Re: Gravity-

On Jun 13, 10:49 am, Khattak <zarm...@xxxxxxxxx> wrote:

<snip repost>

From the link
Under the sub topic “Inside a shell”
“An interesting result occurs when we consider the case in which r <
R, i.e. the point mass m is within the shell. By using symmetry
arguments, it is easy to see that if the point mass is at the exact
center of the shell the force will be zero, but is this true for all
locations within the shell?”

Yes. That is one of the conclusions of the shell theorem. Locations
*outside* the shell will still manifest a gravitational force
indistinguishable from that of an equivale mass located at the center
point of the shell.

Given an earth with a spherically symmetric mass distribution, by
applying both consequences of the shell theorem, we can show that a
point mass m inside the earth at a distance r from the center of the
earth experiences gravitational attraction *only* from that portion of
the earthwithin the radius r. The portion of the earth located at
distances greater than r from the center constitute a hollow
spherically symmetric shell with inner radius r and outer radius R
that exerts no net gravitational influence within its own cavity.

Remember the words “by using symmetry arguments” which means simple
vector forces; this means this law does not obey the simple rule of
vector addition.

That does not follow from your premise. Simple vector forces obey the
law of simpel vector addition whatever the symmetry, if any.

The same vector gravity forces cancel each other if we put an
imaginary earth on earth. So how would apply the law of gravitation in
this case (earth on earth or sun on sun). Here is the detail
Newton derived his law of gravity i.e. F=GMm/R^2 by putting one kg of
sphere on the surface of earth. Now consider if the same sphere is
bigger and bigger till it reach the size of earth. OR simply imagine
earth on earth

Now apply F= GMm/R2 where g = GM/d2. M= m= mass of earth = mass of
imaginary earth, centre to centre distance between two masses =
diameter of earth. So F=GM^2/d^2 which the total mathematical force
modeled by Newton.

But by using symmetry arguments” the vector gravity forces cancel each
other and therefore there is no such net force exists between them.

This means that weight of mass of 1 kg of sphere is start decreasing
by increasing its size on the surface of ground and becomes zero when
it reaches the size of earth.

You are not communicating clearly, perhaps not even with yourself.
Are you referring to the force on one earth upon another, or to the
net force on a test mass located symmetrically between the two?

Therefore the base of whole concepts of Newton law of mathematical
model is wrong even it does not satisfy the rule of vector addition.

Check your math. Better mathematicians than you have validated it
repeatedly. I consider it far more likely that YOU have made an error
than that ALL of the mathematicians and physicists of the past several
centuries have madfe and overlooked an identical error.

The extraneous link talks about
1-      Spherical shell and the point mass outside and inside shell
2-      The force on the point mass at its center

Remember there is difference between “the force on the object” and
“the force between the objects”. Yes, the net force on the point mass
is zero at center of the shell but what will be the force between two
objects or aforementioned rings as per law of gravity?

Treat one of the rings as test mass and calculate the force exerted
upon it by the other ring. The shell theorem applies in the two-
dimensional case as well as the three dimensinoal case.

The planar, symmetric ring is incapable of having a net gravitational
force upon any object wholly within its cavity and residing in the
same plane.

As a corollary, the confined object cannot exert a gravitational
influence upon the ring. The situation was recognized in analyses of
the instability of Larry Niven's Ringworld:
http://en.wikipedia.org/wiki/Ringworld#Instability
http://www.alcyone.com/max/writing/essays/why-niven-rings-are-unstable.html

Tom Davidson
Richmond, VA
.

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