People will tell that there is no air, and this is why we cannot. But if I read on the internet, there is air in space, much less, but still something.

For example:

  • 100 km: $6\ \cdot 10^{-7}$ times as much air as on the surface;
  • 1000 km: $2.5\ \cdot 10^{-14}$ times as much air as on the surface;
  • even on 36000 km there is something ($1.5\ \cdot 10^{-18}$ times as much air as on the surface or 30000 atoms per 1 cubic decimeter).

Talking practically: NASA already built a prototype of the helicopter which will fly on Mars where the atmosphere is the same as 30 km high on The Earth.

if I take some software to calculate propeller thrust like this Propeller Selector and I calculate this very particular practical example. If I take this propeller which cost 50 euro and this engine which cost 9 euro, then I can fly up to 30 km with only 381 watts of power for 8140 rpm. Fly I mean it will produce 1 kg thrust, which enough to hold 29 grams engine, 349 grams propeller and let say 500 grams for the power source (we can even use a solar panel for it). Calculations are shown in the picture below:

enter image description here

and the same configuration will fly (produce same 1 kg thrust) on the surface with only 47 watts of power and just 1000 rpm.

If I go forward, I can calculate that to fly on 100 km I will need 25 meters (1000 inches) diameter propeller and just about 800 watts of power with rotation speed 500 rpm.

enter image description here

I general, we just need big enough propeller, and because of very low air pressure, we will not need a lot of power to make it rotating fast. Also, we can make propeller very thin, because it should not be very strong, because of low pressure as well. Another point, that we can use different propellers for different altitudes, like many stages rockets.

Ultimately, if we make a few kilometer size propeller we can theoretically even fly to Mars and other planets. There are still 30000 atoms per 1 cubic decimeter in space.


Thank you, everyone, for very valuable comments. I will do exact calculations with the real propeller, motor and solar panel to see how high we can go.

The main problems as I understand is the weight of propeller + power source and strength of the material of propeller.


Real calculation of helicopter with solar panels :)

If I take this real propeller which is 6.5 kg and 70 inch diameter, this real motor with 1600W power and these row solar panels with the weight with the weight 4.5 kg for 1350 watt (or 1 kg gives 300 watt) and I add 50% of mass of the solar panel to the mounting of them.

My calculation with the same program (Propeller Selector) shows that it can fly up to 5km taking into account mass of propeller, the mass of motor, the mass of solar panels, the mass of solar panel mounting, max motor power, max propeller rpm, max motor rpm.

If solar panels were 10 times lighter, then 20 km could be reached.

So far, I see only 2 problems:

  1. The weight of the solar panel. If we can make it 10 times lighter, then we can reach much higher altitude.

  2. This program can give wrong results for altitudes higher than 5-10 km.

Other problems from comments:

  • Rotor speed being higher than orbit speed. Calculation shows that it is not necessary and Karman line depends on aircraft weight. So, if we make very lite aircraft/airplane from non-existing/"future invented" material (carbon epoxy, etc.), then Karman line will be higher than 100 km.
  • Shock wave when propeller parts are moving faster than sound. One comment was, that it is not a big problem if air density is very small. So we can start with low speed on the ground and reach high speed propeller when we reach higher altitude.
  • The strength of the material. In my last picture with 25 m (1000 inch) propeller, if I calculate acceleration and g-force it will be 3500g, yes, 10 times more than normal helicopters (Mi-26 helicopter with 32 m propeller and 132 rpm). But, considering that piston inside V8 Formula 1 car engine works with 8500g, then it is kind of feasible). Maybe they do not do it for real helicopters because of sound speed, but as I told before, it does not matter at high altitudes.
  • 4
    $\begingroup$ @OrganicMarble It can be true with high probability, that this program does not work well higher than 10km. But anyway, I calculated from generic momentum conservation law and it gives similar results, we just need big propeller and even 30000 atoms per 1 cubic decimeter can be abough. $\endgroup$
    – Zlelik
    Commented Jun 4, 2018 at 14:59
  • 17
    $\begingroup$ Question 2: Do propellers work outside of the continuum flow regime? Specifically, in the free molecular flow regime? Look up what a Knudsen number is. $\endgroup$ Commented Jun 4, 2018 at 15:10
  • 5
    $\begingroup$ The Karman line would be a hard ceiling - the blade tips would have to be traveling at a rate comparable to orbital velocity to generate any lift. $\endgroup$
    – John Bode
    Commented Jun 4, 2018 at 15:42
  • 6
    $\begingroup$ @OrganicMarble I imagine your propeller will start to look like a turbomolecular pump's inlet at these pressures; a simple wedge to impart downwards velocity $\endgroup$
    – 0xDBFB7
    Commented Jun 4, 2018 at 17:14
  • 6
    $\begingroup$ "0.0000006 times less air" is something like a double negation. It means there is more air because the number is smaller than one. $\endgroup$
    – Nobody
    Commented Jun 5, 2018 at 9:29

7 Answers 7


A 25m diameter rotor has a perimeter of around 78 meters. At that size, at 500rpm, the rotor tips would be going in excess of 1,400mph.

At those kind of speeds, even though it doesn't take much power to get a very light rotor going, there is still an awful lot of force involved which has to be handled by the materials to prevent them literally tearing themselves apart.

  • 50
    $\begingroup$ Not to mention that that's Mach 2, and the shockwaves will do nasty things to your lift. $\endgroup$
    – Hobbes
    Commented Jun 4, 2018 at 15:36
  • 35
    $\begingroup$ This would give a radial acceleration at the tip of some ~34000$ms^{-2}$ (~3500g) - better hope your spacecopter doesn't skip rotor day. $\endgroup$
    – Jack
    Commented Jun 4, 2018 at 15:38
  • 5
    $\begingroup$ @Zlelik No, helicopter blades cannot exceed the speed of sound - even ignoring the massive damage that would cause, it would cause them to lose all lift, losing control of the craft (don't forget that while one blade is supersonic, the opposing blade isn't). Not everything has a linear relationship, you know - you need to understand how things scale. Neither 14 m @ 392 RPM nor 32 m @ 132 RPM give you a good idea o how 25 m @ 500 RPM would behave. $\endgroup$
    – Luaan
    Commented Jun 5, 2018 at 7:49
  • 12
    $\begingroup$ @Zlelik It's normal for helicopter blades to go more than twice as fast as the fastest helicopter blades in the world? Not really following your logic here. $\endgroup$
    – Moyli
    Commented Jun 5, 2018 at 8:20
  • 21
    $\begingroup$ @Zlelik 2 times faster is a big deal in this context. As soon as your blades start to reach the speed of sound, everything about their behaviour changes dramatically, and every one of those changes is going to cause severe problems for any aircraft that wants to remain airborne. $\endgroup$ Commented Jun 5, 2018 at 15:03

At 100 km altitude, you get to the Karman line. This is the altitude where you have to fly at orbital speed to get sufficient lift. This definition is based on the lift equation, which applies to all airfoils including that of a helicopter rotor.

So in a helicopter at 100 km altitude, your blades have to travel at orbital speed (27,000 km/h or 17,000 mph) to generate enough lift.

Because the blades rotate, the inside of the blade travels at a lower speed and the outside travels at a higher speed. Taking the average, the midpoint of your blades would have to revolve at 27,000 km/h.

If you want to get into orbit instead of having to keep spending energy on hovering, you have to fly at 27,000 km/h. When you do that, the advancing blade moves at 54,000 km/h relative to the air. The receding blade moves at 0 km/h relative to the air. The heating effects alone would be enough to melt your blades in a short time.*

I wouldn't want to cope with the aerodynamics of blades that go from 0 to 54,000 km/h twice per rotation, nor with the centrifugal forces in the rotor system.

*: in a rocket launch, the fairing is usually jettisoned at an altitude of ~100 km, when the heating effect drops below 1 kW/m2. The rocket is far below orbital speed at that point (5000 km/h?). Aerodynamic heating scales with the square of the speed, so a helicopter rotor would be subject to 100 times that.

Alternative approach

Approaching from a different angle: at 100 km, atmospheric pressure is 10-7 bar. So your rotor blades need to have 107 times the area to get the amount of lift you need to hover. Your propeller has an area of 32"x1" (roughly), at 100 km that goes to 32 million sq in = 222,000 sq ft is 20,000 m2 is a propeller longer than a Boeing 747. There's no way to make a structure that large within your weight budget. You could increase lift by increasing speed, but then you're back to supersonic propellers.

First principles

The lift of an aerofoil (any aerofoil, including a helicopter rotor) is governed by this equation:

$$L = \tfrac12\rho v^2 S C_L$$

L is the lift force
ρ is the air density
v is the aircraft's speed relative to the air
S is the aircraft's wing area,
$C_L$ is the lift coefficient.

When you go from ground level to 100 km, ρ reduces from 105 Pa to 0.01 Pa (data from atmospheric models discussed here). That means lift also reduces by a factor of 107. You have to compensate for that by either increasing the speed by a factor 103.5 or increasing your wing area by a factor of 107, or a combination of both.

Both inevitably increase weight, which means you need more lift. This is a vicious cycle, and at altitudes far below 100 km you get to a situation where no material in existence is light enough to make your helicopter work.

  • 2
    $\begingroup$ I am not sure if it's fair to consider heating effects from the air resistance. The pressure is very low up there and I believe the actual resistance will be equal to what blades usually resist. Just as the downforce will be the same. $\endgroup$
    – Džuris
    Commented Jun 4, 2018 at 16:43
  • 4
    $\begingroup$ Why would one have to have a forward speef of 25000 mph? A helicopter can have lift without forward speed (and if it has that speed, it already is in orbit). $\endgroup$ Commented Jun 4, 2018 at 18:39
  • 7
    $\begingroup$ The only point of flying at that altitude is to get into orbit, because then you can remain up there without using more fuel. $\endgroup$
    – Hobbes
    Commented Jun 4, 2018 at 19:04
  • 2
    $\begingroup$ @Hobbes see this comment $\endgroup$
    – uhoh
    Commented Jun 5, 2018 at 6:55
  • 2
    $\begingroup$ Not to neglect the heating effects, but the required power to turn a rotor this fast is also shocking at a glance. Each rotor blade does a 360-degree orbital plane change with each revolution of the rotor (since it's at orbital velocity). $\endgroup$
    – Erin Anne
    Commented Jun 6, 2018 at 0:19

Space is really like this (XKCD What if)

enter image description here

In theory a plane could reach most of the way to space, but it won't be able to reach orbital speeds.

Bottom line, it just isn't practical. Maybe someday a helicopter could lift a rocket up high, which would help a bit, but it really just isn't practical. Also, a balloon might just be better in any case, it can go higher and lift more payload.

  • 7
    $\begingroup$ This misses the point of using a helicopter. If you can provide sufficient upward thrust you don't need orbital velocity. Not that I believe any propeller would provide said thrust is space. The same goes for a plane, if it still actually has lift, then it doesn't need to achieve orbit. $\endgroup$
    – Octopus
    Commented Jun 4, 2018 at 15:23
  • 25
    $\begingroup$ @Octopus You need orbital velocity to be in orbit, and not going into orbit is pointless, as anything useful you could be doing is done in orbit. $\endgroup$
    – Polygnome
    Commented Jun 4, 2018 at 16:06
  • 10
    $\begingroup$ It would be cheaper to just use a balloon in any case. I don't think there is anything aside from accuracy that a helicopter can do that a balloon could not do. $\endgroup$
    – PearsonArtPhoto
    Commented Jun 4, 2018 at 17:21
  • 4
    $\begingroup$ @Octopus 100 km is the highest even theoretically you could be via a pure air powered spacecraft (Definition of Karman line). That is too low for most purposes. $\endgroup$
    – PearsonArtPhoto
    Commented Jun 4, 2018 at 17:22
  • 3
    $\begingroup$ @Octopus the fact that an imaginary thing would be useful if it were real does not make it real. $\endgroup$
    – barbecue
    Commented Jun 4, 2018 at 20:03

The propellor needs to be strong enough not to pull itself apart via centrifugal force. If you go through the math, you find that the maximum stress on the propellor blade will be halfway along its length, and will have the value $$ \sigma = \frac{1}{4}\rho L \omega^2, $$ where $\omega$ is the blade's angular velocity (usually measured in radians/second) and $\rho$ is the density of the material (usually measured in kilograms per cubic meter.) Re-arranging this, we find that the specific strength of the propeller material must be $$ \frac{\sigma}{\rho} = \frac{L \omega^2}{4}. $$

For a 12.5-meter length blade rotating at 500 rpm, this works out to be $$ \frac{\sigma}{\rho} \approx 34 \text{ kN}\cdot\mathrm{m/kg}, $$ which is still within the realm of known materials. However, as you get higher, you'll need to increase either the size of the blades, their rotation rate, or (probably both); and eventually you'll get to the point where you have to build your helicopter blades out of unobtanium.

  • 9
    $\begingroup$ At Joe's House of Unobtanium, we're OVERLOADED with space rotors. At low low low prices. That's Joe's!! $\endgroup$
    – zeta-band
    Commented Jun 4, 2018 at 20:57
  • 3
    $\begingroup$ +1 Yay for a science and math-based answer! $\endgroup$
    – uhoh
    Commented Jun 5, 2018 at 1:31

OrganicMarble touched on this in a comment but I think it deserves an answer as well, since the question doesn't stop at the Karman line (approx. 100km if you're really defining it as the height where the velocity required to generate lift exceeds the orbital velocity).

In simplest terms: just because there's a few atoms of gas present at some height doesn't mean it's acting like a gas does at sea level.

Part of the reason a wing (and make no mistake, a helicopter's rotor is absolutely a wing for these purposes) works at all is because air fills in around it. Air rushes in behind it because air molecules fly around and bounce off of each other and fill up space, and so you can keep pushing down air, which in turn pushes up your wing and whatever is attached to it.

This is really aerodynamically important! The wing doesn't have to literally hit a molecule of air to get involved with that molecule, because the molecules are involved with each other. The wing or rotor disc can use a lot of the air around it!

As height increases and the ambient pressure drops (because gravity drags the air down, and there's only so much air around for the other air to essentially stand on top of) that space-filling effect really isn't happening anymore. The air molecules are doing a lot less bouncing (their mean free path is longer) and so once you push them out of the way there's nothing left to push down on. You essentially only interact with the air you hit.

All that is to say this premise

Ultimately, if we make a few kilometer size propeller we can theoretically even fly to Mars and other planets. There are still 30000 atoms per 1 cubic decimeter in space.

is wrong. Once you can't fill in air for the propeller to grab, you're essentially just as likely to bounce your 30,000 atoms off the top than the bottom. This doesn't matter for something like a vacuum pump, because if the molecule bounces off the top eventually it should get another chance to come back to the pump by bouncing off the vacuum chamber's walls. When you're in open space, it means the net force your propeller can create is zero.

If you're an experimentalist, this is trivially true just by observing that the Space Shuttle had giant wings and completely ignored the lift that it generated while it was in Low Earth Orbit. The International Space Station also has giant wings (the solar arrays!) and mostly thinks of those in terms of drag. Think of how many cubic decimeters the ISS has intersected in its decades on orbit! (Lift generation for both is actually more favorable for both of these than for your helicopter, since they move sideways at terrific speeds, thus encountering regions where they haven't already hit all the available air molecules.)

  • $\begingroup$ Nice explanation! $\endgroup$ Commented Jun 6, 2018 at 11:27

We don't fly to space with helicopters because we can't. We would if we could, believe me.

Aside from all the very valid concerns raised by others, the question doesn't properly account for weight*. Maybe your 32 inch diameter propeller weighs 349 grams, but the 10m one certainly doesn't. Oh no! Now 1kg thrust won't lift it at all! So you need more power! So you need more fuel/energy! So you need more thrust...

Solar panels won't solve your problem. They sound great when you don't need a lot of power, but their specific power (Watts per kilogram) doesn't hold up versus something like a jet turbine or rocket engine. Batteries also don't have specific energies (Joules per kilogram) comparable to hydrocarbon fuels yet.

The world's current helicopter altitude record is a bit less than 41,000 feet. Ultimately helicopters just can't put enough power into the air to continue lifting themselves. They all eventually hit a thrust-to-weight ratio of 1, even though a turbine helicopter has vastly more power available than the helicopter you're proposing.

Why does the proposed Mars helicopter work while yours wouldn't? Because it doesn't go very far. The power requirements favor a small helicopter, since the weight drops so much more rapidly than the thrust as it scales down, but the flight times being talked about are about 90 seconds (much like terrestrial drones!) So it couldn't make it from an Earth Sea Level equivalent altitude to the proposed Mars Equivalent altitude; it would run out of energy before it got there.

(*I see in comments on other answers that you've said it's "just a matter of propeller and solar panels weight." It is, but you can't make them arbitrarily light. There's no scaling law that even suggests they'll get dramatically lighter over the next couple decades.)

  • $\begingroup$ Thanks, a good answer. I will do exact calculations with 32 Inch propeller and solar panels to see, how high it can go :) $\endgroup$
    – Zlelik
    Commented Jun 5, 2018 at 7:53
  • 3
    $\begingroup$ The Marscopter also doesn't need as much thrust to lift a given mass through a given density of air, since Mars has much lower gravity than Earth. $\endgroup$
    – Vikki
    Commented Jun 5, 2018 at 20:24
  • $\begingroup$ That's true, but they've also done Mars-atmosphere demo flights on Earth, without tethers, so it doesn't rely on the gravity difference. youtu.be/oOMQOqKRWjU $\endgroup$
    – Erin Anne
    Commented Jun 5, 2018 at 21:11

The propeller would still weigh a lot!

  • You don't want it to bend 90 degree in direction of flight or bend in opposite direction of rotation, which requires some stiffness which does not come cheap in terms of weight. it cannot be very thin.

  • Also I would assume at the base it must me as thick(or similar) as your 81cm propeller base at sea level to support the weight of your aircraft. Let us assume a linear drop in thickness to the tips. Even without calculation I can tell you it will be very heavy.

  • $\begingroup$ Then it is just about propeller weight. We can use many propellers, like thick one for the base, later at 20 km we through it away and start using second thinner and bigger one etc... Or maybe in 10-20 years they will invent something strong enough to make a big propeller. This 81 cm propeller is only 350 grammes. Very light. Also, we can bring just parts to space, like they built International Space Station, and assemble 5 km size propeller there and start flying from there to Mars. $\endgroup$
    – Zlelik
    Commented Jun 4, 2018 at 14:21
  • 7
    $\begingroup$ @Zlelik: You would also need propellant to run the engine with, and you need to lift that as well. To lift that, you're going to need a larger propeller… the tyranny of the rocket equation. Also, every time you think "I found a simple way to solve a problem countless expert have worked on for decades" you should stop and think where you're having vital knowledge problems. $\endgroup$
    – DarkDust
    Commented Jun 4, 2018 at 14:50
  • 1
    $\begingroup$ @DarkDust This is why I asked here, when I found "a simple way to solve the problem". About fuel. Let's use solar panel, then solar energy is 1000 Watt per 1m2. Solar panel has 20% Efficiency and we have 200 Watts electricity from 1m2. Then if we make lite enough solar panel and lite enough propeller we can go up and stay at the same altutude (for example 100 km) for any time. Maybe it is just a matter of propeller and solar panels weight and we just need to wait for 10-20 years, when it will be lite enough. $\endgroup$
    – Zlelik
    Commented Jun 4, 2018 at 14:54
  • 3
    $\begingroup$ @Zlelik, the lightest solar panel I was able to find while looking at portable panels came in at around 22 grams per watt. Hook that up to your proposed 29 gram engine and 349 gram propeller, and you're looking at 996 grams of solar panels to provide the 47 watts you say are required -- somewhat over the 500 grams you can lift at sea level. "If we make it out of unobtanium, it'll work" sounds good, until you actually try to find some unobtanium. $\endgroup$
    – Mark
    Commented Jun 5, 2018 at 0:37
  • $\begingroup$ Let them flex--the outward force way exceeds the lift they generate. You only need to maintain the angle of attack. $\endgroup$ Commented Jun 5, 2018 at 3:54

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