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

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...

perfect!

its been bugging me for a while...

thanks for clearing it up for me

150g would smash a human body into like juice

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.

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.

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

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.

So my question still stands I suppose, where is that ICBM's 150gs coming from?

It can't be from impact because that's only about 5290km which means you have some 20,000km of velocity to go.

So impact deacceleration is actually about 997gs.

The 150g must be when the angular momentum is changing along its trajectory. And acceleration must be around 1.5gs on average the entire trip or else it'd be travelling far swifter than 22,000mph on re-entry.

It'd probably be 1.5 around take-off actually, where it has to fight gravity the most, and it'd probably be around 1+however fast the ICBM is accelerating upon re-entry which can't be too much. I wouldn't think the ICBM would be accelerating much more than 1 or 2gs on its own.

So I think the 150gs has to be from the change in angular momentum. I just don't see where it'd experience so much force since it is not being blasted from the earth at 5292km an hour. In fact it only leaves the earth at about 200-2000mph on its phase from the ground to the outer-atmosphere (hence SDI to kill IBMS before they are going too fast) and so that'd be about .9g if it were not having to fight against earth's gravity. So 1.9gs would the the trust about needed to get to that speed, so on re-entry it'd be around 2.9gs.

Wonder what starwars will have to say about my math ... I'm having to run on fumes of memory

This is in response to all of the message posted in responce to this question.

An ICBM trajectory is indeed parabolic (sort of - the boost doesn't end until near apogee, so not quite - but close) until it begins to re-enter the earths atmosphere.

One thing to keep in mind with all of this: the apogee of a Minuteman III ICBM is 700 miles - whereas only 70 miles or so is required for low-earth orbit - so it is approaching the earth at a fairly steep angle.

Once the RV enters the atmosphere, because it is fairly aerodynamic is does not begin to decelerate rapidly. However, as it approaches the lower atmosphere (still travelling somewhere over 10,000 MPH) aerodynamic drag causes EXTREMELY rapid deceleration (in excess of 150g's actually). It is in fact impossible to have a perfectly parabolic trajectory with any sort of drag effects (with no propulsion)- this non-uniform trajectory only becomes more pronounced when drag only becomes a player in the last 70 miles of a 9,000+ trip.

The basic rules governing re-entry/deceleration and heat is as follows:
Higher Angle: Greater peak decelleration, greater peak heat, lower time in atmosphere
More Aerodynamic: Point of peak deceleration is lower in atmosphere, higher peak heat, less time in atmosphere.

Does that answer the question? I googled it but unfortunately all I could find without endless searching were basic answers like "why does the Space Shuttle get hot"....

One of the most informative (if very corny) sources of information on this is a 1970's or so Dept. of Defense video titled something like "Dr. Parameter".

Here a couple of so-so links:
www.fas.org...
Here is a link with a 60g number (which was about right for a Titan II, Minuteman I - before very good ablatives and smaller warheads became available) www.fas.org...

Oh ... Here we go - a bit technical but definately a fine article:
roger.ecn.purdue.edu...

It answers what you were stating but leaves a new question from what I was thinking, the change in direction (I call it angular momentum I'm not sure if that's acurate).

Because you're always going to feel a force when you're changing direction, the force towards the center of the curve if you're going along a curve, or if it is a point change then you'll still have a force momentarily in the previous path yes?

So what is the amount of force (estimating) that is felt by the payload of an ICBM while going through the changes in its trajectory from say a generally upward motion to a generally downward motion?

At that point it would be post boost (in orbit - an orbit that hits the earth later, but an orbit nonetheless) - a sattelite. It would have a fairly uniform (obviosly not perfect due to changes in radius) acceleration towards the center of the earth. This could never exceed 1g (if a sattelite was at sea level it would experience 1g)

So - using a = G*M(earth)/r^2 , with r = 6378km(radius of the earth at sea level) +1129km (apogee of RV) and accepted values for the rest, I get 7.101 m/s^2.

[edit on 26-6-2004 by Starwars50]

Where's my apology FreeMason? You know, for me being confused about velocity and acceleration and then you sticking your tongue out at me?

I'm not confused. It's called conservation of Energy - read about it. And the only thing that will cause a discrepancy between the kinetic energy at impact and the kinetic energy at take-off is energy lost to drag - on a macro scale that is.