Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm curious why rockets are so big in their size. Since both the gravitational potential one need to overcome in order to put thing into orbit, and the chemical energy burned from the fuel, are proportional to the mass, so if we shrink the rocket size, it would seem to be fine to launch satellites. So why not build small rocket say the size of human? I can imagine small rocket would be easier to manufacture in large quantities and easier to transport. And maybe someone can make a business out of small rocket, carrying one's own satellite.

share|cite|improve this question
up vote 89 down vote accepted

The problem is what Konstantin Tsiolkovsky discovered 100 years ago: as speed increases, the mass required (in fuel) increases exponentially. This relation, specifically, is $$ \Delta v=v_e\ln\left(\frac{m_i}{m_f}\right) $$ where $v_e$ is the exhaust velocity, $m_i$ the initial mass and $m_f$ the final mass.

The above can be rearranged to get $$ m_f=m_ie^{-\Delta v/v_e}\qquad m_i=m_fe^{\Delta v/v_e} $$ or by taking the difference between the two, $$ M_f=1-\frac{m_f}{m_i}=1-e^{-\Delta v/v_e} $$ where $M_f$ is the exhaust mass fraction.

If we assume we are starting from rest to reach 11.2 km/s (i.e., Earth's escape velocity) with a constant $v_e=4$ km/s (typical velocity for NASA rockets), we'd need $$ M_f=1-e^{-11.2/4}=0.939 $$ which means almost 94% of the mass at launch needs to be fuel! If we have a 2000 kg craft (about the size of a car), we would need nearly 31,000 kg of fuel in a craft that size. The liquid propellant has a density similar to water (so 1000 kg/m$^3$), so you'd need an object with a volume of 31.0 m$^3$ to hold it. Our car sized object's interior would be around 3 m$^3$, a factor of 10 too small!

This means we need a bigger craft which means more fuel! And explains why this mass-speed relation has been dubbed "the tyranny of the rocket problem".

This also explains the fact that modern rockets are multi-staged. In an attempt to alleviate the required fuel, once a stage uses all of its fuel, it is released from the rocket and the next stage is ignited (doing this over land is dangerous for obvious reasons, hence NASA launching rockets over water), and the mass of the craft is lowered by the mass of the (empty) stage. More on this can be found at these two Physics.SE posts:

share|cite|improve this answer
The Tsiolkovsky equation in the form you have stated only applies when the net external force is zero (i.e. no gravity). To accurately calculate the $\Delta v$ required, you need to include an additional term $-g(\frac{m_{propell}}{\dot m})$ on the right hand side of the equation. – Asad Saeeduddin Nov 28 '13 at 4:02
@Asad: this is true, but I think it's (mostly) irrelevant to the point that we still need a boat-load of propellant to get ourselves into space, hence large rockets and not person-size ones. – Kyle Kanos Nov 28 '13 at 4:19
@KyleKanos Yes, the gist of your answer is correct. I was taking issue with the calculation you added, which is flawed. Either you need to consider an effective $\Delta v$ which is augmented to approximately account for the retarding effect of gravity as well as the required escape velocity (this is the standard approach) or actually do the calculation taking fuel burn time into account. – Asad Saeeduddin Nov 28 '13 at 4:26
@Asad It might have been easier if Kyle Kanos would have use the $\Delta v$ budget needed to get into low Earth orbit, which is about 9.3 - 10 km/s, but this would still return about the same result. – fibonatic Nov 28 '13 at 4:55
@BЈовић: They usually don't burn oil, it's not efficient enough. But fuel actually isn't that expensive. It's often just a few % of launch costs. – MSalters Nov 28 '13 at 8:45

TL;DR: This answer arrives at roughly the same conclusion as Kyle Kanos', i.e. in addition to payload considerations, the difficulty lies in stuffing a small rocket with a mass of fuel exceeding to the mass of the rocket itself. This answer, however, is more rigorous in how the $\Delta v$ budget is treated.

Developing a relationship between rocket and fuel mass:

Consider the rocket equation, which describes the motion of vehicles that propel themselves by expelling part of their mass with a certain velocity. A simplified version which only takes gravity and thrust into account is given below:

$$ \Delta v(t) = v_e \cdot ln \frac{m_0}{m(t)} - g\left(\frac{m_f}{\dot m}\right) $$ where:

$v_e$ is the effective exhaust velocity,

$m_f$ is the mass of the fuel aboard,

$\dot m$ is the the mass burn rate (constant wrt time),

$m_0$ is the the initial mass of the rocket, and

$m(t)$ is the current mass of the rocket

The value of $\Delta v$ by the time the rocket burns out is fixed by where we're trying to send the rocket, and the mass $m(t)$ at this point will be the mass of the rocket without any fuel, so neither of these quantities is negotiable. The effective exhaust velocity $v_e$ and the rate of mass flow are not in our hands either, since these are a function of the type of engine/propellant available.

All we are left to play with is the initial masses of the rocket fuel $m_f$ and rocket body $m_r$. Let us substitute in the values of $v$ and $m$ at the instant when the rocket escapes gravity, noting that $m_0 = m_f + m_r$:

$$ \begin{align} v_{escape} & = v_e \cdot ln \frac{m_f + m_r}{m_r} - g\left(\frac{m_f}{\dot m}\right)\\ & = v_e \cdot ln(1 + \frac{m_f}{m_r}) - g\left(\frac{m_f}{\dot m}\right) \end{align} $$

Rearranging, we have:

$$ m_r = m_f \cdot \left(exp\left(\frac{v_{esc} + g\left(\frac{m_f}{\dot m}\right)}{v_e}\right) -1\right)^{-1} $$

Note that this is effectively providing $m_r$ as a function of $m_f$, since all the other parameters are fixed by the constraints of the mission and equipment as well as environmental constants. Since the relationship isn't immediately obvious, here is a plot of $m_r$ against $m_f$ for selected values of the constants:

enter image description here

In red, we have a plot of rocket mass versus initial fuel mass, while in blue we have a plot of the ratio of initial fuel mass to total mass. Note that the axis for the blue plot starts at 0.9!! This indicates that regardless of what rocket mass you picked, the net initial mass of your vehicle would have to consist almost entirely of fuel.

So what does this mean?

Filling a vehicle with a mass of fuel exceeding its own is increasingly difficult for small rockets, but not so difficult for much larger rockets (think of how the enclosed volume of a hollow body scales versus mass). This is why making smaller and smaller rockets becomes progressively more difficult.

In addition, a minimum limit on the rocket mass we can choose is imposed by the weight of the payload it must carry, which could be anything from a satellite to a single person.

Upper limit on payload:

A very interesting thing happens near the inflection point of the rocket mass - fuel mass curve. Before the inflection point, adding more fuel allowed us to hoist a larger payload to the desired velocity.

However, somewhere around $4 \cdot 10^6$ kg of fuel mass (for our selected parameter values) we discover that adding more fuel starts to decrease the payload that can be hoisted! What is happening here is that the cost of the additional fuel having to fight against gravity begins to win out against the benefit of having a high fuel to payload mass ratio.

This shows there is a theoretical upper limit to the payload that can be hoisted on Earth using the propellant technology we have available. It is not possible to simply keep increasing the payload and fuel masses in equal proportion in order to lift arbitrarily large loads, as would be suggested by using the Tsiolkovsky equation with no extra terms for gravity.

share|cite|improve this answer
Great answer, with a nice graph to look at too! :) The only thing I feel is missing is the definition of $g$; I would expect it to be the gravitational constant, but it might not be obvious to everyone reading the question out of curiosity. You also don't define $v(t)$, but at least there is a relation there to $v_e$ and $m(t)$ which are both defined in the answer. – Michael Kjörling Nov 28 '13 at 8:20
@MichaelKjörling Thanks! $g$ is in fact the gravitational constant. Are you referring to $\Delta v(t)$, the change in velocity over time $t$? – Asad Saeeduddin Nov 28 '13 at 8:22
Yes, I was referring to $\Delta v(t)$; I know the $\Delta$ part and guessed by $v(t)$ you were referring to velocity at time $t$ (which makes sense). I'm not good enough at this to judge the accuracy of the calculations, but they look about right given what I do know. (The only thing I'm not quite sure I'm buying the example 1 kg/s fuel burn rate; a third stage quite possibly, but first and second? Even so, the answer would still stand IMO, because it really boils down to "rockets are big because you need a metric crapton of fuel to get anywhere, and you need to stuff that fuel somewhere".) – Michael Kjörling Nov 28 '13 at 8:29
@MichaelKjörling Yes, I didn't know what a typical fuel exhaust rate is for conventional fuels, so I just made something up out of thin air. That said, changing the value of $\dot m$ only skews the graph, it doesn't change its shape. – Asad Saeeduddin Nov 28 '13 at 8:31
Yes; like I said, it doesn't change the bottom line. That said, taking Kyle's figure of 31,000 kg of fuel for a 2,000 kg load (unloaded rocket + payload) weight, that would give you a total burn time of close to nine hours. I would imagine (without having checked) for a first and second stage the burn rate being 1-2 orders of magnitude higher. Which by the time you've actually gotten anywhere approximates to "a metric crapton". Of course, with a small rocket, first and second stage burn times may be considerably shorter, but you'll still be burning a lot of fuel to get them off the ground. – Michael Kjörling Nov 28 '13 at 8:34

Because most payloads are quite heavy. I am not sure what kind of payloads you had in mind, I am no expert on this, but I think that most launches contain satellites, which might be heavier then you think, for instance the satellite in this BBC Documentary weighs 6000 kg. And according to Wikipedia, miniaturized satellites weigh less than 500 kg (so heavier is normal). And some of those miniaturized satellites are using excess capacity on larger launch vehicles.

And I think that smaller rockets will experience the turbulence of our atmosphere much violently. Also think of the relatively higher costs in terms of personnel (such as mission control). And I would also expect that certain aspects do not scale linearly in size, but for be this would just be speculation. xxxxxx

share|cite|improve this answer

Consider the problem in the from of a ratio, what is the ratio of mass used to lift the rocket(fuel), to the mass finally put into orbit(cockpit). That proportion will be much the same regarding smaller objects that must be put into orbit. If you use the same ratio or proportion to calculate the needed fuel mass for a small craft, you will find you can't even carry the device holding your fuel. This is also why rockets use stages.

The type of fuel used also has an impact, but those are details that need a new question.

share|cite|improve this answer

Mainly because you need a lot of speed to go into space, and to each that speed, you need to accelerate. If you need a high speed, you will need to accelerate for a long time, thus the need for a large quantity of fuel. You also need to compensate for gravity the whole lift.

There are ways to reduce that fuel requirement, like a horizontal takeoff, you reach a high altitude and then launch, so you keep the engine, but you still need a lot of energy to fight against gravity, and wings can't lift you very high, so that would not be such a good fuel economy, and the plane would still require to be quite big.

share|cite|improve this answer

$ E = mc^2 $

The larger the mass, the more energy can be produced. And we still haven't found any fuel which in small quantities gives the needed amount of energy. I know you will be thinking of nuclear energy; we cannot fit a nuclear reactor inside a rocket with current technology, and even if we can fit it I don't think our existing knowledge of nuclear science is sufficient to ensure accident-free reactors at such velocities.

share|cite|improve this answer
$E=mc^2$ doesn't really apply here. First, I'm not aware of any practical matter-energy conversion process that comes anywhere near close to that (insofar as I know we still haven't figured out how to build matter/antimatter reactors for power generation purposes, and that'd be about the only way to get anywhere near such amounts of energy). Second, if you look at the rocket equation cited in other answers, you'll see that the critical issue is the exhaust velocity. If you can get insane exhaust velocities, each tiny nugget of fuel packs a lot more punch in terms of total system $\Delta v$. – Michael Kjörling Nov 28 '13 at 8:44
We could use the propulsion similar to that of project Orion, but this probably will not be used at take-off due to the nuclear fallout. – fibonatic Nov 28 '13 at 14:15
@fibonatic ...and the fact that you need to worry about nuclear fallout is a pretty good indicator to begin with that you aren't in $E=mc^2$ territory. – Michael Kjörling Dec 2 '13 at 13:36

protected by Qmechanic Dec 16 '14 at 22:55

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.