Re: Electo London Gravity ?


"Significant Zero" <paulpsremove@xxxxxxxxxx> wrote in message news:1120986357.29445.1@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
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> | > > | > | > | > "sue jahn" <susysewnshow@xxxxxxxxxxxx> wrote in message
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> | > > | > | > | > |
> | > > | > | > | > | "sue jahn" <susysewnshow@xxxxxxxxxxxx> wrote in message
> | > > | > | > | > news:42b9c325$0$18649$14726298@xxxxxxxxxxxxxxxxxx
> | > > | > | > 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.

Again, I am not disagreeing about planet sized objects, only
small homgenous objects.



>
> | > >
> | > > |
> | > > | >
> | > > | > So 1/r^2 seems a good ? approximate for above the surface ? Mind
> you the
> | > > | > atmosphere may alter the 1/r^2 slope of the field slightly above
> the
> | > > surface
> | > > | > and seems to fit in nicely with Electric field slopes, Electric
> field
> | > > | > propagation dispersion in the far field and Coulomb at 1/d^2. Have
> not
> | > > had
> | > > | > chance to look up details yet but could near field be fitted into
> r0 to
> | > > r1
> | > > | > as far as gravity, E-fields goes ?
> | > > |
> | > > | It *seems* to work (for chemists?) when you consider everything
> | > > | below the surface to be induced dipoles. It might help to consider
> that
> | > > | the inside of a permanent is a 2 dimensional version with permanent
> | > > dipoles
> | > > | (2 D domains) instead of induced dipoles. (3 D domains)
> | > >
> | > > 2D domains at 1/r^3 and 3D domains at 1/r^2
> | > >
> | > > |
> | > > | I still have not been able to fit 1/r^1
> | > > | Indeed. Finding an approiate URL took a bit of time. Worms
> | > > | and moles must not surf the internet very much, or they have
> | > > | higher aspirations. :o)
> | > >
> | > > 1/r^1 = 1 so anything with 1/r^1 is an inertial ballistic object ? 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?

Sue..

>
>
> |
> | > >
> | > > |
> | > > | > in as that just seems to state it is what it is. ? Sorry about no
> jokes
> | > > at
> | > > | > the moment still a bit pissed off. {:-(
> | > > | Understood. :o(
> | > > |
> | > > | Warm regards,
> | > > | Sue...
> | > > |
> | > > | >
> | > > | > |
> | > > | > | Above the earth's surface, the force diminishes by 1/r^2.
> | > > | > | You can not increase the concentration of mass below your
> | > > | > | toes greater than 100% by moving farther from the center
> | > > | > | after you pass the surface and head toward outer space.
> | > > | > |
> | > > | > | >
> | > > | > | > |
> | > > | > | > | and from the source document:
> | > > | > | > | <<As the non-bonded interaction between atoms and groups
> | > > | > | > | involves less than full formal charge and involves
> polarization
> | > > | > | > | contributions, the distance dependence falls of more quickly
> | > > | > | > | than the 1/r dependence of Coulomb's law.
> | > > | > | >
> | > > | > | > Coulomb's law has a 1/d^2 dependence and an electric field
> has
> | > > 1/r^2
> | > > | > | > dependence AFAIK so where is the 1/r dependence coming from ?
> | > > | > |
> | > > | > | Let's sort this out with the planet earth first. Questions
> above.
> | > > | > |
> | > > | > |
> | > > | > | >
> | > > | > | >
> | > > | > | > | In these more
> | > > | > | > | complicated cases, where the charges can not be represented
> | > > | > | > | by single point locations, the interactions are also less
> | > > isotropic,
> | > > | > | > | falling off not just as a function of distance, but also as
> a
> | > > | > | > | function of orientation: >>
> | > > | > | > |
> | > > | > | > | As the source document involves forces *within* some kind
> | > > | > | > | of fluid or solid, a polymer scientist probaby has little
> need
> | > > | > | > | for the far-field situation. 1/r^2.
> | > > | > | > |
> | > > | > | > |
> | > > | > | > | Sue...
> | > > | > | > | Pardon my careless interchange of d distance and r radius.
> | > > | > | >
> | > > | > | > I thought it made no difference unless you were inside the the
> point
> | > > | > source.
> | > > | > |
> | > > | > | In some cases we are.
> | > > | > |
> | > > | > | Eh! If I knew the Chineese symbol for it I would toss that in
> | > > | > | too just ta keep ya on yer toes. :o)
> | > >
> | > > Culombio ley tiene un 1/ d^ 2 dependencia y un campo eléctrico tiene
> 1/ r^ 2
> | > > ¿dependencia AFAIK así donde está el 1/ r dependencia viene de?
> | > >
> | > > Can you understand that translation ?
> |
> | Absoutely! ¿ bla bal ? is spanish for
> | ~ bla bla ~ which is academic for
> | "I don't know what the f*** is am saying" :o)
>
> ¿Hace ese malo que puedo escribir chistes sucios a usted en español?
>
> BTW I can even write English this just my pc babbling but it can read some
> languages as well{:-)
> | Sue...
> |
> | > >
> | > > | > |
> | > > | > | Sue...
> | > > | > |
> | > > | > | > ?
> | > > | > | >
> | > > | > | > |
> | > > | > | > | > --
> | > > | > | > | > Significant Zero E-field = Electric field, M-field
> =Magnetic
> | > > | > field,
> | > > | > | > two
> | > > | > | > | > unbound field effects
> | > > | > | > | > http://home.freeuk.com/paulps/
> | > > | > | > | > Maybe updates. The spuds, beans and onions are coming up
> | > > nicely.
> | > > | > Ooh
> | > > | > | > | > ah.{:-)
> | > > | > | > | >
> | > > | > | > | >
> | > > | > | > | > | ----
> | > > | > | > | > | Sue...
> | > > | > | > | > |
> | > > | > | > | > |
> | > > | > | > | >
> | > > | > | > | >
> | > > | > | > |
> | > > | > | > |
> | > > | > | >
> | > > | > | >
> | > > | > |
> | > > | > |
> | > > | >
> | > > | >
> | > > |
> | > > |
> | > >
> | > >
> | > >
> | >
> | >
> |
> |
>
>


.

Relevant Pages

  • Re: Electo London Gravity ?
    ... Certainly pressure is a factor for a planet but could we not ... to calculate the gravity slope inside a body of any significant ... | electron cloud half the e- minus would be on you right, ...
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  • Re: Electo London Gravity ?
    ... Certainly pressure is a factor for a planet but could we not ... to calculate the gravity slope inside a body of any significant ... > gravity slopes for various planets. ... > | electron cloud half the e- minus would be on you right, ...
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    ... Certainly pressure is a factor for a planet but could we not ... to calculate the gravity slope inside a body of any significant ... > gravity slopes for various planets. ... > | electron cloud half the e- minus would be on you right, ...
    (sci.physics.relativity)
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