The GRAVITY conspiracy (Part 1)

reply to post by tauristercus





I followed your logic pretty well and what I find is that the element of gravity which Newton found necessary does really remain in your transformed (simplified perhaps) though you have managed to bury it within "constants".

Can we start by recalling the inadequacy of our mathematics? We, at best, deal in approximations knowing that, at some point, math, which is only a language after all, will fail us. Usually by that time a satisfactory conclusion is reached so no matter.

Gravity does not require motion, relative or not, to remain a force. An object in space may be seen as a singularity but only when something else is there as well. Space is a "field" then upon which the singularity of this object has an effect just by being there. The difference between the field in ambience and the field where the object is is measurable. We call this gravity. Again no motion is necessary for this "force" which is only the result of other things to exist. Now we add motion, complex motions in spite of or in addition to or because of gravity and find a way to predict behavior without the considerations of gravity? I feel that "G" is present in your model proportions, indeed that it is foundational to those relationships especially as regards geometric ratios.

Gravity has existed since the smallest elementary particles popped into being. It was the friction through resistance to gravity when particles began interacting with each other that was responsible for ionization and the start up of charged particles. Then also electromagnetism added to the mix bringing a largely resultant force to bear upon, or with, an originating one. Now move up to planets and suns and moons and all the variables involved and devise a way to accurately solve celestial mechanics using geometry, albeit a dynamic one, alone? I think a mistake is evident. I'd like to take a closer look but at this point I am pretty well convinced that "G" is only in disguise. The place I will look is inside of your constants especially velocity. That seems the logical beginning.

What do you think?

tt

reply to post by iismtivu

Venus rotates "counterclockwise". And so does Uranus, though it's hard to tell because it's tilted at close to 90º, sort of laying on its side.

Tilting keeps planets from tumbling? In a sense the planets do "tumble", it's called precession. Not sure how it relates to jet planes though.

I wasn't talking to you when I suggested reading a book but it might not be a bad idea. The planets all revolve around the Sun in the same direction and more or less in the same plane because they all formed in the same protoplanetary disc. The same disc that was determined by the forces of gravity and angular momentum. But, as I pointed out, that plane is not aligned with the equator of the Sun.

What torque are you talking about? Torque is an applied force. What force is applied to the planets other than their own angular momentum and the gravity of other bodies? Torque would come into play if the rate of rotation of the Earth was changing. Well, it is, but very slowly and that is caused by tidal locking effects with the Moon. Gravity again. But the Sun doesn't really produce any torque on the Earth.

I would guess that what we feel and call gravity is bacically current flowing(from the sun) to keep the sun, and earth (and probably the other planets as well) neutrally charged. seeing that everything is made of protons, electrons and neutrons, all would be affected by inductance, and seeing how the sun is putting out millions of volts of electricity (ions) in the form of light, the planet in an effort to stay neutrally charged, has both induced (and created voltage and current) which passes from the sun through the earth and back to the sun and allowing everything to maintain a nuetral charge and accounts for magnetic fields around so called dead planets.

so we have induced current flowing through the earth(the force which is multiplied by the core spinning and creating it's own voltage and current creating our magnetic field) and then we have an atmosphere(causing an equal pressure against everything) and then there is centrifical force, (which is trying to throw everything off) when all are combined we feel what is called gravity.

reply to post by Phage

have you ever seen a gyroscope? or played with one? what happens when you tie a string to the top then set the wheel spinning counter clockwise and then dangle the whole thing by the string and then spin your hand counter clocwise?

the gyro will rotate up 90 degrees, so why dont the planets do the same? and you have yet to point out that while planets dont go arond the sun vertically how is it possible that electrons do that?

yep i was ok at "whats keeps the moon in orbit".. then my head hurt..
but wow most impressive thread send it to nasa
advance maths i like me zzzzzzzzzzzzzzz in my maths class but wow
you have really done your home work...

i kneel before your genius

reply to OP:

Very nice work, but as some posters have already mentioned, you are still using the gravitational constant in your own equations - or at the very least, you are using values that were themselves derived from equations that use the gravitational constant.

I'm not a mathematician, but I get the impression that you are simply restating in a different way the equation that you are trying to replace.

Very well laid out though. Thanks for the post.

Originally posted by Anthony1138
...i read half of your post (more then half) but you draged it on too long...

Love it when this happens. The whole post was equivalent to about 2.5 pages of text in a standard school book. This is why people have no idea what a proper noun is, and why it is to be capitalized, have problems with homonyms, and never learn how double consonants change the pronunciation of words in our beautiful language. I think this one got lucky with the proper use of 'too'...

reply to post by iismtivu

Sure I have. Gyroscopes are cool.

But planets do it too. As I said, it is called precession. In the case of the gyroscope the force is Earth's gravity (which induces torque when the axis of the gyroscope is not vertical). In the case of the Earth, as I said, it is the tidal effects of the Moon (and actually, the Sun which does contribute a bit).

Electrons do not exactly "orbit" the nucleus of the atom. Unlike planets, their "paths" are not determined by gravity and (also unlike planets) their location at a given point in time cannot be predicted.

anyway...

I'm really not very good at math. My life and my world only require me to deal with it at a second grade level, and I'm thankful for that. All that considered, this is probably a silly question, but does this equation also work for elliptical orbits? I assume it does, since the velocity of the orbiting mass changes proportionally to the distance from the mass being orbited. I believe that's a constant.

Constants make math easy? Variables cause great pain and sadness.

Originally posted by RedBird
reply to OP:

Very nice work, but as some posters have already mentioned, you are still using the gravitational constant in your own equations - or at the very least, you are using values that were themselves derived from equations that use the gravitational constant.

I'm not a mathematician, but I get the impression that you are simply restating in a different way the equation that you are trying to replace.

Very well laid out though. Thanks for the post.

I've had a number of people say something similar, in that I've unintentionally "buried" the gravitational constant so that it's no longer visible directly in my alternative equation ... and that it hasn't been removed or dispensed with.

I can understand that it's hard to come to terms with the fact that just perhaps, the gravitational constant may NOT be an actual constant created by nature but one that was artificially introduced to make a particular equation work and produce the "right" sort of answers.

Here's an example of the possible "artificiality" of the gravitational constant and an explanation as to why and how it came about by taking a look at Newton's equation as it would be without the gravitational constant (G) ...

[atsimg]http://files.abovetopsecret.com/images/member/8f3bf92eeed0.jpg[/atsimg]

We can see that the introduction of the gravitational constant was absolutely required by Newton otherwise his notion of an "attracting force" operating between 2 masses would simply not have given the right answers. So from this perspective, we can see that the G constant was an artificial mathematical manipulation.


However, looking at my alternative equation of
[atsimg]http://files.abovetopsecret.com/images/member/637ec13e3bb4.jpg[/atsimg]

there is no hidden or disguised gravitational constant there at all ... and in fact there's no need for it whatsoever as the equation (when multiplied by the mass of the orbiting body) automatically produces results that are already in units of force and therefore doesn't need to be "massaged" as does Newton's.

My equation ONLY uses an integer (4), pi, a distance (r) and the value of K uses only a distance and a time. There is no way to derive a gravitational constant or even to hide one using only distance and time.

Anyone that disagrees is more than welcome to do the work and show how the gravitational constant is "hiding" in the above equation. As I have been saying all along in this thread, from what I can see, the gravitational constant may NOT be a TRUE natural constant but more likely an "artificial" one created to make an equation work correctly.

reply to post by tauristercus

Before you throw Newton out the window, try to compute the force acting on a stationary body without using him (or Einstein if you want to get fancy).

You are working backwards from the observed motion which is exactly what Kepler did. The two "constants" you are using (velocity and radius) are constant because of the force of gravity. G is implicit in your formula because the velocity and radius are dependent on it.

Using Newton, and knowing the mass and distance between two objects, you can compute the force acting upon the orbiting object as well as it's orbital velocity (if it is in orbit). Your method cannot do so. Without knowing the period and distance you cannot calculate the force.

I liked the post, but as several people have pointed out, it suffers from a pretty big flaw.

The value of "K" has to be selected for each solar system. You have essentially made an equation that won't work for arbitrary orbits, but only for orbits around our own sun. The value of K actually replaces the mass of the sun and the gravitational constant. You have removed the mass of the sun by keeping it constant in the equation, and factoring that into your value of K.

Specifically:

K = Gm / 4 (pi)^2

Where "m" in the above equation is the mass of the sun.

Also, you derived K based upon orbital velocity (which took into account both the gravitational constant and the mass of the sun) which won't describe the force exerted on two non-orbiting objects. (That is probably why the force of Pluto is a little off, because it doesn't have a completely constant orbital velocity. Not sure about that.)

The amazing thing about Newton's law is that it applies to all objects, anywhere in the universe, regardless of whether they are in orbit around each other, or not. Contrast that with your formula, which applies only to our planetary system.

Still, I love the fact you made such a cool analysis, so I have to give you credit. Good job

Originally posted by Axial Leader

The value of "K" has to be selected for each solar system. You have essentially made an equation that won't work for arbitrary orbits, but only for orbits around our own sun. The value of K actually replaces the mass of the sun and the gravitational constant. You have removed the mass of the sun by keeping it constant in the equation, and factoring that into your value of K.

Actually K does NOT have the mass of the sun (or any other primary body) factored into it.
K was derived from nothing more than 2 basic values, namely that of a distance and that of a time period. Neither of these two values can be broken down into a mass or gravitational constant.

I liked the post, but as several people have pointed out, it suffers from a pretty big flaw.

The value of "K" has to be selected for each solar system. You have essentially made an equation that won't work for arbitrary orbits, but only for orbits around our own sun.

Actually, I'm pleased to say that my equation is independent of the body being orbited, whether it's the sun, the earth or alpha centauri.

To show this is so, instead of the sun as the primary, lets use the earth as the primary and the moon as the orbiting body.

Again, lets do it Newtons way 1st ...
[atsimg]http://files.abovetopsecret.com/images/member/72b13bacc112.jpg[/atsimg]

Now we'll do it my way ...
[atsimg]http://files.abovetopsecret.com/images/member/fa8c1ac7d54e.jpg[/atsimg]

Again, both give the exact same result which confirms that
my equation correctly calculates the force of attraction but ONLY requires the mass of the orbiting body.

And to just prove that the above using earth instead of the sun wasn't a fluke, here's what I get when I use Saturn and it's moon Titan ...
[atsimg]http://files.abovetopsecret.com/images/member/18b973842031.jpg[/atsimg]

NO gravitational constant and NO primary mass.

Therefore it can be used anywhere in the universe ... just as can Newtons.

Originally posted by Phage
reply to post by tauristercus

 

Your method cannot do so. Without knowing the period and distance you cannot calculate the force.

Without knowing the distance (radius), neither can Newton.

reply to post by Phage

try the experiment i explained above; that's not precession as the wheel is spinning horizontally counter clockwise and you spin the gyro via the string tied to the top of the frame counter clockwise. precession is a wobble in the axial spin of the planet, similar to the wobble of a spinning top when it slows down, i have yet to see the earth's south pole rotate up 90 degrees due to precession!

the combination of forces due to spinning the wheel and also spinning the frame holding the gyro both counter clockwise multiplies the effect, where if you where to spin the wheel counterclockwise and the string clockwise the force would be cancelled out.

Originally posted by Maslo
Instead of second (orbited) mass, you need time to complete orbit, which is a function of second mass and G. Whats the advantage over Newtons formula?

You should watch this video.

vimeo.com...

And then ask yourself if time is that important.

Tauristicus,

The constant G is arrived at empirically - that is by observation and measurement (I had to calculate it in a physics course using two lead balls suspended by horse hair)

Your constant K is no different - it is not derived but observed - arrived at empirically. You cannot know your time of orbit without observation can you?

Or is it obtained using calculations based on the constant G?
You cannot calculate the time of orbit without G.

You see there is no difference.

G, as well as your K, cannot be derived only observed and measured.

Either way you have a formula in which empirical data is required. What's the difference?

I like your thinking but i dare say Newtons formula is more useful as a basis for more complicated orbital dynamics than perfectly circular.


This is true of all physical constants - they are all empirically based (real world observation)

I can understand that it's hard to come to terms with the fact that just perhaps, the gravitational constant may NOT be an actual constant created by nature but one that was artificially introduced to make a particular equation work and produce the "right" sort of answers.

Tauristicus, I have great respect for your contributions here on ATS, but I am disappointed at your attitude on this thread. I concede that my disappointment has no effect on the validity of your claims, nor my or any other criticisms of your OP, But.....
There is no need to imply that any disagreement to your thoughts is merely based in some kind of fear at being unable to come to any terms regarding being "wrong".

Originally posted by tauristercus


However, looking at my alternative equation of
[atsimg]http://files.abovetopsecret.com/images/member/637ec13e3bb4.jpg[/atsimg]

there is no hidden or disguised gravitational constant there at all ... and in fact there's no need for it whatsoever as the equation (when multiplied by the mass of the orbiting body) automatically produces results that are already in units of force and therefore doesn't need to be "massaged" as does Newton's.

You use centripetal acceleration in your reasoning and your equation, do you not?
This is an effect of centripetal force, driven by gravity, including the gravitational constant.

So Yes, there is. It is the constants you mention in your OP, when you substitute 6 into 5. That being Velocity and Radius, both are driven by Gravity and contain the value G expressed in your equation.
The velocity(which we know is variable) and elliptical orbits that Kepler noted, show the Force acting to generate these two constants, and include the G constant. These two constants of velocity and radius are not independent of the system driving it, but consequences of it and expression of these forces.
Mathematically speaking, the numbers match because you are merely describing the forces as the constants of Centripetal acceleration in masses.

A great way to settle this, I guess, is if you would provide ATS with a formula for calculating optimum altitude for Geostationary orbits in satellites, without using the gravitational constant or even the geocentric gravitational constant.

My equation ONLY uses an integer (4), pi, a distance (r) and the value of K uses only a distance and a time. There is no way to derive a gravitational constant or even to hide one using only distance and time.

Like you say, you only need use the interger, pi and K which is its velocity(orbit time) and its radius......BUT you don't know the radius, because you actually need to include certain forces(including the G constant), in order to derive an altitude. An altitude which will then become your radius, that your formula is actually dependent on. This presents somewhat of a paradox.

Once again, I offer my apologies well in advance, if I have misunderstood your posts.

Dont Forget about this ...
Pushing right along ..

Helical Motion
Earth Rotation & Revolution around a moving Sun On youtube (Since Dec 08 2008 )


Our Galaxies Rotation




Interesting from (Universe-review)
Galaxies
universe-review.ca...


Unity of Earth Sun & Moon (Living Cosmos) & Related Stories (hint)
www.livingcosmos.com...

Great Calculations Taur S&F

Nassim Haramein - 3D Solar System, excerpt from Earth Pilgrims

reply to post by atlasastro


The reason his formula gives the same answer is not because it contains the constant G but because both formulas, his and Newtons, are based on empirical data. G in the case of Newton. T, or time of orbit in the case of his formula. (see my post above)

There is no advantage. If we could somehow hook up a force gauge between the planets we could just measure it directly couldn't we? But we can't, so cleverly we make observations and derive the force from those observations and theory. Included in these observations are the constant G. Included in his observations are the orbital time T. What's the difference? One better than the other? I'd say Newtons was more useful.