Re: [Phys-L] apparent weight

[John Denker <jsd@av8n.com>]

On 12/11/2014 10:09 AM, Jeffrey Schnick wrote:
> The person's acceleration is greater than that of the center of the 
> earth at midnight when the person is farther from the sun, and 
> smaller than that of the center of the earth at noon when the person 
> is closer to the sun.

To be charitable, let's pretend that is a statement about
|g| i.e. the scalar magnitude of the acceleration.  As such,
it's more-or-less true, but still not the answer to the 
question about weight.

Weight is defined, operationally, as the indication of an
ideal force-gauge, measuring the force necessary to hold
an otherwise-free object stationary in the chosen reference
frame.  This is completely frame-dependent.  Normally we
choose a frame comoving with the earth. Occasionally we
choose one comoving with the physics department elevator,
or comoving with an aerobatic aircraft, which changes the 
answer temporarily, but I assume that wasn't the point 
of the original question.  Holding the object stationary 
with respect to the sun would be perverse, completely 
inappropriate to the question that was asked.

To repeat:  It is traditional and sensible to weigh
yourself by setting the scale on a /horizontal/ part
of the surface of the /earth/ and then standing on 
the scale.

Therefore what matters is vertical component of the
vector /difference/ between your free-fall acceleration
and the earth's free-fall acceleration.

  Because we are taking the difference, we now have 
  something that is frame-independent.  By choosing a
  uniformly accelerated frame, in accordance with the
  equivalence principle, you can change the acceleration
  of both things (the earth and the test object), but
  you cannot change the difference between the two.

Also, it matters a great deal that the acceleration
g is a vector.  That's because the definition of
"vertical" changes from noon to midnight.
 -- At noon, your acceleration is more sunward i.e.
  more upward, relative to the earth.
 -- At midnight, your acceleration is less sunward
  i.e. once again *more* upward.


> You can check my work here <http://youtu.be/l6RBVHGyRec> and here
> <http://www.anselm.edu/internet/physics/phys-l/apparentWeight.pdf>.

I didn't watch the video, but the .pdf contains numerous
errors.

Starting with the first line:  the term r0 ω0 should be
(r0 + Δr) ω0.  It goes downhill from there.

Also the term involving ωe is not wrong, but it's silly,
because the vertical component thereof is constant over
time and cannot affect the answer.