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

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Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #60 on: March 28, 2019, 08:34:26 PM »
This also raises an interesting question, as to whether an orbit would be possible with a relation ship of the inverse of the 4th power.

It would not matter what inverse power of r the force decreased with, a stable orbit would be possible. It would not even have to decrease as a power of r, it could be any well-behaved decreasing function of r

Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #61 on: March 28, 2019, 08:51:29 PM »
This also raises an interesting question, as to whether an orbit would be possible with a relation ship of the inverse of the 4th power.

It would not matter what inverse power of r the force decreased with, a stable orbit would be possible. It would not even have to decrease as a power of r, it could be any well-behaved decreasing function of r

Sweeet! so magnets on ice might be a rather interesting orbital experiment since we could have two objects of different mass orbiting eachother as a rather interesting visual aide.

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Heavenly Breeze

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Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #62 on: March 28, 2019, 09:14:56 PM »
 Tom Foolery, you surpassed even  JackBlack and rabinos. Bravo, I knock hooves of emotion.
My subtitles don't work, so please tell me what you think about this.
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« Last Edit: March 28, 2019, 09:19:15 PM by Heavenly Breeze »
The earth believes, because magic exists!

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JackBlack

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Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #63 on: March 29, 2019, 02:26:00 AM »
It would not matter what inverse power of r the force decreased with, a stable orbit would be possible. It would not even have to decrease as a power of r, it could be any well-behaved decreasing function of r
Do you have any citation for that?
So far from Rab I have heard it works if the power is greater than or equal to -2, elsewhere I have seen it is true if the power is 2 or -2.

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JackBlack

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Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #64 on: March 29, 2019, 02:32:53 AM »
My subtitles don't work, so please tell me what you think about this.
It is just a typical FE troll ignoring other factors.
Yes, gravity pulls objects towards the centre of Earth, but we also have the centrifugal force to deal with.

Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #65 on: March 29, 2019, 09:22:59 AM »
It would not matter what inverse power of r the force decreased with, a stable orbit would be possible. It would not even have to decrease as a power of r, it could be any well-behaved decreasing function of r
Do you have any citation for that?
So far from Rab I have heard it works if the power is greater than or equal to -2, elsewhere I have seen it is true if the power is 2 or -2.

You can cite me :=)

For a mass m  to orbit earth with mass M at radius R with velocity V, the centrepital force has to equal the attractive force or

(mV^2)/R = GmM/R^2 for the situation where the attractive force decreases as 1/R^2. G is the gravitational constant.

Solve for the Velocity versus radius to get

V^2 = GMR(1/R^2) as the condition for orbit

You can replace the  (1/R^2) with (1/R^3) or (1/R^4) or any other well behaved function f(R) that goes monotonocally to zero faster than 1/R and the equation can be solved (at least numerically) for the Velocity needed for any radius R

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JackBlack

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Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #66 on: March 29, 2019, 02:46:50 PM »
For a mass m  to orbit earth with mass M at radius R with velocity V, the centrepital force has to equal the attractive force or

(mV^2)/R = GmM/R^2 for the situation where the attractive force decreases as 1/R^2. G is the gravitational constant.

Solve for the Velocity versus radius to get

V^2 = GMR(1/R^2) as the condition for orbit

You can replace the  (1/R^2) with (1/R^3) or (1/R^4) or any other well behaved function f(R) that goes monotonocally to zero faster than 1/R and the equation can be solved (at least numerically) for the Velocity needed for any radius R
So you misunderstand what is meant by stability.
You can have an orbit existing with any circularly symmetric force. It can be a nice simple function of r or it can be a horribly complex one complete with discontinuities (as long as you aren't doing it at the discontinuity). If you stick in a value of r you will get some force at that point, you then find the required velocity for that force and radius and you are done.

But this gives a perfectly circular orbit. The issue is what if this orbit is perturbed or you don't start off with a perfect circle, i.e. what if you have a slightly higher or lower tangential velocity, or if you have any radial velocity?
For an unstable orbit, a slight change can result in the orbit collapsing.
A real life example of this is the L1 point, which also links to magnetic levitation.
The L1 point for an orbit is where you have a small body perfectly balanced between 2 larger bodies, e.g. a small satellite between Earth and the sun.
The forces work out to give a system where the object remains in place. For the L1 point this means that as the object orbits the sun, the Earth provides enough force outwards to slow the orbit such that it keeps the same position relative to Earth such that the sun, object and Earth are all aligned.
For magnetic levitation you have it so the magnetic force just cancels out gravity pulling it down so it stays put.
The problem with both is that they are inherently unstable.
If your satellite moves closer to the sun, then it will need a corresponding stronger force from Earth, but as it moves further away from Earth it gets a weaker force from Earth and thus the satellite moves from its position and starts just orbiting the sun disconnected from Earth.
For the magnet example, if it gets closer to the magnet the force increases and it flies up to the magnet; if it gets further away the force decreases and gravity wins and the object falls.

Both of these are intrinsically unstable. While you do have a perfectly balanced point, any slight perturbation will result in it going away from that point.

For a stable orbit the opposite would happen, you have a perfectly balanced point and any slight perturbation (note: slight, not massive), will result in it going back to that point.
For example, with a gravitationally bound orbit around a large body, you have the hypothetical perfectly balanced circular orbit. If you are going too fast then you go to a higher orbit, but also slow down as gravity would then be pulling you back, not just inwards. This results in you reaching a peak where you are going too slow for the orbit and thus you start falling down. Gravity then speeds you up and you end up going back up, oscillating between 2 radii in a stable orbit.


As a simple analogy, a stable orbit corresponds to suspending a long rod from the top, an unstable one would be having it stand vertically on a tiny ball.

Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #67 on: March 29, 2019, 05:11:14 PM »
Well I was not meaning to get into stability analysis, just to show the conditions for a "normal" circular orbit could be met with a wide range of functions for the attractive force.

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rabinoz

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Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #68 on: March 29, 2019, 06:19:10 PM »
Well I was not meaning to get into stability analysis, just to show the conditions for a "normal" circular orbit could be met with a wide range of functions for the attractive force.
A circular orbit that remains bounded when subject to small perturbations is possible for f(r) = -k/rn for n ≥ -2.

Two common ones are n = -2, Newtonian Gravitation, and n = 1, Hooke's Law for a spring.

Re: Is square inverse force orbit possible in reality, and mathematically?
« Reply #69 on: March 29, 2019, 11:00:50 PM »
Tom Foolery, you surpassed even  JackBlack and rabinos. Bravo, I knock hooves of emotion.
Wow thanks... I think...
Quote
My subtitles don't work, so please tell me what you think about this.
Ahh, is that like the old pickup line "Hey I lost my phone number, could I have yours?"


Heavenly Breeze,

I watched your video for you.

First of all, I might not be the best person to answer your questions.

I already did an experiment that proved gravity:



I've also spent plenty of time on a merry go around, like this one:



I can personally confirm for you that I have observed what appears to be gravity, and I have personally experienced the kind of centrifugal force which would cause a globe earth to be oblate.

You can also fill a water balloon with water and hang it on a string with a motor to get it spinning and see if it takes on a generally oblate shape.

I did watch the video, and the guy is totally confused, claiming that if the earth is really an oblate spheroid, then things would fall crosswise if you dropped them.

What he doesn't realize about round earth theory is that the local gravity for any given location causes a level local to that area, and a line perpendicular to level is not even supposed to always go through the center of the earth.

So look, I've observed gravity. I've observed centrifugal force.

The guy in the video has never done the gravity experiment. Probably never been on a merry go around. It would not make sense for me, who has done the gravity experiment, to believe that guy who's never done the gravity experiment.

Would you take advice about using a computer from somebody who'd never seen one in their life and obviously knew nothing about them?

https://vimeo.com/37942209

"Hey Dad, can you help me adjust my email settings on my ipad?"
"Sure can, just let me rinse off the onions."