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After watching the first flight of a rotor-powered craft on Mars, I wonder how similar flying on mars is to earth.
I read:
Mars has a much thinner atmospheric volume compared to Earth. Earth’s atmosphere is over 100 denser than Mars.
If there is less atmosphere to push against, it must be more difficult to fly, how much more difficult is it?
For a rough estimate: just by conservation of momentum, the mass flow rate propelled by the rotor * flow velocity must equal the gravitational force on the vehicle for it to hover. The mass flow rate is directly proportional to flow velocity and density of the surrounding air, so that works out to thrust being proportional to atmospheric density * the square of flow velocity.
The pressure on Mars is about 0.6% of Earth's, but the CO2 atmosphere is denser at any given pressure, making the average surface atmospheric density almost exactly 1% that of Earth's. A helicopter needs to push atmosphere at 10 times the velocity to achieve the same lift, but only needs 0.38 times as much lift in the reduced gravity. Power consumption will be proportional to the square of velocity, so your total power consumption required to hover will be 0.38 times 100...it will take about 40 times as much power to hover.
However, this also means the blades will be moving ~10 times as fast, and are in fact moving at around mach 0.7 in this case. L/D ratios at such high speeds are likely a bit lower, which will increase the power requirements. There are also other effects, like friction losses in a transmission that has to handle a rotation rate 10 times as high, efficiency losses in a motor that must turn faster, etc. In all, 100x is a fairly realistic rough estimate.
Also, this was just for the case of hovering. With all these factors taken into account, you can only exert 0.38 times as much force with your rotor blades. The helicopter will be slower to accelerate, stop, and change direction by the same factor. This is harder to quantify, but means you'll generally take ~2.6 times longer to perform any given maneuver. This easily can get the overall increase in "difficulty" up to 100x.
Any answer to this depends on what you mean by 'difficult'. I'll take a very simple answer: difficulty is power. So how much more difficult is how much more power you need to fly the same helicopter on Mars than on Earth. I'm saying 'the same helicopter' so I can factor out things like blade mass and shape: obviously in real life you wouldn't use the same design, so that would make things more complicated.
There's an expression for lift (this will only be even approximately true for well-subsonic blade velocity) which is
$$L = C_L \frac{\rho v^2}{2}A$$
Where
Well, the lift you need is approximately $mg$ where $g$ is the acceleration due to gravity (less on Mars!) and $m$ is the mass of whatever you want to lift.
So we can rearrange all this to get the ratio between the $v_M$ and $v_E$ -- the blade velocities you need on Mars and Earth:
$$\frac{v_M}{v_E} = \sqrt{\frac{g_M \rho_E}{g_E\rho_M}}$$
And given $g_M/g_E \approx 0.38$ and $\rho_E/\rho_M \approx 100$ we get
$$\frac{v_M}{v_E} \approx 6$$
The energy you need to spin the blades up goes like $v^2$ (let's assume you can use the same blades in Earth's atmosphere, and specifically their masses are the same, and I'll ignore drag while the blades are being spun up). So the energy you need to get the blades up to speed goes is about $36$ times that on Earth.
Once the blades are up to speed you have to deal with drag (and in real life you have to deal with it while spinning them up, but I will ignore that), and there's an equation for drag force:
$$F_D = C_D\frac{\rho v^2}{2}A$$
Where $C_D$ is the drag fudge factor. It's not surprising that drag and lift fo the same way with $v$ and $\rho$ and $A$! But this isn't what we need: we need the drag power -- how much work do we need to do to push the blades through the air with velocity $v$? Well power is force times velocity, so
$$P_D = C_D\frac{\rho v^3}{2}A$$
But now Mars helps us again, because the atmosphere is not very dense. The ratio between the drag power on Mars and on Earth is
$$\frac{\rho_M}{\rho_E}\left(\frac{v_M}{v_E}\right)^3 \approx \frac{6^3}{100} \approx 2.16$$
So, in summary, if I'm right there are two answers here:
So, not one hundred times harder, but quite a lot harder.
Note I worked all this out in a hurry: I may have made more-or-less serious errors.
