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 Bound trajectories check (Read 3649 times)
 frankuitaalst Ultimate Member Great site Posts: 1510 Gender: Bound trajectories check 09/21/14 at 11:14:10   It is known that in a two body problem a planet is bound if the total energy is smaller than zero   ( elliptical orbit ) .   In this case we have : E= Ep+Ek <0 :  mv²/2 - GmM/r <0 . This means the velocity should not be too big .     I now have a system of a planet orbiting  a binary.   And want to have a quick check if the planet is bound to the binary.   Any idea's how I should proceed ?     mv²/2 -GmMtot <0  (vrel to center of mass Mtot )   or :  m Vrel1²/2+ mVrel2²/2 - GmM1/r1 -GmM2/r2 <0   ( two bodies considered ) Back to top IP Logged
 Tony YaBB Administrator Posts: 1060 Gender: Re: Bound trajectories check Reply #1 - 09/22/14 at 06:38:58   Compute the center of mass of the binary.  Pretend all the mass was there, and treat it as a single object.  Now see if energy is negative, or check if ecc < 1 or sma is positive. Back to top IP Logged
 frankuitaalst Ultimate Member Great site Posts: 1510 Gender: Re: Bound trajectories check Reply #2 - 09/23/14 at 13:21:43   Quote from Tony on 09/22/14 at 06:38:58:Compute the center of mass of the binary.  Pretend all the mass was there, and treat it as a single object.  Now see if energy is negative, or check if ecc < 1 or sma is positive. Thanks Tony , you think treating the binary at its COG gives better results than taking account of the two seperate masses ?   I think in general yes it is but :   If COG is used one  does not take into account a change in energy due to a close encounter of one part of the binary .   I'm not very sure in this case .   Is there a way to test  in GravSim the result if I would test the second approach ? Back to top IP Logged
 frankuitaalst Ultimate Member Great site Posts: 1510 Gender: Re: Bound trajectories check Reply #3 - 09/24/14 at 22:41:33   I gave it another thought .   Of couse you are right Tony , at long distances this is indeed the criterium .   I just wondered If at close distances the separate approach could be appropriate . I had the impression that although a body which is in eliptical orbit around a binary can be trown out after a "close " approach to one part of the binary . Back to top IP Logged
 Tony YaBB Administrator Posts: 1060 Gender: Re: Bound trajectories check Reply #4 - 09/25/14 at 18:26:47   I think the secondary bodies have to be about 5x as far as the aphelion of the primary stars for stable orbits. Gravity Simulator can do this.   File > New Create another star around the Sun.     File > New Create a low-mass body around the Sun and choose "barycenter" in the focus object.   There's also a View > View Orbital Energy option, but I think I coded it wrong.  It's not doing what I expect.  You could always output data to view in Excel.  Energy of the system is the sum of U + KE between every body in the system (-GMm/d^2 + 0.5 * m * v^2).  It should be constant. Back to top IP Logged
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