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Re: [Phys-l] Weightless



Joel, responding to me, wrote:

I understand that lots of people (including apparently the folks at NIST) want weight to be the same thing as gravitational force, but I have yet to hear one compelling reason for that.

That's not how I read the following:

"According to NIST (http://physics.nist.gov/Pubs/SP811/sec08.html) - the closest we have in the US to an "official definition":

"In science and technology, the weight of a body in a particular reference frame is defined as the force that gives the body an acceleration equal to the local acceleration of free fall in that reference frame . . ."

Doesn't this say that the weight of an astronaught is zero in the rest frame of the orbiting shuttle, since the local free-fall acceleration is zero in that frame. Therefore the force to give it that acceleration is zero, i.e. weightless??

Yes, it does say that, but it also makes clear that, under this definition, weight is a) a vector and b) frame-dependent. It sounds to me like they are saying weight = gravitational force, but maybe they have a different definition of gravitational force.

In any event, I believe that NIST would say, "the astronauts are weightless in their own reference frame," while I would prefer to say, "the astronauts are weightless."

John Mallinckrodt

Professor of Physics, Cal Poly Pomona
<http://www.csupomona.edu/~ajm>

and

Lead Guitarist, Out-Laws of Physics
<http://outlawsofphysics.com>