Introduction
If you are seriously interested in this question and are willing to devote some time to reading about it, I recommend reading the Atomic Rockets: Engine List page.
It will also discuss the problems you will face as a rocketeer. A partial list is this:
- For planetary launches, your engine must provide a higher
acceleration than the local gravity (e.g. for Earth, it must exceed 1
g). This means you need a high thrust engine.
- To reduce propellant usage, your engine must provide a high specific impulse
($I_{sp}$).
- For most engines these two things are mutually exclusive.
The rocket equation shows that a rocket's total propulsive capability is driven by surprisingly few factors.
$$\Delta v = v_e \cdot \ln{\frac{m_0}{m_f}}$$
- $\Delta v$ - propulsive capability of rocket
- $v_e$ - exhaust velocity of propellant
- $m_0$ - starting mass of rocket
- $m_f$ - final mass of rocket (starting mass minus propellant used)
Stellar Blackhole
Obviously the theoretical limit for anything is the formation of an event horizon (aka Blackhole). This is because the $\Delta v$ requirement exceeds the speed of light and no propellant can exceed that.
You can achieve this formation through many mechanisms. Take a small mass and compress it or keep adding mass to a single object.
A stellar Blackhole forms when several solar masses of matter are brought together under normal conditions (we don't yet know how much mass is required). No amount of fancy rocketry will get you out of a blackhole
3 Stage Chemical Rocket - works up to $M_{Saturn}$
Unlike many types of rocket engines, chemical rockets "burn" their fuel and expel the reaction products as its propellant. This limits your rocket engine to exothermic (energy releasing) reactions.
To keep your $I_{sp}$ as high as possible you must use chemicals that are as low mass as possible. $I_{sp}$ is driven by exhaust velocity, not momentum (increasing propellant velocity decreases propellant usage). So an engine that delivers the same thrust will use less fuel if you expel a low mass at high velocity rather than a high mass at low velocity.
LOX + LH2
The commonly used high performance chemical rocket fuel is liquid oxygen (aka LOX) + liquid hydrogen (LH2). This provides an $I_{sp}$ of about 450 (exhaust velocity of $4,400 \frac{m}{s})$.
LF2 + LH2
An even higher performance fuel would be liquid hydrogen + liquid fluorine. This combination can provide an $I_{sp}$ of about 480 (exhaust velocity of $4,700 \frac{m}{s}$). However, it introduces a host of huge problems:
- Handling the liquid fluorine before launch is tough.
- Keeping hot gaseous hydrofluoric acid from destroying your launch
complex and poisoning the people on the ground is much tougher.
Calculations
For argument's sake if we limit the equation to $\frac{m_0}{m_f} = 10$ (shuttle has a fraction of $\approx 5$ - meaning it is 80% fuel and 20% everything else)
Plugging the provided numbers into the equation would yield the following:
$$\Delta v = 4,700 \frac{m}{s} \cdot \ln{10} = 10,822 \frac{m}{s}$$
Subtract a reasonable atmospheric + gravity drag value ($\approx 20$ % $= 2164 \frac{m}{s}$). This leaves $8,658 \frac{m}{s}$ available for getting to orbit.
Orbital velocity is calculated using this approximation:
$$v_o = \sqrt{\frac{G \cdot M_{planet}}{r}}$$
Now solve for r (and $M_{planet}$):
$$8,658 = \sqrt{\frac{6.7 \cdot 10^{-11} \cdot 23,039 \cdot r^3}{r}} \rightarrow 7.5 \cdot 10^7 = 6.7 \cdot 10^{-11} \cdot 23,039 \cdot r^2$$
$$r^2 = \frac{7.5 \cdot 10^7}{6.7 \cdot 10^{-11} \cdot 23,039} \rightarrow r = \sqrt{\frac{7.5 \cdot 10^7}{1.5 \cdot 10^{-6}}}$$
- $r = 6,971 km$
- $M_{planet} = 7.8 \cdot 10^{24} kg$
Different density planets give different results.
Essentially, the Earth is the limit for single stage chemical rockets.
Staging
But wait a second! Clearly we launch vehicles into space that aren't single stage, so what gives?!
So far we've only discussed doing this as a single stage to orbit. It turns out that by staging a vehicle we actually get better performance and are able to more easily achieve orbit.
How much we actually gain depends upon the number and type of stages. But let's assume we use a 3 stage rocket with each stage has the performance given above. The staging equation is given by:
$$\Delta v = N_{stages} \cdot v_e \cdot \ln{10} \rightarrow \Delta v = 3 \cdot 4,700 \frac{m}{s} \cdot 2.3 = 32,466 \frac{m}{s}$$
All the rest of the numbers stay the same, so solve for r (and $M_{planet}$) again:
$$32,466 = \sqrt{\frac{6.7 \cdot 10^{-11} \cdot 23,039 \cdot r^3}{r}} \rightarrow 1.05 \cdot 10^9 = 6.7 \cdot 10^{-11} \cdot 23,039 \cdot r^2$$
$$r^2 = \frac{1.05 \cdot 10^9}{6.7 \cdot 10^{-11} \cdot 23,039} \rightarrow r = \sqrt{\frac{1.05 \cdot 10^9}{1.5 \cdot 10^{-6}}}$$
- $r = 26,971 km$
- $M_{planet} = 4.1 \cdot 10^{26} kg$
This is almost the mass of Saturn ($5.7 \cdot 10^{26} kg$).
Different density planets give different results.
3 Stage Nuclear Pulse Propulsion - works for all planet masses (up to $170 \cdot M_{Jupiter}$, which is actually a star)
The rocket equation doesn't distinguish between engine type. So you can use exactly the same equations.
According to Atomic Rockets: Engine List, you can expect the optimal performance of a nuclear pulse propulsion engine to be the $100,000 \frac{m}{s}$ design on that page.
If you use that configuration, a nuclear pulse propulsion single stage rocket might be able to launch from a planet 6 times Jupiter's mass (Jupiter's mass $= 1.9 \cdot 10^{27}$, this planet's mass would be $1.2 \cdot 10^{28}$).
A three stage version of this ship would be able to generate about 3 times this $\Delta v$. That would correspond to a planet with the mass of $3.24 \cdot 10^{29} kg$ - about 170 times the mass of Jupiter. However, since a body with a mass above 84 Jupiter masses is a star, we can safely say that a technological civilization could develop nuclear pulse propulsion rocketry to launch into space from any planet.
All planets used in this answer assume an Earth density planet.
