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Remarkably Difficult Newtonian Problem

  1. Dec 6, 2009 #1
    I thought of this problem recently and I was amazed at how difficult it was to solve, given the simplicity of the statement of the problem. I still have no idea how to solve it. The problem is thus: two point masses of 1 kg each are spaced 1 meter apart in a vacuum in space, with no other forces present other than the gravitational force between them. According to Newton's Law of Universal Gravitation, at what time will they meet in the center?
     
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  3. Dec 6, 2009 #2
    Do you know how to solve differential equations? This is a pretty simple problem if you know calculus.... Newtons law of gravity will give you; d^2r/dt^2 = constant/r^2 Use the boundary conditions that the bodies are initally at rest. This will give you the separation distance as a function of time. Invert to find time as a function of r. Assuming you know your starting separation you can find t when r=0 and take the difference to find out how long it takes them to collide.
     
  4. Dec 17, 2009 #3
    Well, I was able to solve the differential equation, but I'm having trouble setting the constants so that the bodies start at rest 1 m apart.
     
    Last edited: Dec 17, 2009
  5. Dec 17, 2009 #4
    Hi.
    x be the distance of a body from the middle point, the equation of motion is m d^2x/dt^2 = - G m^2 / (2x)^2. so x^2 x" = -mG/4 = const. m = 1 [kg], G = 6.67259 E-11 [m^3 s^-2 kg^-1] , x= 0.5 [m] and x'=0 [m/s] at t=0.
    I am interested in the answer. please let me know it.
    Regards
     
    Last edited: Dec 17, 2009
  6. Dec 17, 2009 #5

    cepheid

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    I thought I'd take a shot at it, but I ran into trouble. I'm sure there's some fancy coordinate system you can pick to make the problem easier, but I couldn't figure out what that might be. I decided to use a fixed origin and measure all distances from there. Let's say both masses are to the right of the origin, with mass m1 at x1 and m2 at x2, with x2 > x1 (mass 2 is to the right of mass 1). I'm going to consider it a 1D problem with the positive x direction being to the right.

    Let F21 be the force exerted on mass 1 by mass 2, and vice versa for F12. Then Newton's 2nd Law says that:

    [tex] F_{21} = G\frac{m_1 m_2}{(x_2 - x_1)^2} = m_1 \ddot{x}_1 [/tex]

    [tex] F_{12} = -F_{21} = -G\frac{m_1 m_2}{(x_2 - x_1)^2} = m_2 \ddot{x}_2 [/tex] ​

    Therefore the accelerations are given by:

    [tex] G\frac{m_2}{(x_2 - x_1)^2} = \ddot{x}_1 [/tex]

    [tex]-G\frac{m_1}{(x_2 - x_1)^2} = \ddot{x}_2 [/tex] ​

    If I subtract these two equations, I get:

    [tex] \ddot{x}_2 - \ddot{x}_1 = \frac{d^2}{dt^2}(x_2 - x_1) = -G\frac{m_1 + m_2}{(x_2 - x_1)^2} [/tex] ​

    Making a change of variables, so that r = x2 - x1 is the separation between the masses, I get:

    [tex] \ddot{r} = -G\frac{m_1 + m_2}{r^2} [/tex] ​

    I really have no idea how to solve this second-order non-linear differential equation.
     
  7. Dec 17, 2009 #6

    cepheid

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    It occurred to me that conservation of energy might help here by giving additional info about the first derivatives. Let E be the total energy of the system, U the gravitational potential energy, and T the total kinetic energy. Then:

    [tex] E = U_0 = U + T [/tex]

    [tex] -G\frac{m_1 m_2}{r_0} = -G\frac{m_1 m_2}{r} + \frac{1}{2}m_1\dot{x}_1^2 + \frac{1}{2}m_2\dot{x}_2^2 [/tex]

    [tex] Gm_1 m_2 \left(\frac{1}{r} - \frac{1}{r_0}\right) = \frac{1}{2}m_1\dot{x}_1^2 + \frac{1}{2}m_2\dot{x}_2^2 [/tex] ​

    We also know that:

    [tex] \dot{r} = \dot{x}_2 - \dot{x}_1 = -(|\dot{x}_2| + |\dot{x}_1|) [/tex] ​

    I'm not sure where to go from here though, because I can't see how to use this to get rid of the x's in the kinetic energy terms.
     
  8. Dec 18, 2009 #7

    diazona

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    The trick I've usually seen is to multiply both sides by [itex]\dot{r}[/itex], then you can integrate with respect to time to get a nonlinear first order differential equation. After that, at least for orbital motion, you do something with the angular momentum, which is constant, to get an effective potential... I forget exactly how it goes, I haven't done this in a while.
     
  9. Dec 18, 2009 #8

    rcgldr

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    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 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]

    [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
    https://www.physicsforums.com/showthread.php?t=306442

    [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[/tex]
    [tex]r(1 + u^2) = r_0[/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}{(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_0^\infty \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 ] _0^\infty = \frac{\pi\ {r_0}^{3/2}}{2 \sqrt{2 G (m1 + m2)}} [/tex]

    This matches arildno's method, except I kept m1 + m2 as variables, while he combined the two 1kg masses. I also checked this doing crude numerical integration via a spreadsheet.

    To answer the original post, for an initial distance of 1 meter, t ~= 96136 seconds.
    To check my math, I also tested with 2 meters, where t ~= 271915 seconds
     
    Last edited: Dec 18, 2009
  10. Dec 18, 2009 #9
    this is at least the third time this question is asked.. XD
    and by the way... what is a good source to learn calculus properly..?
     
  11. Dec 18, 2009 #10

    ideasrule

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    Jeff: You absolutely need the integration constants, or else you'll leave out a term in the final equation. Once you add in the constants, 1/2 v2 = G (m1 + m2) / r will become 1/2 v2 = G (m1 + m2) / r - G(m1+m2)/r0. That's something that could have been arrived at using the conservation of energy. If you continue on, you'll find that (1) the next integration becomes harder, and (2) an arcsin term appears in the final equation.
     
  12. Dec 18, 2009 #11
  13. Dec 18, 2009 #12

    rcgldr

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    Which I mentioned in my prior post. It was late so I was going to clean it up today. Using arildno's substitution, I cleaned up my prior post.
    An explanation of arildno's integration by parts trick:

    From an integral table (ignoring the constant here):

    [tex]\int\frac{1}{1+x^{2}}\ dx = tan^{-1}(x)[/tex]

    to integrate by parts instead define u and dv

    [tex]u = \frac{1}{1+x^{2}}[/tex]

    [tex]dv = dx[/tex]

    [tex]du = \frac{-2x}{(1+x^{2})^2}\ dx[/tex]

    [tex]v = x[/tex]

    [tex]\int u\ dv = u v - \int v\ du[/tex]

    [tex]\int\frac{1}{1+x^{2}}\ dx = \frac{x}{1+x^{2}} - \int \frac{-2x^2}{(1+x^{2})^2}\ dx[/tex]

    so

    [tex]\int \frac{x^2}{(1+x^{2})^2}\ dx = \frac{1}{2} \left ( \int\frac{1}{1+x^{2}}\ dx - \frac{x}{1+x^{2}} \right ) = \frac{1}{2} \left ( tan^{-1}(x) - \frac{x}{1+x^{2}} \right ) [/tex]
     
    Last edited: Dec 18, 2009
  14. Dec 20, 2009 #13
    Thanks for your calculation Jeff Reid. 96136 seconds. = 1 day and 2.7 hours , so much short time than I assumed. Astronauts in space trip would find the floating things stick together when he wakes up next morning. I've been wrong in underestimating gravity.
    Regards.
     
  15. Dec 20, 2009 #14

    rcgldr

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    Note that arildno did most of the tricky math in the other thread. I just expanded it to make it easier to follow, and used an alternate approach to get the dv/dr equation.
     
  16. Feb 19, 2011 #15

    rcgldr

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    Sorry to bring up an old thread, but this includes a correction to 3 of the intermediate steps from post #8, which I referred to in a recent thread.

    ...
    [tex]r + r u^2 = r_0 u^2[/tex]
    [tex]r(1 + u^2) = r_0 u^2[/tex]
    ...
    [tex]dr = \frac{2 r_0 u \ du}{(1 + u^2)^2}[/tex]
     
    Last edited: Feb 19, 2011
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