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.
>
> | > >
> | > > |
> | > > | >
> | > > | > 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 ?

You won't like to hear this, but I am not sure protons have gravity either.
Perhap they do if we can detect FQHE.

Regarding R and r
The mathymatishun in you, wants to see a smooth curve but see here:
http://www.physicsclassroom.com/Class/circles/u6l3e5.gif

Missing on that graph is a nearly vertical red line representing
force below the surface. The equation for the entire red line is
probably "uglier than a mud fence covered with toads"
--Mark Twain
http://www.motherearthnews.com/library/1979_September_October/The_Last_Laugh

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...
> | > > | > | > | > |
> | > > | > | > | > |
> | > > | > | > | >
> | > > | > | > | >
> | > > | > | > |
> | > > | > | > |
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> | > > | > | >
> | > > | > |
> | > > | > |
> | > > | >
> | > > | >
> | > > |
> | > > |
> | > >
> | > >
> | > >
> | >
> | >
> |
> |
>
>


.

Relevant Pages

  • Re: Electo London Gravity ?
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  • Re: Electo London Gravity ?
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