Using currently available fuels and technology and an unlimited budget. How big would a rocket without stages have to be to make it to a stable orbit? and how big would a rocket without staged have to be to escape Earth's gravity? "edited" - or for the easier question how big would the rocket have to be if it was massless and only the fuel had mass?

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    $\begingroup$ How massive your payload is? The fuel need depends heavily on the payload (the thing you want to put up there). You should look the Tsiolkovsky rocket equation, and this NASA page can be useful also. $\endgroup$
    – iiqof
    Commented Dec 5, 2013 at 10:38
  • $\begingroup$ Sounds like a perfect question for what-if.xkcd.com :-) . Further, aside from the question of being able to lift your own fuel load, how about a machine which initially is airplane-like and then switches to ballistic? Is dropping wings allowed or would that count as a "stage" ? $\endgroup$ Commented Dec 5, 2013 at 12:29
  • $\begingroup$ The only payload is the rocket and the useful empty space it would contain. No dropping of wings is allowed. I'm just wondering if it is possible at present or do we need something like a 10km high rocket or bigger maybe? $\endgroup$
    – Jitter
    Commented Dec 5, 2013 at 15:44
  • $\begingroup$ There is also the alternative method proposed by Jules Verne, i.e. cannon launch. No need to carry your propellant with you. $\endgroup$ Commented Dec 5, 2013 at 17:57

2 Answers 2


A rocket "without stages" is a one-stage rocket. There have been proposals of this type before. You can often find mentions of SSTO, which means "single stage to orbit". The X-33 was a fairly recent R&D attempt at this. That was a spaceplane and the rocket all in one.

The present question has a different attempt. It is asking about SSTO minimum size with no requirements for reentry, and apparently no payload requirement either. That's difficult to answer in a fully theoretical sense, without just saying "it's zero". If you allow the engineering into the discussion, it seems like there should obviously be an answer.

Further complicating things is the fact that the size of the rocket doesn't necessarily affect the mass fraction of propellant of the tank itself. It is important to look for this number for tanks specifically. NASA gives these examples:

Propellant  Rocket Percent Propellant for Earth Orbit
Solid Rocket     96%
Kerosene-Oxygen  94%
Hypergols        93%
Methane-Oxygen   90%
Hydrogen-Oxygen  83%

There's no obvious academic reason I can't assume an orbital Solid Rocket tank that weighs 1 kg, and is still 96% propellent. As far as tanks go, they physically have a mass proportional to the volume of stuff held in it (to first order..).

Nonetheless, my 1 kg orbital rocket is sure to be highly ineffective. In fact, the Delta v to LEO will increase because air drag and gravity drag will be higher for this rocket. Air drag, mainly. Out of the normal 10 km/s to orbit, around 1 km/s is typically air drag.

A completely accurate way to answer your question would then be to find out when drag gets so large that the tank mass itself satisfies the rocket equation as the payload. To do that, I have to get a metric for air drag. I'll assume it goes about the same speed as other rockets (because decreasing the speed increases gravity drag). That means that this is purely a consequence of surface area to mass ratio.

$$ \Delta v_{drag} \propto \frac{A}{M} \propto \frac{R^2}{R^3} \propto \frac{1}{R} \propto M^{-1/3} \\ \Delta v_{drag} = \Delta v_{nominal} \left( \frac{ M }{ M_{nominal} } \right)^{-1/3} $$

I'm ball-parking things, so I'll assume that the nominal case is a shuttle that has delta v from drag of 1 km/s and weighs 1e6 kg. Now, we formalize the total Delta v and plug it into the rocket equation.

$$ \Delta v = 9 km/s + \Delta v_{drag} \\ v_e = 4 km/s \\ \Delta v = v_\text{e} \ln \frac {m_0} {m_1} $$

Plug and chug...

$$ 9 km/s + 1 km/s \left( \frac{ M }{ 1,000,000 kg } \right)^{-1/3} = 4 km/s \times \ln \frac{1}{0.04} \\ M = 17,179 \text{ kg} $$

The more I look at this, the more I realize this really is your answer. This is a rocket that weighs 17 tonnes. Its delta v to orbit is 12.8 km/s, as opposed to the normal (more favorable) 10 km/s, and this is because of the higher air drag because of its size. It will stand 47 feet tall, and it will deliver its own tank into orbit (which weighs 680 kg). Nothing else.

This is your answer, because you can not go any larger without adding multiple stages.

Is this an efficient way to deliver tanks into orbit? No. However, as people have pointed out before, if we wanted tanks in orbit, we would have used the shuttle external tank, which was 300 m/s short of full orbit when it was dropped into the ocean.

It's still sort of somewhat the most efficient way to deliver a tank to orbit at that size. If you want more tanks in orbit for cheaper, you'll need larger rockets, so that they'll be more efficient. However, I think the problem would reduce to the same thing if you continued to scale it up, and left the rest as empty space. This is a completely insane idea, but it would follow the same mathematics I used here. In other words, you strap a large empty tank onto the solid booster rocket.

Obviously, a better approach would be to take up inflatable modules in a normal rocket. If you absolutely needed rigid tanks, perhaps you could wrap them like Russian dolls around each other, forming a rocket larger than the 17 tonne baseline. Then you would unpack them when you got to orbit. Now you have lots of tanks, and can proceed with your evil plan.

I hereby call our plan "Single Tank To Orbit", abbreviated STTO.

  • $\begingroup$ Thanks for that 17 tonne answer, I guess the extension on this would be what is the magic number for one that could land safely under it's own power;-) I wonder if that's in wolfram? $\endgroup$
    – Jitter
    Commented Dec 5, 2013 at 18:23

Couple of additional points to AlanSE answer

  1. You can experiment with rocket equation by yourself using WolframAlpha. Just plug in your own values for final mass (payload), exhaust velocity etc.

  2. Lockheed Martin X-33 was supposed to be technology demonstrator for VentureStar single-stage-to-orbit spaceplane. So you can use its design specifications to compare with estimates from rocket equation. VentureStar was expected to deliver 20 tonnes of payload to low-earth orbit, and then safely return back to the surface. The mass at launch would have been 1000 tonnes.

  3. One possibility to avoid the tyranny of the rocket equation is to use the atmospheric oxygen as oxidizer, that is to use air-breathing jet engines to provide at least part of delta-v. The scramjet technology could provide the speeds in the atmosphere up to Mach 12-17, thus reducing the delta-v needed for the final acceleration into orbit.


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