Re: Free fall

Luigi Fortunati il 24/03/2024 07:29:00 ha scritto:

[[Mod. note -- Perhaps.
>
That is, let's call your 1-gram body "A", and your 2-gram body "B".
We can think of B as a pair of one-gram halves (call them "B1" and "B2")
glued together.  The question is, does the presence of B1 change the
electromagnetic (EM) field at B2's location, or vice versa, by an
amount large enough that we need to care about it?  If *not*,
then the EM force acting on B will be the sum of
(a) the EM force acting on B1 alone (i.e., if B2 were NOT there), and
(b) the EM force acting on B2 alone (i.e., if B1 were NOT there).
Assuming that B is small enough that the EM field doesn't vary
significantly across B's diameter, we should have (a) = (b), so in
this case the EM force acting on B should be twice the EM force
acting on A.
>
But, if the presence of B1 *does* change the EM field at B2's location
by a significant amount, then the EM force acting on B will *not*
equal the sum of (a) and (b) above, and the EM force acting on B will
*not* be twice the EM force acting on A.

This thing you say also applies to the gravitational field: if the presence of B1 changes the gravitational field at the position of B2 by a significant amount, then the gravitational force acting on B is not equal to the sum of (a) and (b ) and the gravitational force acting on B is not double the gravitational force acting on A.

The proportionality between force (whether gravitational or EM) and mass depends on each individual particle because it is precisely the particle that intercepts its share of the field force (whether gravitational or EM).

Obviously (and to avoid any misunderstanding) the comparison must be made between equal materials and, that is, if the material of the first body is XYZ, that of the second body is also XYZ (if the material of the first is iron, that of the second is also iron , if one is wood the other is also wood).

Luigi Fortunati