Re: Weight and reference frame

["John S. Denker" <jsd@MONMOUTH.COM>]

At 12:35 PM 9/25/01 -0500, Rick Tarara wrote in part:
>  the 'weight is the gravitational force due to the nearest planet'
> camp doesn't agree that the astronauts are weightless.

Is anybody really arguing for that point of view?  The counter-arguments are:
  1) It doesn't meet the criterion Ken Fox suggested ("I like to
communicate with my students").  The students think the astronauts are
weightless.  The astronauts think so, too.
  2) It doesn't comport with modern physics, specifically the principle
that gravitation at a given point is equivalent to an accelerated reference
frame.
  3) It is spectacularly pathological if you are midway between two planets.


> If ALL books refrained from identifying the gravitational force
> on an object (mg) as weight

All books?  ALL books?!!  If we wait until ALL books are perfect, we'll
never accomplish anything.

Even more importantly: identifying weight=mg is not a problem at all.  It
would become a problem if we foolishly insisted that g be solely due to the
nearest planet.  That is, it would be problematic to identify
weight=GMm/r^2, for reasons enumerated above.

The recommended modern-physics approach is to define g operationally as the
acceleration of a freely-falling object relative to the chosen reference
frame.  For astronauts, g=0.  In the lab, |g|=9.8 m/s^2 or thereabouts.
  http://cires.colorado.edu/~bilham/Absgdata.html

The tabulated data shows that taking g=GM/r^2 is a reasonable
approximation, but it isn't exact, and it certainly isn't a definition.

Extra credit:  Note that the tabulated values are given to ten sig digs,
but there is considerable variability in the 3rd digit.  Does this
variability look random, or can you spot a trend?