How will hovercraft work on Mars? The facts are:


*

*On Mars atmosphere pressure is way much lower than on Earth.

*To hover hovercraft blows air under itself to create air cushion. This air cushion as I understand must have enough pressure to lift hovercraft and in the same time this pressure must be higher than atmosphere pressure to create lifting force.
My specific question is: given the pressure on Mars and on Earth will the same model of hovercraft need more or less "engine power" to be lifted than on Earth? By engine power I mean power of air blowing engine not propulsion.
I'm deducing that less engine power than on Earth but please give some calculation example. Also please correct if I made any mistake above:).
 A: Parameters
The surface gravity of Mars is ~0.376 g, where g ~ 9.81 $m/s^{2}$ for Earth.  The surface pressure of the atmosphere on Mars is ~0.636 kPa, which is roughly 0.63% of Earth's atmospheric pressure (i.e., ~101.325 kPa at sea level).  The density of air at STP on Earth is ~1.2 $kg/m^{3}$, compared to Mars at ~0.020 $kg/m^{3}$.
Background
Typical hovercrafts make use of an impeller, or a type of axial fan.  We will assume the hovercraft's altitude is in steady state, and the only differences are in the atmospheric pressure and density.
The thrust force magnitude (assume one dimensional for now) is just given by:
$$
F = \frac{ 1 }{ 2 } \rho \ A_{disc} \ \left( C_{in}^{2} - C_{out}^{2} \right) = \frac{ 1 }{ 2 } \rho \ A_{disc} \ \Delta C^{2}
$$
where $\rho$ is the atmospheric density, $A_{disc}$ is the area of the propeller disc, $C_{in}$ is the inflow speed at air intake, and $C_{out}$ is the outflow speed at the exhaust.
Application
If everything were equal, then the thrust ratios between Earth and Mars would just be the ratio of their respective atmospheric densities (Note: $\rho_{E}/\rho_{M}$ ~ 60).  The force of gravity ratio is $F_{gE}/F_{gM}$ ~ 2.67.  Thus, ratio of thrust-to-weight ratios is $ratios_{E}/ratios_{M}$ ~ 22.6.
Since we cannot change $\rho_{M}$, we must increase either $A_{disc}$ or $\Delta C^{2}$ by a factor of ~ 22.6 to get an equivalent performance.
Caveats/Notes
Things I did not account for include, but are not limited to:


*

*differences in lift and drag on the fan blades that would arise due to the difference in pressure and density between the two atmospheres

*the efficiency of the fan

*shape of fan blades and its affect on the engine performance

*effect of thin atmosphere on fuel combustion

*etc.

A: This is more of an engineering question than physics, I think. 
Anyway, think about it in terms of force. To hover, you have to provide a downward force to lift the craft upwards, then hold it at, say a metre above the ground. Force is mass by acceleration, and mass is density by volume. As the atmospheric density on Mars is very low, it seems unlikely, even though gravity is reduced compared to Earth, that hovercraft would be used on Mars. 


*

*The engine would need to be more powerful than on Earth, to spin the impeller fan  faster than on Earth, so as to pull in more low density "air" per unit time. 

*The impeller fan would have to be much greater in diameter compared to Earth fans to generate sufficient volume in a given time,
to increase the effective mass of working fluid sufficiently to generate any reasonable downforce. A larger impeller implies more engine power would be required.

