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vf321 Group Title

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?

  • 2 years ago
  • 2 years ago

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

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

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

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

    • 2 years ago
  6. vf321 Group Title
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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.

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

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

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

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

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

    • 2 years ago
  12. vf321 Group Title
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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...

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

    • 2 years ago
  14. experimentX Group Title
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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 ...

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

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

    • 2 years ago
  17. experimentX Group Title
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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.

    • 2 years ago
  18. vf321 Group Title
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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?

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

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

    • 2 years ago
  21. experimentX Group Title
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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.

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

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

    • 2 years ago
  24. experimentX Group Title
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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.

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

    • 2 years ago
  26. vf321 Group Title
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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.

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

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

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

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

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

    • 2 years ago
  32. experimentX Group Title
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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.

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

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

    • 2 years ago
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