From: utzoo!kcarroll
Newsgroups: net.physics
Title: CG is  CM 
Article-I.D.: utzoo.2356
Posted: Tue Aug 10 11:40:36 1982
Received: Tue Aug 10 11:40:36 1982

   Aha! I think I see! Center-of-mass is  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, ). However, unless the
expanding gas-cloud has a spherically symmetric density-distribution,
the CG of the cloud does  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