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Re: Weight



I see nothing odd. In both cases, I would measure the weight as the
gravitational attraction of the earth on the object. The _apparent_ weight
at the equator will be slightly less due to the earth's rotation and in
orbit the _apparent_ weight will be zero due to the free fall condition,
but in each case the 'weight' would be GMeMo/R^2 where Me is the mass of
the earth and Mo is the mass of the object. If you want a spring scale to
accurately measure 'weight' then you need to define weight as the apparent
weight, not the gravitational attraction. The vast majority of
introductory texts choose the gravitational attraction--which of course
doesn't make it gospel, but near enough for me.

Rick
----------
From: James Mclean <jmclean@chem.ucsd.edu>

It seems odd to me that you suggest using "weight" in a way that agrees
with every-day experience on Earth, but that once you are up on the space
shuttle you want to ignore simple experience.

I think the answer lies in the last sentence of your first paragraph. On
earth, the apparent centrifugal force is negligible; nonetheless, it
technically should clearly be included in what we experience everyday as
"weight". If is is also included on the space shuttle, then we find the
astronauts to be "weightless."