Is square inverse force orbit possible in reality, and mathematically?

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I just heard in another thread of a different topic that gravitational orbit is not mathematically possible due to it's square inverse component.

Aside from the "Bumble bee cannot mathematically fly" type of joke, I am exploring the two questions:

1: Is gravitational orbiting mathematically possible

2: Is any orbit physically possible under a force which is squarely inverse to the distance.

There do seem to be orbit simulators which indicate that orbits are mathematically possible with an inversely square gravitational force.

I'm also trying to think of ways to physically demonstrate an orbit with a square inverse attractive force between two bodies.

The only idea I've had so far is a large magnet mounted solidly and a small magnet hanging above it on a long strong string.

The attractive force between the two magnets should be essentially squarely inverse to the distance, so it should essentially orbit. It should be able to take on any number of the real orbits that the planets are said to adopt.

I'm sure some folks will respond with 32 pages of formulas, which I'm afraid I've been kind of glossing over.

But if you good people could also do either or both of the following, I would be most grateful:

1: Explain how an inverse  square force orbit is mathematically impossible *in your own words* and in simple terms, to show that you actually understand it.

2: Come up with suggestions how we could physically demonstrate whether a square inverse force orbit is possible in the real world.

EDIT:
Had another idea:
Take a large pane of glass, put a large magnet under it, and a large weight with a strong magnet on top, with ice cube feet.
Since magnetic attractive force also reduces with the square of the distance, and ice cubes slide real nicely on wet glass, we should be able to see of the free moving weight can orbit on any of the traditional orbital paths.
« Last Edit: March 17, 2019, 08:37:11 PM by Tom Foolery »

No it isn't possible,because orbital motion requires an infinite series of higher order forces whereas gravitation is postulated to be a mere acceleration.  http://septclues.com/TYCHO_SSSS/Gopi%20Papers/Celestial%20Dynamics%20and%20Rotational%20Forces.pdf

In textbooks, the derivation of the equations for orbital motion stop at the first derivative but due to the nature of orbital motion an infinite number of non-vanishing derivatives exist for orbital motion.  Gravitation (as postulated) provides only first order acceleration in the radial direction.  Read the above paper for the derivation.

« Last Edit: March 17, 2019, 04:01:07 PM by GDg43SA »
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No it isn't possible,because orbital motion requires and infinite series of higher order forces whereas gravitation is postulated to be a mere acceleration.  http://septclues.com/TYCHO_SSSS/Gopi%20Papers/Celestial%20Dynamics%20and%20Rotational%20Forces.pdf

In textbooks, the derivation of the equations for orbital motion stop at the first derivative but due to the nature of orbital motion and infinite number of non-vanishing derivatives exist for orbital motion.  Gravitation (as postulated) provides only first order acceleration in the radial direction.  Read the above paper for the derivation.

Is that paper or similar published in any scientific journals by chance?  Septclues.com is a conspiracy kook website pushing fringe science and conspiracy theories.

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JackBlack

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I would say yes it is possible. The nature of the force involved (i.e. what kind of pattern it follows) is irrelevant.
For a hypothetical perfectly circular orbit, any force directed towards the centre of rotation, as long as it has the appropriate magnitude is enough.

As for demonstrating them in reality, all the satellites which exist do a pretty good job of that.
Otherwise the simplest thing I can think of are ions.
Have a positive and negative ion (or charged object) orbiting each other while in free fall.

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JackBlack

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No it isn't possible,because orbital motion requires and infinite series of higher order forces whereas gravitation is postulated to be a mere acceleration.
As has been explained to you repeatedly, these higher order "forces" are not forces. They are merely the result of changes in acceleration.

These higher order derivatives existing simply means the acceleration is changing which is simply due to the motion of the changing the direction to the central force, not that you need something to provide a force to cause it.

So I will say it again:
Jerk is simply how acceleration changes. You don't need a force to supply the jerk.

If you wish for anyone to take that claim seriously (i.e. accept it) you will need far more than a baseless assertion.

Gravitation (as postulated) provides only first order acceleration in the radial direction.  Read the above paper for the derivation.
The motion of the object in the orbit provides the rest.
And I am not calling that pile of nonsense a paper. It is a blog article, nothing more.

Quote
These higher order derivatives existing simply means the acceleration is changing which is simply due to the motion of the changing the direction to the central force, not that you need something to provide a force to cause it.
No the acceleration is NOT changing, the position of the satellite is moving but the acceleration is constant at a given radius in polar/transverse coordinates under Newtonian assumptions. 

Quote
As has been explained to you repeatedly, these higher order "forces" are not forces. They are merely the result of changes in acceleration.
You explained nothing.  You asserted something that violates basic tenets of physics and logic.  In order for the acceleration of a body to change there must be a second order force to change it and on and on for higher and higher order forces.

You're just throwing words together that you don't understand.  The equations for elliptical and circular orbits contain infinite nonvanishing derivatives in the radial and transverse directions.  It follows that there MUST be higher order forces (snaps, crackles, pops ad infinitum) in order to keep the object on an elliptical or circular orbit.  If those higher order forces don't exist the minor body will eventually crash into the major body, regardless of the initial tangential velocity of the object because objects move in the direction of whatever net force and net higher order forces acting upon them thus no matter what the initial tangential velocity of the body it will eventually collide with the source of the force that is:  the radius of the orbit will diminish continually if only a centripetal force is acting on said body.  This is a mathematical necessity:  you can't just arbitrarily ignore the higher derivatives of a motion by fiat.

Quote
So I will say it again:
Jerk is simply how acceleration changes. You don't need a force to supply the jerk.
You can say it again and again, but that doesn't make it any less of a lie.  Of course you don't need a "force  to supply the jerk" if by force you mean "first order force":  you need a second order force to supply the jerk, a third order force to supply the snap, a fourth order force for the crackle etc. etc.  If those higher order forces do not exist as mathematically necessitated then the body will not conform to the assumed path.
« Last Edit: March 17, 2019, 04:31:29 PM by GDg43SA »
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No it isn't possible,because orbital motion requires and infinite series of higher order forces whereas gravitation is postulated to be a mere acceleration.  http://septclues.com/TYCHO_SSSS/Gopi%20Papers/Celestial%20Dynamics%20and%20Rotational%20Forces.pdf

In textbooks, the derivation of the equations for orbital motion stop at the first derivative but due to the nature of orbital motion and infinite number of non-vanishing derivatives exist for orbital motion.  Gravitation (as postulated) provides only first order acceleration in the radial direction.  Read the above paper for the derivation.

Is that paper or similar published in any scientific journals by chance?  Septclues.com is a conspiracy kook website pushing fringe science and conspiracy theories.
Ad hominem.  Read the paper and critique it, or else GTFO.
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Jackblack and rabinoz:  I will grant you the indulgence of one last chance to enjoy the privilege of my time: refute the mathematics presented in the paper I linked to above:  If you continue to spout nonsensical word-salad gibberish you will be put back on my ignore list.  So put up or shut up:  refute the math or go away.
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NotSoSkeptical

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No it isn't possible,because orbital motion requires and infinite series of higher order forces whereas gravitation is postulated to be a mere acceleration.  http://septclues.com/TYCHO_SSSS/Gopi%20Papers/Celestial%20Dynamics%20and%20Rotational%20Forces.pdf

In textbooks, the derivation of the equations for orbital motion stop at the first derivative but due to the nature of orbital motion and infinite number of non-vanishing derivatives exist for orbital motion.  Gravitation (as postulated) provides only first order acceleration in the radial direction.  Read the above paper for the derivation.

Is that paper or similar published in any scientific journals by chance?  Septclues.com is a conspiracy kook website pushing fringe science and conspiracy theories.
Ad hominem.  Read the paper and critique it, or else GTFO.

I inferred from your postings that you are retarted.  Since you post links to conspiracy theory websites, this confirms that you are retarted. 


That's ad hominem.

Thanks for playing.

As for your "paper", we already have a blog pimper, we don't need another. 


As for the topic of discussion, is there a valid reason to why they wouldn't be.
Rabinoz RIP

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rabinoz

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Jackblack and rabinoz:  I will grant you the indulgence of one last chance to enjoy the privilege of my time: refute the mathematics presented in the paper I linked to above:  If you continue to spout nonsensical word-salad gibberish you will be put back on my ignore list.  So put up or shut up:  refute the math or go away.
Oh, what a total joke you are with your pious holier than thou attitude!
You will grant us lowly creatures the indulgence of one last chance to enjoy the privilege of your, oh so valuable, time".
Yes, your majesty ::), no your majesty ::), anything you say your majesty ::). What a useless prig! 

But why shouldn't we "continue to spout nonsensical word-salad gibberish"? All that paper does is to "continue to spout nonsensical word-salad gibberish".

But I couldn’t care less about buying any silly indulgences from a troll like you. It's much better to be ignored.

But what math?
The higher derivatives all exist simply because the direction of the velocity vector is continually changing.

But they not what defines this uniform circular motion about a point.
That definition is simply rotation with a constant angular velocity, ω, at a constant distance, R from the centre, C.

PS If you don't accept this central 1/r2 what keeps the moon moving as is observed?

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rabinoz

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Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #10 on: March 17, 2019, 07:23:51 PM »
Ad hominem.  Read the paper and critique it, or else GTFO.

I inferred from your postings that you are retarted.  Since you post links to conspiracy theory websites, this confirms that you are retarted. 

An ignorant newbie like GDg43SA might not know the meaning of "retarted". Maybe you should have explained that.
Quote from: Urban Dictionary
retarted
A retarded person's way of spelling the word "retarded."
Though I do not like describing it like that because it casts a slur of retarded people who are that way through no fault of their own and deserve our understanding.

But we should realise that it is impossible to refute the beliefs of conspiracy theorists. All evidence simply strengthens their belief in conspiracy.

Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #11 on: March 17, 2019, 08:34:17 PM »
No it isn't possible,because orbital motion requires an infinite series of higher order forces
Could you elaborate -- in your own words, so I know you understand it -- what would happen? Are you saying it would be impossible for something to orbit without either crashing into the center mass or flying off away from it if the attractive force was related to the inverse of the square of the distance?
Quote
whereas gravitation is postulated to be a mere acceleration.
I might be confused.. But isn't gravity a force? Acceleration is the result of a force (Any force) acting on a mass over time. Whether one newton of force is exerted from gravity or a rocket motor, the acceleration is the same for a given mass.
Quote
http://septclues.com/TYCHO_SSSS/Gopi%20Papers/Celestial%20Dynamics%20and%20Rotational%20Forces.pdf

In textbooks, the derivation of the equations for orbital motion stop at the first derivative but due to the nature of orbital motion an infinite number of non-vanishing derivatives exist for orbital motion.  Gravitation (as postulated) provides only first order acceleration in the radial direction.  Read the above paper for the derivation.
Well considering your earlier math blunders and the oddness of your current claims, and then your high reliance on septclues.com which seems, shall we say, just a tad questionable as a reliable source of information, I must admit to being a little skeptical of reliability of your claims.

I guess we need to come up with a project I can build to test whether one object can orbit the other with an attractive force that diminishes with the square of the distance.

After all, just because *you* can't come up with math that says it works, that doesn't mean that it can't actually work.
Lots of things worked before math came up with it's kind approval.

Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #12 on: March 17, 2019, 08:38:18 PM »
Had another idea:
Take a large pane of glass, put a large magnet under it, and a large weight with a strong magnet on top, with ice cube feet.
Since magnetic attractive force also reduces with the square of the distance, and ice cubes slide real nicely on wet glass, we should be able to see of the free moving weight can orbit on any of the traditional orbital paths.

That should essentially be good, right? I mean there's mass, there's square inverse attractive relationship, and there's freedom of movement.

Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #13 on: March 17, 2019, 10:33:07 PM »
Quote
These higher order derivatives existing simply means the acceleration is changing which is simply due to the motion of the changing the direction to the central force, not that you need something to provide a force to cause it.
No the acceleration is NOT changing, the position of the satellite is moving but the acceleration is constant at a given radius in polar/transverse coordinates under Newtonian assumptions. 
This is wrong. For uniform circular motion, the magnitude of the acceleration is constant, but the acceleration vector is constantly changing.

Moreover your article has numerous errors, one of which is not recognizing the time dependence of the basis vectors and coordinates therefore have gotten incorrect equations of motion on page 3.
Second of which is you assumed your object was rotating uniformly then went on to talk about acceleration magnitude changing, like what? When you are in uniform circular motion the magnitude of the inwards acceleration is constant.

----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Edit: Just too add onto your paper that you posted.

You don't need a "an infinite series of higher-order forces" to remain circular, this is wrong and what you have stated is not a series.

Although you can have higher order directives of acceleration, this measures the higher order rates of change of the acceleration due to the central force, it does not imply you need extra forces as you have proposed.

That is a simply lack of understanding of newtons second law, coordinate systems and rates of change so I have attached further readings for you.

https://en.wikipedia.org/wiki/Polar_coordinate_system
https://en.wikipedia.org/wiki/Circular_motion
« Last Edit: March 18, 2019, 03:17:23 AM by cravingTD »

Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #14 on: March 18, 2019, 02:42:08 AM »
1: Is gravitational orbiting mathematically possible

Short answer is yes.

Slightly longer answer. There are a set of laws called keplars laws of planetary motion describes a satellites or planets orbit as elliptical.

We can derive these laws under a set of assumptions, a caveat being that these assumptions arn't strictly true in reality but they give a good approximation for the most part (too my knowledge) and do show that gravitational orbits are possible.
« Last Edit: March 18, 2019, 04:17:37 AM by cravingTD »

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JackBlack

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Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #15 on: March 18, 2019, 03:12:16 AM »
No the acceleration is NOT changing
The acceleration is a vector. If the direction changes, that means the acceleration changes.
If the acceleration didn't change then that would mean there is no jerk or any higher order terms.

You explained nothing.  You asserted something that violates basic tenets of physics and logic.
There you go projecting again.

Once more:
Forces are what causes objects to accelerate. You apply a force, and then dependent upon the mass, the object accelerates.
Your higher order "forces" are not forces, they are simply a measure of the change in acceleration.

You are the one asserting non-physical and illogical nonsense claiming that there must be some real, physical cause of these magic higher order "forces".
The burden of proof is upon you to show that these higher order "forces" are needed when simply physics clearly indicates that objects are accelerated based upon forces, and that these higher order "forces" are merely measures of changes in the actual forces and resultant acceleration.

You're just throwing words together that you don't understand.  The equations for elliptical and circular orbits contain infinite nonvanishing derivatives in the radial and transverse directions.  It follows that there MUST be higher order forces (snaps, crackles, pops ad infinitum) in order to keep the object on an elliptical or circular orbit.
No, it is you who seems to be throwing together words.
These higher order derivatives are mearly how the acceleration is changing. They don't correspond to any real force.

If those higher order forces don't exist the minor body will eventually crash into the major body, regardless of the initial tangential velocity of the object because objects move in the direction of whatever net force
NO!
Explain how the force acting on the object doesn't accelerate it based upon what that force is and keep it following the circular path?
Can you do that? NO!
Objects accelerate based upon the real force acting on it, not some magical higher order force.

This is a mathematical necessity
It is a complete absurdity based upon completely ignoring reality.

you need a second order force to supply the jerk
Why?
So far that is just a baseless claim.
Why is a second order force needed?
That would be akin to claiming you need a lower order force to provide the velocity, and that if you don't have that the object will magically stop, and an even lower order force to provide the position, and without that the object will magically fly to 0.
It is pure nonsense.

refute the mathematics presented in the paper I linked to above
The math in the pile of crap (again, not a paper) is not the issue. The physics is.
The existence of higher order derivatives doesn't magically mean you need a higher order force to supply them.

I will give you another chance (and probably plenty more) to justify your delusional claims that you magically need these higher order forces in reality rather than them simply being the product of the changing acceleration on the object due to its changing position.

Start by explaining just what you think a force which is constant in magnitude and always pointed towards the centre of rotation would do to an object initially on a tangential trajectory with a velocity such that |F|=|v|2/r, clearly explaining why it would do that. Don't just say it will magically crash, make sure you explain why.

Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #16 on: March 18, 2019, 03:37:49 AM »
The only idea I've had so far is a large magnet mounted solidly and a small magnet hanging above it on a long strong string.

The attractive force between the two magnets should be essentially squarely inverse to the distance, so it should essentially orbit. It should be able to take on any number of the real orbits that the planets are said to adopt.

EDIT:
Had another idea:
Take a large pane of glass, put a large magnet under it, and a large weight with a strong magnet on top, with ice cube feet.
Since magnetic attractive force also reduces with the square of the distance, and ice cubes slide real nicely on wet glass, we should be able to see of the free moving weight can orbit on any of the traditional orbital paths.

My problem with your magnetic orbiting idea is that the force between two magnets don't have a inverse square relationship so they would not have a orbit your trying to replicate.
« Last Edit: March 18, 2019, 03:39:28 AM by cravingTD »

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NotSoSkeptical

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Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #17 on: March 18, 2019, 06:17:37 AM »
Ad hominem.  Read the paper and critique it, or else GTFO.

I inferred from your postings that you are retarted.  Since you post links to conspiracy theory websites, this confirms that you are retarted. 

An ignorant newbie like GDg43SA might not know the meaning of "retarted". Maybe you should have explained that.
Quote from: Urban Dictionary
retarted
A retarded person's way of spelling the word "retarded."
Though I do not like describing it like that because it casts a slur of retarded people who are that way through no fault of their own and deserve our understanding.

But we should realise that it is impossible to refute the beliefs of conspiracy theorists. All evidence simply strengthens their belief in conspiracy.

I will assume that he figured it out, because he added me to his ignore list without posting a reply.



I wish I could actually provide some good discussion to this, but my knowledge of this is limited.  If I understand it correctly, isn't it similar to having a ball attached to a string and then spinning it in a circle.  The ball tries to pull away due to the centripetal force and the string acts as the gravity pulling it in.  A very basic and rudimentary example, but that's the general gist.
Rabinoz RIP

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Zaphod

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Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #18 on: March 18, 2019, 07:52:27 AM »
Feynman Lost Lecture

This is a really good short video on elliptical orbits based on a lecture by the great Richard Feynman, who was teaching Newtonian mechanics.




Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #19 on: March 18, 2019, 09:05:38 AM »
Quote
These higher order derivatives existing simply means the acceleration is changing which is simply due to the motion of the changing the direction to the central force, not that you need something to provide a force to cause it.
No the acceleration is NOT changing, the position of the satellite is moving but the acceleration is constant at a given radius in polar/transverse coordinates under Newtonian assumptions. 
This is wrong. For uniform circular motion, the magnitude of the acceleration is constant, but the acceleration vector is constantly changing.
Wrong.  In polar coordinates the acceleration is of constant magnitude and direction at a given radius(radially inward.)  You must confusing Cartesian with polar coordinates. 

Quote
Moreover your article has numerous errors, one of which is not recognizing the time dependence of the basis vectors and coordinates therefore have gotten incorrect equations of motion on page 3.
Do you not realize that the dot notation means derivative with respect to time?  And what are you talking about "time dependent coordinates."  The motion of the object is dependent upon time, not the coordinate system unless you want to needlessly complicate things.
Quote
Second of which is you assumed your object was rotating uniformly then went on to talk about acceleration magnitude changing, like what? When you are in uniform circular motion the magnitude of the inwards acceleration is constant.
Yes, the acceleration is constant in the Newtonian system in magnitude and direction.  But, as demonstrated a constant inward acceleration is not able to support a circular or elliptical orbit.  You have not challenged the derivation but merely asserted that you disagree with it.

Quote
You don't need a "an infinite series of higher-order forces" to remain circular, this is wrong and what you have stated is not a series.
Okay, it is a sequence, not a series.  Nice quibble.  The author is not a native English speaker.


Quote
Although you can have higher order directives of acceleration, this measures the higher order rates of change of the acceleration due to the central force, it does not imply you need extra forces as you have proposed.
There can be no "higher order rates of change of the acceleration due to the central force" if there is only a non-varying central force as Newtonian mechanics asserts.  There can be no effects without causes.


Quote
That is a simply lack of understanding of newtons second law, coordinate systems and rates of change so I have attached further readings for you.

https://en.wikipedia.org/wiki/Polar_coordinate_system
https://en.wikipedia.org/wiki/Circular_motion
Such irony!  And Newton's second law isn't pertinent here, we are talking about DERIVATIVES of Newton's second law.
« Last Edit: March 18, 2019, 10:46:26 AM by GDg43SA »
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Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #20 on: March 18, 2019, 09:57:42 AM »

Quote
Could you elaborate -- in your own words, so I know you understand it -- what would happen? Are you saying it would be impossible for something to orbit without either crashing into the center mass or flying off away from it
Yes, I am saying that.  An object will move in the direction of the net force acting upon it if the net force does not vary with time.
whereas gravitation is postulated to be a mere acceleration.
[/quote]
Quote
I might be confused.. But isn't gravity a force? Acceleration is the result of a force (Any force) acting on a mass over time. Whether one newton of force is exerted from gravity or a rocket motor, the acceleration is the same for a given mass.
Yes, gravity is (asserted to be) a force.  A better wording would have been "whereas gravitation is postulated to merely provide an acceleration and no higher order rates of change of acceleration." 

Quote
Well considering your earlier math blunders and the oddness of your current claims, and then your high reliance on septclues.com which seems, shall we say, just a tad questionable as a reliable source of information, I must admit to being a little skeptical of reliability of your claims.
What "math blunders?"  Using the wrong radius for your Cavendish experiment and calculating the acceleration of the larger mass rather than the smaller mass?  The first was your fault for not providing that information initially, the second was a mistake made in haste.  I guarantee that my math skills are more than adequate to assess this matter.  Furthermore, your passive-aggressive tone is very annoying.  Cut it out.
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Mikey T.

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Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #21 on: March 18, 2019, 12:07:02 PM »
Pedantic.  Classic troll tactics.  Claim stupid shit, link to someone else's incorrect garbage, put those trying to correct you on ignore. GD you are not worth my time to insult you further.  Queue change to ignore list and schoolyard level insults.  Not like the small minded ego of the truly idiotic ones could handle mild critiques of there utterances without lashing out.

Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #22 on: March 18, 2019, 12:51:29 PM »
So this “infinite derivatives of higher order forces” business must also apply to all rotating machinery, right?

If that means planets should career off into interstellar space, does it also mean spinning shafts should tear out of their bearings, turbine blades rip out of housings, etc, etc?

Doesn’t sound right to me.


Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #23 on: March 18, 2019, 01:17:22 PM »
So this “infinite derivatives of higher order forces” business must also apply to all rotating machinery, right?

If that means planets should career off into interstellar space, does it also mean spinning shafts should tear out of their bearings, turbine blades rip out of housings, etc, etc?

Doesn’t sound right to me.
This is a poor analogy.  A rigid body undergoing torque (turbine, spinning shafts) is not the same as a body orbiting around another point due to a central force directed from that point.
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JackBlack

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Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #24 on: March 18, 2019, 01:38:30 PM »
Wrong.  In polar coordinates the acceleration is of constant magnitude and direction at a given radius(radially inward.)  You must confusing Cartesian with polar coordinates.
You mean you are intentionally using dishonest coordinates to pretend it is constant to cover up your earlier mistake?
Like I said, if it is constant, there are no higher derivatives.
The derivative of a constant is 0. That means no jerk and your entire claim falls to pieces.

But even with polar coordinates, the vector is changing direction.
With polar coordinates you have a magnitude and a direction, which will be expressed as a magnitude and direction rather than an x and y component (dealing with 2D).
When the orbiting object has a position of (r,0), the acceleration vector is (a,π). When the position is (r,π), the acceleration vector is (a,0).
Are you really going to claim that (a,π) and (a,0) are the same?

The frame it remains constant in is a rotating reference frame, in which the position remains constant and there are 2 forces, an inwards force pulling the object towards the centre, and an outwards centrifugal force pulling it to the edge. This keeps the position constant and all derivatives of it as 0. This means no jerk.

So what is it going to be?
Is the acceleration changing, or is it static and thus there is no jerk?

But, as demonstrated a constant inward acceleration is not able to support a circular or elliptical orbit.
WHERE?
You keep asserting this but you are yet to provide any justification of it.
Yes, the blog post you linked does show there are higher order derivatives due to this inwards acceleration.
It even shows it produces a circular orbit.
It in no way disproves it.
Instead it just jumps to the completely unsupported conclusion that because jerk is a thing it magically can't do it.
There is absolutely no basis for it at all.

There can be no "higher order rates of change of the acceleration due to the central force" if there is only a non-varying central force as Newtonian mechanics asserts.  There can be no effects without causes.
So you admit your blog post is garbage?

Such irony!  And Newton's second law isn't pertinent here, we are talking about DERIVATIVES of Newton's second law.
No, that is what is extremely important.
It indicates that if you have a force acting on an object then it

Yes, I am saying that.  An object will move in the direction of the net force acting upon it if the net force does not vary with time.
Try actually addressing the question.
This object isn't just sitting there having a net force on it. It is already moving.
What happens?
We can easily see that the force applied will accelerate it towards the centre of the orbit, but the sideways motion will mean that it can't just go straight down and instead remains the same distance, circling.

Can you explain what you think should happen and why.
gravitation is postulated to merely provide an acceleration and no higher order rates of change of acceleration.
Again, it is the motion of the object which provides them.
If the object was stationary (and held stationary) then the acceleration from gravity would be constant.
But it isn't. It moves. As it moves, it can change the magnitude (if it moves radially) and the direction (if it moves tangentially). This is all that is required to provide these higher order terms.
You don't need magical higher order forces.

Using the wrong radius for your Cavendish experiment
Even when using the 4 m, you don't get anywhere near the number you provided. You would need roughly 15 m, which is a massive blunder.
So either you pulled a number out of thin air, or you just failed at a simple calculation.


The first was your fault
No, it was entirely your fault for making such a horribly flawed assumption. There was no reason at all for you to use roughly 15 m.
You could estimate it from the footage and masses provided like I did, or you could have asked.

I guarantee that my math skills are more than adequate to assess this matter.
Your math skills do seem to be quite poor.
You are claiming the acceleration/force is constant, but also changing by virtue of having derivatives.
As a nice simple question, what is the derivative of a constant?

More importantly, what about your physics skills which seem to be severely lacking, with you repeatedly claiming you need magical higher order forces?

This is a poor analogy.  A rigid body undergoing torque (turbine, spinning shafts) is not the same as a body orbiting around another point due to a central force directed from that point.

Then how about a bunch of objects on a string, setup so they are equally spaced around a circle with an even number of them.
That way there is no net force on the shaft, just a torque to keep it spinning.
According to you, all the objects should magically fly to the centre.
We can even have it set up to be on a spring so you can't complain that the string gets loose and lets the balls fly back out, or you can attach a load cell to each string/object to confirm it is constantly being pulled towards the centre.

The spring version already exists in the form of a centrifugal clutch.
At low speed, the spring is providing a force greater than the centripetal force and the clutch pulls inwards. At a very specific speed it is balanced and would remain in place. At a higher speeds it goes outwards even though the only force acting on it is the force (not torque) from the springs pushing it inwards.

Now again, can you use your allegedly adequate math skills to explain exactly what will happen to a object experiencing a force directed towards the origin with a magnitude of |F|=mv2/r to produce an acceleration of |a|=v2/r.
Can you provide a formula (in either Cartesian or polar coordinates) to describe the path taken, preferably in parametric form where t is the time in seconds?

For someone as allegedly skilled as you, this should be no problem.

*

JackBlack

  • 18586
Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #25 on: March 18, 2019, 01:40:15 PM »
And if all that post was too long for you, this is the important part which really gets to the issue of the thread:
Can you use your more than allegedly adequate math skills to explain exactly what will happen to a object experiencing a force directed towards the origin with a magnitude of |F|=mv2/r to produce an acceleration of |a|=v2/r also directed towards the centre?
Can you provide a formula (in either Cartesian or polar coordinates) to describe the path taken, preferably in parametric form where t is the time in seconds?

For someone as allegedly skilled as you, this should be no problem.

*

JackBlack

  • 18586
Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #26 on: March 18, 2019, 01:55:18 PM »
Had another idea:
Take a large pane of glass, put a large magnet under it, and a large weight with a strong magnet on top, with ice cube feet.
Since magnetic attractive force also reduces with the square of the distance, and ice cubes slide real nicely on wet glass, we should be able to see of the free moving weight can orbit on any of the traditional orbital paths.

That should essentially be good, right? I mean there's mass, there's square inverse attractive relationship, and there's freedom of movement.
It should be hypothetically possible to set it up, but I'm not sure on how stable it is, because magnets, while you can approximate them as a collection of magnetic monopoles, don't actually follow the inverse square law as they are dipoles. There is also the issue of the magnet trying to tip over.

Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #27 on: March 18, 2019, 02:04:36 PM »
This is wrong. For uniform circular motion, the magnitude of the acceleration is constant, but the acceleration vector is constantly changing.
Wrong.  In polar coordinates the acceleration is of constant magnitude and direction at a given radius(radially inward.)  You must confusing Cartesian with polar coordinates. 
Please follow to the new link which could clarify this for you.
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Moreover your article has numerous errors, one of which is not recognizing the time dependence of the basis vectors and coordinates therefore have gotten incorrect equations of motion on page 3.
Do you not realize that the dot notation means derivative with respect to time?  And what are you talking about "time dependent coordinates."  The motion of the object is dependent upon time, not the coordinate system unless you want to needlessly complicate things.
Yes I am fully aware of what the dot means, I will post a more clearer link that I suggest you read.

Roughly speaking the components of the basis vectors in polar coordinates change depending on where you are so we, they are not like coordinates such as Cartesian that are constant. So they are taken into account when you take time derivatives. Do you understand this?

This is a good demonstration of what I mean as I don't want to fill the thread with calculations.
http://www.iitg.ac.in/physics/fac/saurabh/ph101/Lecture2-SBasu.pdf
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Second of which is you assumed your object was rotating uniformly then went on to talk about acceleration magnitude changing, like what? When you are in uniform circular motion the magnitude of the inwards acceleration is constant.
Yes, the acceleration is constant in the Newtonian system in magnitude and direction.  But, as demonstrated a constant inward acceleration is not able to support a circular or elliptical orbit.  You have not challenged the derivation but merely asserted that you disagree with it.
I have stated the definition of uniform circular motion, which you stated your system to be undergoing.
 
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You don't need a "an infinite series of higher-order forces" to remain circular, this is wrong and what you have stated is not a series.
Okay, it is a sequence, not a series.  Nice quibble.  The author is not a native English speaker.
Its not just a quibble, but a important distinction if this were a scientific paper.

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That is a simply lack of understanding of newtons second law, coordinate systems and rates of change so I have attached further readings for you.

https://en.wikipedia.org/wiki/Polar_coordinate_system
https://en.wikipedia.org/wiki/Circular_motion
Such irony!  And Newton's second law isn't pertinent here, we are talking about DERIVATIVES of Newton's second law.
You are talking about derivatives of newtons second law, which is in fact talking about newtons second law...

Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #28 on: March 18, 2019, 02:25:21 PM »
So this “infinite derivatives of higher order forces” business must also apply to all rotating machinery, right?

If that means planets should career off into interstellar space, does it also mean spinning shafts should tear out of their bearings, turbine blades rip out of housings, etc, etc?

Doesn’t sound right to me.
This is a poor analogy.  A rigid body undergoing torque (turbine, spinning shafts) is not the same as a body orbiting around another point due to a central force directed from that point.

It’s not a analogy.

You are talking about these higher orders fundamentally as a result of the circular motion, so it must hold for all rotating systems.

In a rigid system, any additional forces would introduce coresponding stresses and strains.

All of which would be a surprise to tens if not hundreds of thousands of engineers who carefully design such systems.

Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #29 on: March 18, 2019, 05:35:46 PM »

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Could you elaborate -- in your own words, so I know you understand it -- what would happen? Are you saying it would be impossible for something to orbit without either crashing into the center mass or flying off away from it
Yes, I am saying that.  An object will move in the direction of the net force acting upon it if the net force does not vary with time.
Oh? And if the force does vary with time, I suppose the object will what, move away from it? move crosswise? not move at all?
Seriously, you're causing me to question my sanity and your sanity. One of them is breached and I can't tell which one. But I got a hunch it's not entirely mine.
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whereas gravitation is postulated to be a mere acceleration.
Now wait a second. If you say that it's distance and mass dependent force, fine.
But just because a force changes depending on other parameters doesn't mean it's not a force.
Even the force of a spring changes with distance. Lots of forces depend on other things for their exact value and direction.

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I might be confused.. But isn't gravity a force? Acceleration is the result of a force (Any force) acting on a mass over time. Whether one newton of force is exerted from gravity or a rocket motor, the acceleration is the same for a given mass.
Yes, gravity is (asserted to be) a force.  A better wording would have been "whereas gravitation is postulated to merely provide an acceleration and no higher order rates of change of acceleration." 
Sorry my friend, we're losing ground.
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Well considering your earlier math blunders and the oddness of your current claims, and then your high reliance on septclues.com which seems, shall we say, just a tad questionable as a reliable source of information, I must admit to being a little skeptical of reliability of your claims.
What "math blunders?"  Using the wrong radius for your Cavendish experiment and calculating the acceleration of the larger mass rather than the smaller mass?  The first was your fault for not providing that information initially, the second was a mistake made in haste.  I guarantee that my math skills are more than adequate to assess this matter.  Furthermore, your passive-aggressive tone is very annoying.  Cut it out.
OK where should I start. Now look, I don't have time right now to go through every post and recreate a court-worthy account of your exact blunders.
If you want me to, I'll be glad to do it another time. But from memory:

Your blunders: First calculating acceleration for the heavy mass. And telling *ME* I was all wet. Doh? The wrong mass? That's like this guy that is hired to drive a thousand miles. When he arrives, he's late. It took him 3 times as long as it should have. Boss says "What happend?" He says "Oh not much, I got a little lost."  Going 1000 miles in the wrong direction then having to drive 2000 miles back to the correct destination is not a small blunder.

And then I'd asked you what the velocity should be after 10 seconds, and instead you calculated the distance after first 6 minutes, but didn't tell me what the units for the number was.
Then you assumed some absurd radius - which you can't pin on me because you knew the info was missing - don't you know that if you're missing an input parameter you need to ask? You can't just guess at it! That's a blunder for sure. Especially an input which affects it by the square of the distance!

And now you're stuck on trying to contort gravity into something other than a force.

Not to mention that you told me my experiment looked wrong before you even did any math at all.

Can you see why I would not be inclined to place a strong confidence in information you provide to me?