The recent question about Cessna reentering from ISSCessna reentering from ISS got the answers that all imply a rapid drop.
But from what I know, air drag is proportional:
- to square of airspeed
- to air density
- to attack surface (with all the fancy caveats of aerodynamics).
Meanwhile, lift of a plane is proportional
- to square of airspeed
- to air density
- to wing area (with angle of attack and all the fancy caveats of aerodynamics).
That means if an airplane in air experiences X newtons of air drag, its wings are capable of producing Y newtons of lift, with Y being usually considerably higher than X, and mostly proportional. The actual ceiling limit comes from the engine performance which has not enough air to push against.
Is there, then, a specific reason why a plane couldn't enter from the sufficiently low, circular orbit, without need for heatshields? The moment it starts experiencing significant drag - that means considerable energy is being dissipated - wouldn't the same speed and air density that create that drag, contribute equally to its lift, allowing it to maintain altitude where the power dissipation would be kept in check? In other words, instead of dropping steeply into thick atmosphere like most reentry vehicles do, couldn't it just glide down, losing altitude very slowly, to keep both drag and lift to survivable/usable values? Or am I missing something here?
