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vf321

  • 2 years ago

With the initial conditions for a two mass system m1 and m2, where neither mass is significantly larger than the other, the resulting DE is unsolvable, right? Assume that they both start at rest: (see DE below) Do things get simpler if I give a starting velocity to both masses? What if I look at position relative to the com of the system?

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  1. vf321
    • 2 years ago
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    The DE for the two masses in 2D space in general:\[m_1\frac{d^2\vec r_1}{dt^2}=G\frac{m_1 m_2 (\vec r_2-\vec r_1)}{|\vec r_2-\vec r_1|^3}\](switch necessary subscripts for m2) Making the assumption that they're not moving at first, we know movement will be linear. Make the linear axis "x": \[m_1\frac{d^2x_1}{dt^2}=G\frac{m_1 m_2 (x_2-x_1)}{|x_2-x_1|^3}=\pm G\frac{m_1m_2}{(x_2-x_1)^2}\]So we get a system of DEs: \[m_1\frac{d^2x_1}{dt^2}=\pm G\frac{m_1m_2}{(x_2-x_1)^2}\]\[m_2\frac{d^2x_1}{dt^2}=\mp G\frac{m_1m_2}{(x_2-x_1)^2}\]Still unsolvable to my knowledge...

  2. vf321
    • 2 years ago
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    EDIT: last equation should be\[m_2\frac{d^2x_2}{dt^2}=\mp G\frac{m_1m_2}{(x_2-x_1)^2}\]

  3. experimentX
    • 2 years ago
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    consider one of the mass to be at rest and another is accelerating.

  4. vf321
    • 2 years ago
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    You can't do that though! m_2 is accelerating too!

  5. experimentX
    • 2 years ago
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    well if you want a dynamic system then ... it's going to be a system of second order DE.

  6. vf321
    • 2 years ago
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    Yeah... But that's the problem. The DE seems to be unsolvable. My question is asking if there's some clever symmetry (like the com or something) that we can use to get around the DE.

  7. experimentX
    • 2 years ago
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    not sure ... first order system is fairly complicated.

  8. experimentX
    • 2 years ago
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    plus ... this is non linear :(( looks like there must be some other approach.

  9. vf321
    • 2 years ago
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    Yeah but its not a question from a textbook... I just made this up. There might not be an approach :(.

  10. experimentX
    • 2 years ago
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    let me try to ask mathematica. do you want general solution or ... do you have some boundary condition?

  11. experimentX
    • 2 years ago
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    didn't get any solution from mathematica.

  12. vf321
    • 2 years ago
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    I think I have something that might help simplify a bit (N's 3rd law): \[m_2\frac{d^2x_2}{dt^2}=-m_1\frac{d^2x_1}{dt^2}\]\[\int_0^tm_2\frac{d^2x_2}{dt^2}dt=\int_0^t-m_1\frac{d^2x_1}{dt^2}dt\]\[m_2(\frac{dx_2}{dt}-v_{i2})=-m_1(\frac{dx_1}{dt}-v_{i1})\]\[v_{i1}, v_{i2}=0 m/s\]\[\int_0^tm_2\frac{dx_2}{dt}dt=\int_0^t-m_1\frac{dx_1}{dt}dt\]\[m_2(x_2(t)-x_{i2})=-m_1(x_1(t)-x_{i1})\]\[x_2(t)=\frac{-m_1(x_1(t)-x_{i1})}{m_2}+x_{i2}\]That seems right. If it is, then we don't have a system of DE's anymore...

  13. vf321
    • 2 years ago
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    I think we get: \[m_1\frac{d^2x_1}{dt^2}=\pm G\frac{m_1m_2}{(-(m_1x_1/m_2)-x_1+c_1)^2}\]\[c_1=\frac{m_1x_{i1}}{m_2}+x_{i2}\]

  14. experimentX
    • 2 years ago
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    let dx/dt = v ... dv/dt = dv/dx *dx/dt = v dv/dx not sure if you integrate it wrt dt ...

  15. vf321
    • 2 years ago
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    is \[\int v(t)dt=x(t)\]?

  16. vf321
    • 2 years ago
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    Cause that's all i did.

  17. experimentX
    • 2 years ago
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    this doesn't solve x(t) at all ... we need x as explicit function of time. sorry ... i think integration method would work.

  18. vf321
    • 2 years ago
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    It doesn't solve for x(t) as a function of time, but it makes x_2 represented as x_1. This way we have one DE to solve instead of a system, right?

  19. experimentX
    • 2 years ago
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    http://en.wikipedia.org/wiki/Two-body_problem i remember doing this kinda (central force) problem using reduced mass :((

  20. experimentX
    • 2 years ago
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    something like |dw:1345580217154:dw|

  21. experimentX
    • 2 years ago
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    still this system is interesting ... right now i'm working on first order systems. Maybe i'll start PDE right after it. If i encounter second order system ... i'll try to update.

  22. experimentX
    • 2 years ago
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    i think using symmetricity we can simplify this problem for two identical bodies. |dw:1345580536599:dw|

  23. experimentX
    • 2 years ago
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    for non identical bodies , the acceleration produced is equal to their ratio of masses |dw:1345580770508:dw|

  24. experimentX
    • 2 years ago
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    |dw:1345580931804:dw| choosing initial condition, and choosing a suitable coordinate system, i think .... this is pretty easy to solve x(t) as explicit function of time.

  25. experimentX
    • 2 years ago
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    I'm not sure for general case ... with non rest IC.

  26. vf321
    • 2 years ago
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    Okay thx i looked at the wiki link. I'm on to something on my own and I'll see if that works.

  27. vf321
    • 2 years ago
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    OK no one else is helping at physics so I'll repost the problem in the math section...

  28. vf321
    • 2 years ago
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    If any1's interested: http://math.stackexchange.com/questions/185593/solving-the-de-for-a-two-body-system

  29. experimentX
    • 2 years ago
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    try asking this instead. How to solve for general solution of system of this differential equation? |dw:1345676009886:dw|

  30. experimentX
    • 2 years ago
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    more over this is just 1D case ... for general 2d case the system of DE could me much worse.

  31. vf321
    • 2 years ago
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    Yeah I know it's 1D I was trying just to tackle this, let alone the 2D.

  32. experimentX
    • 2 years ago
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    i guess CM frame of reference is a better choice to solve problem. later we can translate origin to any other choice after we find x and y. Try to simulate this graphically.

  33. experimentX
    • 2 years ago
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    since we know, the distance between the objects from CM with be at constant proportion of their masses.

  34. vf321
    • 2 years ago
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    It's okay, I'll just do it out on my own using the CM. Don't worry about it.

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