Re: Gravity-
- From: Khattak <zarmewa@xxxxxxxxx>
- Date: Sat, 13 Jun 2009 13:32:17 -0700 (PDT)
On Jun 13, 9:32 am, "tadc...@xxxxxxxxxxx" <tadc...@xxxxxxxxxxx> wrote:
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#Instabilityhttp://www.alcyone.com/max/writing/essays/why-niven-rings-are-unstabl...
Tom Davidson
Richmond, VA- Hide quoted text -
- Show quoted text -
You are not communicating clearly, perhaps not even with yourself.
You might have gone through in hasty but it’s quit lucidly mentioned.
By the way this was my question that there is difference between “the
force on the object” and “the force between the objects” Anyway
symmetric arguments (vector addition) is used in shell theorem instead
of measuring center to center distance between two masses as in law of
gravity while in case of earth on earth, both the mathematical model
and the symmetric arguments (vector addition concept) of gravities
show different results. Further I never mentioned any test mass
between two earths.
ALL of the mathematicians and physicists of the past several centuries
have madfe and overlooked an identical error.
This is your thinking not theirs and fair criticism is my right.
The net force on the point mass at the center of shell is zero but it
will still be there forever because of equal forces from shell on its
all sides so F is not zero between the two masses.
Furthermore,
Since g=GM/R^2 and doesn’t even depend upon the test or falling mass
therefore atoms and molecule of AIR have also masses, so why do they
not fall on the earth, e.g a blimp of helium. Newton didn’t mention
any temperature or pressure in his gravity equation.
.
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