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Homework Help: Time for two bodies to collide under gravity.

  1. Jul 25, 2008 #1
    1. The problem statement, all variables and given/known data

    Two bodies mass [itex]m_1[/itex] and [itex]m_2[/itex] are a distance [itex]x[/itex] apart. How long do they take to collide as a result of their gravitational attraction to each other.

    2. Relevant equations

    This question was asked in another thread which I found in a search. However the final answer was never posted, and I am trying to work it out.

    This is a post from the thread.

    3. The attempt at a solution

    I then get that:

    [tex] \int v dv = -G(m_1+m_2) \int x^{-2} dx\\ [/tex]

    [tex]\frac{v^2}{2} = \frac{G(m_1+m_2)}{x} + c\\ [/tex]

    [tex]\frac{dx}{dt} = \sqrt{\frac{2G(m_1+m_2)}{x} + c}


    However from here I am not sure how to get x as a function of t. I think it is explained later in the thread (for reference, https://www.physicsforums.com/showthread.php?p=1791110#post1791110) but I do not understand.

    I tried to integrate both sides with respect to t, but then I couldn't rearrange it to get x on its own on one side.

    Thanks for the help :)

    (edited many times to get latex right, first time I've done it)
    Last edited: Jul 25, 2008
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  3. Jul 25, 2008 #2


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    You don't want to. The integration of the separable differential equation

    dt = \frac{dx}{\sqrt{\frac{2G(m_1+m_2)}{x} + c} }[/tex]

    (don't forget to evaluate the arbitrary constant from the previous integration first!)

    gives you t as a function of x . Remember that it is t that you want to evaluate, so if we call T the time to collision, then we would write

    T = [tex]t |^{T}_{0} [/tex] = (integral function of x)[tex]|^{0}_{x_o}[/tex] .

    [Note: if the evaluation threatens to be undefined for x = 0, use the sum of the radii of the two objects instead: x = [tex]R_1 + R_2[/tex] .]

    General gravitational infall problems often lead to results which are very difficult to invert into the form of radius as a function of time.
    Last edited: Jul 25, 2008
  4. Jul 25, 2008 #3
    You could use dimensional analysis to determine a good ball-park figure that will differ from the real answer by some constant factor. To my knowledge the following applies to two or more objects initially at rest and oriented in an evenly-spaced manner about the circumference of a circle.

    [tex]t = \sqrt{\frac{R^{3}}{GM}}[/tex]

    where R is the radius of the circle, M is the total mass of the system, and G is the gravitational constant.

    This above equation can be determined through trial-and-error by manipulating the variables accordingly so that the units would cancel.

    [tex]R = [L][/tex]
    [tex]M = [M][/tex]
    [tex]G = [M^{-1}L^{3}T^{-2}][/tex]
    [tex]t = [T][/tex]
    Last edited: Jul 25, 2008
  5. Jul 25, 2008 #4


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    One doesn't have to find this by trial-and-error with units; dimensional analysis will lead directly to it by setting

    [tex]t \propto G^{\alpha}M^{\beta}R^{\gamma} [/tex]

    and solving for the values of the exponents that yield units of time only.

    Unfortunately, the result is so generic that it can be applied to all sorts of gravitational interactions with a considerable range in the dimensionless constant. (This expression is, after all, a form of Kepler's Third Law.) It does, however, suggest the way to answer the question in the case where, say, m1 = M >> m2 . We then have the "radial infall problem" for a small body falling onto a(n infinitely) massive body, the time for which is half the orbital period of a (degenerate) elliptical orbit having a semi-major axis of half the distance to fall (since the apofocus is the starting distance x and the perifocus is the center of the massive body). We can then just use Kepler's Third Law to say

    [tex]T^2 = \frac{4 \pi^2 }{GM} \cdot R^3
    = \frac{4 \pi^2 }{GM} \cdot (\frac{x}{2})^3
    = \frac{\pi^2 }{2GM} \cdot x^3 [/tex]

    The infall time is then half of this "period", or

    [tex] \tau = \frac{T}{2} = \frac{\pi }{\sqrt{8GM }} \cdot x^{3/2} [/tex]

    However, marmoset's problem appears to call for the general result, with no particular relationship between the two masses; so the above result can only be a limit at one extreme.
  6. Jul 25, 2008 #5
    I understand the limitations of dimensional analysis. I liked your explanation, but wouldn't marmoset's generalized result be of this form though, except that M = m1 + m2 instead? I thought by applying dimensional analysis, marmoset would have a target for the final form of his derivation. Perhaps this is only true if m1 = m2? Perhaps I didn't think this through as fully as I should have.
  7. Jul 25, 2008 #6


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    The basic form with only one mass M in it assumes that the massive body has negligible acceleration. In the general problem, what actually happens is that both masses fall toward each other, so the treatment of the motion of these bodies is not as simple (unfortunately) as just assigning M as the sum of the masses. Dimensionally, the answer will have the form you describe, but the dimensionless ratio will involve some combination of the two masses. (I should take some time and work that integral through now...)
  8. Jul 25, 2008 #7
    I don't know how to do that integral, the wolfram online integrator gives this bad boy:


    I think I would work out c from setting my expression for [itex]\frac{dx}{dt}[/itex] equal to zero when [itex]x = x_0[/itex]. After that, the [itex]t |^{T}_{0} [/itex] is just T, the time taken for collision which is what I want to find, so T is equal to the horrible integral evaluated between [itex]x_0[/itex] and 0.

    Is this the right way to do it? And is there a cleaner way to do the problem from scratch than this method?

    Thanks for the help.
    Last edited: Jul 25, 2008
  9. Jul 25, 2008 #8


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    I was proposing to work from this:

    \frac{v^2}{2} = \frac{G(m_1+m_2)}{x} + c\\

    to get

    0 = \frac{G(m_1+m_2)}{x_o} + c\\

    which leads to

    \frac{dx}{dt} = \sqrt{\frac{2G(m_1+m_2)}{x} - \frac{2G(m_1+m_2)}{x_o}}[/tex]

    The separated differential equation now looks like

    \sqrt{2G(m_1+m_2)} \cdot dt = \frac{dx}{\sqrt{\frac{1}{x} - \frac{1}{x_o} } }

    which may be a bit easier to deal with...

    EDIT: I believe the typical method for dealing with this integration is to use the substitution

    [tex]u = \frac{1}{x} \rightarrow du = -\frac{1}{x^2} dx \rightarrow dx = -\frac{1}{u^2} du [/tex]
    Last edited: Jul 25, 2008
  10. Jul 26, 2008 #9


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    Well, maybe I haven't either... Trying out the integration, I still get the factor G(m1 + m2) , so maybe that should be there. (I'd thought that it should end up as the "reduced mass". OTOH, I haven't checked that non-inertial description thoroughly: it looks OK, but perhaps I should check some references...)

    I've looked at the other extreme, for which m1 = m2 = M. We can use an argument similar to the one I gave earlier, with one adjustment. The two equal masses will fall toward each other at equal accelerations and meet at the midpoint of the initial separation, x. So each now acts as if it is on a "radial infall" over a distance (x/2). Since each acceleration is initially [tex]\frac{GM}{(\frac{x}{2})^2} = \frac{G \cdot 4M}{x^2} [/tex], each mass behaves as if it is falling toward a central effective mass of 4M .

    The degenerate ellipse for each mass here has an apofocal distance of (x/2) and a perifocal distance of 0, so the "semi-major axis" is (x/4). Applying Kepler again gives us

    [tex]T^2 = \frac{4 \pi^2 }{G \cdot 4M} \cdot R^3 = \frac{\pi^2 }{GM} \cdot (\frac{x}{4})^3 = \frac{\pi^2 }{64GM} \cdot x^3

    and the infall time to the midpoint is given by

    \tau = \frac{T}{2} = \frac{\pi }{16 \sqrt{GM }} \cdot x^{3/2}
    [/tex] .

    This is why I think the dimensionless constant ought to have some dependence on the ratio of the masses.
    Last edited: Jul 26, 2008
  11. Jul 26, 2008 #10
    Aren't the bodies moving towards eachother in a straight line? If yes, how can you possibly end up with a circle with radius R?

    From the other thread:
    You can find t as a function of x, which should be good enough for you to calculate the time it takes for the two bodies to collide. (This of course only works as it is if the initial velocity of both bodies is 0. Otherwise you might need to change the resulting formula so you can add an initial velocity.)
    This formula for t as function of x is a bit large... I used Mathematica to find the integral mentioned, so it might not be as simple as possible.

    I found that:
    [tex]t=\int \frac 1 {\sqrt{\frac{2G(m_1+m_2)}{x}-\frac{2G(m_1+m_2)}{x_o}}}\,dx[/tex]

    [tex]= \frac{1}{\sqrt{2G(m_1+m_2)}} \int \frac {1}{\sqrt{\frac{1}{x}-\frac{1}{x_o}}} \,dx[/tex]

    [tex]= \frac{\frac{1}{2}\sqrt{x_o^3}\arctan{(\frac{\sqrt{x_o}\sqrt{\frac{1}{x}-\frac{1}{x_o}}(2x-x_o)}{2(x-x_o)})}+x_o x \sqrt{\frac{1}{x}-\frac{1}{x_o}}+C}{\sqrt{2G(m_1+m_2)}} [/tex]

    [tex]= \frac{\frac{1}{2}\sqrt{x_o^3}\arctan{(\frac{\sqrt{x_o}\sqrt{\frac{1}{x}-\frac{1}{x_o}}(2x-x_o)}{2(x-x_o)})}+x_o x \sqrt{\frac{1}{x}-\frac{1}{x_o}}+\frac{1}{4}\pi\sqrt{x_o^3}}{\sqrt{2G(m_1+m_2)}}
  12. Jul 26, 2008 #11
    I guess, though, that (Special) Relativity also has a part in this...
    I may be wrong, but I thought that because the bodies could get very high velocities... Especially when the bodies start getting very close to eachother, their velocities increase greatly.
    Of course, if the bodies are quite large (if they have a large radius), they may never achieve those high velocities, for they would have collided before their centre of mass get close enough together. Still...
    Depending on the situation, the bodies can also have a big mass.
    Now, I don't know much about (Special) Relativity, but it does affect the situation, right?
  13. Jul 26, 2008 #12
    The objects can be initially oriented around a circle although their motion is a straight line, such as in the current example--perhaps, you can think of this as using geometry to simplify the physical situation. If the masses are equal, the point of collision should be the origin (center of mass of the system); if the masses differ, then the system's center of mass will be displaced depending on the magnitude of their difference.

    If we could precisely place (no error whatsoever) three equal masses in an evenly-spaced manner about the circumference of a circle, we would find that all three masses will converge at the origin of the circle. Their motions are still straight lines, but instead occur in two dimensions (three "points" determine a plane).
  14. Jul 26, 2008 #13
    I think that the absence of the "reduced mass" ratio is fine in this context, since the objects don't actually orbit each other. If they did, then one object would likely be considered the origin of our system, and then we would need the reduced mass to accurately describe the system.
  15. Jul 26, 2008 #14
    It depends. In general, we would find that the time value is large (days, months) since gravity is a feeble force. Since velocity = distance / time, the initial distance between the objects must be extremely large (many orders of magnitude greater than time) to cause the objects to approach relativistic speeds.

    However, if we changed the scenario so that the objects have some charge density and are opposite in polarity and wanted to determine the time to collision using the electromagnetic force, relativistic speeds are likely to more prevalent than the gravitational scenario. This is because the electromagnetic force is some 10^30 times stronger than gravitational force.
  16. Jul 26, 2008 #15


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    R is just a variable name, carried over from the quoted post. In that context, it comes from the term radial vector, the vector pointing from one mass (chosen as the origin) to the other. In the form of Kepler's Third Law I've been writing, that "R" is really the semi-major axis of the orbit; I suppose I should have switched over to "a" to avoid confusion.

    I guess I'm old-fashioned and avoid computer integrators (especially having seen some of the messes they create); I used Gradshteyn and Ryzhik, the monster compendium of series and integrals, which is the major reference people go to -- if it's been integrated, it's probably in there somewhere. I'll have to post it later, since I'm not where my notes are, but their result is along those lines. It simplifies considerably since we are integrating from x = x_0 down to x = 0: all but one of the terms goes to zero and the remaining term has the requisite (x_0)^(3/2) . (I still want to check to see if I have the dimensionless constant right...)

    On another point, I overdid the fours in my result for m1 = m2; I believe that should just be

    T^2 = \frac{4 \pi^2 }{G \cdot M} \cdot R^3 = \frac{4 \pi^2 }{GM} \cdot (\frac{x}{4})^3 = \frac{\pi^2 }{16GM} \cdot x^3

    and [tex] \tau = \frac{T}{2} = \frac{\pi }{8 \sqrt{GM }} \cdot x^{3/2} [/tex]

    and ignore my remark about 4M being the effective mass.

    You can estimate whether relativity would be an issue right from the start: the final velocity would be given classically as

    [tex]\frac{v_{f}^2}{2} = G(m_1+m_2) \cdot [ \frac{1}{x_f} - \frac{1}{x_o}] [/tex]

    (choose some small non-zero radius reasonable for the objects involved)

    If the final velocity is larger than, say, 0.01 c, we probably need to invoke relativity theory. If the mean density of the system (total mass/current volume) reaches something like 10^4 kg/m^3, we probably should be working with general relativity.
  17. Jul 26, 2008 #16

    D H

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    Looking at this problem as a degenerate orbit does provide a nice and easy way to arrive at the collision time. Suppose that instead of a zero initial velocity the initial relative velocity is normal to the line between the two masses and has a small non-zero magnitude. This places the masses in an elliptical orbit about each other. The period of an elliptic orbit is

    [tex]T = 2\pi \sqrt{\frac {a^3}{G(m_1+m_2)}[/tex]

    where [itex]a[/itex] is the semi-major axis length. The time taken to move from apofocus to perifocus is half the orbital period:

    [tex]t_c = \pi \sqrt{\frac {a^3}{G(m_1+m_2)}[/tex]

    As the magnitude of the initial velocity goes to zero, the perifocus goes to zero, so the semi-major axis becomes half the initial separation. The semi-major axis length is the mean of the perifocus and apofocus separation distances. In the limit of zero initial velocity, the perifocus separation distance goes to zero and thus

    [tex]t_c \to \pi \sqrt{\frac {x_0^{\;3}}{8G(m_1+m_2)}[/tex]

    where [itex]x_0[/itex] is the initial separation distance.
    Last edited: Jul 26, 2008
  18. Jul 27, 2008 #17
    You're right, and I somewhat misunderstood what was said before; to be honest I didn't quite know the meaning of every word as well (English is not my native language)...

    Anyway, for what I understand now, is that
    [tex]t_c \to \pi \sqrt{\frac {x_0^{\;3}}{8G(m_1+m_2)}[/tex]
    gives the time it would take the two bodies to "get together", so the time it takes to get from x=x0 to x=0 m.
    I guess this solution should be good enough for Marmoset...
    Of course, if the bodies have some volume/dimensions, those should be taken in consideration as well I think, but I guess that's obvious.
  19. Jul 27, 2008 #18

    D H

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    In which case one must revert to the original form of the problem.

    I'll start with

    [tex]\frac{dx}{dt} = -\sqrt{2G(m_1+m_2)\left(\frac 1 x - \frac 1 x_0\right)}[/tex]

    Applying the substitutions

    k &\equiv G(m_1+m_2) \\
    u &\equiv \frac {x_0} x


    [tex]\sqrt{\frac{x_0^3}{2k}}\; \frac 1 {u^2\sqrt{u - 1}}\;\frac{du}{dt} = 1[/tex]

    Solving for [itex]t_f[/itex], the time taking to go from [itex]u=u_0=1[/itex] to [itex]u=u_f[/itex],

    t_f &=
    \sqrt{\frac{x_0^3}{2k}}\,\int_1^{u_f} \frac{du}{u^2\sqrt{u - 1}} \\
    &= \sqrt{\frac{x_0^3}{2k}}\,
    \left.\left(\frac{\sqrt{u-1}}u+\tan^{-1}\sqrt{u-1}\right)\right|_1^{u_f} \\
    &= \sqrt{\frac{x_0^3}{2k}}\,
    \left(\frac{\sqrt{u_f-1}}{u_f}+\tan^{-1}\sqrt{u_f-1}\right) \\
    &= \sqrt{\frac{x_0^3}{2G(m_1+m_2)}}\,

    In the limit [itex]x_f \to 0[/itex], the radical term vanishes and the inverse tangent term goes to [itex]\pi/2[/itex], and thus

    \lim_{x_f\to0} t_f =\pi \sqrt{\frac{x_0^3}{8G(m_1+m_2)}}

    which is the same result as in post #16.
  20. Jul 27, 2008 #19


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    That is true, but for most of the applications where this result would be used, the size of the two bodies is negligible compared to the initial separation distance. Since this is a time integral, the proportion of time spent in falling from, say, the sum of the radii of the two bodies, x = R1 + R2, to a separation x = 0 to the total infall time to x = 0 is negligible, so the result may be reasonably used as an approximation for the time to collision of finite bodies, rather than "mass points".

    Thank you for presenting the complete integration, D H. I had found the same result, but wanted to think it over more because I wasn't believing in the dimensionless constant. I think now that I was just confusing things in the labeling for the m1 = m2 case, as I described it in posts #9 and 15. (I also carried the integration to x = 0, or [tex]u = \infty[/tex], and found that one term requires L'Hopital, which nonetheless takes it to zero.) So I learned something from this thread...
    Last edited: Jul 27, 2008
  21. Aug 3, 2008 #20
    Sorry to bump this thread, but I have been away for the past week and I would like to thank everyone who has helped in this thread. You have all been great : ) I cannot do the integration, but I will look at the clever way about letting the initial velocity tend to zero.
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