## r^-5 "Orbital Decay"?

What did you draw?
testtubegames
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### Re: r^-5 "Orbital Decay"?

A Random Player wrote:Not necessarily.. The universe is isotropic (I think) and infinite, so one will be pulled from all sides the same.. But how would the universe know which "side" of the infinity has more mass?
Yes, classically things would be very divergent on such a big scale. Though once we get to those scales, I suppose we need to turn our Bizarro-Newtonian-Gravity into some Bizarro-General-Relativity. Without thinking too hard about what all that means... I'd guess all the extra gravity would give us a 'Big Crunch' pretty quickly. Or, at least something close to a crunch... since as the universe got smaller, gravity would get weaker and weaker.

On that topic, since spacetime would curve *more* the further away you are from a mass... every object would have an event horizon if you go out far enough.

Thank goodness this is just a 2D simulation!

19683
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### Re: r^-5 "Orbital Decay"?

Gravity would actually be r^-1 in a 2-dimensional universe, just as it is r^-2 in a 3-d universe, r^-3 in 4-d and so on. However, a 2-d universe would be too simple for life, and a 4-d universe would be unstable.

Do you think there could be r^-2.5 or other decimals?
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robly18
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### Re: r^-5 "Orbital Decay"?

19683 wrote:a 4-d universe would be unstable.
Really? Please do explain why, I'm interested.
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19683
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### Re: r^-5 "Orbital Decay"?

robly18 wrote:
19683 wrote:a 4-d universe would be unstable.
Really? Please do explain why, I'm interested.
Because forces would follow an inverse cube (r^-3) force, there would be no stable orbits. Because of this, planets would fall into or escape from their stars. Also, electrons in atoms would fall into their nuclei. Without stable orbits, all large scale structures, including life, would either collapse or fly apart.

On a sub-microscopic scale however things might be different. I believe it would have something to do with a 4-dimensional Schrodinger equation. Any ideas?
Binomial Theorem: ((a+b)^n)= sum k=0->k=n((n!(a^(n-k))(b^k))/(k!(n-k)!))

robly18
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### Re: r^-5 "Orbital Decay"?

19683 wrote:Because forces would follow an inverse cube (r^-3) force, there would be no stable orbits.
How come? It sounds easy to me, to make an orbit with an inverse cube force...

Also, why would electrons fall into their nuclei?
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testtubegames
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### Re: r^-5 "Orbital Decay"?

Neat stuff! I wasn't thinking about dimensions when I added that feature... I was just looking to simply tweak the laws of physics and see what happens. But you're absolutely right.

The simulator was envisioned as a 2D slice of a 3D world, but of course, it could just as well be a 2D slice of a 4D world, or more. Very cool! As for fractional dimensions... I suppose perhaps in a fractal universe? Fractals can be described as having non-integer dimensions, IIRC. Though now I feel a bit like I'm writing the plot to a star trek episode...

And Robly, the inverse-cube force is indeed the dividing line between stable and unstable orbits. So while circular orbits are possible, they're highly unstable. If you try to make a circular orbit with r^-3.1, it lasts for a bit, then veers off. (In theory, it should so the same for r^-3, but I got bored watching and waiting )

Classically, the effect would the same for electrical forces -- hence why electrons wouldn't have stable orbits either. But as you say, 19683, we'd really need quantum mechanics to discuss that properly. And my extra-dimensional quantum mechanics is a bit rusty!

robly18
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### Re: r^-5 "Orbital Decay"?

testtubegames wrote: And Robly, the inverse-cube force is indeed the dividing line between stable and unstable orbits. So while circular orbits are possible, they're highly unstable. If you try to make a circular orbit with r^-3.1, it lasts for a bit, then veers off. (In theory, it should so the same for r^-3, but I got bored watching and waiting )
I thought that was do to rounding errors?
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testtubegames
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### Re: r^-5 "Orbital Decay"?

robly18 wrote:I thought that was do to rounding errors?
It's both, together. If the circular orbit were stable (like in r^-2), rounding errors don't matter... the orbit self-corrects. And if there were no rounding errors, no perturbations, the stable vs unstable distinction wouldn't matter.

It's like the proverbial pencil standing on it's tip. It is a highly unstable state. In an ideal world (with no rounding errors, nothing disturbing it, perfectly balanced to *all* the decimal places) it could stay standing that way forever, the fact that it's unstable would be irrelevant. In the real world, not so much.

So it is precisely because there *are* rounding errors in the simulation that the stable vs unstable distinction is important.

(And just to be clear, 'rounding errors' aren't a bug or anything. It's just that -to riff off you signature- while .999... equals 1, .9999999999999 most certainly does not equal 1)

A Random Player
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### Re: r^-5 "Orbital Decay"?

robly18 wrote:
testtubegames wrote: And Robly, the inverse-cube force is indeed the dividing line between stable and unstable orbits. So while circular orbits are possible, they're highly unstable. If you try to make a circular orbit with r^-3.1, it lasts for a bit, then veers off. (In theory, it should so the same for r^-3, but I got bored watching and waiting )
I thought that was do to rounding errors?
In theory, a perfect circular orbit would not decay, but because of rounding errors here, the orbit is not actually perfect, and decays.

Edit: Ninja'd. I agree with Andy though.
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19683
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### Re: r^-5 "Orbital Decay"?

robly18 wrote:
testtubegames wrote: And Robly, the inverse-cube force is indeed the dividing line between stable and unstable orbits. So while circular orbits are possible, they're highly unstable. If you try to make a circular orbit with r^-3.1, it lasts for a bit, then veers off. (In theory, it should so the same for r^-3, but I got bored watching and waiting )
I thought that was do to rounding errors?
The rounding errors are what cause it to go slightly off, but after that the instability takes it way off course. If the orbit was truly stable, a small error can't throw it off course.

In a real orbit, small external forces would cause the slight offset.
Binomial Theorem: ((a+b)^n)= sum k=0->k=n((n!(a^(n-k))(b^k))/(k!(n-k)!))