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Re: [Phys-l] When is g = GM/r^2?



Hi all-
Good point! However the statement is true no matter where you put your origin of coordinates. It's jsut that the relevant coordinate is the earth-moon distance. M_earth + M_moon is new to me, too.
Don't forget to set up the companion exquation for the non-accelertion of the center-of-mass.
Regards,
Jack

On Sun, 26 Nov 2006, John Mallinckrodt wrote:

Note to all: I would be particularly interested in critiques of the
following as it represents a recent evolution of my own thinking.

The current thread has stimulated me to do a lot of thinking about
how to construct (and then, hopefully, begin to teach) a world view
that is fully compatible with all of the results of nonrelativistic
physics (in which space is Euclidean and time is absolute) AND the
principle of equivalence in which g is frame dependent and determined
operationally by the local acceleration of a freely falling object.
I think I have managed to do this.

The final piece of the puzzle for me had to do with the problem of
how to sensibly reconcile the Newtonian formula for g -- i.e., sum
over i of -GMi/ri^2 ri_hat -- with the gravitational field
determined operationally by observing a freely falling object. The
answer is motivated by the following curious observation:

The acceleration of the Moon as determined in a nonrotating
reference frame attached to the Earth is approximately
G(M_earth+M_moon)/r_earth-to-moon^2, toward the Earth.

That's right, M_earth PLUS M_moon. It's easy to check for yourself
that this is true, but I wonder how many have noticed this fact
before. I hadn't. (BTW, the formula is approximate only because I
have ignored the influence of other bodies like the Sun, which, in
fact, turn out to have a fairly significant effect.)

Thus, in the Earth-based reference frame, the gravitational field
vector at the position of the Moon is in the same direction as but
larger than the value (apparently) predicted by the Newtonian
formula. This fact can be understood by noting that, in the center
of mass frame, the Earth has an acceleration G*M_moon/r_earth-to-
moon^2 toward the Moon. So the added term can be understood as the
ordinary correction due to the acceleration of the Earth-based
reference frame.

Accordingly, I propose the following conceptually simple reconciliation:

In the center of mass frame of a set of isolated* point
particles, the gravitational field vector at any position
p is given by

g_CM = G * sum over i (Rpi_hat * Mi/Rpi^2) .

where Mi is the mass of the ith particle,
Rpi is the distance from position p to particle i, and
Rpi_hat is a unit vector pointing toward particle i from
position p.

To find the gravitational field at the same position in any
other frame, one must, as always, correct by subtracting the
acceleration of that frame relative to the center of mass
frame.

In particular, the gravitational field at the same position in
a nonrotating reference frame attached to the jth particle is

g_j = g_CM -
G* sum over all i except j (Rji_hat Mi/Rji^2) .

where Rji is the distance from particle j to particle i and
Rji_hat is a unit vector pointing toward particle i from
particle j.

* Note: Practically speaking, the condition of "isolation"
requires that no particle and no position of interest be
close enough to any other particles massive enough that
including them in the calculations above significantly
affects the results.

All of this is simply to say that, to determine g at any position P
from a frame attached to any body B, we just use the standard
Newtonian formula and then correct it by subtracting the acceleration
of B that is induced by the other relevant massive bodies.

John Mallinckrodt

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

and

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