It's worth noting that, if the drag is managed correctly, and entry speed is relatively slow, one does NOT need heat shielding other than the (minimal) needed to reduce blackbody losses.
See also Space Ship 1's flights, and SpaceX and ULA reusable booster programs...all of which cross the space barrier, and return without a reentry heat shield. (SS1 on a high drag configuration, SpaceX and ULA both using thrust to prevent excessive friction heating.)
Correct. What a lot of people do not realize is that the heat is not truly caused by 're-entry'. It is actually caused by aerobraking.
As I said earlier, an object at an altitude of 200km in a circular orbit travels at around 24 times the speed of sounds. This is because at 200km the force of gravity is still something like 98% of the surface of the planet. As a result in order to stay in orbit an object has to move so fast that if it traveled in a perfectly straight line (and if the surface of the Earth was perfectly smooth) the surface of the Earth would drop away by nearly 5 meters just because of its curvature.
On the other hand an object dropped from 200km above the surface of Earth would only reach a maximum speed of a bit over 6 times the speed of sound, discounting the fact that its drag would slow it down. When you add in the effect of drag the top speed is unlikely to be much more than 2-3 times the speed of sound (Felix Baumgartner approached Mach 1, I believe, during his jump, but the human body is hardly that aerodynamic a shape).
The basic formula for atmospheric heating involves v
2 x d where v is the velocity of the object and d is the density of the atmosphere. This means that an object travelling at 24 times the speed of sound will generate 16 times as much heat as an object traveling at 6 times the speed of sound, assuming the same atmospheric density.
All of which is a really long way of saying that deorbiting objects generate way more heat than objects simply entering from space. But things like meteorites hit our atmosphere and burn up and they aren't deorbiting, so how does that work? That occurs because the meteorite is zipping through space at even higher velocities than orbiting objects.
I would imagine that for approaching an Earth-like planet the physics would run something like this:
The approaching ship would begin braking well outside the atmosphere. How far out would be a factor of how fast the ship was approaching before braking and how strong the maneuver drives are. The target is to be travelling at about 4.5 times the speed of sound when you reach 200 km above the surface.
At that speed a ship would only generate about 3.5% of the energy of a deorbiting ship. 4.5 times the speed of sound would still generate a lot of heat at normal atmospheric pressure but at 200 km the density of the atmosphere is so low that the ship probably would have no issues.
The ship would continue to decelerate at about 1/2 a G. It will take a ship around 5 minutes to decelerate from 4.5 times the speed of sound to 0 at 1/2 G and that will coincidentally cover around 200 km.
I chose 1/2 G because that gives ships a good safety margin. If they find they are coming in a little bit 'hot' they can just increase thrust briefly to adjust. Additionally if they use a greater thrust then they will hit the atmosphere at higher speeds.
My guess is that decelerating from 4.5 times the speed of sound there won't be any issue with the ship travelling too fast when it reaches the richer atmosphere at around 50 km (at which point its speed is a little over 2 x the speed of sound). In the case of a ship that is too fragile to handle these forces the rate of deceleration could be reduced even more, although this would require an even lower speed at 200 km and would extend the time necessary to cover the 200 km and land. However, since the original figure was 5 minutes you are still looking at relatively short amounts of time in order to land. Ships would be spending far more time just maneuvering so that they are above the landing site and moving into the landing pattern.