View Full Version : Cosmological.... no topic


Zarkov
03-20-04, 10:53 PM
This thread is started by Zarkov

You may post relevant or non relevant science

But any slanderous or nasty comments are not welcome..... and maybe those responsible should be totally ignored by all other posters, until their unwelcome posts are deleted.

In this thread, Cosmology, old and new is the theme

Past history threads for this stream are... Invisible light, Physics without Einstein, Electrodynamic spin gravity theory, Ether of magnetism... and any other threads I have posted it.

In this thread these are all valid topics for discussion......

Zarkov
03-21-04, 04:44 PM
from 1100f

>> Where did you get the observation that rv^2 is constant all the time?
(third time that I am asking)

Calculations from observed orbital parameters. You can do it yourself !!!!!!

( Gcentral spin-perihelion X Gcentral spin-aphelion )^0.5 = Gcentral spin-inertial-circle

The values of rv^2 for each planet point to the Gcentral spin-sun value = 1.328 X 10^11 km^3 sec^-2..within a reasonable tolerance..... except Pluto, which is no longer under a purely 1/r^2 force.

Gcentral spin is an energy value and closely related to the Poynting energy vector.

Pete
03-22-04, 12:26 AM
Hi Zarkov,
The question is asking about specific observations of radius and velocity of planets. Not about gcentral spin, space-time curvature, newtonian gravity, or any other theory-based notions. Simply distance from the Sun, and velocity relative to the Sun.

Calculations from observed orbital parameters. You can do it yourself !!!!!!
That's what I've done. I'm afraid I have some bad news for you.

You maintain that observations suggest that rv<sup>2</sup> is constant for all planets at all times. Easy enough to check, if you accept the observations of orbital radius and velocity.

Do you have a preferred source for this data? I'm using NASA Goddard Space Center (http://nssdc.gsfc.nasa.gov/planetary/factsheet/index.html).
If you disagree with their observations, please provide an alternative source.

Note that I've left out the outer planets, because not all the specific parameters we are interested in have yet been directly observed.

All results are given to four significant figures, which may be stretching the limitations of the models and available figures.

<table border=1><tr><th>Planet and position</th><th>v (km/s)</th><th>r (km)</th><th>r.v<sup>2</sup> (km<sup>3</sup>/s<sup>2</sup>)</th></td><tr><th>Mercury at aphelion</th><td>38.86</td><td>69.82x10<sup>6</sup></td><td><b>1.054x10<sup>11</sup></b></td><tr><th>Mercury at perihelion</th><td>58.98</td><td>46.00x10<sup>6</sup></td><td><b>1.600x10<sup>11</sup></b></td><tr><th>Venus at aphelion</th><td>34.79</td><td>108.94x10<sup>6</sup></td><td><b>1.319x10<sup>11</sup></b></td><tr><th>Venus at perihelion</th><td>35.26</td><td>107.48x10<sup>6</sup></td><td><b>1.336x10<sup>11</sup></b></td><tr><th>Earth at aphelion</th><td>29.29</td><td>152.10x10<sup>6</sup></td><td><b>1.305x10<sup>11</sup></b></td><tr><th>Earth at perihelion</th><td>30.29</td><td>147.09x10<sup>6</sup></td><td><b>1.350x10<sup>11</sup></b></td><tr><th>Mars at aphelion</th><td>21.97</td><td>249.23x10<sup>6</sup></td><td><b>1.203x10<sup>11</sup></b></td><tr><th>Mars at perihelion</th><td>26.50</td><td>206.62x10<sup>6</sup></td><td><b>1.451x10<sup>11</sup></b></td><tr><th>Jupiter at aphelion</th><td>12.44</td><td>816.62x10<sup>6</sup></td><td><b>1.264x10<sup>11</sup></b></td><tr><th>Jupiter at perihelion</th><td>13.72</td><td>740.52x10<sup>6</sup></td><td><b>1.394x10<sup>11</sup></b></td><tr><th>Saturn at aphelion</th><td>9.09</td><td>1514.50x10<sup>6</sup></td><td><b>1.251x10<sup>11</sup></b></td><tr><th>Saturn at perihelion</th><td>10.18</td><td>1352.55x10<sup>6</sup></td><td><b>1.402x10<sup>11</sup></b></td></table>

There appears to be a problem. Observations do not match your theory to more than one significant figure!

Now, let's try the formula I proposed earlier (adjusted for units and significant figures).
Rememberthat I said this formula should give a constant for one planet at all points on its orbit. Each planet will have its own constant value (as the table shows, more distant planets have smaller values).

I derived this formula from theory - If the theory I used was correctly derived from observations, then my formula should closely match those observations:

<table border=1><tr><th>Planet and position</th><th>v (km/s)</th><th>r (km)</th><th>v<sup>2</sup> - 2.6545x10<sup>11</sup>/r</th></td><tr><th>Mercury at aphelion</th><td>38.86</td><td>69.82x10<sup>6</sup></td><td><b>-2292</b></td><tr><th>Mercury at perihelion</th><td>58.98</td><td>46.00x10<sup>6</sup></td><td><b>-2292</b></td><tr><th>Venus at aphelion</th><td>34.79</td><td>108.94x10<sup>6</sup></td><td><b>-1226</b></td><tr><th>Venus at perihelion</th><td>35.26</td><td>107.48x10<sup>6</sup></td><td><b>-1226</b></td><tr><th>Earth at aphelion</th><td>29.29</td><td>152.10x10<sup>6</sup></td><td><b>-887.3</b></td><tr><th>Earth at perihelion</th><td>30.29</td><td>147.09x10<sup>6</sup></td><td><b>-887.2</b></td><tr><th>Mars at aphelion</th><td>21.97</td><td>249.23x10<sup>6</sup></td><td><b>-582.4</b></td><tr><th>Mars at perihelion</th><td>26.50</td><td>206.62x10<sup>6</sup></td><td><b>-582.5</b></td><tr><th>Jupiter at aphelion</th><td>12.44</td><td>816.62x10<sup>6</sup></td><td><b>-170.3</b></td><tr><th>Jupiter at perihelion</th><td>13.72</td><td>740.52x10<sup>6</sup></td><td><b>-170.2</b></td><tr><th>Saturn at aphelion</th><td>9.09</td><td>1514.50x10<sup>6</sup></td><td><b>-92.64</b></td><tr><th>Saturn at perihelion</th><td>10.18</td><td>1352.55x10<sup>6</sup></td><td><b>-92.63</b></td></table>

Draw your own conclusions.

Pete
03-22-04, 12:41 AM
In case you're wondering, the constant in that formula is 2GM, where G is the Gravitational constant, and M is the mass of the Sun.

Zarkov
03-22-04, 03:20 AM
Hi Pete.... well you are in for a suprise :)

Lets look at your Gcentral spin values.... Below I have calculated the square root of the product to give the Gcentral spin-sun.... see formula below.

Mercury... 1.298 X 10^11
Venus...... 1.327 X 10^11
Earth....... 1.327 X 10^11
Mars........ 1.321 X 10^11
Jupiter..... 1.327 X 10^11
Uranus..... 1.324 X 10^11

Close enough for large distances
As I said
( Gcentral spin-perihelion X Gcentral spin-aphelion )^0.5 = Gcentral spin-inertial-circle

The above are Gcentral spin from the planet in an inertial state, a circular orbit, reflected in the Sun, the centre of spin.


Both Pluto and Mercury are not under a pure 1/r^2 law.
Sedna is almost 1/r^3.
:)


Now this would be interesting to you

Gcentral spin / r^2 = gravity value.

To calculate the Gcentral spin for any specific planet you need to take the data from the parameters of the natural/artificial satellites of that planet.

All the formulae involving Gcentral spin apply to each system.... however they only specifically apply to the spin system you looking at.

eg

Gcentral spin-earth = 3.99 X 10^5

Gcentral spin-earth / r^2 = g at that radius on Earth

Gcentral spin-sun / r^2 = g at that radius on the Sun

This structure is reflected throughout the Universe.

Thus each system is preferred centred, nothing out there is relative... since the whole system of the Universe is completely and intimately linked.

PS Pete, I must look at the site you used.... I have never seen it.... {looked at it, yep a condensed version of my data sources, NASA

I am looking for the raw value of the advance of the perhelion of each of the planets... know any site where I may get this data.

macx
03-22-04, 05:23 AM
Happy Birthday Mr Zarkov.

Trust you are well.

:)

macx

blackholesun
03-22-04, 01:50 PM
Zarkov....since no "Gspin" values have been observed, you can't say they are used in observation....other than that, you're just making up the values of Gcentral spin because you haven't OBSERVED anything.

Pete
03-22-04, 07:27 PM
Your logic re Gcentral spin is not sound, it is specific theory oriented.
The specific theory seems to be inconsistent - you said earlier that "rv^2 for any system is a constant, Gcentral spin is unique to the spinning system. All the planets give the same Gcentral spin, that of the Sun's system." Yet now you maintain that the planets give different Gcentral spins.

To hell with theory, Zarkov, I'm interested in observations.

I've listed the radius and velocity of each planet at two different times on each orbit. Simple, direct observations.

Your prediction that rv<sup>2</sup> is constant for all planets at all times has failed. Your prediction does not match observations.

My prediction that v<sup>2</sup> - 2GM/r is contant for a planet at all times has succeeded. My preduction does match observations.

I care not for any theory that does not produce predictions that match observations.

Zarkov, you said earlier "If a theory takes no notice of observed results of experiments then it is doomed." I have given you observed results. Are you changing your theory to match these results?

Zarkov
03-22-04, 09:21 PM
>> My prediction that v2 - 2GM/r is contant for a planet at all times has succeeded. My preduction does match observations.

errrr, show me again, I failed to appreciate your first attempt

>> Your prediction that rv2 is constant for all planets at all times has failed. Your prediction does not match observations.

I am wondering about you Pete, did you not see that Gcentral spin-sun is constant, it is derived from all the planets?????

I do not understand you

Pete
03-22-04, 11:56 PM
I am wondering about you Pete, did you not see that Gcentral spin-sun is constant, it is derived from all the planets?????
Zarkov,
You gave several figures for Gcentral spin-sun, between 1.298 X 10<sup>11</sup> and 1.327 X 10<sup>11</sup>. That's not a constant!

You said that r.v<sup>2</sup> is constant for all planets at all times, but observational values vary between 1.054x10<sup>11</sup> and 1.600x10<sup>11</sup>. That's not a constant!

errrr, show me again, I failed to appreciate your first attempt
The post is still there. Which part don't you understand?

Zarkov
03-23-04, 02:36 AM
Pete, I give up, your logic is confused..... Gcentral spin-sun = 1.328 X10^11

Are you quibbling over a few parts in 10^8. Cosmological distances and velocities are imprecise in the first place

I told you Pluto and Mercury are not totally 1/r^2 controlled, the value they give for the Sun is predictably biased.... if are you trolling then I will not answer your posts anymore, if not your understanding certainly leaves a lot to be desired.

I said this thread is for serious discussion.... I am not interested in trolls.

Pete
03-23-04, 07:00 AM
Pete, I give up, your logic is confused..... Gcentral spin-sun = 1.328 X10^11

Plus or minus 7x10<sup>8</sup> (even ignoring the planets that don't fit your formula :rolleyes: )

Are you quibbling over a few parts in 10^8. Cosmological distances and velocities are imprecise in the first place
It's a few 10<sup>8</sup>s in 10<sup>11</sup> - that's a few parts in a thousand. Not good!

Actually, I'm quibbling over one part in five.
That's the variation in rv<sup>2</sup> which you said was constant.

My formula, on the other hand, matches observations to better than one part in 10,000, for all planets, comets, asteroids, and satellites. When adjustments are made for the gravity of the planets, not just the Sun, the match is even better.

Pete
03-23-04, 07:35 AM
On the topic of Gcentral spin...
If you really want to come up with a constant figure, you're better off averaging the radius and velocity, then calculating rv<sup>2</sup> (because rv<sup>2</sup> is constant for circular orbits, like I showed you before).

It's still not great (I don't know if it should be - I suspect that an averaged elliptical orbit is not the same as a circular orbit), but it's better than your geometric mean of rv<sup>2</sup> at perihelion and aphelion. At least it doesn't need an exception for Mercury and Pluto ;) .

<table border=1><tr><td></td><td>Mercury</td><td>Venus</td><td>Earth</td><td>Mars</td><td>Jupiter</td><td>Saturn</td><td>Uranus</td><td>Neptune</td><td>Pluto</td></tr><tr><td>Average radius (10<sup>6</sup>km)</td><td>57.91</td><td>108.21</td><td>149.60</td><td>227.92</td><td>778.57</td><td>1433.53</td><td>2872.46</td><td>4495.06</td><td>5869.66</td></tr><tr><td>Average velocity (km/s)</td><td>47.87</td><td>35.02</td><td>29.78</td><td>24.13</td><td>13.07</td><td>9.69</td><td>6.81</td><td>5.43</td><td>4.72</td></tr><tr><td>rv<sup>2</sup> (10<sup>11</sup>km<sup>3</sup>/s<sup>2</sup>)</td><td>1.3270</td><td>1.3271</td><td>1.3267</td><td>1.3271</td><td>1.3300</td><td>1.3460</td><td>1.3321</td><td>1.3254</td><td>1.3077</td></tr></table>

Zarkov
03-23-04, 10:35 PM
Pete, that is what I am on about, you brought up the perihelion and aphelion.... all I showed there was that the relation still holds but as a square root of the product.
Actually this is better than artificial averages.

>> (because rv2 is constant for circular orbits, like I showed you before).

Glad you agree on this as it is a relationship that applies to all cosmological spinning systems.

Good work :)

Zarkov
03-24-04, 03:10 AM
Now what does this result mean?
given
( Gcentral spin-perihelion X Gcentral spin-aphelion )^0.5 = Gcentral spin-inertial-circle

The inertial circular orbit obeys a 1/r^2 force law.
The values of Gcentral spin-aphelion and Gcentral spin-perihelion are different to the Gencentral spin-inertial circle.
OK ?
Therefore at all points in the orbit where the observed Gcentral spin does not equal the Gcentral spin-inertial, the force law is not exactly 1/r^2

In Electrodynamic spin gravity theory this is due to the 1/r and 1/r^3 force components not being in harmonic resonance.

The observation above shows that simple attraction of matter for matter is not the mode of action, because the force law relation varies from the expected 1/r^2 law.

The orbit of a planet is controlled by the interaction of a magnetic-electric force, which respectively obey 1/r^3 and 1/r force relations.
The resulting expressed force therefore varies between 1/r to 1/r^3,

The closer to a circular orbit a planet attains the closer the force law is to 1/r^2.

But no planet has a circular orbit, so nothing is inertial.

Pete
03-24-04, 02:16 PM
My formula, on the other hand, matches observations to better than one part in 10,000, for all planets, comets, asteroids, and satellites. When adjustments are made for the gravity of the planets, not just the Sun, the match is even better.

When I say "my formula", of course, I mean the not-particularly-interesting equation derived directly from the law of conservation of energy, and the formulas for kinetic energy and gravitational potential energy. I naturally don't intend to claim it as an ingenious breakthrough.

Zarkov
03-24-04, 04:05 PM
OK, back up

>> My formula, on the other hand, matches observations to better than one part in 10,000, for all planets, comets, asteroids, and satellites. When adjustments are made for the gravity of the planets, not just the Sun, the match is even better.

where is this formula you speak of ?

is this it "v2 - 2.6545x1011/r"

where your results range from -2292 to -92 ???????????????/

Pete
03-26-04, 06:01 AM
You're not paying attention, Zarkov. I get the felling that you are not giving my posts the attention that you demand for your own.

Please review exactly what I said that the formula predicted:
Remember that I said this formula should give a constant for one planet at all points on its orbit. Each planet will have its own constant value (as the table shows, more distant planets have smaller values).

So for Mercury, the number is -2292 at all points in its orbit.
For Saturn, the number is -92.6 at all points in its orbit.
etc...

Zarkov
03-26-04, 06:12 AM
mmmh, I wasn't paying attention.

So what does that formula mean?

Zarkov
03-26-04, 06:18 AM
I have re-read
so you are saying v^2-2GM/r = a local constant

is that correct?

Zarkov
03-26-04, 06:26 AM
GM in the case you showed should equal 1.328 X10^11 (or NASA's figure 1.327 X 10^11)
if it represents the Gnewton X Mass of the Sun

Pete
03-26-04, 06:26 AM
It means that the sum of an orbiting body's kinetic energy and gravitational potential energy doesn't change (much).

v<sup>2</sup> - 2GM/r is constant for a particular orbiting body. Not for all bodies.

Zarkov
03-26-04, 06:34 AM
Oh I see

but v = ( Gcentral spin / r )1/2 ………(19)

V^2 = GM / r

so we have v^2 (observed velocity of planet squared)
equals twice the predicted velocity of the planet squared ......

Logic needed here.

Pete
03-26-04, 06:38 AM
GM in the case you showed should equal 1.328 X10^11 (or NASA's figure 1.327 X 10^11)
if it represents the Gnewton X Mass of the Sun

That's right. According to the best figures I could quickly find, 2GM = 2.6545x10<sup>11</sup>

Actually, this sort of calculation is most accurate using units of AU for distance, and solar masses for weight. I believe that those units are required when doing precise calculations for the cumulative affects of all planets on each other (eg for precession). The value of G in SI units is embarrassingly imprecise!

Gaussian gravitational constant (http://scienceworld.wolfram.com/physics/GaussianGravitationalConstant.html) k = &radic;G
http://scienceworld.wolfram.com/physics/gimg51.gif

Pete
03-26-04, 06:43 AM
Oh I see

but v = ( Gcentral spin / r )1/2 ………(19)

V^2 = GM / r

so we have v^2 (observed velocity of planet squared)
equals twice the predicted velocity of the planet squared ......

Logic needed here.

Some skill at maths may help.
Try again!

Zarkov, you use Gcentral spin in odd places. If you mean GM, why not just say GM? If you mean v<sup>2</sup>/r, why not say so?

Zarkov
03-26-04, 06:44 AM
Gnewton and mass are contrived units... meaningless in cosmology, and only leads astronomers into incorrect errors and assumptions.

Best understand G only measures geomagnetic-electric energy content per cubic kg of matter.

G may not exportable to other systems, but so far I have no proof of this.

Pete
03-26-04, 06:48 AM
Circular orbit:

v<sup>2</sup> = GM/r....................(1)
v<sup>2</sup> - 2GM/r = K.............(2)
v<sup>2</sup>-2v<sup>2</sup> = K................(1 and 2)
-v<sup>2</sup> = K
v = constant

Therefore, v is constant for a circular orbit. No surprise there.
This also demonstrates that the identity v<sup>2</sup> = GM/r only holds for circular orbits.

Zarkov
03-26-04, 06:54 AM
>> Zarkov, you use Gcentral spin in odd places. If you mean GM, why not just say GM? If you mean v2/r, why not say so?

OK Gcentral spin for the centre of the spin system is derived from satellite parameters = G times Mass of planet

eg Gcentral spin derived from each planet = Gcentral spin-sun = 1.328 km^3 / sec^-2
so only the mass of the Sun can be calculated from GM

If you want Gcentral spin-earth for example you need to calculate rv^2 using the parameters of Earth satellites.... they all point back to Earth centre..
just as the planets point back to the Sun centre

So to get the mass of the Earth you use
Gcentral spin-earth = G times mass of earth......

There is a repeating logic to it all

Zarkov
03-26-04, 06:58 AM
>> This also demonstrates that the identity v2 = GM/r only holds for circular orbits.

Yes, all the logic that has been presented is for unrealcircular inertial orbits..

REAL precessing ellipses require a non quadratic equation, involving 1/r and 1/ r^3

Pete
03-26-04, 07:04 AM
Well, all your logic so far is for the trivial circular case.

The formula I presented gave a specific prediction for how velocity and radius are related in a non-circular orbit. That was the whole point! The prediction turned out to have a startling match with observations (I was only expecting 3 sig figures). And based solely on a 1/r<sup>2</sup> force relationship... go figure.

Zarkov
03-26-04, 05:37 PM
>> The formula I presented gave a specific prediction for how velocity and radius are related in a non-circular orbit

mmmh, all you have done is inadvertably used
( Gcentral spin-aphelion X Gcentral spin-perihelion )^.5 = Gcentral spin-inertial
so that v'^2 = v^2 moderated by r' / r remember rv^2 = the GM you used.

That formulae is just a useless individual application and of no predictive value, sorry.

Now the purpose of the exercise we have just gone through is to show that there is a absolute prefered repeating pattern (and frame of reference) to the structure of Universe.... and the formula derived in one specific frame of reference is applicable to other proper frames of reference.

It is interesting when two interacting spin frames are analysed, such as the Earth's field spin being driven by the Sun's field spin ... an example of toroidal and poloidal interacting spin fields.

>> Well, all your logic so far is for the trivial circular case.

This is not a trivial analysis, because the predictive values for a circle are also valid in an ellipse. Remember as I have said, terrestrial gravity is caused by (almost circular for Earth) field spin whereas planetary "gravity" is caused by oppositely opposed forces, (electric/magnetic) and the orbital path taken is the driven resultant.

Newton's analysis was for the "trivial case" of inertial circles and it is the most successful analysis, used by NASA.

Now the analysis of 'actual' planetary orbits is enlightening.... for instance the value of experienced gravity on Mercury doubles in one complete orbit.......but you can only calculate this from the inertial and actual parameters and constants..... using the formula derived from what you call the "trivial case" !!!

Pete
03-27-04, 06:19 AM
mmmh, all you have done is inadvertably used
( Gcentral spin-aphelion X Gcentral spin-perihelion )^.5 = Gcentral spin-inertial
so that v'^2 = v^2 moderated by r' / r remember rv^2 = the GM you used.
Wrong again. Perhaps you should check your algebra.

This is not a trivial analysis, because the predictive values for a circle are also valid in an ellipse.
Not so, as you have ably demonstrated! That's why you get such poor results for your predictions, and why you need to make exceptions for the elliptical orbits of Mercury and Pluto.

Newton's analysis was for the "trivial case" of inertial circles
Oh dear! You need to brush up on your history as well, I see. I suggest you spend a little time at a library, reading up on what Newton's analysis actually involved.

Now the analysis of 'actual' planetary orbits is enlightening.... for instance the value of experienced gravity on Mercury doubles in one complete orbit.......but you can only calculate this from the inertial and actual parameters and constants..... using the formula derived from what you call the "trivial case"
You have things backwards, as usual. Merecury's acceleration can be calculated directly from observing its orbit. It can also be calculated directly from Newton's universal law of gravitation. Neither of these are derived from circular orbits.

Zarkov
03-27-04, 03:24 PM
>> v2 = GM/r....................(1)
v2 - 2GM/r = K.............(2)
v2-2v2 = K................(1 and 2)

OK in (1), v is the theoretical orbital velocity of any satellite of a central body of mass M at distance r

in (2), lets look at 2GM/r. This is equal to the (escape velocity)^2, or V^2 of an object projected at radius r from a central body of mass M.
Vescape in this case is Vsun at distance r.

So the equation you posted is, in words...

[The orbital velocity(satellite)]squared minus [the escape velocity(Sun) at satellite distance]squared equals a constant.

You see how this equation is totally trivial
Now the escape velocity = 2^.5 times the orbital velocity at r from a central source of spin.

V[escape] = (2)^.5 X v[satellite at r]

Your original v (in (2) above) was the observed orbital velocity of a satellite of the Sun.
The r in that equation is the distance to that satellite.
The r and v values used in you calculation were for aphelion and perihelion of the satellite.

so we have v[satellite]^2 - { (2)^.5 X v[satellite] }^2 = k

so v^2 ( 1 - 2 ) =k

therefore v^2 = -k

The above is for a circular orbit

But because you are working in squares the vector resultant of (Gcentral spin -aphelion) X ( Gcentral spin-perihelion) = ( Gcentral spin-inertial )^.5 ( which you used as 1.327 X 10^11 is inadvertedly used in the relation.

The discrepency between Gcentral spin-aphelion and Gcentral spin-perhelion causes an equivalent factor that multiplies k, when perihelion or aphelion parameters are used

The equation you presented is trivial, meaningless and represents unsound orbital mechanics logic.



As far as accuracy of Gcentral spin is concerned, the distance of the planets from the Sun, and their orbital velocities are not precisely known, and even with this uncertainty Gcentral spin-sun, derived from aphelion or perihelion or inertial considerations yeilds the same result to a few parts in a thousand.

Moreover, the same Gcentral spin principles apply to planets and their satellites, galaxy centres and their satellites.... it is a universal principle for the Universe.


:)

Zarkov
03-30-04, 11:19 PM
OK, the structure of the Universe is now laid out.

It is interesting to calculate the magnetic component of 'gravity' at the surface of the Sun, and on the surface of each of the planets.

It turns out that the value is approximately (within one order of magnitude) the same on the surface of each cosmic body including the Sun!!!!!.

This observation begs the question..

Is the Solar System bathed in an intergalactic magnetic field?
since it is improbable that a planet would have the same magnetic relation to its radius as the Sun.

Or alternately, it makes me wonder just "what is magnetism?"
We know that electric charge can be shielded but magnetic charge can not.

It appears that Sedna's orbit (the new planetary body orbiting way past Pluto ) is almost entirely controlled by magnetic forces because of the observed eccentricity of its orbit.

Our Solar System holds many more surprises IMO.