Originally posted by The Bandit
Ok heres my thoughts and ill try to make as much sense as i can...
the earth rotates at (x) speed and the problems facing re-entry is correctly entering the atmosphere right?
well if the shuttle (for example) was to travel at the same speed as the earths rotation, would it not hit the atmosphere at 0 speed and therefore 'drop' through the atmosphere?
or..
if the shuttle was travelling at 0, and then 'dropped' into the atmposphere, would it not pass through it then?
i know these are 2 totally different theories, but i am at a loss!
can anyone help?
Not quite. Most of the heat produced is because of deceleration in the up/down axis, not forward/backward. Simply put, you have about 70 miles of air to get through (and want to hit the ground with a low velocity), so your going to have to decelerate via drag (heat). (to dump the potential energy from falling 70 miles+)
To complicate matters, with a human payload there is a certain maximum g-load that the re-entry vehicle can withstand (and thus it needs a shallow re-entry angle). For example, an ICBM re-entry vehicle has a very steep re-entry path and heat is not a huge problem (as it is on the shuttle). But it undergoes about 150g's of peak deceleration - not good for human cargo.
It theoretically would be possible to stop the vertical deceleration, however then the backward drag would become an issue. My guess is (without running through any calculations) that this would create more of a problem than it would fix. (It would also keep you in the hot shell longer, which in itself is the major problem of a steep re-entry angle).
Hope this helps...


therefore roughly you will only receive 1g regardless of how fast you are going if you let earth do the work.
Wonder what starwars will have to say about my math ... I'm having to run on fumes of memory 