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Re: [Phys-L] apparent weight

The centripetal acceleration of a person at the equator due only to the Earth's rotation is about 3.4 cm/s^2 while the acceleration of the Earth in its orbit about the Sun is about 0.6 cm/s^2. So the "classical acceleration" that you refer to of the person in the cartoon picture is ~2.8 cm/s^2 AWAY from the Sun at noon and 4.0 cm/s^2 TOWARD the Sun at midnight. None of that, of course, bears on the question of differences in weight due to tidal effects.

On Dec 11, 2014, at 12:20 PM, Jeffrey Schnick wrote:

With respect to a non-rotating reference frame in which the center of mass of the sun is fixed at the origin, a person standing on a scale at the equator has a classical acceleration that is roughly directed toward the sun both at noon and at midnight--it would be straight at the sun if the equatorial plane were coplanar with the orbital plane. That acceleration is not g. It is certainly not weight, and it certainly is not an answer to the question about weight. John Clement said that the magnitude of that acceleration is greater at noon than it is at midnight. I said it is just the other way round.

-----Original Message-----
From: Phys-l [mailto:phys-l-bounces@www.phys-l.org] On Behalf Of John
Denker
Sent: Thursday, December 11, 2014 1:02 PM
To: Phys-L@Phys-L.org
Subject: Re: [Phys-L] apparent weight

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.
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