\Delta m \\ \end{aligned}$

⚠️ Here, I’m treating all my velocity terms as positive, then adding negative signs as needed to indicate the exhaust velocity is opposite the direction of rocket motion. If you want, you can treat your velocities as vectors that can be either positive or negative (in other words, moving any negative signs into the terms themselves) – just make sure you stay consistent to avoid confusing yourself.

Canceling out terms and rearranging gives us

(1) $m\Delta v - \Delta m\Delta v- v_{e}\Delta m = 0$

Now I’m going to do something crazy. I’m going to claim that we can just ignore the $\Delta m \Delta v$ term. Just cross it out and keep going like it was never there.

✋🏻 🛑 But wait you say! That’s illegal. You’re absolutely right. For the equation in front of us, simply ignoring term would just make this wrong. However, we’re trying to derive the behavior of real life rockets, for which the $m, v_e$ are far larger than $\Delta m, \Delta v$ , in which case we can approximate this equation by just ignoring this really small term.

If that sounded really hand-wavy, that’s because it is! We’re deriving it like this to maximize our physical understanding of the scenario, especially if you’re learning this for the first time. The more rigorous explanation can be found in the FAQ here.

Rearranging the terms from Eq. 1 then gives us

(2) $\frac{\Delta v}{\Delta m} = \frac{v_{e}}{m}$

Let’s stop and think about what Eq. 2 tells us.

On the left-hand side, we have a rate $\Delta v/\Delta m$ : the change in the rocket’s forward velocity $\Delta v$ when a mass $\Delta m$ is expelled toward the back. We can think of this rate as how effective the rocket engine is at accelerating the rocket, as a higher $\Delta v/\Delta m$ means that the rocket’s velocity increases more when the mass is ejected.

The right-side side tells us what this rate depends on:

As the exhaust velocity increases, this rate increases. The engine is more effective at pushing the mass out the back, which means (thanks to conservation of momentum) that the engine is more effective at accelerating the rocket forward.
As the mass of the rocket decreases, this rate increases: the engine is more effective at accelerating the rocket because the rocket is lighter, so by Newton’s Second Law, the force from the ejected mass causes a greater acceleration.

While Eq. 1 is only an approximation (remember that we just ignored the $\Delta m \Delta v$ term), for the case of rockets, it turns out to be a very effective approximation, and one that strongly aligns with our intuition.

Summing up all the small changes in mass and velocity

But Eq. 1 only tells us this relationship between mass and velocity for a small time interval. To understand how mass and velocity evolve over the entire time period, let’s sum up the changes in mass and the changes in velocity.

Let’s re-arrange Eq. 2 to arrive at

$\frac{\Delta v}{v_{e}} = \frac{\Delta m }{m}$

Next, we have to make a subtle but crucial change from $\Delta m \to - \Delta m$ .

Up to now, we’ve been looking at the relationship between the ejected mass $\Delta m$ , the mass of the rocket $m$ , and the change in the velocity of the rocket $\Delta v$ .
However, we want to sum up the changes in the mass of the rocket and relate it to the changes in the velocity of the rocket, not the mass of the ejected mass. We do know that if the ejected mass is $\Delta m$ , then the rocket has lost mass $- \Delta m$ , so let’s go ahead and make this replacement.

$\frac{\Delta v}{v_{e}} = \frac{-\Delta m }{m}$

✏️ Don’t worry if this didn’t seem obvious to you, because it’s not! The first time I derived this, I didn’t add this negative sign, and my final rocket equation showed that less fuel was required to achieve a higher velocity change, which told me I had messed up somewhere. This is why it’s important to interpret your setup as you go, and do a sanity check at the end – does your answer line up with what you think should happen?

Now, let’s sum up the small changes in velocity and mass over the entire time interval.

(3) $\sum^{t}_{t=0} \frac{\Delta v}{v{e}} = \sum^{t}_{t=0} \frac{-\Delta m }{m} \Rightarrow \sum^{v_f}_{v_i} \frac{\Delta v}{v_{e}} = \sum^{m_f}_{m_i} \frac{-\Delta m }{m}$

So far, our derivation has assumed that $\Delta t$ is small enough for us to treat the mass as constant in that time interval. But this is just an approximation. Unless…we take $\Delta t$ to be infinitesimally small. Looking at both sides of Eq. 3, what if we took the limit of $\Delta t \to 0$ (and therefore $\Delta m, \Delta v \to 0$ )?

This doesn’t have to be true – this is just a hypothesis that we have that, as it turns out, yields a rocket equation that makes sense.

$\lim_{\Delta v \to 0}\sum^{v_f}_{v_i} \frac{\Delta v}{v_{e}} = \lim_{\Delta m \to 0}\sum^{m_f}_{m_i} \frac{-\Delta m }{m}$

👀 This is the definition of an integral! That’s not an accident. We’ve set up a physical scenario where we’re summing up a variable in very small amounts, then taking the limit as that small amount goes to zero. That’s exactly what an integral does, and that’s why we can go from $\Delta m$ and $\Delta v$ to $dm$ and $dv$ and apply the rules of single-variable calculus.

(4) $\frac{1}{v_{e}}\int^{v_f}_{v_i} dv = -\int^{m_f}_{m_i} \frac{1}{m} dm$

Using the integration rules from single-variable calculus gives us

$\Rightarrow \frac{1}{v_e}(v_f-v_i) = \text{log }\frac{m_i}{m_f}$

The change in velocity $v_f - v_i$ is known as delta v, or $\Delta V$ .

Although most conventions use $\Delta v$ to denote “delta-v”, we will use $\Delta V \equiv v_f - v_i$ to denote the total change in velocity to avoid confusion with our previous usage of $\Delta v$ , which we defined as the small but finite change in rocket velocity when a small but finite mass is ejected.

(5) $\Delta V = v_e \text{log }\frac{m_i}{m_f} \Leftrightarrow \frac{m_i}{m_f} = e^{\Delta V / v_e}$

🚀 This is the rocket equation! 🚀 Here’s what the rocket equation tells us:

As the delta-v requirement increases, we need a larger $m_i/m_f$ ratio, which means we need more fuel mass initially to get the rocket to the target velocity.
A more effective engine has a higher exhaust velocity, which means a smaller $m_i/m_f$ ratio is needed to reach the target velocity, which means less fuel is needed.

Is this how rockets work?

Sort of. For a rocket with no external forces acting on it, this equation can be an effective description of how a rocket’s velocity changes with with its mass over time.
However, if a rocket is affected by atmospheric drag (for example, if launching from Earth’s surface) or gravity (for example, if flying close enough to any large celestial body such as Earth or the Moon), we would need to incorporate these forces into the derivation.
- We derived the basic rocket equation without gravity or drag by starting with conservation of momentum.
- Another way of writing conservation of momentum is $F = \frac{dp}{dt} = 0$ . By including the relevant forces, we can use the same method to arrive at a similar result (although the integrals might not as pretty).