It’s not an escape orbit, it’s just a more eccentric orbit (unless it is much higher). Still, you are correct that my second solution will not work (see my second edit).
I started solving the trajectory for an exponentially decaying air density and a drag force that scales linearly with density and quadratically with velocity, but I did not immediately see the solution to the resulting differential equation, nor did I see a clever trick for avoiding the calculation. I’ll look at it again later.
Your first idea should be elaborated. But it is quite sound.
Your second idea is wrong. Above geostationary orbit and not moving relative to the Earth surface, means an escape orbit.
Still, I would prefer a non—atmosphere solution. But yes, your first idea is also good albeit a little undeveloped.
It’s not an escape orbit, it’s just a more eccentric orbit (unless it is much higher). Still, you are correct that my second solution will not work (see my second edit).
I started solving the trajectory for an exponentially decaying air density and a drag force that scales linearly with density and quadratically with velocity, but I did not immediately see the solution to the resulting differential equation, nor did I see a clever trick for avoiding the calculation. I’ll look at it again later.
You are absolutely right. It CAN be an escape orbit, if it is high enough. But it may also not be an escape orbit.
You are right. Still, not a good solution.