I know Earth's atmosphere is a show stopper for rail gun launch to orbit from Earth's surface.

The vacuum of the Moon is often cited by lunar enthusiasts who point out lunar mass drivers could provide a lot of delta V.

But how about a rail up Olympus Mons? At $0.02 \ kg/m^3$ Mars surface atmosphere is already rarefied. And at 22 km, Olympus Mons is a very tall mountain. At 3.5 km/s, Mars orbital velocity is a little less than half that of earth.

Is the air at the top of Olympus Mons thin enough to allow rail gun launches to orbit?


2 Answers 2


I've been wondering this for years, not knowing how to approach the problem.

Then at a NasaSpaceFlight thread a poster using the name R7 told me about dynamic pressure, also known as q.

$q \ = \ 0.5*\rho*v^2$

q is measured in pascals. R7 told me the expression Max-Q is the maximum dynamic pressure a space craft endures during it's flight.

Page 109 of Introduction to Rocket Science and Engineering by Travis S. Taylor says the Shuttle's Max-Q was at an altitude of 11 km at a speed of 442 m/s. Max-Q was about 35,000 pascals.

What's the density of Martian air at the top of Olympus Mons? According to the Mars Fact Sheet the density of Mars surface atmosphere is $ 0.02 kg/m^3$ and the scale height is 11.1 km. Since Olympus Mons is 22 km high, it seems to me the the density at the mountain top would be $e^{-22/11.1}*0.02kg/m^3$ where e is Euler's number, about 2.72. That gives a density of about $ 0.0028 kg/m^3 $. Mars orbital speed at 22 km altitude is about 3,540 m/s. Plugging these numbers into

$q \ = \ 0.5*\rho*v^2$

I get a dynamic pressure is about 17,268 pascals, or about half of the shuttle's Max-Q.

So at this point I believe Mars' atmosphere wouldn't be a showstopper for an Olympus Mons rail gun.

  • $\begingroup$ Dynamic pressure relates to the size of your structural problems (the acceleration loads in the mass driver will likely be a bigger concern than the aerodynamic loading at the muzzle). The other key parameter to consider is stagnation temperature, to assess the magnitude of your thermal problems. $\endgroup$ Commented Feb 4, 2015 at 12:12
  • 1
    $\begingroup$ @DanielChisholm Olympus Mons is about 600 km in diameter. Assuming a 300 km track, I get a little over 2 g's over 3 minutes to achieve 3.5 km/s. I followed the link but I still don't know what's meant by stagnation temperature. $\endgroup$
    – HopDavid
    Commented Feb 4, 2015 at 23:29
  • 1
    $\begingroup$ You'll need to add another factor: the extra speed you need to compensate for atmospheric resistance on the way up. $\endgroup$
    – Hobbes
    Commented Feb 5, 2015 at 13:53

First, let's broaden this question to mean "mass-driver" instead of "railgun" since a railgun might not be the best kind of mass driver for Mars. Mars is a dusty planet and the dust could interfere with the electrical contact between the armature and the rails.

To determine if the air on Mars is thin enough we can calculate the velocity we need to achieve at the exit of the mass driver with


Where 'G' is the gravitational constant, 'M' is the mass of Mars, 'r' is the distance from the center of Mars to the mass driver exit, and 'a' is the semimajor axis of the elliptical orbit you will be on after you launch. 'a' is also the average of the periareion (Mars equivalent of perigee) and apoareion (Mars equivalent of apogee) of your elliptical orbit. enter image description here So, for example, if you wanted to get to a 1000 km orbit and the mass driver's muzzle was at the top of Olympus Mons (21.229 km) then 'a' would be $$(3389.5+1000+3389.5+21.229)/2 = 3900 km$$

and 'r' would be $$3389.5+21.229 = 3411 km$$

Mars's mass is 6.39E+23 kg and G is 6.67e-11 N⋅m2⋅kg−2. Plugging these numbers in you should arrive at a muzzle velocity of 3639 m/s. Add a little to account for atmospheric drag and subtract 242 m/s if you're taking advantage of Mars' rotation. Then you still need to do a circularization burn at the apoareion to get into a circular orbit. For this, you will need a rocket with a delta-v of 100.1 m/s.

Your drag will be $$DynamicPressure = 0.5 * dragCoefficient * airDensity * v^2$$

If you make a very long pointy nose cone, the drag coefficient will be quite low - perhaps as little as 0.035 (see Figure 5.7 in this report). enter image description here

Air density we can obtain from HopDavid's answer above. He estimated 0.0028kg/m3 for the top of Olympus Mons. So plugging these numbers in we get, roughly

$$DynamicPressure = 0.0028 * 0.035 * (3639-242)^2 = 1131 N/m^2$$

... which is not very much. A relatively small rocket could easily provide an offsetting force so that atmospheric drag wouldn't slow down the vehicle. For reference, a single RS-25 (Space Shuttle main engine) generates 2.279 MN of force, about 2000X the aerodynamic drag force per square meter.

Therefore, aerodynamically speaking, launching from the top of Olympus Mons with a mass driver is absolutely possible.

As the density of the atmosphere at the surface of Mars is 0.02 kg/m2, the dynamic pressure at the surface would be 8079 N/m^2; therefore, launching from the surface (technically called the Mars datum surface, or average surface elevation) is also possible.

Let's also consider the broader question of whether it makes sense compared to the alternative of using rockets.

The mass driver approach has the following advantages:

  1. Rockets would probably kick up a lot of dust on Mars, which might trigger additional cleanup costs such as having to clean off solar panels and windows around the launch site.
  2. Making rocket fuel, say methane and liquid oxygen, will consume a lot of energy, and energy is hard to generate on Mars. On Earth, oxygen and methane can be extracted from the air and oil wells and do not need to be synthesized. Pushing off the planet with a mass driver is significantly more energy efficient than manufacturing propellant and then using most of it to overcome the physics of the rocket equation.

However, a mass driver has the following disadvantages:

  1. It has a higher up-front cost to build. This would be especially true if most of its mass had to be shipped from Earth. If most of the parts can be manufactured on Mars this is less of a concern.
  2. A mass driver is less aim-able than a rocket. This might not be an issue if your goal is always to launch in roughly the same direction, for example, on a trajectory that will take you back to Earth.
  3. It might not be easy to set up a mass driver close to where you want to locate your settlement - at least not unless you are prepared to do a lot of tunneling and bridge building. Suitable locations for rocket launch pads will be easier to find.

A mass driver on Mars will make more sense once we have established a small colony on Mars. A powerful one could generate a lot of delta-v making it cheaper, faster, and more convenient for people to return from Mars to Earth. Having a fast-return option might be a deciding factor for people who are contemplating whether they want to visit and perhaps settle permanently on Mars. If you can grow your colony faster, that's worth a lot.

A mass driver on Mars might help humanity expand into the rest of the solar system since Mars is further out from the sun and its gravity well is smaller. It might one day be considered critical infrastructure for humanity's journey to the furthest reaches of the solar system and perhaps beyond.

  • $\begingroup$ The linked video is long on why people should spend money on a gun launch system and light on practical details for earth and had little to say about Mars - suggest pulling the actual hard details out of the video and looking at how they would apply on Mars, or at least timestamping the proposal part. $\endgroup$ Commented Nov 28, 2023 at 8:02
  • 1
    $\begingroup$ "That is a myth perpetuated by certain people who are interested in defending the status quo of launching payloads with chemical rockets." That is quite an accusation. I would ask for evidence for this assertion, but I won't bother because I know there is none, because this claim doesn't pass the giggle test. It implies that the top space engineers and leadership in the U.S., Russia, China, India, etc. have all been bamboozled by "certain people" who have managed to perpetuate a hoax that rail gun launching is inferior to chemical rockets for launching payloads from Earth. $\endgroup$ Commented Nov 29, 2023 at 14:16
  • $\begingroup$ Evidence provided. And, the engineers haven't been fooled. If you download nap.nationalacademies.org/catalog/13354/…, and look at the chart on page 17, item 1.5.1, you'll see that "Ground Launch Assist" is tied for the highest score on "Technical Risk and Reasonableness". What I think is holding us back from developing ground launch assist infrastructure is not physics, science, or math, but an institutional bias toward more proven solutions that happen to be more synergetic with national defense priorities. $\endgroup$
    – phil1008
    Commented Nov 29, 2023 at 21:44
  • 1
    $\begingroup$ phil1008 - I didn't ask for evidence that some people have advocated railguns or evidence that some people (like Elon) have dismissed its viability. I asked for evidence of your claim that certain people are perpetuating falsehoods about railguns because they "are interested in defending the status quo of launch payloads with chemical rockets", which means they are purposely lying. If you had put "I suspect" in front of that statement then it would have been clear that you are just voicing your conspiratorial suspicions, but you didn't add that qualifier and instead presented it as a fact $\endgroup$ Commented Nov 30, 2023 at 0:47
  • 1
    $\begingroup$ phil1008 - in the NRC link that you provided (I assume you mean Figure D.1 on page 108), 28 out of the 32 items on that page received the same identical score for Technical Risk and Reasonableness, so yeah I guess you could say ground launch assist is in a 28-way tie for first. But that's talking about railgun launch assist, which is sort of the same category as air launch. This discussion is about using a railgun as the sole method of putting spacecraft into orbit, and I'm pretty sure that is what Elon was commenting on also. $\endgroup$ Commented Nov 30, 2023 at 1:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.