Re: shell theorem
- From: Dave Typinski <nospam@xxxxxxxxxx>
- Date: Mon, 10 Nov 2008 22:42:48 -0500
"Androcles" <Headmaster@xxxxxxxxxxxxxxxx> wrote:
"winston71" <donkey71@xxxxxxxxxxxxx> wrote in message
news:82688d95-e246-47b2-b5c1-13e6a31708c0@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Nov 11, 1:48 am, eric gisse <jowr.pi.nos...@xxxxxxxxx> wrote:
On Mon, 10 Nov 2008 15:30:03 -0800 (PST), winston71
<donke...@xxxxxxxxxxxxx> wrote:
Hello,
From wikipedia "1. A spherically symmetric body affects external
objects gravitationally as though all of its mass were concentrated at
a point at its center."
I don't understand the integration. I tried to make a computer
simulation (million random points represent a sphere, there is a test
point outside the sphere, calculate and add each gravity force
vector), but the results came out wrong; total force is minimum when
sphere radius touches test point, and gets stronger as sphere radius
gets smaller (number of sphere points is constant and test point
coordinates is constant). Did anyone try such a simulation and got
good results? Thanks.
Divergence theorem.
Look it up.
Sorry I don't understand that either. At least can you clarify for me
what "spherically symmetric" means?
==========================================
All it means is that for every point within the sphere there is
a corresponding point diametically opposite - a mirror image.
==========================================
Would a mass system of 6 points
which are at the center of a cube's faces be spherically symmetric?
==========================================
Yes... and so would the eight vertices (corners) of the cube.
A little qualification with that is in order, I think. Spherical
symmetry is not the only requirement. The distribution also has to be
continuous.
==========================================<snip>
I could add just 6 vectors and see where my simulation went wrong..
==========================================
You could.
He could add them, but it still woudn't come out right. Those six or
eight points would be common to a sphere, but they would not act as a
point mass.
For example, if one does the acceleration vector addition for six unit
masses lying at points +/- 1 on all three axes relative to a point at,
say, (2,3,3) in space, the vector sum is not the same as what you get
if all six points were at the origin.
Trivial proof: inside a spherically symmetric shell, the net force of
gravity is zero. One could easily pick a point in space, say (0.8,
0.8, 0.8). That's well inside the "shell" described by the six
points, yet it's obvious that those six points would produce a net
acceleration at (0.8, 0.8, 0.8) since they all lie on the origin side
of that point.
Which is all by way of saying that the derivation commonly used to
show that the force due to gravity of a spherical distribution is the
same as the force generated were it were a point mass at the center of
the distribution assumes that the distribution is continuous.
So, yes, those six points are a spherically symmetric distribution,
but they are not a spherically symmetric continuous distribution.
Therefore, they're not gravitationally equivalent to a point mass at
the center of the distribution.
--
Dave
.
- References:
- shell theorem
- From: winston71
- Re: shell theorem
- From: eric gisse
- Re: shell theorem
- From: winston71
- Re: shell theorem
- From: Androcles
- shell theorem
- Prev by Date: Re: shell theorem
- Next by Date: Re: Physicists of today should be called abstractists instead.
- Previous by thread: Re: shell theorem
- Next by thread: Re: shell theorem
- Index(es):
Relevant Pages
|
