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Objects falling towarda each other.

  1. Dec 16, 2009 #1
    1. The problem statement, all variables and given/known data
    m1 and m2 only interact gravitationally. They are initially at rest and a distance L apart. They where then released. Where and when will they collide?


    2. Relevant equations
    [tex]F_g=G\frac{m_1m_2}{x^2}[/tex]

    3. The attempt at a solution
    For the where part, my argument is that initially at the CM frame, the total momentum is zero, therefore, they must collide at the origin of the CM frame. I think this is not a good argument.

    For the time part, I have no clue. I mechanics is really weak. Despite being a physics major I am not required to take any classical mechanics course to graduate.
     
  2. jcsd
  3. Dec 16, 2009 #2

    tiny-tim

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    Hi E92M3! :smile:
    It is an excellent agrument! :approve:

    Have a mince pie! :smile:
    Stay with the CM frame.

    Hints:

    i] if the distance of mass 1 from the CM is x, the distance of mass 2 is … ?

    ii] special trick: a = dv/dt = dv/dx dx/dt (chain rule) = v dv/dx. :wink:
     
  4. Dec 16, 2009 #3
    Just to make sure, is this because the frame of the masses are accelerating and therefore not inertial frames so newtons' laws don't apply?

    The distance of mass 2 from the origin of the CM frame is initially L-x. But since they both move towards the origin I don't know what it will be at a later time.

    Well I can look at mass 2 (it is define to be the mass on the negative x-axis initially):
    [tex]F_2=G\frac{m_1m_2}{(x_1-x_2)^2}[/tex]
    Here, x1 is the distance from the CM to m1 and x2 is the distance from the CM to m2. They are both functions of t.

    Following the hint I can say:
    [tex]a_2=G\frac{m_1}{(x_1-x_2)^2}=v_2\frac{dv_2}{dx_2}[/tex]
    [tex]Gm_1\int\frac{dx_2}{(x_1-x_2)^2}=\int v_2 dv_2[/tex]
    [tex]\frac{-Gm_1}{x_1-x_2}=\frac{v_2^2}{2}+C[/tex]

    At t=0, v2=0 and x2=x1-L:

    [tex]\frac{-Gm_1}{L}=C[/tex]
    [tex]\frac{-Gm_1}{x_1-x_2}=\frac{v_2^2}{2}+\frac{-Gm_1}{L}[/tex]

    I'm kindda still stuck. This is a differential equation, but it is not one that I can solve readily.
     
  5. Dec 16, 2009 #4

    tiny-tim

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    Hi E92M3! :smile:
    No, it's because there are no external forces, so (Newton's first law) the CM has uniform velocity … which, in the CM frame, means it stays where it is.

    Good ol' Newton's laws always apply (in an inertial frame).
    (just to make it clear, "the CM frame" means the inertial frame in which, at one particular time, the CM is stationary: Newton's first law then tells you that it will always be stationary in that frame)
    You're missing the point … the CM is stationary, so the ratio of the distances will always be m1/m2

    so use that to write x2 as a function of x1 in the force equation. :wink:
     
  6. Dec 16, 2009 #5
    Ok, I think I'm in deep trouble. I am taking an astronomy course and final is in 3 days. While doing the past papers, I found out that I can do any of the classical mechanical questions.

    Here's what I think,

    [tex]m_1x_1=m_2x_2[/tex]
    therefore:
    [tex]x_1=\frac{m_2}{m_1}x_2[/tex]

    Putting this into the force I get:
    [tex]F_2=G\frac{m_1m_2}{x_2^2(\frac{m_2}{m_1}-1)^2}[/tex]

    Following the same procedure as I did before I get:
    [tex]\frac{-Gm_1}{x_2(\frac{m_2}{m_1}-1)^2}=\frac{v_2^2}{2}-\frac{Gm_1}{(\frac{m_2}{m_1}-1)L}[/tex]

    Once again I ended up with a differential equation I can't solve.
     
  7. Dec 16, 2009 #6

    tiny-tim

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    That's ok, except you need a + in the denominator, not a -, don't you? :wink:

    Now write v = dx/dt = f(x), so dx/f(x) = dt, and integrate. :smile:
     
  8. Dec 16, 2009 #7
    Sorry, I am really not getting this. I guess of I want to keep the minus sign I need to write:
    [tex]m_1x_1+m_2x_2=0[/tex]
    Since I defined x1 and x2 to be the vector pointing from the origin of the CM frame to the 2 masses. But I really don't know what to do next. I can modify this to be:
    [tex]
    \frac{-Gm_1}{x_2(-\frac{m_2}{m_1}-1)^2}=\frac{v_2^2}{2}-\frac{Gm_1}{(-\frac{m_2}{m_1}-1)L}
    [/tex]

    Am I correct here? Is my integral valid?
    [tex]
    Gm_1\int\frac{dx_2}{(x_1-x_2)^2}=\int v_2 dv_2
    [/tex]
     
  9. Dec 16, 2009 #8

    tiny-tim

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    Hi E92M3! :smile:

    If we make x1 and x2 both positive, then r = x1 + x2 = (m1 + m2)x2/m1.
    Where did dv come from? :confused:

    The trick is to write v = dx/dt, the dx is then joined by the function of x on the other side (it'll be inside a √, because you started with v), while the dt goes to the other side, on its own. :wink:
     
  10. Dec 16, 2009 #9
    Wait what is r here? It seems like r is the separation between the two masses am I correct? Then how did you get this:r = x1 + x2 = (m1 + m2)x2/m1?

    I got the dv from the chain rule:
    [tex]a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}[/tex]



    [tex]
    a_2=G\frac{m_1}{(x_1-x_2)^2}=v_2\frac{dv_2}{dx_2}
    [/tex]

    Then I integrated both sides:

    [tex]

    Gm_1\int\frac{dx_2}{(x_1-x_2)^2}=\int v_2 dv_2

    [/tex]

    And applied the initial condition that initially v=0 and they are separated by L to get the integration constant.
     
  11. Dec 16, 2009 #10

    tiny-tim

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    ah, now I see it in full, it's ok :smile:
    if x1 and x2 are both positive, then r = x1 + x2 = m2x2/m1 + x2 :wink:
     
  12. Dec 16, 2009 #11
    Ok I set them both to positive then I get this after the integration:
    [tex]

    \frac{-Gm_1}{x_2(\frac{m_2}{m_1}+1)^2}=\frac{v_2^2}{2}-\frac{Gm_1}{(\frac{m_2}{m_1}+1)^2L}

    [/tex]
    This is not something I can solve though.
     
  13. Dec 16, 2009 #12

    tiny-tim

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    As I said before …
     
  14. Dec 16, 2009 #13
    So... this?
    [tex]


    \frac{-Gm_1}{x_2(\frac{m_2}{m_1}+1)^2}=\frac{v_2^2}{2}-\frac{Gm_1}{(\frac{m_2}{m_1}+1)^2L}


    [/tex]
    [tex]


    \frac{Gm_1}{(\frac{m_2}{m_1}+1)^2L}-\frac{Gm_1}{x_2(\frac{m_2}{m_1}+1)^2}=\frac{v_2^2}{2}


    [/tex]
    [tex]


    \sqrt{2}\sqrt{\frac{Gm_1}{(\frac{m_2}{m_1}+1)^2L}-\frac{Gm_1}{x_2(\frac{m_2}{m_1}+1)^2}}=v


    [/tex]
    [tex]\sqrt{\frac{2Gm_1}{(\frac{m_2}{m_1}+1)^2}}\sqrt{\frac{1}{L}-\frac{1}{x_2}}=\frac{dx_2}{dt}[/tex]
    [tex]\sqrt{\frac{2Gm_1}{(\frac{m_2}{m_1}+1)^2}}\int dt=\int \frac{1}{\sqrt{\frac{1}{L}-\frac{1}{x_2}}} dx_2[/tex]
    [tex]t\sqrt{\frac{2Gm_1}{(\frac{m_2}{m_1}+1)^2}}=\int \sqrt{\frac{Lx_2}{x_2-L}} dx_2[/tex]
    This is not a form of integral that I recognize. I tried a few tricks but they don't really work.
     
    Last edited: Dec 16, 2009
  15. Dec 17, 2009 #14

    tiny-tim

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    Hi E92M3! :smile:

    (just got up :zzz: …)
    Try a trig or hyperbolic trig substitution (if you like, substitute x = y2 or y2 + C first) :smile:

    (btw, I think we missed a minus in the original equation … r'' = -F :wink:)
     
  16. Dec 17, 2009 #15
    Thanks!
    I think I got:
    [tex]t=\frac{\pi L^{3/2}}{2(\frac{m_2}{m_1}+1)^{3/2}\sqrt{2Gm_1}}[/tex]
    I think that this is the solution. I'm dead if the test consists of this king of question.
     
  17. Dec 17, 2009 #16

    tiny-tim

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    I'm not convinced …

    I haven't worked it through, but I think it's going to involve an inverse trig …

    if you want me to check it, you'd better write it out in full. :smile:
     
  18. Dec 17, 2009 #17
    well, the units worked out though. anyways:

    [tex]F_2=\frac{Gm_1m_2}{(x_1+x_2)^2}[/tex]
    [tex]a_2=\frac{Gm_1}{(x_1+x_2)^2}=\frac{Gm_1}{x_2^2(\frac{m_2}{m_1}+1)^2}[/tex]
    note:
    [tex]m_1x_1=m_2x_2[/tex]
    [tex]a=v\frac{dv}{dx}[/tex]
    [tex]\frac{Gm_1}{(\frac{m_2}{m_1}+1)^2}/int \frac{1}{x_2^2}dx_2=\int v_2 dv_2[/tex]
    I used the condition that their initial velocity is zero and the initial separation is L to solve for the integration constant.
    [tex]\frac{Gm_1}{(\frac{m_2}{m_1}+1)^2} /left ( \frac{1}{L}-\frac{1}{x_2(\frac{m_2}{m_1}+1)} \right )=\frac{v_2^2}{2}[/tex]
    Now I set v=dx/dt then integrate:
    [tex]\sqrt{\frac{2GM_1}{\frac{m_2}{m_1}+1}} \int dt= \int \sqrt{\frac{x_2L(\frac{m_2}{m_1}+1)}{x_2(\frac{m_2}{m_1}+1)-L}}dx_2[/tex]
    Now I let u=x2(m2/m1+1):
    [tex]t=\frac{1}{(\frac{m_2}{m_1}+1)^{3/2}sqrt{2GM_1}} \int\sqrt{\frac{uL}{L-u}}[/tex]
    To find the limits of integration, u=x1+x2, then u is from L to 0. This allows me to flip the sign in the bottom.
    [tex]t=\frac{1}{(\frac{m_2}{m_1}+1)^{3/2}\sqrt{2GM_1}} \int_0^L \sqrt{{\frac{uL}{u-L}}}=t=\frac{1}{(\frac{m_2}{m_1}+1)^{3/2}\sqrt{2GM_1}}\frac{\pi L^{3/2}}{2}[/tex]
     
  19. Dec 17, 2009 #18

    tiny-tim

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    No, you can't just flip the sign and make negative square roots.

    And I don't understand the last bit, are you saying ∫√(u/(L - u)) du is proportional to u3/2 ?

    Try u = y2.
     
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