Why are rockets so big?
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:
- Why do rockets have multiple stages?
- Why do rockets jettison fuel tanks?
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:
- $\Delta v(t_{escape})$ is simply our desired escape velocity (assuming the rocket starts from rest), which is dictated by where we're trying to send the rocket
- $m(t_{escape})$ will optimally be the mass of the rocket without any fuel
- 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.
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.
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.