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Non constant accelleration equation(s)

  1. Sep 11, 2012 #1
    Hey guys, it's been awhile since I've been in the class room and I'm a bit fuzzy on non-constant acceleration problems.

    This actually started with a something of an amorous note sent to a lady friend, with the general idea that there is always a (small) gravitational force experienced between two people and if you make everything else disappear, the two will always(eventually) find each other.

    So, that of course made the geek in my curious to derive the equation describing the length of time(and final velocity out of morbid curiosity) that it would take two objects with space, with no other gravitational influences to "find each other"(intersect).

    I started looking at the basic kinematic equations but soon found myself out my depth when looking at the complexity introduced by acceleration not being constant.


    The note actually was a hit.
  2. jcsd
  3. Sep 11, 2012 #2
    Yea the acceleration is going to depend on the distance between the two, which obviously changes, I started looking at the differential equations for it but they are fairly nasty. I do remember during my general physics class however that there is a very easy way to do this with energy methods. The only thing I can't remember off the top of my head is I know there is something odd about the sign convention with the potential energy, and I don't have my physics book near. Anyone care to help us out?
  4. Sep 11, 2012 #3


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    [strike]It's actually a differential equation. The general solution to it is orbital motion. But for a special case of no orbital angular momentum, it's pretty straight forward. The attractive force is GMm/r². That gives you dr/dt=-G(M+m)/r². By rearranging terms and integrating, you have the solution [itex]t=-\frac{r^3}{3G(M+m)}+C[/itex]. Setting boundary condition t=0 at r=R, you get your answer [itex]T=\frac{R^3}{3G(M+m)}[/itex]. Here, T is time it takes for bodies to meet, R is initial distance, and M and m are masses of two bodies. The size is assumed to be negligible compared to initial separation distance.[/strike]

    Please disregard this silliness.
    Last edited: Sep 11, 2012
  5. Sep 11, 2012 #4
    There won't be any terminal velocity, the velocity will be growing till the impact. You could estimate the impact velocity, though, it is fairly simple for the conservation of energy.

    Time, however, is problematic. If you integrate the equation of motion, you end up with a hairy expression which is anything but intuitive.
  6. Sep 11, 2012 #5
    That should be dv/dt=-G(M+m)/r². Which changes a lot.
  7. Sep 11, 2012 #6
    Yes at the moment im stuck at integrating [tex]\frac{dr}{\sqrt{2m_1G(\frac{1}{R}-\frac{1}{r})}}[/tex]
    Last edited by a moderator: Sep 11, 2012
  8. Sep 11, 2012 #7
    It should be 1/r - 1/R. Potential energy is MINUS GMm/r.
  9. Sep 11, 2012 #8
    Hmmm I'll have to go double check, I wasn't looking at my work when I wrote it out. Yea I thought I remembered potential being negative, and it's supposed to be 0 at infinity, no?
  10. Sep 11, 2012 #9


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    Ha, ha. I must be tired. Good catch, voko.

    Well, in that case, 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. [itex]T=\pi\sqrt{\frac{(R/2)^3}{G(M+m)}}[/itex]. How is that?
  11. Sep 11, 2012 #10
    If I remember correctly, any derivation of the third law rests on the assumption that the angular momentum is not zero. Well, it may be that it still turns out to be true even in this case. Let me think about it.
  12. Sep 11, 2012 #11
    Yes, Kepler's third law holds even in this case. Using conservation of energy: [tex]
    \frac{m\dot{r}^2} {2} - \frac {GMm} {r} = - \frac {GMm} {R}

    \\ \dot{r} = - \sqrt {2GM (\frac 1 r - \frac 1 R) }

    \\ \int_R^0 \frac {dr} {\sqrt {\frac 1 r - \frac 1 R}} = - \sqrt {2GM} T

    \\ \cos \theta = \sqrt {r/R}
    \\ \theta = \arccos \sqrt {r/R}
    \\ d\theta = - \frac 1 {\sqrt {1 - r/R} } \frac 1 {2R\sqrt {r/R} }
    = - \frac 1 {\sqrt {1/r - 1/R} } \frac 1 { 2 \sqrt {R} r }

    \\ \cos^2 \theta d \theta = - r/R \frac 1 {\sqrt {1/r - 1/R} } \frac 1 { 2 \sqrt {R} r }
    = - \frac 1 {2R^{3/2}} \frac 1 {\sqrt {1/r - 1/R} }

    \\ \int_R^0 \frac {dr} {\sqrt {\frac 1 r - \frac 1 R}} = -2R^{3/2} \int_0^{\pi/2} \cos^2 \theta d \theta
    = -2R^{3/2} \frac 1 2 \left[ \theta + \sin \theta \cos \theta \right]_0^{\pi/2}

    = -R^{3/2} \frac {\pi} {2} = - \sqrt {2GM} T

    \\ T = {\pi} \sqrt {\frac {(R/2)^3}{GM}}

    One thing to note, though. The bodies were assumed to be point masses. If we take them to have some finite size, then we have to integrate to r > 0, so we won't have everything nicely cancelling at the upper limit of integration. Instead, we will have, as I said, some hairy mess of square roots and trigonometric and inverse trigonometric functions.
  13. Sep 11, 2012 #12
    Hmmmm, I didn't approach the problem from an energy standpoint, I started with:


    Then I continued with solving the differential equation but I still end up with the reverse 1/R-1/r....
  14. Sep 12, 2012 #13
    The acceleration should be be negative, because the force is pulling the object toward r = 0.
  15. Sep 12, 2012 #14


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    There are multiple ways to come to the same integral.

    For any initial separation significant larger than the size of the objects, the finite final distance is negligible.
    The final velocity is can be calculated with energy conservation. For simplicity, I'll assume that both masses are same, both objects are spherical with radius r/2 and initial separation R.
    With R>>r, m≈80kg and r=50cm, this gives ~0.1mm/s. Basically every movement will give you escape velocity ;(.
  16. Sep 12, 2012 #15
    I did not dare to spell out this assumption :)
  17. Sep 12, 2012 #16


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    Well, it is a very bad approximation (especially together with m=80kg), but otherwise we would need simulations.

    Oh, concerning the time: That should depend on the reduced mass ##\mu=\frac{Mm}{M+m}##, not on the mass of one object only. The approximation that one mass is negligible compared to the other is mean ;).
  18. Sep 12, 2012 #17


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    The math for point objects was done in a previous thread, in post #8, with a minor correction for intermediate steps in post #15, although there is one line with a latex issue (latex was updated since this thread was made). alrildno figured out a different substitution to handle the integral, which was used in this thread.


    arildno did the math for 2 finite sized objects, each with the same mass, in this thread, but almost all the lines of latex are messed up, except the final result in post #4 is ok.


    There was another thread where all the latex issues were cleaned up, but I couldn't find it.
    Last edited: Sep 12, 2012
  19. Sep 12, 2012 #18
    True. I derived the equation assuming M >> m. Assuming both masses are comparable, the potential energy term become ## - \frac {GMm} {\mu r} ## and ## - \frac {GMm} {\mu R} ##, where ## \mu = 1 + \frac m M ## and ## r ## and ## R ## are the distances of mass ## m ## from the center of mass, where the masses will collide.

    This basically means we have to replace ## M \rightarrow \frac M {\mu } = M + m ## throughout, thus getting ## T = \pi \sqrt {\frac {(R/2)^3} {G(M + m)} } ##.
  20. Sep 12, 2012 #19


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    Reposting the math from the previous thread here with the corrections:

    v= dr/dt

    a = dv/dt

    multiply by dv/dt by dr/dr:

    a = (dr dv)/(dt dr) = v dv/dr

    This gets you to the first step:

    v dv/dr = -G (m1 + m2) / r2

    v dv = -G (m1 + m2) dr / r2

    for v = 0 at r = r0 you get:

    [tex]1/2 v^2 = G (m_1 + m_2) / r - G (m_1 + m_2) / r_0 [/tex]

    [tex]v = ± \sqrt{ 2 G (m_1 + m_2) / r - 2 G (m_1 + m_2) / r_0} [/tex]
    chosing the negative root since r decreases over time:

    [tex]v = \frac{dr}{dt} = - \sqrt{ \frac{2 G r_0 (m_1 + m_2) - 2 G r (m_1 + m_2)} {r\ r_0}} [/tex]

    [tex]\frac{- \sqrt{r_0 \ r} \ dr} {\sqrt{ 2 G r_0 (m_1 + m_2) - 2 G r (m_1 + m_2)}} = dt[/tex]

    [tex] {- \sqrt{ \frac{r_0}{2 G (m_1 + m_2)}}} \ \ \sqrt{\frac{r}{r_0 - r}} \ dr = dt[/tex]

    Using arildno's method from this thread

    [tex]u = \sqrt{\frac{r}{r_0-r}}[/tex]

    [tex]u^2 = {\frac{r}{r_0-r}}[/tex]
    [tex]r = r_0 u^2 - ru^2[/tex]
    [tex]r + r u^2 = r_0 u^2[/tex]
    [tex]r(1 + u^2) = r_0 u^2[/tex]
    [tex]r = \frac{r_0 u^2}{1 + u^2} = \frac{r_0 + r_0 u^2 - r_0}{1 + u^2} = \frac{ r_0(1 + u^2) - r_0}{1 + u^2} = r_0 - \frac{r_0}{1 + u^2}[/tex]

    [tex]dr = \frac{2 r_0 u \ du}{(1 + u^2)^2}[/tex]

    at r = r0, u = ∞, at r = 0, u = 0.

    [tex]t = \frac{-2 {r_0}^{3/2}}{\sqrt{2 G (m1 + m2)}} \ \int_\infty^0 \frac{u^{2}}{(1+u^{2})^{2}}du= \frac{-2 {r_0}^{3/2}}{\sqrt{2 G (m1 + m2)}} \left [ \frac{1}{2} \left (\tan^{-1}(u)-\frac{u}{1+u^{2}} \right ) \right ] _\infty^0 = \frac{\pi\ {r_0}^{3/2}}{2 \sqrt{2 G (m1 + m2)}} [/tex]

    For finite size objects, the upper limit in the integral would need to be changed from u = 0 to the value of u that corresponds to r at the point of contact between objects.
    Last edited: Sep 12, 2012
  21. Sep 12, 2012 #20
    Both methods seem to have the same result, which is reassuring.
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