r^(-2.9)

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19683
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Joined: Wed Jun 05, 2013 12:15 pm

r^(-2.9)

Postby 19683 » Wed Mar 09, 2016 6:32 pm

I was playing with various gravity laws and found that r^(-2.9) was very strange. The orbit spirals inward, and it seems like it will collapse, but then it swings around the star several times and the comes back out. Why does it behave like this?
Binomial Theorem: ((a+b)^n)= sum k=0->k=n((n!(a^(n-k))(b^k))/(k!(n-k)!))

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testtubegames
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Re: r^(-2.9)

Postby testtubegames » Thu Mar 10, 2016 2:57 pm

That's really weird, isn't it?

I'd noticed this with General Relativity on, too. But showing it with an even simpler force law (like r^-2.9) helps cut through some of the mystery around it.

Looking at it, I think what we're seeing is just precession. So just as an orbit in r^-2.1 will slowly advance forward each cycle, like so:
Screen Shot 2016-03-10 at 1.51.49 PM.png
Screen Shot 2016-03-10 at 1.51.49 PM.png (154.18 KiB) Viewed 5910 times

An orbit in r^-2.9 will do that same thing, but to a greater extreme.

It's as if the asteroid is tracing out an ellipse, but an ellipse that's also spinning around the star. That makes it look like an inward spiral followed by an outward spiral.

Make sense?

19683
Posts: 151
Joined: Wed Jun 05, 2013 12:15 pm

Re: r^(-2.9)

Postby 19683 » Thu Mar 10, 2016 5:03 pm

Yeah, I think I see what you mean. So the orbit itself precesses faster than the planet is moving and that makes it go around several times before getting farther again. I wonder what force laws can produce this effect, like if r^(-2.8) or r^(-2.5) will work.
Binomial Theorem: ((a+b)^n)= sum k=0->k=n((n!(a^(n-k))(b^k))/(k!(n-k)!))


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