Ever since Jules Verne wrote "From the Earth to the Moon" space guns have long been proposed as a method of launching objects into space. And for non-human, non-fragile payloads they make sense as the payload doesn't have to carry much of it's fuel with it. But apart from a few experiments no successful space gun has been developed.

So I'm wondering, what technological/engineering hurdles that need to be overcome before we can see a working and efficient space gun?

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    $\begingroup$ Related: space.stackexchange.com/questions/815/… $\endgroup$ – Everyone Oct 10 '13 at 17:05
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    $\begingroup$ There is a better answer on physics.SE at physics.stackexchange.com/questions/35139/… than the ones here. Even better than mine :) $\endgroup$ – Chris Mueller Mar 6 '14 at 20:36
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    $\begingroup$ There is a very practical geo-political problem. Chiefly, not getting your space gun blown up by any country in range. And for a space gun that is every country. $\endgroup$ – C. Towne Springer May 16 '14 at 0:41
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    $\begingroup$ J Verne and H G Wells wrote scifi using space guns. Garret P Serviss wrote a follow up to War of the Worlds where magnetic propulsion was described. It's funny that rockets were not thought of in the 19th century, given that firework does fly high. The artillerist Conrad Haas suggested rocket flights to space in the 1550s. $\endgroup$ – LocalFluff May 16 '14 at 13:00

Scaling is probably the biggest issue.

First of all, you always need an upper stage, as a gun cannot get you into a reasonable orbit. It can get you high enough, and maybe fast enough, but not into an orbit.

So now your gun has to be large enough to launch a payload, including at least some kind of upper stage.

Once you scale up to those kind of sizes to deliver meaninful payloads, these guns get VERY large and much more complex than would seem.

There are many issues that come up with scaling. As things get larger, their frontal area must get longer, air resistance gets higher, and now more energy is needed to push all the way through the atmosphere, with sufficient velocity at the end, to be close enough to orbital velocity so that a small upper stage can circularize the orbit.

But as you trade off upper stage size/payload/initial speed, the initial G load gets higher and higher in order to deliver a useful payload.

  • $\begingroup$ And how much reusability would you get with a gun? The big German guns' caliber increased with each shot, resulting in a very short lifespan. $\endgroup$ – Don Branson Oct 10 '13 at 17:42
  • $\begingroup$ You might want to add in the problem of aerodynamics - if you get all or most of your speed at ground-level, you need to go at orbital velocities in full-atmosphere. $\endgroup$ – john3103 Oct 10 '13 at 20:08
  • $\begingroup$ @john3103 That is just a scaling issue. You need to get going fast enough to get through the atmosphere, to get through the atmosphere sort of a tautology. Just go faster. Maybe that never works but that math is beyond me. $\endgroup$ – geoffc Oct 10 '13 at 21:39
  • $\begingroup$ Going faster is not a solution to the atmosphere. Going through material will stop a mass by the time it's displaced it's own mass regardless of it's initial velocity. If you want to punch through the atmosphere you need something that weighs well over 14.7 pounds per square inch if you're going straight up, even more if you're on an angle. Realistically, this means a BIG craft. $\endgroup$ – Loren Pechtel Feb 28 '14 at 2:37

The biggest problems with a space gun are inherent the simple nature of ballistics.

We can use railguns to accelerate projectiles very fast; much faster than explosive propellants can do over the same distance. The current record is 33 MJ; that's a one-kilo projectile propelled at approximately escape velocity (8127 m/s). So, theoretically, the most powerful railgun we have (currently in the hands of the Navy Labs, who are developing the technology for a new class of warship) could put a 1-kilo object into space.

However, there are those problems I mentioned. First, in the Navy railgun, this amount of energy is currently imparted to the projectile over a distance of approximately 12 meters, in a timespan of 10 milliseconds. An 8000m/s velocity change in a 1kg projectile over .01 seconds equals an acceleration force of 800,000 Newtons, which is about 81,632.65 G-forces. Humans can survive about 14 Gs when lying flat on their back or stomach, and 4 Gs is the maximum we like submitting our astronauts to in a seated position, since they're not all fighter pilots in prime condition able to handle 6-8 Gs for long maneuvers.

To get the same delta-V with acceleration limited to 40m/s2 (just over 4G) would require 200 seconds, during which time you would have traveled 800,000 meters (the required length of the barrel of the railgun). 800,000 meters is just shy of 500 miles; you're talking about engineering a track from Dallas to El Paso, TX, with millimeter tolerances for deviation per kilometer of track from being perfectly straight. It would be the biggest and most expensive single civil engineering project the human race has ever undertaken; bigger by far than the current bullet train systems, bigger than the CERN supercollider, bigger than the U.S. interstate project.

Second, at the instant you leave the barrel, you're travelling as fast as you ever will. Drag from air resistance will slow you down, and the air is thickest right at the surface of the Earth, where it's cheapest to build this 500-mile track. So as soon as you leave the barrel, you're blasted with Mach-25 winds that will instantly begin slowing you down below escape velocity. And if it happens to be raining in Dallas (remember you want to travel west to east, in the direction of the Earth's rotation, to take advantage of the extra 465 m/s of angular velocity), the raindrops will shatter the windshield. We mitigate this with modern rockets by limiting our acceleration and velocity until we exit the troposphere, at which point we throttle up the rocket to maximum power as the air thins. Our railgun, on the other hand, would have to be built on about a 7* incline, with the muzzle about 100,000 feet in the air, in order for the projectile to avoid the worst of the air drag as it exits the barrel. So now, you're taking what's already the most expensive project in history and adding to it the task of building a ramp 36 times taller than the Burj Khalifa, and that's the shorter dimension, by far. We're now also working against gravity as we accelerate, requiring us to add an additional 1.225m/s2 to our railgun's acceleration just to account for gravity (we've already been assuming that air drag and other friction within the barrel is negligible).

Lastly, 33MJ is our current world record for imparting kinetic energy into a projectile via magrail without completely destroying it, and that's only enough to get one kilo up to escape velocity. The Space Shuttle orbiter, empty, weighs 130,000 kg. The lightest vehicle we've ever put into orbit that is capable of supporting human life was the Mercury capsule at 1,400kg; a railgun capable of putting this capsule in orbit would have to be able to impart, in the ideal, not one millijoule less than 46.2GJ of kinetic energy to the capsule. In the grand scheme of things that's not much; it's about the yearly electricity consumption of the average refrigerator. However, that's after all inefficiencies and losses, which when you're dealing with electromagnetic inductors can be several orders of magnitude more than the energy delivered.

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    $\begingroup$ So we take all the mass of Antarctica, pile it up in the Sahara to make the largest pyramid in human history, and put a rail gun on the slope! Easy! $\endgroup$ – Stephen Collings May 29 '14 at 20:53
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    $\begingroup$ The Navy's railgun comparison is a little unfair. Part of its design challenge was how to accelerate a projectile to 8,000+M/S in a comparatively short barrel. You stated the record at that time was 33MJ for a 1KG projectile over a timespan of 10 milliseconds in a 12M barrel. Just saying, in many ways, it would have been far easier put 33MJ in a 1KG projectile over a 12KM barrel. You would have roughly the same energy requirements, but power requirements drop by a factor of 1000 and put far less stress on the gun. But I guess those issues return when you scale the payload up to 1000KG $\endgroup$ – Keith Reynolds Feb 6 '16 at 22:39
  • $\begingroup$ Certainly, the project is daunting if one does not consider a hybrid system. But a Hybrid chemical and either a rail or linear magnetic motor gun, along with non human payload, makes the time in the barrel shortened to 3 seconds when accelerating at 6.8G over 300 meters and an exit velocity of 200m/s, considerably less air resistance compared than launching at 8KM/s on magnets alone. Considering how much chemical fuel is spent during the first few seconds, considerable saving might be realized with a hybrid approach. $\endgroup$ – Keith Reynolds Feb 6 '16 at 23:32

Since we've been fielding a lot of questions on this subject lately, I just want to chime in with some of the rules-of-thumb that seem very natural to me. Unlike rockets, a gun sees the maximum density of the atmosphere at its fastest point. So provided you understand the necessity to circularize an orbit, and you understand that high accelerations need hardened equipment, the viability of many proposals can be easily ruled out envelope calculations for the drag.

Here's a helpful metric to look at the atmosphere with: the mass-thickness of the atmosphere is the mass per unit area, looking straight up into the sky at sea-level.

$$ \frac{ \text{Mass of Atmosphere} }{ \text{Area} } := \mu_{atm} \approx 10 \frac{ \text{tons} }{ m^2 } $$

In a naive sense, look at a bullet straight-on. Divide its mass by the area (same area it presents to the atmosphere). A more accurate approach would be to do some manipulation of the drag equation. We're not interested strictly on the force on the projectile. Alternatively, I'll consider the velocity lost due to its trip through the atmosphere, $\Delta v$ here. I obtained the following from the drag equation, under the assumption that the total velocity is notably greater than the loss in the atmosphere (if not, it's nonviable anyway).

$$ \frac{ \Delta v }{ v } = \frac{1}{2} \frac{C_d \epsilon \mu_{atm}}{ \rho D f \sin{( \theta)} } $$

Due to practical considerations, if this ratio is about 1, you don't have a chance. It will burn up in the atmosphere, and even if not, you can't produce those speeds. Speeds near orbital have never been demonstrated. So if you can't push this ratio well below 1, the idea is out the window. I'll go over all the terms below, dividing them into more-or-less into things that are completely impossible to push beyond a certain window, and things that have not inherent limitation.

Hard limits:

  • C_d the drag coefficient for bullets are in the neighborhood of 0.2 for well engineered projectiles. You can easily push it a little bit higher or a little lower, but there is no where close to an order of magnitude of wiggle room. The high mach numbers make it an even tighter range for different shapes.
  • epsilon, is the correction for a thinner atmosphere at higher altitudes. For Mt. Everest the factor is something like 0.3. I would believe a factor of 0.5 for placing it on a high mountain. Obviously sea launches face the full factor of 1.
  • rho, the average density of the projectile is constrained by your propellant if you're going to circularize its orbit. This is very low, probably around 1.0 specific gravity. If you're doing some other scheme (like a Rotovator), you could pack it full of steel payload, getting to 7.0 or 8.0 at theoretical maximum. Unless you're sending Uranium or something like that. I'll give it 2.0 for a practical payload.
  • f, the length to diameter ratio of the projectile is limited by aerodynamic considerations. Rockets tend to be very slender, but they have active control systems. Fins can help push the envelope here. But even with that, I'm calling a "hard" limit of about 10.
  • sin(theta), this factor works against you, and the best case scenario is 1. This would apply for suborbital vertical launches. For orbital launches, you risk defeating the point of a gun in the first place if this angle isn't low enough. You could shoot something straight up and then burn the necessary 7.8 km/s, but this would make a very poor mass ratio, and probably can't be engineered to withstand the gun acceleration. For Quichlaunch, I would give this a value of 0.5.

Soft limits:

  • D, is the diameter, which serves as a metric for the overall size of the payload. How large can it be? How large of a gun can you construct?

So we have to use our (only) soft limit in order to engineer around the other hard limits. I'll take a velocity loss ratio of 0.2 for now. You could stretch this, but not by much. You actually come out with a set of parameters like:

$$ D \approx 4.5 cm \\ M \approx 11.7 kg $$

This doesn't sound shockingly huge. But if you change it to a sea-level launch, the required mass goes up by a factor of 8, due to the fact that mass scales with D^3. I was also extremely generous with the diameter to length factor. A bullet shape that has a factor of 5 would be, again, 8 times the mass. So if we're looking t something like a practical sea launch to ultimately reach orbit, the minimum mass is closer to the scale of 750 kg.

You can see here how the minimum size depends strongly on the assumptions of the type of system you're using. Nonetheless, 10-750 kg is still a starting range for the minimum bullet mass needed for the system to ever work.

But the bullet size and speed also set a limit on the minimum size of the gun itself. If you can get pressures on the order of 50,000 psi, then the 750 kg case (Quicklaunch) at 8 km/s would need a volume of about 110 m^3. That's big.

Compare, a nuclear reactor pressure vessel is probably about 2-3 times the volume of that. It's also about 1/20th the pressure. And it costs upwards of $100 million.

Straightforward material requirements for a pressure vessel are proportional to the (pressure)x(volume) product. So things are not looking good for our space gun, although the economics of heavy forgings are far from simple. On the other hand, a minimalist space gun mounted on the slope of a tall mountain might have a hard cost minimum below the $10 million scale.

Or you could just buy a Falcon 9 flight, for somewhere around $50 million. Of course you could get better economics from the space gun if you used it enough times to amortize the cost sufficiently. However, the total yearly demand for payload into LEO is only about 240 tons. The Falcon 9 can carry 10 tons in one shot. Only a small fraction of the payloads could be substituted by the space gun (hardened equipment). So maybe there would be enough demand for the space gun to replace 1 or 2 rocket launch equivalents. On top of that, the risks are huge. Global flight frequency would have to be much larger for people to seriously put the necessary capital into this.

  • $\begingroup$ I'm a little confused about your talk of pressure vessels. Why would the gun need to withstand any significant amount of pressure? I guess you must be assuming a gun powered by explosives? In reality a railgun would probably be a much better way to achieve the sort of velocities needed for this feat. $\endgroup$ – Ajedi32 Mar 22 at 19:16

I think there is another, more fundamental problem which hasn't been mentioned by the other answers. The drag on a body due to friction with the air increases quadratically with velocity. The drag force is given by (taken from this Wikipedia article) $$ F_D=\frac12\rho C_D A v^2, $$ where

  • $\rho$ is the density of the fluid.
  • $C_D$ is the drag coefficient which admittedly can decrease with velocity.
  • $A$ is the cross-sectional area.
  • $v$ is the velocity of the object relative to the fluid.

This is a problem because a gun would need to impart all of the energy necessary to get the object into orbit right at the beginning, meaning that all of the energy would need to be stored in the objects velocity rather than the chemical energy of its boosters. This would significantly increase the total amount of energy necessary due to the quadratic dependence of the drag force.

  • $\begingroup$ To be fair, air resistance / drag was mentioned in both previously existing answers, but OK I guess there's no harm done in emphasizing it more. ;) $\endgroup$ – TildalWave Feb 27 '14 at 13:39
  • $\begingroup$ Indeed; I honestly missed the 4th paragraph of @KeithS's answer which makes some of the same points. My point about the quadratic dependence wasn't mentioned elsewhere though. $\endgroup$ – Chris Mueller Feb 27 '14 at 13:43
  • $\begingroup$ What effect would such an orbital gun shot with tons of payload have on its near surrounding? Something like that of a small asteroid impact? $\endgroup$ – LocalFluff Feb 27 '14 at 18:32
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    $\begingroup$ @LocalFluff Good question. We can make some arguments just by symmetry. Lets neglect air effects, assume that the asteroid comes in at escape velocity and we launch the payload at escape velocity, and assume that they have the same mass. The total impulse would be roughly equal because they happen over similar timescales. The only difference then would be the footprint of the gun versus the footprint of the asteroid. I guess the footprint of the gun needs to be very large :) $\endgroup$ – Chris Mueller Feb 27 '14 at 18:46
  • $\begingroup$ Thanks for linking to an informative Wikipedia article. Also in that article is the power associated with drag, which scales with velocity cubed. At 8 km/s through earth's troposphere the ship must endure a lot of thermal watts per second. The shooting stars we see at night typically burn up in the mesosphere about 70 km in much more rarefied atmosphere. The spacecraft would need an extremely robust thermal protection system. $\endgroup$ – HopDavid May 16 '14 at 15:16

Several posts have mentioned atmospheric drag. Drag slows the spacecraft. Traveling 8 km/s through the troposphere also induces heat and stress. The spacecraft would need a sturdy structure and thermal protection system or it'd burn up and/or crumple like beer can.

An important factor would be the flight path angle as the space craft exits the muzzle. If the spacecraft departs at a 0 degree flight path angle (in other words, horizontal), it must travel a long ways through the atmosphere. A huge column of air would be displaced:

enter image description here

That part of the path traveling through the atmosphere is colored red.

Over a trip of around 8000 kilometers the ship endures slowing resistance, heat and pressure.

If the flight path angle is close to 90 degrees (in other words, near vertical), the trip through the atmosphere is a lot shorter. The ship endures extreme atmospheric friction for only about 100 kilometers. But then the orbit would be an elongated ellipse that would quickly return and crash into earth's surface:

enter image description here

A typical rocket starts with a nearly vertical ascent. As the air grows thinner, it leans more towards the horizontal. It doesn't do the major horizontal burn until the ship is well above most the atmosphere. This isn't an option for space guns.

The notion is much more viable for airless worlds like Luna or Ceres.


Fundamentally, the issues are that a functional spacecraft is only engineered for between 3 G's and 30 G's; typical rockets only to about 100 Gs. (3.5 for the shuttle, 30-60 for ICBMs.) A reasonable launch tube would require an exit velocity in excess of 7.8km/s (the LEO speed). Adding 1 km/s for atmospheric losses, that gives a need for about 8.8km/s; for simplicity, let's round up to 9km/s. Note that a 180 km high suborbital flight was attained by the US Navy HARP project, with a 3.6 km/s launch velocity and near vertical launch.

9 km/s, at a reasonably robust 30 G's, is a hair over 30.6 seconds. That requires (using d=0.5AT^2) 137644 m... almost 138 km. (The HARP space gun was 41m long, and launched at 3600m/s, near vertical, and around 18500 G.)

Peak electronics safe launch is around 15500 G's (used on artillery shells), but designing spacecraft for that is highly unlikely. At that acceleration, and using the same 9m/s target, it's 274 meters long, and in the barrel for 0.06 seconds.

A more reasonable benchmark is the 100 G's that is routinely survived in momentary impact events - it's a good benchmark for frame survival of relatively thin-skinned craft. (After all, most cars survive this level relatively intact. It's also the acceleration of the Sprint Interceptor Missile.) At 100 G's, that's 9.2 seconds in the tube, and about 41.5 km long.

A human-safe launch is 12 G's... for a target of 9 km/s, it needs to be 345 km long, and is 70.6 seconds in the tube.

As can be seen, the length of the tube is a huge limit. Essentially, one has to launch a small missile - it doesn't need a whole lot of delta-V, just enuough to raise the perigee - and thus the reasonable level is a high acceleration missile. At that it's impractical, but doable. (Designs for this have been worked up by various engineering schools as sample problems.) The problem being that the gas can only accelerate so fast, and to maintain a consistent thrust requires adding additional combustion along the tube.

It's possible now, but impractical. It would be monumental architecture, unusable for humans, but a practical method of delivering certain categories of payload to orbit - bulk fluids, such as fuel, water, and air, and raw materials for construction, such as girders. It would require on-orbit tugs. Any failure has no abort mode, as well, since the payload is unpowered in flight.


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