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-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu
[mailto:phys-l-bounces@carnot.physics.buffalo.edu] On Behalf
Of David Bowman
Sent: Monday, November 27, 2006 9:34 AM
To: Forum for Physics Educators
Subject: Re: [Phys-l] When is g = GM/r^2?
Regarding John M.'s reconciliation of Newtonian gravitation
theory with the Equivalence Principle:
(BTW, the formula is approximate only because Ihave ignored
the influence of other bodies like the Sun, which, in fact,turn out to
have a fairly significant effect.)
That may be somewhat of an understatement in that the
acceleration of the moon (along with the earth) toward the
sun in a frame in which the Sun is at rest is about 2.17
times *greater* than the acceleration of the moon relative to
a frame in which the earth is at rest. In fact the orbit of
the moon is concave toward the sun at all points of its
orbit, regardless of lunar phase, as seen in a frame in which
the Sun is at rest.
Thus, in the Earth-based reference frame, the gravitational fieldNewtonian formula.
vector at the position of the Moon is in the same direction as but
larger than the value (apparently) predicted by the
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.
In light of my observation above, how about saying that the
condition of 'isolation' requires that the magnitude of the
*change* in the external gravitational field (produced by all
ignored bodies outside the collection of interest for the
system) across the relevant spatial extent of the motions of
the bodies in their mutual CM-at-rest frame, be an
insignificant fraction of the magnitude of the internal
gravitational field produced by the constituent particles of
that system?
standard Newtonian
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
formula and then correct it by subtracting the accelerationof B that
is induced by the other relevant massive bodies.
John, what is your opinion on also making a correction for
the acceleration of the CM of the system relative to a frame
that includes the other ignored bodies in the calculation of
the CM which is to be taken as effectively not accelerating?
John Mallinckrodt
Professor of Physics, Cal Poly Pomona
<http://www.csupomona.edu/~ajm <http://www.csupomona.edu/~ajm> >
and
Lead Guitarist, Out-Laws of Physics
<http://outlawsofphysics.com <http://outlawsofphysics.com> >
David Bowman
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