A few things i dont understand about re-entry...

I didn't know ICBMs really had any deacceleration forces, they fly in at 22,000 mph so where the hell do they pick up the "g-force"?

FreeMason

I'm not real "up" on ICBM's. So here is my question...

Do ICBM's come in at a constant velocity? Or do they operate basically on a lob-lolly trajectory? (i.e. accelerate to altitude and then no more propulsion but decelerate to the point they just "fall" in to their target?)

That point is important to your question.

Originally posted by Valhall
Do ICBM's come in at a constant velocity? Or do they operate basically on a lob-lolly trajectory? (i.e. accelerate to altitude and then no more propulsion but decelerate to the point they just "fall" in to their target?)

yeah, they're lobbed... they fly a parabolic arc to their destination. that's how it's easy to determine where and when one will hit.

Thanks CKK...

Well, then, the g's coming down will pretty much match (with a bit of discrepancy) the g's going up.

Actually off hand I'm not too sure about that myself, but if it were as you described it would be only 1g.

Remember earth's gravity is 1g therefore roughly you will only receive 1g regardless of how fast you are going if you let earth do the work.

So if the ICBM takes that path, the only times it can experience more than 1g is during lift-off when it is accelerating. Because as you describe its re-entry would be "gravity" powered and thus its deacceleration would be caused by gravity and its reacceleration towards a target and such.

But I don't think it is, I think it is a fully powered flight, in a parabolic sense but that it just keeps acellerating so maybe (thought just hit me) the 150g is at the moment when the missile changes its direction.

Instead of going up it goes down, of course it wouldn't be a sharp on the dime change of direction...but that whole segment of the flight-path would experience Xg due to centripital force.

I mean (more stating for myself) imagine an egg in the missile...as it accelerates upwards perpindicular to the earth it experiences only one direction of force, towards the earth, as it goes at an arch, it experiences forward acceleration and gravitational excelleration.

So as it changes directions (towards earth instead of orbit) that little egg would still be heading towards orbit, thus experiencing the g forces caused by the change in direction of acceleration. From orbit towards earth.

(Answered his own question)

So I guess by the physics used, I've also answered your question and no the ICBM does not work on a "lob-lolly" trajectory because then the forces would never exceed 1g.

However something changing its velocity from 22,000mph towards orbit, to 22,000 mph towards the earth is going to have a major deaccelerating effect on what ever objects still wanting to go towards that orbital trajectory...

Right?

Just like when a car suddenly turns and you're thrown against the door...centripital force

That would be a big negatory good buddy. A trajectory will fall (neglecting reduction in speeds due to drag) at the same "acceleration" that it is lofted.

So, if the ICBM ascends at XX g's, it will descend at XX+/-% g's.

It will be in the same ball-park.

No no no...not negatory

If something is hurled at say 15,000mph at 45degrees upward...it'd have accelerated 15,000mph/s. Which is a HUGE number of G's.

Now let's say that happend, on the lob-lolly trajectory you'd let gravity handle the rest.

Deacceleration is -9.8m/s^2 because that's roughly what it is for earth's gravity.

Deacceleration would take place until the velocity (15,000mph) reaches 0, and then acceleraton would begin at 9.8m/s^2 (or second per second...whatever I don't care about the accuracy of the formula at the moment ) until impact. Which would thus stop the object and thus end the velocity...which would also have a deacceleration of say it was going 15,000mph before impact...so it would deaccelerate at -15,000mph/s.

But the acceleration before impact would always be 1g, and the deacceleration on the way up would be -1g.

Unless something like a rocket were powering that acceleration//deacceleration.

I think, you're confusing Velocity with Acceleration valhall

You can be going 1 billion miles a second but if you are experiencing no acceleration you will float around like you weren't moving

Originally posted by Valhall

Originally posted by Starwars50[/i

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...

No, there would be no problem. To stop the vertical deceleration you simply maintain the orbital speed required to maintain that orbit. If you stop descent, and you maintain orbit, there is no problem as long as you have sufficient fuel.

And that is what negates the option of a gradual spiralling descent, sufficient fuel. But there would be no problem with "keeping you in the hot shell longer" there is no hot shell if you are traveling at the correct orbital velocity for a given orbit.

The problem will lie that in order to create power there must be some sort of atmosphere. If that is the case there will be aerodynamic heating of the airframe.

The simplest way to think of it is that you still have the same amount of energy to dissipate - in addition to whatever energy you impart attempting to keep the altitude.