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Re: [Phys-l] Weightless

If I define weight as the reading on a scale, as I do, then there is no ambiguity. One might confuse oneself with explanations, but the definition is unambiguous.

The scale reading in an orbiting satellite is zero. The astronaut is therefore weightless by definition.

Now make a free-body diagram of the orbiting vehicle. There is a single force acing on the vehicle, graviyy, pointed toward the earth's center. Show this force with a red-arrow. The vehicle is accelerating in the direction of the force. Show this acceleration with a green arrow, pointed in the direction of the red arrow, but of an arbitrarily different length.

Recognize that your free-body diagram is the same as any free-boady diagram that you would draw for an object in free fall.

The astronaut is in free fall, therefore the scale-reading is zero, and the astronaut is weightless.
Regards,
Jack


On Mon, 20 Nov 2006, Anthony Lapinski wrote:

True weight = w = mg (acting downward toward Earth's center.

Saying that scales read your "weight" has ambiguity (in certain
situations). Again, take one up in the orbiting shuttle. All scales read
zero. Then if the astronauts have no weight, what force keeps them
orbiting the Earth? Very confusing.

But if the scale measures the apparent weight (normal force, or support
force), then the orbiting astronauts are "apparently" weightless. There is
much gravity acting on them (around 8.7 m/s2, compared to 9.8 on the
Earth's surface). Many kids think g = 0 up there! And it is this force
(weight = mg, where g = 8.7) which provides the centripetal force
necessary for the shuttle (and astronauts) to orbit. Less confusing.

Forum for Physics Educators <phys-l@carnot.physics.buffalo.edu> writes:
But I don't know what you mean by "true weight". I am strictly an
operational physicist. "Weight" is the reading on a scale. That's a
definition that has no, as far as I can see, ambiguity. It may require
extrapolation to the very large and very small, but there is a regime
where it can be straightforwardly applied.
Regards,
Jack


On Mon, 20 Nov 2006, Anthony Lapinski wrote:

People in their everyday experience think a bathroom scale measures
their
downward weight. And so when the scale reads zero for the astronauts,
they
think g = 0! Lots of confusion and misconceptions with this with ALL
people.

I guess one could never have a true weight equal to zero unless you are
"very far" from all objects (so that g = 0).

We as teachers will never reach any consensus on this topic. What works
for one might not work for someone else. To me, bathroom scales are
common
devices, and very useful for illustrating ideas about free body diagrams
in elevators, N3L for objects colliding, and "apparent" weightlessness.

Forum for Physics Educators <phys-l@carnot.physics.buffalo.edu> writes:
But if an astronaut isn't "weightless," then what is? Isn't the word
of absolutely no use if we can't apply it to astronauts? Weight
implies "heaviness" and heaviness implies (to me) a tendency to stay
"pinned by gravity" to some surface. If a person (or thing?) finds
that she (or it?) has absolutely no such tendency, then it seems to
me that she has every right to say that she is "weightless" and that
we ought to agree with and abide by her assessment.

There's certainly no problem justifying any of this from a Newtonian
viewpoint. It's simply a matter of deciding what you want to mean by
"weight." I don't see any good reason to require that it mean the
same thing as "gravitational force." Indeed, it seems to me a
genuine disadvantage.

Anthony Lapinski wrote:

I agree. This is why I discuss bathroom scales and apparent weight. If
someone says there is gravity acting on the astronauts AND they are
also
weightless, it ads to more confusion...

[Rick Tarara wrote:]
In the 'g' thread, we've (yet again) opened up the definition of
weight
debate. While I see certain pedagogical advantages to either of
the two
main approaches, I would ask if the proponents of saying that one IS
weightless while in the space station can explain that from a
Newtonian
viewpoint? Seems to me that there is only one force acting on the
person.
If we call that the gravitational force due to the earth, then what
(again
in the Newtonian viewpoint) is weight? Isn't this gravitational
force a
component of your weight? But it is the only component here in the
Newtonian view and is not zero!

John Mallinckrodt

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

and

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



_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l


_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l


--
"Trust me. I have a lot of experience at this."
General Custer's unremembered message to his men,
just before leading them into the Little Big Horn Valley



_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l


_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l


--
"Trust me. I have a lot of experience at this."
General Custer's unremembered message to his men,
just before leading them into the Little Big Horn Valley