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From: bet@ecsvax.UUCP (Bennett E. Todd III)
Newsgroups: net.puzzle
Subject: Re: someone asked why weightless ...
Message-ID: <971@ecsvax.UUCP>
Date: Fri, 20-Dec-85 15:51:21 EST
Article-I.D.: ecsvax.971
Posted: Fri Dec 20 15:51:21 1985
Date-Received: Sun, 22-Dec-85 01:30:04 EST
References: <96@decwrl.DEC.COM> <970@ecsvax.UUCP>
Reply-To: duccpc!bet@ecsvax.UUCP (Bennett E. Todd III)
Distribution: net
Organization: Duke University Computation Center
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In article <970@ecsvax.UUCP> dgary@ecsvax.UUCP (D Gary Grady) writes:
>If you're in the space shuttle orbiting 300 miles up, the acceleration
>of gravity is very little less than it is at the surface.  If weight =
>mass X acceleration, how can you be "weightless"?

First off, let us define acceleration as the second derivative of
position with respect to time. So what's position? You know, spatial
coordinates. There's where you get into trouble in this one -- an object
in free fall is weightless *in it's local reference frame* -- a falling
rock wouldn't normally be called weightless; the taken-for-granted
reference frame for falling objects is the "outside" world, against
which they are falling. Space ships, by finding a ballistic trajectory
which doesn't intersect the surface of the Earth, can keep falling for a
good while, and with their size, make an interesting reference frame in
their own right. Thus "weightlessness".

On the other hand, the gravitation attraction of the Earth is indeed
still important; if it weren't then the craft wouldn't orbit, it would
head elsewhere to frolic. So in the reference frame of the Earth the
satellite is not in fact weightless; it weighs something less than what
it does on the surface of the Earth, and its weight is its mass times
the acceleration it is suffering to keep it in orbit.

-Bennett
-- 

"Hypocrisy is the vaseline of social intercourse." (Who said that?)

Bennett Todd -- Duke Computation Center, Durham, NC 27706-7756; (919) 684-3695
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