Re: Electo London Gravity ?


"sue jahn" <susysewnshow@xxxxxxxxxxxx> wrote in message
news:42d15b5a$0$18642$14726298@xxxxxxxxxxxxxxxxxx
|
| "Significant Zero" <paulpsremove@xxxxxxxxxx> wrote in message
news:1121013880.18182.1@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
| >
| snip
| > | > | > > | > | > | To your point. We should have no ambiguity with the
| > | > far-field
| > | > | > > effects.
| > | > | > > | > | > | Might we clear up the matter with the statements?::
| > | > | > > | > | > | Gravitational force diminishes by 1/r^2 above the
earths
| > | > surface.
| > | > | > > | > | > | Gravitational force diminishes by 1/r^1 below the
earths
| > | > surface.
| > | > | > > | > | >
| > | > | > > | > | > I thought that below the surface it fell to zero at
the
| > center
| > | > of
| > | > | > > the
| > | > | > > | > | > partical at the center of the earth but still followed
a
| > 1/r^2
| > | > as
| > | > | > > far as
| > | > | > > | > the
| > | > | > > | > | > feet of a person in a lift was concerned ?
| > | > | > > | > |
| > | > | > > | > | That is what I get for trying to speak a foreign
language.
| > | > | > > | > |
| > | > | > > | > | Graviational force is maximum at the earth's surface
because
| > | > | > > | > | 100% of it's mass is concentrated below your toes..
| > | > | > > | > |
| > | > | > > | > | Gravitational force is zero at the earth's center
because
| > any
| > | > | > > | > | plane cut through a test mass has 50% of the earths mass
| > | > | > > | > | on either side.
| > | > | > > | > |
| > | > | > > | > | Force is directly proportional to the distance from
center.
| > | > | > > | > | Is 1/r or 1/r^1 the way to say that ? My spelling is
much
| > | > better
| > | > | > > | > | than my math. :o)
| > | > | > > | >
| > | > | > > | > It is claimed at
| > | > | > > http://hyperphysics.phy-astr.gsu.edu/hbase/forces/isq.html
| > | > | > > | > that Gravity source strength equals 4 pi GM = I_s and the
| > surface
| > | > | > > intensity
| > | > | > > | > is 4 pi GM/4 pi r^2 so 4 pi GM/4 pi r^2 = I_e = GM/r^2 =
g so
| > g =
| > | > | > > 9.80665 m
| > | > | > > | > s ^-2 = I_e at r_e which I hope means the same as m s-2
with ^
| > | > meaning a
| > | > | > > | > superscript exponent to you and others and in long hand
meters
| > per
| > | > | > > second
| > | > | > > | > per second ? So twice the center to surface radius from
the
| > center
| > | > is
| > | > | > > 2r_e
| > | > | > > | > and the intensity there is I_2r_e = g_e/4 and three times
the
| > | > center to
| > | > | > > | > surface radius is 3r_e so at 3r_e we have I_3r_e = g_e/9
| > | > | > > | >
| > | > | > > | > So the gravitational field intensity at
| > | > | > > | > 0r = GM ? so this should be zero intensity and zero mass
but
| > for
| > | > | > > calculating g you
| > | > | > >
| > | > | > > So could we agree that (phy-astr.gsu.edu) information on the
| > source
| > | > | > > intensity being 4 pi GM at the centre of the earth as being
| > | > misleading?
| > | > | >
| > | > | > Yes... I'll agree that it is oversimplified.
| > | > | >
| > | > | > > and
| > | > | > > that it is in fact zero at the dynamic centre
| > | > | >
| > | > | > It is dynamic, whether induced or atomic domains.
| > | > | >
| > | > | > > but may have a positive value
| > | > | > > at the average or mathmatical centre as this would be a more
| > useful
| > | > value in
| > | > | > > some cases?
| > | > | >
| > | > | > By average or mathmatical centre I think you mead geometric
centre.
| > | > | >
| > | > | > For true spherical bodies:
| > | > | > The earth-moon barycentre does not coincide with the earth's
| > | > | > geometric center. If you were at the earth's geometric center
| > | > | > you would fall toward the moon.
| > | > | >
| > | > | >
| > | > | > > Perhaps if the average centre of mass of an isotropic field
| > generator
| > | > was
| > | > | > > expressed something as 1/0r+1 = GM or something perhaps.
| > | > | > > | > 1/2r = GM/1/2r^2 ?
| > | > | >
| > | > | > What seems to be missing there is R and r
| > | > | > See:
| > | > | > 1/2 (R^2 - r^2) = 1/2 R^2 (1 - (r/R)2^)
| > | > | >
| > | > | > Gotta break here 'cause I see homework below
| > | > | > =======
| > | > | >
| > | > | > Sue...
| > | > | >
| > | > | > >
| > | > | > > This should be intensity for1/2r = g/x with x being some
planet
| > | > density
| > | > | > > factor that
| > | > | > > for earth is about 1 and is about 2 for a small hydrogen cloud
of
| > | > about
| > | > | > > earth
| > | > | > > volume with g taking the appropriate value. ?
| > | > | > Sheeshs? I haven't a clue. <O)
| > | > | > >
| > | > | > > | > 1r = g = 9.80665 ms^-2 = GM/r^2
| > | > | > > | > 2r = GM/2r^2 = 2.4516625 m s^-2
| > | > | > > | > 3r = GM/3r^2 = 1.0896277 m s ^-2
| > | > | > > |
| > | > | > > | No... That does not look right for 1/2 and 0 radius.
| > | > | > > | If I ascii-ised this wrong
| > | > | > > | 1/2 (R^2 - r^2) = 1/2 R^2 (1 - (r/R)2^)
| > | > | > > | it is about 2/3 down the page:
| > | > | > > | http://cseligman.com/text/planets/integration.htm
| > | > | > > |
| > | > | > >
| > | > | > > w_r = p_r = (3/8 pi G)g_R^2(1-(r/R)^2) with w being weight and
p
| > the
| > | > | > > pressure per unit area at r distance from the centre of mass
being
| > | > only true
| > | > | > > for a uniform density case like the small hydrogen cloud I
| > mentioned
| > | > above.
| > | > |
| > | > | Pressure? Density ? Are you sure we are ready to consider that. I
gave
| > the
| > | > | link only because I couldn't find 1/2 (R^2 - r^2) = 1/2 R^2 (1 -
| > (r/R)2^)
| > | > | very redily. Certainly pressure is a factor for a planet but could
we
| > not
| > | > ignore
| > | > | it for something the size of a Cavendish balance ?
| > | > |
| > | >
| > | > Fraid not, to calculate the gravity slope inside a body of any
| > significant
| > | > mass you have to integrate the mass at many levels to obtain a slope
| > like an
| > | > upside down tick as far as that paper goes anyway. Trudge through
"The
| > | > internal Pressure of Planets" and near the end there is the
internal
| > | > gravity slopes for various planets.
| > |
| > | I am not sure I agree with that, when I compare a cannon ball in
| > | a hollow at the earth's center with a canon ball surrounded by
| > | molten iron. (Corning an de Beers are helping with the obvious heat
| > | problems)
| > |
| > | One ball is under tremendous pressure, the other is not. They
| > | are the same mass. If incompressable, they are the same density.
| >
| > But thats it when is something incompressable and we are into
singularities
| > and other stuff but its not at the center the dificulty with 1/r^2 is
its
| > near the surface.
|
| "Something" doesn't have to be at the center, to appear at the
| barycentre. There is nothing at a binary star's barycenter but
| that is where things in the farfield fall to.
|
| > |
| > | Again, I am not disagreeing about planet sized objects, only
| > | small homgenous objects.
| >
| > If its true for planets why would it not be true for particles and small
| > objects although the effect may be immeasurably small in these cases ?
|
| First, I think you and I need to get on the same planet... or star
| system at the very least. The pressure at the barycenter of a
| rotating double star should be very near the same as empty
| space... but that is where things *fall to* from a distance.
| Are you quite sure you did not find some old hydraulic and
| Deep sea books at the thrift store and you just want to put them
| to good use ? :o)
|

Ok if you want to consider the case of binarys and how the field gradient
looks in their case we have to either consider the average mass about the
barycenter or take a vector of their period. Allthough the average mass is
hollow and not isotropic we could consider it through the plane of its
rotation and say that when we can draw a line throught the centers of the
two masses we have a gravity slope line as per a solid planet, an inverted
tick or something simlair. Other views will have there own slope lines and
the pressure profile will not match a solid planet at all.?



| >
| snip
| ?
| > so
| > | > I
| > | > | > > don't know why that guy was quoting Coloumbe at 1/r below. ?
| > | > |
| > | > | It is the same as gravity. If you were a proton in the center of
an
| > | > | electron cloud half the e- minus would be on you right, half on
your
| > left.
| > | > | If you were on the surface of the e- cloud, 100% the electrons
would
| > | > | be on one side of you. Same as gravity, 1/r^1 inside the cloud.
| > | > | 1/r^2 as you move away from the cloud.
| > | >
| > | > Its not 1/r^1 see the internal gravity slope graph I mention above
its
| > more
| > | > complex than that, inside a planet anyway. I will thing about your
| > example
| > | > as I don't think that equates to the field law slope inside a
planet. I
| > | > think your example at particle level to equate to a planet would
have to
| > be
| > | > stating the coulomb force experienced by an electron as it burrowed
into
| > the
| > | > centre of a proton and to make the gravity and coulomb examples
parallel
| > you
| > | > would have to be talking about the forces experienced by a proton as
it
| > | > burrowed its way to a cloud of electrons which would not be 1/r but
a
| > | > complex slope as per gravity so the bloke who quoted 1/r below has
not
| > | > thought about it enough ?
| > |
| > | I have to stick up for that bloke just now because his polymer
| > calculations
| > | look right. Let's see if we can eliminate the pressure and density
| > calculations by
| > | modeling small stuff first. Does that work?
| >
| > I'll go with you but I don't think there will be any experimental data
as to
| > the measure of gravity at the particle level ?
|
| Ya mean Van der Waal and London and BEC experiments don't count ?

Van der Waal and London are the subtractive components that bring coulomb
more inline with the force of gravity?

|
| > and as far as the forces
| > experienced by a proton as it traverses a cloud of electrons might be a
bit
| > outside current capabilities but I'll go with ideas and try and see what
| > makes sense.
|
| I was just illustrating how when you are *within* the collection of
entities
| you divide and conquer. Whether you are surrounded by galaxies or
| surrounded by fishing sinkers, the gravity is zero. If all the galaxies or
| all the fishing sinkers are to your left, then you will be pulled to the
left.
|

Yes but I thought we were talking about the gravity slope inside source as
not a 1/r^2 field effect and no hint of a 1/r field
The field slope inside a gravity source follows a complex law of pressure
and mass distribution.

| Sue...
| ============
| >>>
| > | > | > > | > | > | > http://home.freeuk.com/paulps/
| > | > | > > | > | > | > Maybe updates. The spuds, beans and onions are
coming
| >
| >
|
|


.

Relevant Pages

  • Re: Electo London Gravity ?
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  • Re: Electo London Gravity ?
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  • Re: Electo London Gravity ?
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  • Re: Electo London Gravity ?
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  • Re: Electo London Gravity ?
    ... Certainly pressure is a factor for a planet but could we ... to calculate the gravity slope inside a body of any ... |> gravity slopes for various planets. ... |> stating the coulomb force experienced by an electron as it burrowed into ...
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