hint: Apparently the Tsiolkovsky rocket equation does not actually say that you can launch a conventional rocket into orbit around an arbitrarily large and massive body.

I'm looking for a number based on scaling the earth radius and maintaining the same average density. Must attain LEO, which also gets faster as the planet grows. Don Pettit's Tyranny mentioned in this nice answer is fun, but does not present enough math.

On this Earth, rockets barely work. Payloads can only be a few percent of the total mass for LEO, and less than one percent for deep space.

If we define slightly heavier Earths, say Earth1.1, Earth1.2... where the radii were 1.1, 1.2, etc. times that of Earth and the masses were 1.13, 1.23, etc. times the Earth's mass (in other words same average density, same "iron/rock ratio") what happens? Is there some point where chemical rockets simply will no longer be able to put things in space, or does the payload mass simply become ridiculously tiny? If there is a cut-off, is it different for LEO and deep space?

For our purposes, let's not explore alternative or hybrid launch systems or boost systems (such as balloons, planes, laser beams, space elevators etc.). Just stick to chemical propellant rockets.

edit: here is a guide. So for a scaling factor $f$:

$$ r = f r_{earth} $$ $$ m = f^3 m_{earth} $$ $$ g = G \frac{m}{r^2} = \frac{f^3}{f^2}g_{earth} = f g_{earth} $$ $$ H = \frac{kT}{gm_{molecule}} = f^{-1}H_{earth} $$

We catch a little break here. Assuming same surface atmosphere composition, temperature and pressure (STP), the scale height H actually decreases with increasing $f$. (If we were "world builders" we should probably increase pressure to get more oxygen needed for moving in the higher gravity, but that's a different Stack Exchange.)

As far as LEO altitude is concerned (thanks @Lex for catching that) one might define it as the same number of scale heights as would be on Earth. That's not really so useful because the density profiles of the bits of the atmosphere responsible for drag (Thermosphere and Exosphere are affected by many phenomenon, including the solar wind, and don't scale at all like the lower layers. Nonetheless for historical reasons I'll leave the following, as it is not essential to the question:

$$ h_{LEO} = h_{LEOearth} \frac{H}{H_{earth}} = f^{-1} h_{LEOearth} $$ $$ v_{LEO}=fv_{LEOearth} $$

The LEO period is independent of the size of a planet, if the average density is fixed. However, the velocity of LEO does scale with radius!

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    $\begingroup$ Related: On a Super-Earth 1.5x the volume and mass of Earth, would our rocket technology allow us to reach orbit?. Look at Russell Borogove's answer at the bottom. $\endgroup$
    – kim holder
    Commented Mar 9, 2016 at 15:53
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    $\begingroup$ True, but the main thing was that, in theory, any gravity can be overcome if you put enough stages in your humongous rocket. So then the question turns into a matter of what is practical. I don't know if there is a way to define the question to put a useful limit on that. The limits of structural materials? The GDP of the planet? Not being facetious, btw. $\endgroup$
    – kim holder
    Commented Mar 9, 2016 at 16:06
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – kim holder
    Commented Mar 9, 2016 at 16:33
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    $\begingroup$ I still think the issue is what point we will assign as being the boundary of 'ridiculous'. Say, if there was a beautiful garden world that was a moon of this planet, maybe they would expend extraordinary effort to get just one rocket there. $\endgroup$
    – kim holder
    Commented Jul 23, 2016 at 15:05
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    $\begingroup$ Yeah, but that is the thing - as i understand it, there is no such number, theoretically. Any number of stages can be placed on a rocket, and each one increases the gravity that can be overcome - by ever smaller amounts, but that amount is never zero. I suppose there is a point where the amount is so small it is less than the variance introduced by wind and temperature and such... maybe that could be a limit... $\endgroup$
    – kim holder
    Commented Jul 23, 2016 at 15:27

6 Answers 6


Because linear increases in delta-v require exponential increases in mass, small changes to the assumptions you make about fuel tank structural mass and engine thrust-to-weight ratio start to make very large changes in the final size of the rocket.

For example, if you're getting off a 3.6g planet with a 7-stage rocket, the difference between 88% fuel fraction and 92% fuel fraction yields about a 10:1 difference in the total mass of the rocket.

So I don't think it's really reasonable to talk about ultimate theoretical limits; too many engineering factors are involved.

Locking down a lot of variables, I can tell you what kind of rocket you'd need for a given surface g, though. Let's make these assumptions:

  • We are placing 1 ton of payload into low planetary orbit.
  • Required delta-v to reach orbit, including atmospheric and gravity losses, is 10,000m/s per surface g. Seems to hold for Earth, Mars, and the "Earthtoo" which was discussed in another Q/A.
  • We can build rocket stages of arbitrary size, with a tankage propellant fraction of 90%; the rocket stage mass is the tank mass plus the engine mass -- ullage rockets, interstage, etc. is all handwaved out.
  • We have an infinite supply of Apollo-era rocket engines: RL-10, J-2, M-1, H-1, and F-1.
  • First-stage TWR at ignition must be at least 1.2 (relative to local gravity)
  • Middle-stage TWR at ignition must be at least 0.8
  • Final-stage TWR at ignition must be at least 0.5

Given those assumptions, here is a table of surface gravity, stage count, first-stage engines, and total rocket mass.

Surface                         First        Total       Saturn V 
Gravity   Stages                Stage      Mass, t     Equivalent
 0.5           2             1x RL-10          4.5
 1.0           3             1x   H-1         49.4          0.02
 1.5           3             1x   F-1        249.2           0.1
 2.0           4             5x   F-1       1329.0           0.5
 2.5           5            40x   F-1       8500.9             3
 3.0           6           274x   F-1      50722.2            17
 3.5           7          2069x   F-1     331430.9           100
 4.0           8         20422x   F-1    2836598.4           950
 4.5           8        392098x   F-1   47 million         15000
 5.0           9    3.5 million   F-1  391 million        130000
 6.0          11    400 million   F-1   38 billion      millions
10.0          18        2.88e19   F-1      1.65e21  quadrillions

Up above 10g, something really interesting happens that is kind of a theoretical limit. The mass of the rocket reaches a measurable fraction of the mass of the entire planet it's launching from.

At 10.3g, rocket mass is 0.035 of the mass of the planet. 10.4g, rocket mass is one fifth of the mass of the planet. This doesn't actually alter the ∆v requirement -- we're going into orbit around the rocket/planet barycenter! At 10.47g, the rocket is the planet, and we're... just... chewing it up entirely, pulverizing it in a dust cloud expanding at 4km/s.

These extreme conclusions appear to be corroborated by this independently derived paper, which explores some other related aspects of super-Earth-based chemical rockets.

Another consideration recently brought up by user @uhoh is that as the linear scale of a given rocket stage increases, its mass, and thus the required thrust force to lift it, goes up by the cube of the scale, but the area available at the base of the rocket to mount engines goes up only by the square of the scale; this problem is made even worse here by the increasing surface gravity. The Saturn V was just about at the point where this relation starts to become problematic; the outboard engines on its first stage are mounted at the very edge of the stage in order to make room for their nozzles to gimbal.

Solid rockets don't have the same dimensional constraints, and have very good thrust-to-weight and thrust-to-cost ratios, so they're probably more likely to be used in lower stages for these very large rockets.

Stages much larger than the Saturn V first stage would need to address this with some combination of being shorter and squatter, or compromising engine gimbal range, or mounting engines in pods surrounding the tankage, and there might be fairly hard engineering limits at some point for those reasons. At the 3g mark, for example, the 274 first-stage engines would require a stage about 90 meters in diameter and 9 meters tall, at which point the engineering inefficiencies associated with the fuel tank proportions will be becoming serious.

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    $\begingroup$ Awesome. I'm happy now. :) $\endgroup$
    – kim holder
    Commented Jul 30, 2016 at 21:01
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    $\begingroup$ @kimholder me too - thank you very much for your support and interest! I'm glad we go to the bottom of this. $\endgroup$
    – uhoh
    Commented Jul 30, 2016 at 22:18
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    $\begingroup$ Thanks you very much Russell Borogove for your systematic approach and persistence. We can call this The Tyranny of the Exponential! Very nice. $\endgroup$
    – uhoh
    Commented Jul 30, 2016 at 22:22
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    $\begingroup$ From that article: "On worlds with a surface gravity of >10g0, a sizable fraction of the planet needs to be used up as chemical fuel per launch, limiting the total number of flights" -- good corroboration! $\endgroup$ Commented Apr 5, 2018 at 17:42
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    $\begingroup$ This is like an XKCD what-if, the everybody dies kind. Love it. $\endgroup$
    – GdD
    Commented May 9, 2018 at 15:22

First, let us look at the rocket equation:

$$\Delta v=\ln \left(\frac{m_0}{m_f}\right)v_e$$

That tells how much a rocket can change its velocity (the $\Delta v$). The requirements for reaching a higher velocity for a minimal orbit would increase on your heavier Earth. (For constant density it is proportional to the radius.)

How can we increase the $\Delta v$ of the rocket to keep up? We can increase the exhaust velocity, $v_e$, of the engine, but that cut-off is around 5000 m/s for chemical engines. The other thing we could do is increasing the mass ratio of the rocket $\left(\frac{m_0}{m_f}\right)$. That is problematic too, as we can not really make the fuel tanks out of soap bubbles. Staging is the option left, you could place a big rocket under a small rocket to get a little more change in velocity. Then you are getting a linear benefit for an exponential expense.

As an example, the Saturn V rocket got into LEO (~9000 m/s), sent a payload towards the Moon (3120 m/s), the service module slowed the stack into LMO (820 m/s), and finally the LM landed and took off again (2*1720 m/s). There are still some unused fuel left in the service module then, so let us just call the total $\Delta v$ of the Saturn V/Apollo 17 km/s. That is less than the requirements for a 2x radius Earth. The Apollo program was pretty expensive [citation needed], so it may take a while before a nation of a 2x Earth world attempts to go into orbit. The limit is, as you state, the ridiculously low payload ratio.

Another consideration is the increased surface gravity. (That scales linearly with diameter at constant density). That requires the rocket to have a higher thrust to weight ratio, and that will increase the dry mass, reducing the possible $\Delta v$. (It also increases gravity losses, but that is mostly compensated by the lower scale height of the planet, reducing drag losses).

Eventually, the gravity is so high that even the most powerful engine can not lift itself from the ground. That at least is a definitive limit.

A more theoretical consideration, is $\Delta v$ requirements actually a finite limit?

Surprisingly, it is not. Remember what I said about staging earlier: "you are getting a linear benefit for an exponential expense". But there is not limit to what we can expend! Consider the following scenario: We add more and more stages at the bottom of the rocket, each of them has the same mass as all the stages on top of it. Then burning each of them gives the same mass ratio between before and after, therefore each of them are supplying the same amount of $\Delta v$. To add 10 times that amount, you need 10 stages each doubling the mass. To add 100 times that amount, you need to double a hundred times. The mass grows ridiculously fast, even doubling 10 times are over a thousand times more. But why should we stop :)

But can we really continue to add exponentially larger stages for ever?

After a while, other problems show up. For instance: Rockets are long and thin, to minimize drag. That shape can not be kept for very large rockets. The reason not is the square cube law. Conserving the same dimensional proportions, a rocket twice the height has 8 times more mass. But the base area of the rocket has only increased 4 times. That means that each unit of area has to support more mass. Sooner or later, even the strongest materials must give up, and you must give up the traditional rocket shape in favour of a wider base. That adds a lot to the drag! Problems like that are going to continue to show up:

"More mass means more problems, exponentially more mass means exponentially more problems."


A modern design, larger rocket than the Saturn V, with modifications to increase the T/W ratio could probably make it to orbit on a 2x radius, 8x mass Earth. That is a feasibility limit, rockets that are ridiculously much larger may have a few km/s extra $\Delta v$, but that does not alter the numbers a lot. In theory though, rockets can grow until the drag stops them, or the engines can no longer lift even themselves.

Or perhaps you at some point want to use the available resources of the planet to launch a single rocket to orbit.

  • $\begingroup$ Thanks! Can you summarize as a number in the format described in the question? Are you saying that a Saturn V can give you "LEO" on an earth with 2x radius and 8x mass (e.g. earth_2.0)?? I don't think that's going to work! I'm looking for the limits. The discussion of things to consider is great, but let's get to a single number, or possibly two - one for LEO one for "deep space". $\endgroup$
    – uhoh
    Commented Mar 9, 2016 at 15:05
  • $\begingroup$ OK, when you say "That is less than the requirements for a 2x radius Earth." can you show that using math, not words? What are the requirements for a 2x radius (8x mass) earth, numerically, and why does a Saturn V satisfy it? If there are design changes (T/W) how much? 10%? 10X? Is there a chance this can be done with a simulator program or even Kerbal? $\endgroup$
    – uhoh
    Commented Mar 9, 2016 at 16:08
  • $\begingroup$ @uhoh Don't be too fast on that, especially when a bounty is involved. I agree this is a good answer, but sometimes even better answers come along. Bounties exist to stimulate that kind of interest. $\endgroup$
    – kim holder
    Commented Jul 25, 2016 at 16:23
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    $\begingroup$ I was interested by @RussellBorogove's comment in The Pod Bay about the limitations on engine T/W ratios. That seems to be something that can be given a theoretical limit if only chemical engines are considered. $\endgroup$
    – kim holder
    Commented Jul 25, 2016 at 16:27
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    $\begingroup$ @kimholder I can try to write something about that too. $\endgroup$ Commented Jul 25, 2016 at 16:34

note: I've accepted an answer 2.5 years ago. This paper was published recently so I thought I would add this supplemental answer since it may be an interesting reference for future readers.

The Space.com article No Way Out? Aliens on 'Super-Earth' Planets May Be Trapped by Gravity links to Michael Hippke's ArXiv preprint Spaceflight from Super-Earths is difficult.

While the calculation is based on escape velocity rather than LSEO (Low Super-Earth Orbit) the conclusion is similar, the problem is exponential and it gets really difficult quickly.

The author uses the example of the planet Keppler-20b (see also here), and although there is some uncertainty, the planet's size is roughly 1.9 that of earth, and it's mass is almost 10 times that of Earth.

For a mass ratio of 83, the minimum rocket (1 t to $v_{esc}$) would carry 9,000 t of fuel on Kepler-20b, which is 3× larger than a Saturn V (which lifted 45 t). To lift a more useful payload of 6.2 t as required for the James Webb Space Telescope on Kepler-20 b, the fuel mass would increase to 55,000 t, about the mass of the largest ocean battleships. For a classical Apollo moon mission (45 t), the rocket would need to be considerably larger, ∼ 400,000 t. This is of order the mass of the Pyramid of Cheops, and is probably a realistic limit for chemical rockets regarding cost constraints. (emphasis added)

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    $\begingroup$ More stages that's all. $\endgroup$
    – Muze
    Commented Jan 1, 2019 at 20:25
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    $\begingroup$ @Muze: I was taken aback to discover that indeed the linked paper makes no attempt to consider additional stages beyond a simplistic two-stage Falcon-9-alike. (It also makes the distressing blunder of conflating the linear mass increase from a payload increase with the exponential mass increase of a velocity increase.) $\endgroup$ Commented Jan 2, 2019 at 10:48
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    $\begingroup$ Actually, the best quote from this paper is "On worlds with a surface gravity of $\geq$ 10g, a sizable fraction of the planet would need to be used up as chemical fuel per launch, limiting the total number of flights." $\endgroup$ Commented Jan 13, 2020 at 6:59
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    $\begingroup$ "limiting the total number of flights" Unless you make your flights with escape velocity, in which case you will soon not be troubled by a high gravity any more. :) $\endgroup$
    – Suma
    Commented Feb 5, 2021 at 8:47

Not a planetological exposition in sight so, I'll add my two cents to this rather theoretical discussion.

Amongst exoplanetologists, the consensus has emerged that 1.6 Earth radii and 5 Earth masses is likely to be the upper limit to rocky planets. Simulations have shown that above these figures, the bodies develop increasingly Mini-Neptune like characteristics. This means very thick Helium Hydrogen atmospheres and crushing surface pressure.

Also since Michael Hippke's slightly whimsical paper was referenced in one of the answers it seems appropriate to mention Ocean worlds at Super Earth masses. Ocean worlds present a host of habitability hurdles including a paucity of certain life critical elements like phosphorus, lack of volcanism, no water rock interface due to high pressure ice on the marine floor and others. These conditions will likely limit or even prevent the establishment of the vibrant prebiotic chemical environments that are necessary for biogenesis.

If the first assumption holds true, the highest gravity on a potentially habitable world will not exceed approximately 2.5g.(edit: and thus making it not quite so difficult to reach orbit with chemical rockets as would have been the case with a higher g value)

  • $\begingroup$ Thanks for your answer! I can't guess what "increasingly Mini-Neptune like characteristics" means or what "Hippke's paper" is. Without any linked sources or cited references in your post it's hard to know what this means or judge your answer's veracity. Is it possible to support this with some links and/or citations? Thanks, and Welcome to Space! $\endgroup$
    – uhoh
    Commented Feb 16, 2020 at 3:40
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    $\begingroup$ Where in your answer is the question about rockets addressed? $\endgroup$ Commented Feb 16, 2020 at 4:20
  • $\begingroup$ Hi, uhoh! Thanks for the greeting. I edited for a little clarity and added a couple random expository footnotes. There are thousands of papers on these topics. $\endgroup$ Commented Feb 17, 2020 at 5:36
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    $\begingroup$ @ Organic Marble, the primary impediment to reaching orbit discussed here is high gravity. Thus, I think a comment on the likely upper limit of habitable worlds gravitational strength is warranted. As a side note greater aerodynamic forces at higher atmospheric pressures than exist on Earth also deserves attention, but as an aerospace engineer you're likely better equipped to write about that than I am. Cheers. $\endgroup$ Commented Feb 17, 2020 at 16:04
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    $\begingroup$ "1.6 Earth radii and 5 Earth masses" It seems en.wikipedia.org/wiki/Kepler-20b mentioned in another anwer is a counterexample to this, assumed to be a rocky world with 9 M⊕ and 1.9 R⊕ (still Wolfram Alpha tells me its surface gravity should be around 2.5g) $\endgroup$
    – Suma
    Commented Feb 5, 2021 at 8:52

Great answers have been given, but one of the major themes is they assume a fixed wet to dry mass ratio of 10:1 (ish). The justification is:

  • You need to fix this as: there are no meaningful answers with without a value and, which value is subject to engineering nuances, which are difficult to handle.

  • 10:1 is a good pick. (We can't do much better than this and still have everything work so it seems sensible to stick at this)

The problem is that's the limit of what we can make work on earth. A lot of the dry mass of a rocket is either:

  • directly related to the thrust-to-mass ratio (i.e. number/size of engines)

  • indirectly related to TMR (i.e. supports the structural loads)

Note, to keep gravity loses equivalent in practice the accelerations needed, hence TMR, is linear with the surface gravity. Hence so is a part of the the wet/dry mass ratio.

Once we take that into consideration things look a lot bleaker for the high g super-earths getting something into orbit using chemical rockets.

The actual numbers here are a little difficult to know, but if 5g world leads to a rocket with a w/d mass ratio of 5 to 1 (which I think is about right but...), you're staring down the barrel of a $10^{20}$t type figure for launch mass. To put that into perspective, the 'moon rocket' is no longer a good comparison. That's the mass of the moon it got to.

Theoretical limit? I'd say so.

At that mass things start taking a turn for the 'XKCD'. Forget the practical issues they're clearly long gone at "moon-sized-anything". We hit cold hard theoretical limits. You start having to deal with your own gravity.

Firstly those practical issues are big ones even if we laugh a little 'engineering' problems (like money, and where we might find $10^{19}$t of aerospace grade materials). For example that's the sort of size that when you're made out something solid and are already floating in space under 0G, you deforming under you own gravity into a ball. Trying to make that out of mostly liquid fuel and subject it to 5-10g..., you're not staying the shape you started. Doesn't matter what mass-ratio 'hit' you are willing to take. But we've got this far, we aren't going to let a lack of unobtainium stop us.

No the real hard limit is how being so heavy effects your exhaust velocity. At the risk of getting too meta here if you're heavy enough, its difficult to get things to come apart from you. It applies to planet sized rockets as much as it does planets.

If you're a few million kilos, your 'exhaust velocity' is the velocity you can get your propellant to get to. If you have more mass than the moon, your propellant will have lost a lot of momentum by the time it's left your gravitational influence. And this is the fate our rocket meets. LOX/H2 has an exhaust velocity of about $4,400ms^{-1}$, about as good as we can do. Let's just say our moon-sized rocket has the density of the moon too, and so has a similar escape velocity of $2,380ms^{-1}$. Then the useful exhaust velocity of our rocket (initial less escape) is less than half. hence half the delta-v. You won't be going to space to day.

"Ok", I hear you say, "that just mean's you can't go to space in that rocket.", "How about a bigger one?". Well "No". This is another one of those "Even if everything sort of worked as before, you want go twice as fast, which is going to be a lot more mass." type problems. Except now we really can't just take the "make in 10 orders of magnitude bigger" approach. Apart from the fact that our rocket is now lot bigger than the planet which means we couldn't possibly construct it, now we have no chance of using chemical rockets to propel us anywhere. To gain any momentum we need to chuck something out of our gravity well, and the exhaust velocity of chemical rockets don't make when we are this big. We are now truly stuck.

But wait: directly the exhaust doesn't make it out, but I wonder if you could try different way of getting mass out of a very deep gravity well. Shouldn't be too hard. Even if it was only a little bit, we could always just scale it up...

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    $\begingroup$ Try making your answer less verbose and heavier on facts. As written, it doesn't answer the question either. $\endgroup$ Commented Feb 1, 2019 at 18:26
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    $\begingroup$ @OrganicMarble thank's for your interest (and down-vote). To start with not answering the question: you're right I didn't give a hard answer. The problem is this is really a grey area. How big really just equates to how many issues are you're willing to magic away. My answer really only says unless your willing to ignore conservation of mass, 5 times as big will be a problem. I think that's interesting. $\endgroup$
    – drjpizzle
    Commented Feb 1, 2019 at 18:33
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    $\begingroup$ @OrganicMarble To answer your "being heavy lowers exhaust velocity, not making sense" comment sorry, perhaps I explained it badly. Maybe this image would help. Imagine throwing a ball tethered to your hand by a strand of elastic. How thick the elastic is will effect how much momentum you can ultimately put into it for a given throw. $\endgroup$
    – drjpizzle
    Commented Feb 1, 2019 at 18:41
  • $\begingroup$ I can't see how this is an answer to the question, or even supplements other answers. This reads like a conversation or a blogpost. Can you add a tl;dr at the beginning summarizing the key points of your answer? The conversational approach doesn't really fit the Stack Exchange Question/Answer format. $\endgroup$
    – uhoh
    Commented Feb 3, 2019 at 0:48

On a practical engineering side of things. Ultimately you are limited by exhaust velocity. In theory you can always just make a bigger engine, bigger tanks, etc. Ridiculously expensive, but possible. This would seem to set the real limit to material strength. Material strength is likely to give out before the gravity wells pull exceeds the exhaust velocity of even moderately modern fuels.

For example, LF+LOX typically has an exhaust velocity of around 4,400 m/s. Which will fight up to 448 G of gravity. Literally more than the sun. Practically however much less than that. So size of the planet itself presents no real deal killers, it just makes the payloads mass fraction very VERY low.

At some point though other technologies, like nuclear bomb drives (https://en.wikipedia.org/wiki/Project_Orion_(nuclear_propulsion)), become the only feasible affordable way off the planet.

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    $\begingroup$ I'm not sure how you are connecting exhaust velocity with "fighting gravity". It is true that a rocket with an exhaust velocity of 4400 m/s could only lift about 1 seconds worth of fuel supply against 448g, but that's not really directly relevant. You need to use the rocket equation and formulae for the escape or orbital velocities of planets of various sizes and densities to get a sensible answer. $\endgroup$ Commented Jan 1, 2019 at 18:43
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    $\begingroup$ A real rocket will never get close to the 1 second point. Payload for this calculation includes the engine and fuel tanks as well, which will always have nonzero mass. the 448 G's is literally the point where the fuel cant even lift itself off the ground. That is a jet of exhaust moving at that velocity would not even be able to escape the engine if pointed straight up., let alone lift any additional fuel $\endgroup$ Commented Mar 18, 2019 at 21:08

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