2013-11-28 02:00:58 8 Comments

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.

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## 6 comments

## @Asad Saeeduddin 2013-11-28 05:08:03

TL;DR:This answer arrives at roughly the same conclusion as Kyle Kanos' answer, i.e. in addition to payload considerations, the difficulty lies in stuffing a small rocket with a mass of fuel exceeding the mass of the rocket itself. This answer, however, is more rigorous in how the $\Delta v$ budget is treated.The rocket equation:Consider the Tsiolkovsky 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 (constant) 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 with respect to time), $m_0$ is the the initial mass of the rocket and $m(t)$ is the current mass of the rocket.

Note that this is essentially a momentum exchange equation: you have a finite amount of momentum available from expulsion of fuel, which you must spend on increasing the velocity of the rocket + remaining fuel system,

as well as overcoming gravity(i.e. dragging the planet ever so slightly). A form of the Tsiolkovsky equation that does not take this into account (as in the other answer) will give you non-physical results.Constrained variables:Now, what can we play with in this equation? Assuming $t_{escape}$ is the time at which the rocket escapes Earth's gravity:

This means none of these quantities are negotiable; we are constrained by the demands of the mission and the available technology.

Developing a relationship between rocket and fuel mass: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 $m_0 = m_f + m_r$:

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

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:

In red, we have a plot of rocket mass versus initial fuel mass, while in blue we have a plot of the

ratioof initial fuel mass to total mass. Note that the axis for the blue plotstarts 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

decreasethe 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.

## @dmckee --- ex-moderator kitten 2018-06-07 20:16:51

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## @fibonatic 2013-11-28 02:35:51

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

## @Kyle Kanos 2013-11-28 03:46:54

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

biggercraft which meansmorefuel! 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:

## @Asad Saeeduddin 2013-11-28 04:02:53

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.

## @Kyle Kanos 2013-11-28 04:19:12

@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.

## @Asad Saeeduddin 2013-11-28 04:26:22

@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.## @fibonatic 2013-11-28 04:55:12

@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.

## @Asad Saeeduddin 2013-11-28 05:49:08

@fibonatic The delta v budget you are quoting is only a reasonable approximation for rockets with similar burn time to actual rockets. Since this question is specifically about rockets that can be very small, using the gravity inclusive delta v budget for a large rocket will yield poor results.

## @BЈовић 2013-11-28 07:18:45

holy crap! that means they are burning oil like it is nothing, just to get some junk up there. And they did it lots of times. What kind of maniac is needed to form nasa?!?

## @MSalters 2013-11-28 08:45:38

@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.

## @AJMansfield 2013-11-28 19:51:57

@BЈовић For a better idea of what type of fuel gets used, see the Wikipedia pages for Solid-fuel Rockets and for Liquid Rocket Propellants.

## @David Hammen 2014-11-11 09:22:27

@MSalters - They oftentimes do burn oil. The first stage of the Saturn V rocket used RP-1, a highly refined kerosene, to launch men to the Moon. RP-1 with liquid oxygen as the oxidizer is very widely used as a propellant.

## @Kyle Kanos 2017-01-25 11:02:08

Might the few downvoters mention what they think is wrong with this answer?

## @user77220 2017-08-21 00:01:59

"as speed increases, the mass required (in fuel) increases exponentially", this is wrong assumption. It depends on the type of fuel that are used for rocket propulsion. Are you talking about liquid fuel, solid fuel or solar fuel. All have different capacities.

## @my2cts 2019-06-21 16:47:24

Kyle I found my answer at physics.stackexchange.com/questions/487194/… deleted . I formally object. The reason was that not enough explanation was given, only a link to arxiv.org. You know that arxiv.org is a very stable site. It has been consistent for over 20 years. Also I should be given the opportunity to explain my answer in more detail. Please undelete. Apologies for the cross post. I saw no other means.

## @Kyle Kanos 2019-06-21 18:06:33

@my2cts physics.meta.stackexchange.com

## @jokoon 2013-11-28 10:37:05

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.

## @user34882 2013-11-28 08:09:27

$ 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.

## @a CVn 2013-11-28 08:44:24

$E=mc^2$ 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 $\Delta v$.

## @fibonatic 2013-11-28 14:15:07

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.

## @a CVn 2013-12-02 13:36:03

@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.

## @jean 2018-06-07 12:35:28

We can put it on an airplane en.wikipedia.org/wiki/Nuclear-powered_aircraft

## @Luke Burgess 2013-11-28 02:22:13

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.

## @lurscher 2017-12-07 14:33:31

this is the correct answer. Also, you need to factor the fact that atmospheric drag grows as the square power of width, while total fuel mass grows with the third power, even assuming constant fuel to dry mass ratio