How to calculate the lowest possible altitude a satellite can orbit at due to aerodynamic heating if provided with a sufficient propulsion system?

Question

If a satellite is equipped with a propulsion system which is enough for compensating the local drag and maintaining the orbit, then aerodynamic heating would be the limiting factor for attaining the lowest possible altitude.

How can one calculate or at least estimate the limiting altitude for a given satellite? What major parameters or aspects of the satellite are required?

Answer 1

Following @JCRM's lead: rocket horsepower questions to the rescue! See this answer and this answer for derivations and explanations.

Power

If we assume that most of the kinetic energy of the air molecules striking the spacecraft is converted to heat (perhaps it's more like half or 2/3) then we can use the concept of "rocket power" which is really just the kinetic energy of the gas leaving a spacecraft calculated in the frame of the spacecraft.

$$\frac{dE}{dt} = P = \frac{v^2}{2}\frac{dm}{dt}$$

$\frac{dm}{dt}$ would be the mass of air encountered per unit time and is the density times the velocity time the area $\rho v A$.

If our drag shield were a plate of metal held "into the wind" thermally shielded and on insulating posts maintained at a temperature $T$ of 1000 Kelvin (about 730C) it could dissipate about $\sigma A T^4$ by thermal radiation assuming a shock wave hasn't formed in front that is so dense it starts radiating back and blocking radiation out. If that were the case then you'll need to absorb heat in front and re-radiate it out the back using a circulating liquid to transfer the heat, which sounds hard and also sounds like someone may have thought of this in the past.

$$P = \sigma A T^4 = \frac{v^2}{2}\frac{dm}{dt} = \frac{v^2}{2} \rho v A$$

$$P = \sigma A T^4 = \frac{1}{2} \rho v^3 A.$$

I'm leaving the drag coefficient equal to one, otherwise that is what Wikipedia gets as well. Solving for density;

$$\rho = \frac{2 \sigma T^4}{v^3}.$$

The Stefan Boltzmann constant $\sigma$ is about 5.67E-08 W m^-2 K^-4.

Put in 1000 K and 7800 m/s for example and we get roughly 2E-07 kg/m^3 or (also roughly) 2E-07 bar which puts it roughly at (found here) the Karman line at 100 km which makes @JCRM's comment about this being another "Karman plane question" either eerily prescient or profoundly insightful!

What thrust is needed?

Since force is just power divided by velocity we remove one power of $v$ to get

$$F = \frac{1}{2} \rho v^2 A.$$

At 2E-07 kg/m^2 that's 12 Newtons which is much bigger than you could easily do with solar-electric on a spacecraft with a 1 square meter cross section orbiting at 100 km. You'll need a conventional thruster and so you'll run out of propellant quickly.

I'll leave it as an exercise for the reader to calculate the thruster's horsepower ;-)