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From: cjh@petsd.UUCP (Chris Henrich)
Newsgroups: net.sf-lovers
Subject: Re: Gravity on an Integral Tree
Message-ID: <355@petsd.UUCP>
Date: Mon, 24-Sep-84 18:58:43 EDT
Article-I.D.: petsd.355
Posted: Mon Sep 24 18:58:43 1984
Date-Received: Wed, 26-Sep-84 07:23:39 EDT
Organization: Perkin-Elmer DSG, Tinton Falls, N.J.
Lines: 35

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>  In Larry Nivens book 'The Integral Tree' I don't understand the 'gravity'. In
>the tree tufts there is 'gravity' and in the mid-trunk area there is zero g. How is
>this possible. The tree doesn't rotate end for end, one end is always toward Voy.
>Can anyone out there explain it in a relatively simple manner?

     In a word, "tide".
     An integral tree, or any other reasonably small object in
orbit, moves as if its mass were concentrated at the "center
of mass." (Halfway up the trunk.) If it is in a circular
orbit, then the gravitational force at that distance from 
Voy is equal to the centripetal force needed to keep the
object in that circular path.
     At the inner tuft, which is closer to Voy, the
gravitational attraction of the star is greater, and the
centripetal acceleration needed to keep in that orbit is less.
So a person in the tuft feels a pull towards Voy.  Likewise,
in the outer tuft, the gravitational attraction is less, and
the centripetal acceleration needed to keep up with the
integral tree is greater, so a person there feels a pull
outwards. The integral tree itself is under tremendous
tension, which explains why it can come apart if it is weakened
at the middle.
     This force is called "tidal stress" because the same
mechanism explains the tides of Earth's oceans: the major
gravitational attractor being the Moon.
     
Regards,
Chris

--
Full-Name:  Christopher J. Henrich
UUCP:       ..!{decvax,ucbvax,ihnp4}!vax135!petsd!cjh
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