The question I'm trying to answer seemed simple: how hard would it be to fly on a planet with lower gravity but also thinner atmosphere compared to Earth. If the answer could hint me at how much different an airplane designed to fly there would look like.

I know atmospheric pressure, atmospheric composition (and hence molar mass) and temperature at the surface of the hypothetical planet. However I have a problem with determining lift and drag coefficients. The NASA site says this coefficients are dependent on viscosity and compressibility of air, the form of the aircraft and angle of attack. My first thought was to separate the part of the coefficients that is dependent on the aircraft from atmospheric parameters. However, I have trouble finding a formula for it. This page says that for certain condition lift coefficient is $$ C_l=2\pi \alpha $$ But I'm not sure if this approximation is true for different atmosphere.

L/D Ratio and Mars Aircraft may be relevant.

Also, can I assume that lift to drag, or maximum lift to drag, is the same in any atmosphere, and if so under what conditions?

  • 2
    $\begingroup$ Obligatory xkcd-what-if link. $\endgroup$
    – Qmechanic
    Sep 18, 2015 at 11:50
  • $\begingroup$ @count_to_10: Pressure at surface is 0.456 atm or 46.2042 kPa. Gravity is 0.84g. Temperature is 278.15K and density is 0.5842914 kg/m^3. $\endgroup$
    – Shiolle
    Sep 18, 2015 at 12:18
  • $\begingroup$ I will be honest with you, I am not the greatest fan of speculative questions, I don't mean any offence to you by that, but this page: en.wikipedia.org/wiki/Lift-to-drag_ratio seems to suggest that the answer to can I assume that lift to drag, or maximum lift to drag, is the same in any atmosphere is yes, as I can't see an immediate density term jumping out at me in the L/D equation. $\endgroup$
    – user81619
    Sep 18, 2015 at 12:23
  • $\begingroup$ @count_to_10 Thank you for your time. I've read the entry on Wikipedia about lift to drag and wanted to link it here, but you need a certain amount of karma before you can post more than 2 links per post. I just thought that maybe I'm missing something, since both coefficients depend on atmosphere properties. $\endgroup$
    – Shiolle
    Sep 18, 2015 at 12:38
  • $\begingroup$ I took the liberty of putting in the links, I hope that's ok, you should be able to delete them if you don't want them. $\endgroup$
    – user81619
    Sep 18, 2015 at 12:43

3 Answers 3


Physics should not be different on other planets, so the same laws apply as on earth. Only the results of an optimization might look unfamiliar. See here for an answer on Aviation SE on a Mars solar aircraft.

The lift slope equation you found is only valid for slender bodies, like fuselages and fuel tanks, and once wing span becomes a sizable fraction of length, more complicated equations will be needed, and Mach number effects must be considered, too. See here for a more elaborate answer.

Generally, to fly like on earth would mean that the ratio of dynamic pressure and mass is the same. Then you would use the same aircraft as on earth (provided the other planet's atmosphere contains enough, but not too much oxygen for the engine to function).

Dynamic pressure $q$ is the product of the square of airspeed $v$ and air density $\rho$: $$q = \frac{\rho\cdot v^2}{2}$$

Lift is dynamic pressure times wing area $S$ and lift coefficient $c_L$ and must be equal to weight, that is the product of mass $m$ and the local gravitational acceleration $g$: $$L = q\cdot S\cdot c_L = m\cdot g$$

The lift coefficient is a measure how much lift can be created by a given wing area and can reach values of up to 3 in case of a landing airliner. Then the wing uses all kinds of high-lift devices (slats, slotted flaps), and once those are put away, the lift coefficient of an airliner is at about 0.5. For observation aircraft, less speed is required, and a normal lift coefficient for them would be 1.2. I see no reason why this number should be different just because the atmosphere is different.

The most important number would be the Reynolds number $Re$. It is the ratio of inertial to viscous forces in a flow and is affected by the dimensions of your plane (on earth we use the wing chord $l$) and the density and dynamic viscosity $\mu$ of your planet's atmosphere. $$Re = \frac{v\cdot\rho\cdot l}{\mu}$$

Lower Reynolds numbers will translate into higher friction drag, which depresses the maximum achievable lift-to-drag ratio. Gliders fly at Reynolds numbers between 1,000,000 and 3,000,000 and airliners can easily achieve 50,000,000. When you need to optimize for a more gooey atmosphere, your wings will become less slender than on earth, because you will enlarge wing chord $l$ to keep $Re$ up.

Once you need speed to get the weight lifted, the Mach number $Ma$ might become important. Generally, subsonic flight is the most efficient, and it has a natural limit at $Ma^2 \cdot c_L = 0.4$. This is what can be achieved with todays technology. The speed of sound in a gas is mainly a function of temperature - Mach 1 on Mars is 238 m/s.

The first part of an airplane to hit a Mach limit are the propeller tips. Maybe you need to have several slow-spinning, small propellers than one big, honking propeller which would provide the best efficiency as long as its tips are well below Mach 1.

Last, you need to know the number of atoms per gas molecule. Air is dominated by two-atomic molecules, but maybe your planet has an atmosphere like early earth with lots of carbon dioxide. This will affect the ratio of specific heats $\kappa$ - this means the rate of heating and cooling with compression and expansion of the gas might be different than on earth. This will come into play when you approach or exceed Mach 1.

  • $\begingroup$ Actually less gravity can affect lift. Especially when gravity helps generate lift. $\endgroup$
    – Ethan
    Sep 19, 2015 at 19:05

This site is an awesome resource.

From my own knowledge, airplanes can fly with a wide range of weights (gravities), at a wide range of altitudes (densities), at a wide range of speeds.

One thing to be aware of is the Coffin corner where stall speed (the minimum speed at which a wing can support the load placed on it) approaches critical Mach number. (Speed of sound decreases with temperature.) A way around this is to transition to supersonic flight.


The surface of Mars is 0.095 PSI. The highest flying drone on Earth flew at 13000 feet with specialized propellers. The PSI of 13000 feet on earth is 2.39 PSI. The temp of 5.5 F. The temp found at 13000 feet. This is Calculator I used.


The calculator stops at 20000 meters. Or 20 km up. Temp minus 50c. Still maintaining a PSI of .25. This is well below the altitude that Mars if equivalent to Earth. That is to say 20 km into Earths atmosphere has much greater air density then Mars surface Atmosphere.

The NASA calculator for lift I saw was just an annoying equation. This Calculator is much more complete. https://www.omnicalculator.com/physics/lift-coefficient

Its unlikely that Ingenuity is capable of lift on Mars. I have not finished all my calculations. Even with .89 gravity of Earth. 4x2×2 ft carbon fiber blades. Weighing 1.4 kg. With 2500 RPM.


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