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Re: "acceleration due to gravity"



Being of Leigh's persuasion in this thread, after having been persuaded from
a close but not isomorphic position in past threads; I'd add that one must
be very careful to not equate mg with GMm/r^2 with this viewpoint! Which I
think is what Rick Tarara is saying when he refers to "staying in the
Newtonian box".

I'd disagree that Leigh's viewpoint is necessarily the "modern viewpoint".
I'd suggest that it is fully in the Newtonian box; but is done in such a way
that it will minimize difficulties when one jumps boxes later on; and is
consistent with "vulgar" terminology as well; a pedagogical advantage in my
opinion.

Joel

-----Original Message-----
From: phys-l@lists.nau.edu: Forum for Physics Educators
[mailto:PHYS-L@lists.nau.edu]On Behalf Of John Mallinckrodt
Sent: Friday, January 26, 2001 7:59 AM
To: PHYS-L@lists.nau.edu
Subject: Re: "acceleration due to gravity"


On Thu, 25 Jan 2001, Leigh Palmer wrote:

My point has been missed entirely. I did not talk about corrections
at all. In my picture g is the quantity by which one multiplies the
mass to find what I call the weight.

I think Leigh is too conservative here: g is the quantity by which
one multiplies the mass to find what *everyone* generally calls
the weight. It's far simpler and more practical this way, but
more importantly, it's the *only* way to insure that g is well
defined and to insure consistency. As he notes you can not
determine which part is "truly gravitational" and which part is
"merely due to noninertial effects" anyway. Indeed, the question
itself is meaningless from a practical point of view even within
the context of Newtonian mechanics.

John Mallinckrodt mailto:ajm@csupomona.edu
Cal Poly Pomona http://www.csupomona.edu/~ajm