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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.... |
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