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Re: [Phys-l] definition of gravity

If you choose to use a rotating reference frame (as we do on the surface of the earth) then we include the v^2/r term in our 'definition' of g.

It affects the direction and the magnitude.

.
At 7:16 PM -0500 11/7/11, Robert Cohen wrote:
I've thought about JD's response below and am a little confused. So it
is *incorrect* to say that GM/r^2 is equal to the strength of the
gravitational field (of the object of mass M at the distance r from its
center)?

Robert A. Cohen, Department of Physics, East Stroudsburg University
570.422.3428 rcohen@esu.edu http://www.esu.edu/~bbq

-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu
[mailto:phys-l-bounces@carnot.physics.buffalo.edu] On Behalf Of John
Denker
Sent: Thursday, November 03, 2011 4:55 PM
To: Forum for Physics Educators
Subject: [Phys-l] definition of gravity

On 11/03/2011 01:10 PM, Robert Cohen wrote:
When introducing F=mg, is g the gravitational field or is it the local

"acceleration of gravity" (acceleration of an object in the local
frame when the only force acting is gravity)? In other words, does mg

include the centrifugal force associated with the rotating frame of
reference? Might this impact how you introduce the material?

That's an interesting, important question.

... when the only force acting is gravity ...

I'm pretty sure what those words were intended to mean, namely that
"gravity" was intended to be calculated in accordance with the law of
universal gravitation:

"gravity" = G M / r^2 [1]

However, whether or not I'm right about that, I wish to use a different
definition. I recommend *defining* "gravity" to be the acceleration of
the chosen reference frame (as measured by reference to freely-falling
objects).

This is pretty much required for consistency with a modern (post-1915)
understanding of what gravity is, including the equivalence principle.
It is also required for common-sense practical applications such as
architecture.

Let's be clear: The thing that we calculate using equation [1] must not
be considered "the" gravity (except in a few special cases). It is
often the dominant contribution to the gravity, but the other
contributions are quite significant in ordinary real-world applications.

I write:

g_I = G M / r^2 [2]

where the LHS is emphatically not g but rather g_I, which is only one
contribution to g.

For the next level of detail on this, see
http://www.av8n.com/physics/weight.htm

gravity one way in the chapter on experiments in the lab frame
(implicitly including all contributions to the frame-
acceleration) and define it another way in the chapter on cosmology
(including only g_I).

It may be that g is numerically equal to g_I for cosmology, if/when we
choose to use a nonrotating reference frame ...
but that's an equation, not a definition. It's a choice, not a law of
nature. For 99% of the practical applications students (and other
folks) see in real life, g is only roughly approximated by g_I.

==================

To summarize:

When introducing F=mg, is g the gravitational field or is it the
local "acceleration of gravity" (acceleration of an object in the
local frame when the only force acting is gravity)?

Yes, both. Those are the same thing, according to the
recommended definitions.

In other words,
does mg include the centrifugal force associated with the rotating
frame of reference? Might this impact how you introduce the
material?

The centrifugal force is *included* in the definition of g,
and in the definition of gravity.
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