Re: Hale-Bopp THEN and NOW

Article: <5f1grb$3fo@dfw-ixnews11.ix.netcom.com>
From: saquo@ix.netcom.com(Nancy )
Subject: Re: Hale-Bopp THEN and NOW
Date: 26 Feb 1997 14:20:27 GMT

In article: <ESjMkmABB2EzEwUX@spacetime.demon.co.uk> Chris Lawrence writes:
> ZetaTalk[TM] writes:
>> If the comet has slowed and has the Sun more to it's back
>> than to its side, WHY would the comet NOT start falling
>> back toward the Sun, directly.
>
> BECAUSE it already has a component of velocity tangential
> to the direct path from it to the sun. ... The way in which an
> oribiting body moves is a vector sum of it's current velocity
> and that velocity which any forces on it (i.e. sun) creates.
> Chris Lawrence <chris@spacetime.demon.co.uk>

(Begin ZetaTalk[TM])
Examine that tangential path. It has various components. A curve is a COMBINATION of a forward motion and a sideways tug, always. There is no such thing as a curved motion, except as a combination of forces, which you may call vectors if you insist. Is this not the case, that in vector analysis, you must draw TWO vectors to equate to a curve? Given that, draw the comet on the long stretch away from the Sun as vectors.

1. Picture the comet with the Sun to its left as it exits the Solar System. It is drifting slightly to the right, as your human scientists know, and which they describe as following an ellipse (as though the comet were dutifully drawing geometry). Draw the vectors at this point, as a straight line from the center of the Sun on out into space, which will be a line parallel to the path the exiting comet is taking.
2. For the second vector, draw a second line, this one pushing to the right, as this is the direction the comet is moving in, away from the first vector drawn from the center of the Sun. As you see comets exit, they are at all times moving to the SIDE, toward the wide end of the nose cone of the ellipse you observe when it is within your Solar System.

Now, given that there is NO vector moving the comet to the left, how does it get way over there for its re-entry into the Solar System?

(End ZetaTalk[TM])

In article: <ESjMkmABB2EzEwUX@spacetime.demon.co.uk> Chris Lawrence writes:
> ZetaTalk[TM] wrote:
>> WHY would it move increasingly toward the side instead?
>> Motion is in a straight line unless perturbations occur ...
>> WHY would it drift to the side over this dramatic swath
>> of space
>
> If you whirl a rock round on a string, there is an acceleration
> towards the end of the string you're holding (if there wasn't
> the rock would not move in a circle) and this continuously
> changes the velocity of the rock. That doesn't mean that
> this acceleration causes the rock to smash into your hand,
> does it?
> Chris Lawrence <chris@spacetime.demon.co.uk>

(Begin ZetaTalk[TM])
Where the twirling rock on the end of a string is often used as an analogy for planets in orbit around the Sun, there are several significant differences. The planets are not bound to the Sun in the same manner as the rock is bound to the arm, by the length of string. The string keeps the rock from moving AWAY, where a planet is free to do so. Likewise, the rock is not moving about in a circle on its own, it is being THROWN, continuously by a tugging arm. Freeze the arm that the string is attached to, and see how long the rock stays in motion!

The comet exiting from the Solar System is likewise not bound to the Sun except as the sum of the forces affecting it may equate to that. Thus, one must examine these forces, and analyze their effect, to understand why the cosmos works as it does. What happens when the swinging arm that sets a rock on the end of a string, LETS LOOSE of that string? Does the rock continue in a curve? Does the rock circle round and return after a few years? It moves in a straight line, away from the arm that THREW it, until gravity brings it to the Earth and the friction of the atmosphere slows it. Plonk. No perfect ellipse.
(End ZetaTalk[TM])