# Is there an upper limit to a rocket's size/payload? [duplicate]

This question already has an answer here:

It's obvious that as you increase the payload of a rocket, you need to also increase the amount of rocket fuel. Yet, as you increase the amount of rocket fuel, you are also increasing the overall mass of the rocket and, therefore, require even more fuel to get the rocket off the ground.

From this observation, one could potentially conclude that as you double the payload of a rocket, you must roughly quadruple its fuel (I am making an assumption here, so correct me if I'm wrong on that one). With that said, is there a theoretical/practical upper limit to the size and payload of a chemical rocket?

## marked as duplicate by Kyle Kanos, Danu, BMS, John Rennie, Qmechanic♦Jul 6 '14 at 17:09

• This may be helpful: what-if.xkcd.com/85 – xebtl Jul 5 '14 at 19:08
• How much money, engineering talent and fuel do you have? – dmckee Jul 5 '14 at 19:34
• @dmckee suppose a country has unlimited resources. Would they ever reach a maximum size? – Jason Jul 5 '14 at 21:01
• If they have unlimited engineering talent they'll recognize an efficiency limit before they hit a physical limit. They'll get to the point where they say, "this is the biggest and heaviest thing we need to lift, build me the right rocket to do it" and they won't muck around with anything bigger than that. – dmckee Jul 6 '14 at 0:38
• possible duplicate of Why are rockets so big? – Kyle Kanos Jul 6 '14 at 14:38

In the idealised case, the answer to this is slightly surprising. The fact that the mass of a rocket must include the mass of its fuel is embodied in the rocket equation, $$\Delta v = v_e \ln\frac{m_i}{m_f},$$ where $m_i$ is the initial mass of the rocket (including fuel, payload and everything else), and $m_f$ is the final mass, including the payload but much less fuel. $v_e$ is the effective exhaust velocity, which we might as well assume stays fixed for a given type of rocket, and $\Delta v$ is essentially the velocity change required to reach escape velocity, which we'll also assume stays constant.

The above equation does not include the acceleration due to gravity, which is of course an important factor. This is because (as is usually done) it's included in the $\Delta v$ term, which includes the velocity you lose to gravitational acceleration as the rocket ascends. You can put in the gravitational acceleration explicitly and the result doesn't change, as I'll show below.

Rearranging the rocket equation gives us $$m_i = m_f e^{\Delta v/v_e},$$ which tells us the amount of fuel (the majority of $m_i$) we need to lift a mass $m_f$. You can see that this is exponential in $\Delta v$, meaning that if we want to go a little bit faster we need a much bigger rocket. This is called "the tyranny of the rocket equation."

In this case we don't want to go faster, we just want to send more stuff, i.e. we want to increase $m_f$. But the equation is not exponential in $m_f$, it's linear. Therefore if we ignore any changes in rocket design that would be needed to increase its size, we can conclude that if you want to double the payload, you only need to double the size of the rocket, not quadruple it.

If we want to do this more precisely, we should include gravitational acceleration in the rocket equation. As per this answer by Asad to another question, this gives us $$\Delta v = v_e ln \frac{m_i}{m_f} - g\left(\frac{m_f}{\dot m}\right),$$ where $g$ is acceleration due to gravity and $\dot m$ is the rate at which fuel is burned, which we assume is constant over time. According to the reasoning in Asad's answer, we end up with $$m_i = m_f \left(\exp\left(\frac{\Delta v + g\left(\frac{m_f}{\dot m}\right)}{v_e}\right) -1\right)^{-1},$$ where $\Delta v$ is now the true escape velocity rather than the effective escape velocity. In Asad's answer, he assumes that $\dot m$ stays constant as you change $m_f$, and he concludes that there is a strong limit to the size of a rocket. But in fact if you were going to make a rocket twice the size, it wouldn't make sense to keep $\dot m$ the same. To take it to an extreme, imagine building something the size of a Saturn V that burns fuel at the same rate as a hobby rocket. It obviously wouldn't be able to lift itself off the launch pad, and nobody would consider building such a design. So let's instead assume that the burn rate is proportional to the size of the rocket. This means that $\frac{m_f}{\dot m}$ is a constant, and the equation as a whole is still of the form $$m_i = m_f \times \text{a constant},$$ so it's linear in $m_f$.

In fact none of this is really all that surprising after all, because if you want to send twice the mass you could always just use two rockets of the original size. By just strapping those rockets next to each other you've got one of twice the size that can send twice the payload. Moreover, it burns fuel at twice the rate, just as I assumed above. There's no reason that wouldn't work in principle. (Though in practice it would be another matter of course!)

If the equation had been exponential in $m_f$ then there would have been a point at which increasing the payload mass would require an unreasonable amount of extra fuel, and that would have imposed a strong practical limit on rocket size. But since it's linear this doesn't really happen. The limits on rocket size are not due to an exponential increase in propellant mass, but to the engineering challenges in building a structure of that size and complexity that won't fail under the violent conditions of a rocket launch.

These include factors to do with the way the strength of a structure scales with its size and (I imagine) practical issues involved in getting fuel where it needs to be at the right time. In this respect the factors that limit the size of rockets are quite similar to the factors that limit the size of buildings.

• Fascinating! I never considered using two rockets. When you put it that way it seems so intuitive! Even more fascinating is the exponential relationship between dV and the mass of fuel. – Jason Jul 6 '14 at 5:32
• This is actually wrong, there is an upper limit. Please see Asad's answer to the question Why are rockets so big. The basic problem is that you are not considering the ever-important gravitational acceleration in the rocket equation. – Kyle Kanos Jul 6 '14 at 14:41
• @KyleKanos thanks, you're right that I should have included the gravitational acceleration in the equation. But unless I misunderstood it on my quick reading, I don't really agree with Asad's answer. He assumes the burn rate remains constant as the rocket size changes, which seems a weird assumption to me. If you let it scale proportionally to the mass you still end up with $m_f$ times a constant, as in my answer. – Nathaniel Jul 6 '14 at 15:47
• @KyleKanos Thank you for pointing that out. I didn't see that when reading the other question. – Jason Jul 6 '14 at 15:55
• @KyleKanos I've fixed the answer so that it explains the issue with gravitational acceleration properly, and also explains why Asad's answer to the linked question is wrong (and yours is right). – Nathaniel Jul 9 '14 at 3:23

The upper limit is the size of the planet. Consider that you could mount a linear accelerator or "mass driver" pointing upward and use the material the planet is made of as reaction mass. Now drive off to explore the cosmos. By the time you whittle down the planet to an asteroid sized rock, you will be going at a pretty good clip!