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How long for 2 particles to collide due to gravity?

  1. Jun 26, 2013 #1
    1. The problem statement, all variables and given/known data

    Two particles (masses m1, m2) are released from rest a distance D apart in space. How long until they collide?

    2. Relevant equations

    The force between the particles is [itex]F_G(t)=\frac{Gm_1m_2}{r(t)^2}[/itex].

    The center of mass is located a distance [itex]r_1(t)=\frac{m_2}{m_1+m_2}r(t)[/itex] from particle 1 and [itex]r_2(t)=\frac{m_1}{m_1+m_2}r(t)[/itex] from particle 2. Note that [itex]r_1(t)+r_2(t)=r(t)[/itex].

    3. The attempt at a solution

    The collision will take place at the center of mass. Particle 1 needs to traverse a total distance of [itex]d_1=\frac{m_2}{m_1+m_2}D[/itex] to reach the COM. Its acceleration over this distance is
    [itex]\ddot r_1(t)=\frac{m_2}{m_1+m_2}\ddot r(t)=\frac{Gm_2}{r(t)^2}[/itex].​

    OR! Another approach I thought of was to use potential and kinetic energy:
    [itex]=-\frac{Gm_1m_2}{r(t)}+\frac{1}{2}m_1m_2\dot r(t)^2[/itex].​

    But either way, I wind up with a differential equation that's totally inhumane. I feel like I'm missing something here. Open to suggestion on how to further either of these two approaches, or other approaches entirely.
  2. jcsd
  3. Jun 26, 2013 #2
    Using energy is definitely the right way to go about it. Start off with the fact that the center of mass is 0 potential energy for both objects. You should be able to find a useful relationship between [itex]m_1, m_2, r_1[/itex] and [itex] r_2[/itex]. Next you will probably want to compare the kinetic energy of the two objects when the collide and the potential energy they begin with. We know kinetic energy has velocity in it. We also know we can relate distance, time and velocitz to each other. That is where you will get the amount of time it takes for them to collide.

    Hope that helps.
    Last edited: Jun 26, 2013
  4. Jun 26, 2013 #3
    Sorry double posted by accident
  5. Jun 26, 2013 #4


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    This problem has been posted at least twice before, but assuming you want to solve it yourself, start off by determining the rate of acceleration of each particle versus r(t), and subtract the accelerations (consider one of the accelerations to be negative, so you're really adding the magnitudes of accelerations) to determine the rate of acceleration of closure between the particles, and use that as the second derivative of r(t) (which will be negative since it's a rate of closure). This will end up with a tricky integral, but if you get that far, I or someone else can offer help.
    Last edited: Jun 26, 2013
  6. Jun 26, 2013 #5
    Thanks for the reply, rcgldr. And sorry for the duplicate question but all the words in this problem are so common to physics problems that it was really hard to search for.

    Anyhow, particle 1 feels a force of
    and so it accelerates toward the COM at a rate of
    [itex]\ddot r_1(t)=\frac{F_G(t)}{m_1}=\frac{Gm_2}{r(t)^2}[/itex].​
    Having established the direction of particle 1's motion as positive, particle 2 then feels the equal and opposite force of
    and accelerates at
    [itex]\ddot r_2(t)=\frac{F_G(t)}{m_2}=-\frac{Gm_1}{r(t)^2}[/itex].​
    The distance between the particles is
    so the acceleration of closure is
    [itex]\ddot r(t)=\ddot r_1(t)+\ddot r_2(t)=\frac{G}{r(t)^2}(m_2-m_1)[/itex].​
    You say this acceleration is necessarily negative, but it looks to me like its sign would be based on which particle is more massive. So am I right so far? If so, I'm once again stuck with an intractable differential equation: I even tried feeding it to Wolfram|Alpha and I just got back a message saying “Standard computation time exceeded.”
  7. Jun 26, 2013 #6

    Let [itex]v(t)=\frac{dr}{dt}[/itex]. Then [itex]\ddot r(t)=\frac{dv}{dt}=\frac{dv}{dr}\cdot\frac{dr}{dt}=v\frac{dv}{dr}[/itex].

    So our equation becomes [itex]v\frac{dv}{dr}=\frac{G(m_2-m_1)}{r}[/itex], which is separable to [itex]\int v dv = G(m_2-m_1)\int\frac{1}{r}dr[/itex].

    Got to go to work now but this seems pretty promising.
  8. Jun 26, 2013 #7


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    A minor change is needed. Assume m2 is to the right of m1, position of m1 is x1(t) (left of COM) and position of m2 is x2(t) (right of COM). For the integration constant, at t = 0, v = 0, r = r0.


    [itex]\ddot r_1(t)=\frac{F_G(t)}{m_1}=\frac{Gm_2}{r(t)^2}[/itex]

    [itex]\ddot r_2(t)=\frac{F_G(t)}{m_2}=-\frac{Gm_1}{r(t)^2}[/itex]

    [itex]\ddot r(t)=\ddot r_2(t)-\ddot r_1(t)=-\frac{G}{r(t)^2}(m_1+m_2)[/itex].


    [itex]\int v dv = -G(m_1+m_2)\int\frac{1}{r^2}dr[/itex].

    Last edited: Jun 26, 2013
  9. Jun 26, 2013 #8
    Thank you, thank you. Of COURSE r(t)=x2(t)−x1(t). It's always the little things that get me. Can't take the time to finish the problem right now because I'm at work, but you got me over the hurdle, I think.
  10. Jul 7, 2013 #9
    Alas, I spoke too soon. I finally returned to this problem to find myself still stuck. When we left off, I had this integral
    [itex]\displaystyle\int_0^{v(t)} v \,dv = -G (m_1 + m_2) \displaystyle\int_{r_0}^{r(t)} \dfrac {1} {r^2} \,dr[/itex]​
    which I can evaluate to get
    [itex]\dfrac 1 2 v(t)^2 = G (m_1 + m_2) \bigg( \dfrac {1} {r(t)} - \dfrac {1} {r_0} \bigg)[/itex].​

    But where does that leave me, in terms of my original project of finding the time to collision? If I put back dr/dt in place of v, I have
    [itex]\dfrac {dr}{dt} = \bigg[ 2G (m_1 + m_2) \bigg( \dfrac {1} {r(t)} - \dfrac {1} {r_0} \bigg) \bigg]^{1/2}[/itex],​
    but I decidedly have no idea what to do when faced with the new integral that leaves me with:
    [itex]\displaystyle\int_0^{T_{coll}} dt = \displaystyle\int_{r_0}^0 \bigg[ 2G (m_1 + m_2) \bigg( \dfrac {1} {r} - \frac {1} {r_0} \bigg) \bigg]^{-1/2} dr[/itex].​

    Maybe there's just some trick to this integral that I'm missing, but it feels like I'm going in circles and like there must be some easier way to go about this.
  11. Jul 7, 2013 #10


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    You can rearrange the right side to:

    [itex]\displaystyle\int_0^{t_{coll}} dt = \displaystyle \sqrt{\frac{r_0}{2G (m_1 + m_2)}} \int_{r_0}^0 \sqrt{\frac{r}{r_0 - r}} \ dr[/itex]​

    Then substitute

    [itex]\displaystyle u = \sqrt{\frac{r}{r_0 - r}}[/itex]​


    [itex]\displaystyle u^2 = {\frac{r}{r_0 - r}}[/itex]​

    Then solve for r and dr in terms of u and du, and change the limits of the integral for u. You'll end up with an integral that can be solved more easily (integral table lookup, and integration by parts).
    Last edited: Jul 7, 2013
  12. Jul 7, 2013 #11


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    or use a trigonometric substitution (sorry for butting in). I don't know if there is an 'easy way' to do this problem...
  13. Jul 7, 2013 #12


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    Assuming this isn't homework, perhaps this thread should be moved to general or classic physic section?
  14. Jul 8, 2013 #13

    rude man

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    Has anyone come up with an answer? I only arrived at a solution for D = 1m, which was

    T = (π/2)/k seconds where
    k = (1+m1/m2)√(Gm1m2/M) and
    M = (m1/2)(1 + m1/m2).

    So for m1 = m2 = m and D = 1m,
    T = (π/2)/2√(mG).

    (The units are consistent since "π/2" has dimensions of L3/2 here.)
  15. Jul 8, 2013 #14


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    This problem has appeared at least twice before. For r_0 = 1 meter, m1 = m2 = 1kg, t_collision ~= 96136 seconds. Waiting for the OP to try and work out the final integral.
    Last edited: Jul 9, 2013
  16. Jul 9, 2013 #15

    rude man

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    It's not a function of m1 and m2? Not what I got, and doesn't appear that way from your post #10 either ...

    I wound up with the same integral and, after sorting out my k coefficient, found it to be the same as yours.

    But T has to be a function of m1 and m2 so your number 96136 s. must be for certain values of m1 and m2.

    That second integral is not trivial - Wolfram alpha refused to evaluate it for me unless I paid - so I got T(D=1m) which was free.
    Last edited: Jul 9, 2013
  17. Jul 9, 2013 #16


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    It is, I missed an edit on my previous post where I set m1 = m2 = 1kg, it's fixed now.

    As far as I know, arildno was the first member here to have figured out the u = sqrt(...) substitution as a way to solve the second integral. What's left to do after my previous post is to figure out how to convert dr into a function of du, and change the limits of the integral based on u instead of r. Once this is done, the integral based on u and du will be solvable, including masses that are not point masses.
  18. Jul 9, 2013 #17
    The trig sub was what I used in one of those earlier cases mentioned :)

    In that same thread K^2 made this remark: "I'm going to cheat. This is still an orbital motion problem. By Keppler's Laws, period depends on semi-major axis only, and what we are looking for is half the orbital period." So much for the easy way.
  19. Jul 9, 2013 #18


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    the difficult bit with the trig sub is getting the limits of the integral right.
  20. Jul 9, 2013 #19
    Frankly, I do not remember that was a problem. Guessing the sub and doing the requisite algebra was more of a bother.
  21. Jul 9, 2013 #20


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    I think I have always been a bit unsure about converting limits of integrals, especially for sinusoid functions, which are not 1-to-1. I mean, how can you know what value of theta to use as the limit (because it does matter for the answer), but there is no way to tell which value you should use because the sinusoid is not a 1-to-1 function... you see what I mean? kinda tricky.
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