spaceship one

Ron Wanttaja wrote:

Like I said on an earlier post, I don't have much background on re-entry
physics. But I think it's possible to deorbit going slowly at a fairly
shallow angle...you just have to time the deorbit burn properly.


Me either, Ron. And I don't think my Holiday Inn Express line is gonna
work on this one, either.

I thought I'd try to see if I could at least set the problem up.
Hey, it's only calculus, right? But I found nothing there I could get
hold of except the (obvious) arrogance of ignorance. Humbling...


For a body in motion, the first derivative gives the rate of change
in position per unit time (i.e.: speed = rate of change of position
per second)

If the object is accelerating, the second derivative gives the rate
at which the first derivative is changing, or the rate of change in
speed (acceleration = rate of change in position per second per second).

Third derivative gives the rate of change in acceleration (what the
physics guys call 'jerk' = rate of change of position per second per
second per second). Like the way an old car jerks if there is too
sudden a change is how it is accelerating.

Quoting Martin Gardner, "Beyond the third, higher order derivatives are
seldom needed. This testifies to the fortunate fact that the universe
seems to favor simplicity in it's fundamental laws".

BUT
Simplicity is relative.

On orbit, our ship is in steady state unaccellerated motion, right?
Well, not exactly.

Due to the curved path of the orbit there is an 'outward' centrifugal
force that is exactly opposed by the opposite 'inward' centripetal force
(of gravity).

So our steady state 'unaccelerated' motion is actually a _second_
derivative from the straight line path (ASSUMING the orbit path is
perfectly circular?).

therefore

Adding an acceleration to our _forward_ motion (second and third
derivatives) causes an immediate third derivative reaction of the
orbital
altitude, i.e.: motion inward (if slowing) or outward (if speeding up).

If I'm not too badly mistaken, we are up to the SIXTH derivative,
and still haven't accounted for any deviation that would result if the
acceleration vector is not EXACTLY aligned with the true orbital path
in both pitch and yaw.

Taking those into account, we are looking at the TWELFTH derivative
just to predict what's going to happen when we try to change speed.

If we are off in pitch, I think the end result would be an oscillation
in the the orbital path. Think about an AC electrical signal imposed on
a DC carrier.

If we were thrusting straight 'outward', the thrust pushes us to a highe
altitude that our orbital velocity will not be able to maintain.

As soon as the thrust is removed, and momentum decays, we will drop back
down, gaining inward momentum on the way, which will cause an
'undershoot'
of our previous altitude, which will again bleed off momentum until we
go back 'up', and over shoot again.

There is probably going to be a fairly strong damping effect that will
eventually (sota) stabilize at the original altitude, but I haven't a
clue
how to set THAT one up...



Sheesh! Rocket Science....