Why are rockets so big?

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.

• How much fuel do you need to drive your car at 28,968 kilometers per hour against headwinds as strong as gravity? 200mph winds!!!! powerful telecoms sattellites weigh 1000 KG. they put the maximum amount of satellites in each payload, including small ones, and if you tried to save energy by going slower, you would have to fight gravitty for longer. the trick is to get it out as fast as possible above 30 miles high. a GPS sat weighs 2080 KG. – com.prehensible Oct 17 '17 at 10:39

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:

• 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
• @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
• @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. – David Hammen Nov 11 '14 at 9:22

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:

1. $\Delta v(t_{escape})$ is simply our desired escape velocity (assuming the rocket starts from rest), which is dictated by where we're trying to send the rocket
2. $m(t_{escape})$ will optimally be the mass of the rocket without any fuel
3. The effective exhaust velocity $v_e$ and the rate of mass flow $\dot m$ are a function of the type of engine/propellant available

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 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.

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.

• Comments are not for extended discussion; this conversation has been moved to chat. – dmckee Jun 7 '18 at 20:16

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.

• 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 – lurscher Dec 7 '17 at 14:33

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

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.

$E = mc^2$

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.

• $E=mc^2$ 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 $\Delta v$. – a CVn 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. – a CVn Dec 2 '13 at 13:36
• We can put it on an airplane en.wikipedia.org/wiki/Nuclear-powered_aircraft – jean Jun 7 '18 at 12:35

protected by Qmechanic♦Dec 16 '14 at 22:55

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