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


-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu
[mailto:phys-l-bounces@carnot.physics.buffalo.edu] On Behalf
Of Jeffrey Schnick
Sent: Monday, November 27, 2006 10:10 AM
To: Forum for Physics Educators
Subject: Re: [Phys-l] When is g = GM/r^2?



-----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 I
have 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.

This factor of 2.17 and the concavity of the moon's orbit
relative to the sun are interesting. Thanks for pointing
them out. But, I think that if you did the problem as a
three body problem in the center of mass of the sun frame and
then subtracted the earth's acceleration in that frame from
the moon's acceleration in that frame to get the acceleration
of the moon relative to the earth, the only difference from
John M.'s solution would be due to the non-uniform nature of
the sun's gravitational field in the region of the earth-moon
system. These tidal effects are small compared to the
acceleration of either body relative to the sun. I think
that John M. accurately characterized them when he said that
they "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.

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?


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, 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?

If I understand John's hybrid model correctly, he has chosen
a reference frame in which the center of mass is actually not
accelerating. I think that if you put an ideal accelerometer
at rest at the location of the center of mass, (a) it would
stay there, and

I don't really think it would stay there unless in were
massless/unphysical.

(b) it would read zero.
Rather than make a correction, in order to obtain greater
precision in John's model, e.g. for the earth/moon problem,
one should just include more bodies in both the determination
of the center of mass and in the sums he described.


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
_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l


Jeff Schnick
_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l