Challenge: Lagrange Points

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19683
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Challenge: Lagrange Points

Post by 19683 »

Try to put an asteroid at each Lagrange point of a star-planet system. Try to make them as stable as possible.
Binomial Theorem: ((a+b)^n)= sum k=0->k=n((n!(a^(n-k))(b^k))/(k!(n-k)!))
Nicbudd
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Re: Challenge: Lagrange Points

Post by Nicbudd »

I think I have done it. It might be a little off

Code: Select all

Gravity Fun at TestTubeGames.com: [ForceG: 1,Qual: 1,Zoom: 1,xSet: 35,ySet: 23], [x0: -55,y0: -47,vx: 0,vy: 0,t0: 0,who: 1,m: 1000], [x0: -28,y0: 25,vx: 2.28,vy: -0.85,t0: 227.7,who: 2,m: 100], [x0: 1,y0: -46,vx: 0.03,vy: -1.77,t0: 464.4,who: 3,m: 0]
Chemi
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Re: Challenge: Lagrange Points

Post by Chemi »

Code: Select all

Gravity Fun at TestTubeGames.com: [ForceG: -2,Qual: 1,Zoom: 1,xSet: 0,ySet: 0], [x0: 3,y0: -13,vx: 0,vy: 0,t0: 0,who: 1,m: 1626], [x0: -150,y0: -13,vx: 0,vy: 3.26,t0: 0,who: 2,m: 42], [x0: -70,y0: 121,vx: 2.87,vy: 1.56,t0: 296.1,who: 3,m: 0], [x0: -77,y0: -140,vx: -2.78,vy: 1.75,t0: 296.1,who: 3,m: 0]
Okay? I placed it at L4 and L5 point!
A Random Player
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Re: Challenge: Lagrange Points

Post by A Random Player »

Nicbudd wrote:I think I have done it. It might be a little off

Code: Select all

Gravity Fun at TestTubeGames.com: [ForceG: 1,Qual: 1,Zoom: 1,xSet: 35,ySet: 23], [x0: -55,y0: -47,vx: 0,vy: 0,t0: 0,who: 1,m: 1000], [x0: -28,y0: 25,vx: 2.28,vy: -0.85,t0: 227.7,who: 2,m: 100], [x0: 1,y0: -46,vx: 0.03,vy: -1.77,t0: 464.4,who: 3,m: 0]
That's in r^1 gravity :P

---

So, I've tried this several times, but I never got it to work. My most recent attempt went like this:

Begin by using a simple orbit:

Code: Select all

Gravity Fun at TestTubeGames.com: [ForceG: -2,Qual: 1,Zoom: 1,xSet: 0,ySet: 0], [x0: 0,y0: 0,vx: 0,vy: 0,t0: 0,who: 1,m: 1000], [x0: 100,y0: 0,vx: 0,vy: -3.1622776601683795,t0: 0,who: 2,m: 100]
Find the radius that makes the adjusted orbital period equal to that of the planet:
2pi*r/sqrt(r*(1000/r^2+100/(r-100)^2))=2pi*sqrt(100^3/1000)
Explanation: 1000/r^2+100/(r-100)^2 is the force/acceleration toward the center of the system. By a=v^2/r, sqrt(r*(1000/r^2+100/(r-100)^2)) is the orbital velocity, and with C=2pi*r and d=vt, 2pi*r/sqrt(r*(1000/r^2+100/(r-100)^2)) (the LHS) is the period. 2pi*sqrt(100^3/1000) is the orbital period of the planet.
This gets us r=135.23367415675359103. It's rather straightforward to get the velocity from there, and to generate the resulting code:

Code: Select all

Gravity Fun at TestTubeGames.com: [ForceG: -2,Qual: 1,Zoom: 1,xSet: 0,ySet: 0], [x0: 0,y0: 0,vx: 0,vy: 0,t0: 0,who: 1,m: 1000], [x0: 100,y0: 0,vx: 0,vy: -3.1622776601683795,t0: 0,who: 2,m: 100], [x0: 135.23367415675359103,y0: 0,vx: 0,vy: -4.2764642668839177,t0: 0,who: 3,m: 100]
But running this has the asteroid escaping. Where am I going wrong with this?
$1 = 100¢ = (10¢)^2 = ($0.10)^2 = $0.01 = 1¢ [1]
Always check your units or you will have no money!
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robly18
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Re: Challenge: Lagrange Points

Post by robly18 »

I feel you, Random. I think it has to do with the gravitational constant :\
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A Random Player
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Re: Challenge: Lagrange Points

Post by A Random Player »

robly18 wrote:I feel you, Random. I think it has to do with the gravitational constant :\
I don't think that's very likely.. G is equal to 1 uzp^3/(μ*pf^2) (1 Unit-zoom pixel^3/(Mass unit * Pseudoframe^2), totally valid unit in games! :P), and the coordinates in the code are expressed in the same units, so I think it's probably a problem with the math.
$1 = 100¢ = (10¢)^2 = ($0.10)^2 = $0.01 = 1¢ [1]
Always check your units or you will have no money!
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robly18
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Re: Challenge: Lagrange Points

Post by robly18 »

I'm not sure... I did that too, and it failed for me too. What are the odds of both of us messing up?
Convincing people that 0.9999... = 1 since 2012
A Random Player
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Re: Challenge: Lagrange Points

Post by A Random Player »

robly18 wrote:I'm not sure... I did that too, and it failed for me too. What are the odds of both of us messing up?
Probability of you messing up * Probability of me messing up, from general multiplication rule? ;)

I'm going to think about this some more..
$1 = 100¢ = (10¢)^2 = ($0.10)^2 = $0.01 = 1¢ [1]
Always check your units or you will have no money!
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testtubegames
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Re: Challenge: Lagrange Points

Post by testtubegames »

This is a curious one. I took a look over the math... and I couldn't find any issues. I worked through it independently, got the same answer... checked references, got the same answer. And the asteroid still stubbornly zooms off into space, instead of staying nicely in L1.

So if it isn't an issue of doing the math wrong, I figure it has to come down to stability. This is, after all, a Lagrange point that is unstable. "But" you may object "the asteroid doesn't even look like it is *trying* to orbit next to the planet." That's what I thought, at least, until I realized that the stability may well be connected to the masses of the objects. As in -- Lagrange points might only be moderately unstable when you're talking about the Earth, Sun, and Moon, because the ratio of masses between the earth and sun is so very small (instead of .1, as in your example, the ratio is more like .00001). It turns out, as you'd expect, the stability (and amount of unstability) does hinge upon these types of issues. (for more on that, check out this mathy site).

I tried to make a system that was a bit more realistic, then. With the ratio of the masses at .001, and the distance increased to 1000 instead of 100, to boot.

Code: Select all

Gravity Fun at TestTubeGames.com: [ForceG: -2,Qual: 1,Zoom: 0.12,xSet: 0,ySet: 0], [x0: 0,y0: 0,vx: 0,vy: 0,t0: 0,who: 1,m: 1000], [x0: 1000,y0: 0,vx: 0,vy: -1,t0: 0,who: 2,m: 1], [x0: 1070.900553804888,y0: 0,vx: 0,vy: -1.07090055380487,t0: 0,who: 3,m: 0]
The asteroid still breaks free from the lagrange point in less than a rotation (heck, less than 1/4 a rotation!), but it's certainly better. Anyone want to take a crack at making it better? I imagine decreasing the mass even further is the way to go (we've got a couple orders of magnitude still before we reach the earth-sun case)
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robly18
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Re: Challenge: Lagrange Points

Post by robly18 »

I say we just give up and do this in r^1. That should prove easier.
Convincing people that 0.9999... = 1 since 2012
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