[net.physics] CG is <not> CM

kcarroll (08/10/82)

   Aha! I think I see! Center-of-mass is <not> the same as
center-of-gravity!  I agree that the path that the CM of a
supernova follows is the same as the path followed by the pre-nova
star (neglecting friction effects with surrounding gas-clouds, and
electro-magnetic interaction with the rest of the galaxy; let's
just say, <by Newtonian physics>). However, unless the
expanding gas-cloud has a spherically symmetric density-distribution,
the CG of the cloud does <not> follow that path. In fact, it seems
that "center of gravity" is a nearly meaningless fiction,
except in the case of spherically-symmetric bodies. 
   Consider 2 bodies, A and B. A is a point-mass, while B has
some volume. Bring A and B close together; they will be attracted
towards each other, by gravitational force. Now, the force on A
will be the integral over all the mass in B, of the force on A
due to a differential portion of B's mass. The direction of the
force acting on A will be the direction of the vector sum of all 
the differential forces on A due to infinitesimal portions of
B's mass. Supposedly, this direction is the direction from A to the
center of gravity of B. Right?
   Now consider the two cases shown below:

CASE 1:                        B

           A


CASE 2:    B1                                      B2

           A

In case 1 B is a single, compact body. The direction of the
gravitational force on A due to B points pretty well directly
towards B. In case 2, B has been split up into 2 (say equal) masses,
B1 and B2. The center of mass of this system is the same as for
that  of case 1. However, considering that the force of gravity
varies with the inverse square of the distance between the interacting
bodies, the force on A due to B1 (pointing in the direction of B1)
will be much stronger than the force on A due to B2 (towards B2). Hence,
the net force on A due to (B1 and B2) will point almost towards
B1. The apparent CG of B has moved to the left. Note that if A
had originally been placed over to the right, then the CG would have
appeared to have shifted to the right, instead. CG is thus a function
of the position of the observer; it doesn't seem to be the "center"
of the B system at all, does it?
(thanks Wedekind.ES@PARC-MAX, and others, for clearing this up)


Kieran A. Carroll
...decvax!utzoo!kcarroll

G:wing (08/12/82)

I better hope that CG is not fake.  That's how pilots balance their aircraft!

rhm (08/12/82)

I would like to apologize for having used "center of gravity" instead of
"center of mass" in a moment of frenzied typing. utzoo!kcarroll is quite
correct in pointing out that CG is essentially meaningless.

G:wing (08/12/82)

As a followup to my posting, on Earth we must use CG because of gravity.

thomas (08/12/82)

G:wing's articles indicate a common misconception about Center of Gravity
vs. Center of Mass.  (Seems to parallel the weight vs mass problem.)
What we commonly call "Center of Gravity" is REALLY the center of mass.
"Center of Gravity" refers to the point in space towards which an object is 
attracted by the gravitational attraction of the object under consideration.
As an earlier article pointed out, if the attracting object is not spherical,
the "Center of Gravity" can depend on the position of the attracted object.

=Spencer

G:wing (08/13/82)

Do take notice that on Earth it is a LITTLE more difficult to find the true
center of mass than the apparent center of gravity...