Using the simple momentum theory, the thrust generated by a rotor is:
$ T = \sqrt[3]{2 \rho A P^2} $
where $A$ is the rotor area, $\rho$ is the density and $P$ the power needed to spin the rotor. Since Ingenuity uses two coaxial contra-rotating rotors, this value has to be doubled:
$ T = 2 \cdot \sqrt[3]{2 \rho A P^2} $
The highest thrust is normally required for hovering where it simply equals the weight:
$ T = W = mg $
where:
- $m$ is the mass
- $g$ is the gravity
As a proof, let's introduce on the right part of the equation the values for Ingenuity and let's see if we get the correct thrust i.e. weight on the left. So:
- $\rho = 0.01 kg/m^3$ Note: density on Mars changes according to season from 0.01 to 0.02; we use the lowest value in order to be conservative
- $A = \pi \cdot 0.6^2 m^2$
- $P = 350W$
Considering the martian gravity of $g_{mars} = 3.72m/s^2$ this returns a mass of $1.88kg$ which matches quite perfectly the official mass of $1.8kg$. So, even on Mars, Earth's physics is right 😅
Now, according to the question we have to bring Ingenuity on Earth at some $30km$ height and see if it can still hover. Well, the right part of the equation remains the same since at $30km$ height the density is the same as on Mars. But the left part doesn't match anymore because the gravity is now $g_{earth} = 9.5m/s^2$ i.e. some 2.5 times higher than on Mars. So no, at 30km height, gravity is too high on Earth to let Ingenuity hover.
Anyway if we go down to some $16km$ height then the density rise to $0.15kg/m^3$ giving again a perfect match between thrust and weight. Obviously other aerodynamic or structural issues have not been considered and could play an important role.