Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A physics problem, simple or subtle?

  1. Jul 13, 2004 #1
    We all know (I hope) that the standard Newtonian formula for the gravitational attraction of two (point like) masses is:

    F = G M*m/r^2 where G is the Gravitational Constant - M is the mass of

    one object, m is the mass of the other and r is their separation.

    So far so good.

    Now imagine two point masses as above, separated by an initial distance 2r (it helps the algebra to make it 2r) and that they are initially at rest.

    They are then 'let go' and allowed to move together by gravity alone. I could couch this in some fancy language about assuming the space is maximally symmetric blah blah blah - but the initial conditions are as stated above - it doesn't need GR or such.

    Now the question is:

    How long does it take them to collide?

    A friend asked me this and I couldn't figure it out, can you?
  2. jcsd
  3. Jul 13, 2004 #2


    User Avatar
    Science Advisor
    Homework Helper

    [tex]F_1 = \frac{GM_1M_2}{(2R - x_1 - x_2)^2}[/tex]

    Where [itex]F_1[/itex] is the force on object 1, [itex]M_1[/itex] and [itex]M_2[/itex] are the masses of objects 1 and 2, respectively, [itex]2R[/itex] is their initial separation, and [itex]x_1[/itex] is the distance towards the center that object 1 has travelled from its initial position (and you can guess [itex]x_2[/itex]). Now, divide both sides by [itex]M_1[/itex]

    [tex]a_1 = \frac{GM_2}{(2R - x_1 - x_2)^2}[/tex]

    [tex]x_1'' = \frac{GM_2}{(2R - x_1 - x_2)^2}[/tex]

    You can make a similar equation for [itex]x_2''[/itex], and maybe you can get two differential equations you can solve.
  4. Jul 13, 2004 #3
    if you have the force which you have calculated and you have the mass then surely you could use F=MA

    REGARDS jamie
  5. Jul 13, 2004 #4
    yes, once you calculate force, then both objects will have that force on eachoter. The distance they will need to ravel is r, since they will both meet at midway. use the formula F = ma, calculate a for one body. The sub that a, and r into the formula:
    d = vot + 1/2at^2
    this becomes: r = 1/2at^2
    therefore t = sqrt(2r/a)
    or if you wanna do it the long way
    t = sqrt(2r/(G M*m/r^2)/m))
  6. Jul 13, 2004 #5


    User Avatar
    Science Advisor
    Homework Helper


    Why would both objects meet in the middle? If one has much greater mass, then it will accelerate much less, and not make it to the middle. It was not specified that the masses are the same.
  7. Jul 13, 2004 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Best to view them from the inertial frame in which the masses were initially at rest. They will of course meet at their center of mass.
  8. Jul 14, 2004 #7
    That's the math, what about the physics? What is the mechanism that causes the masses to move toward each other is what I'd like to know. GR's answer is space-time curvature no? So how does that work? Let's see, the masses start off by distorting space-time. Then what? If they were initially at rest, what would cause them to accellerate? Say the balls are on the edges of the grand canyon (representing the space-time curvature), if nobody pushes them why would they fall in? Now, if space has some properties like it can attract matter it might be a different story. Say perhaps not only does matter attract space and vice vera but space can have different densities so that in the presence of two masses, the density of space in between them becomes greater (more effective mass), the attraction between matter and space bringing the masses together. On a second thought, this probably begs the question since we then have to propse a mechansim by which space attracts matter.

    Another mechanism I've read is that these masses exchange force particles called gravitons that moves them toward each other. This seems pretty strange also- I can't picture how anything with forward momentum could hit something and cause it to move towards the collision. Still easier to picture than space-time curvature though.

    Then there's the Le Sage "pushing gravity" theories that explain this acceleration as a result of a shortage of "pushing particles" in between the masses (shadowing effect) causing the two masses to be pushed toward each other. This is probably the easiest to picture since transfer of momentum is very familiar to us.

    Any comments on these or any other bright ideas out there?
    Last edited: Jul 14, 2004
  9. Jul 14, 2004 #8


    User Avatar
    Staff Emeritus
    Gold Member
    Dearly Missed

    The masses cause the curvature of spacetime to change, in that new geometry the new shortest paths give the masses a natural, zero-effort path to approach each other, and while they approach they continue to chage the curvature, changing the geodesics, changing their paths, it's dynamic. But if they are prevented from following the path, like a weight on a cliff then they won't follow it. Duh! General Relativity is not bone stupid.
  10. Jul 14, 2004 #9


    User Avatar
    Science Advisor
    Gold Member

    I'd take a slightly different approach, to Doc Al and say it's best to view them from a non-inertial frame, as (without actually attempting solving the problem) an inertial frame is fibne for finding out where they collide, but you'd have to do a little bit of extra work to find the equation that is trivially found in a non-inertial refernce frame.

    The equation you want is a = -G(M + m)/r^2 (which the accelartion of the object which don't choose to be stationery in your non-inertial refernce frame). This is a seconnd order differential equation, which you must solve in order to find t.
  11. Jul 14, 2004 #10


    User Avatar
    Science Advisor
    Gold Member

    That's wrong, remember the accelration is not constant.
  12. Jul 14, 2004 #11


    User Avatar
    Science Advisor

    hehe, the joys of the 2 body problem.. In this case, its easy b/c its not really a 2 body problem, but more like a 1.5 body problem (in difficulty) =) Its quite a bit tougher if you specify different initial conditions..
  13. Jul 14, 2004 #12

    Doc Al

    User Avatar

    Staff: Mentor

    two roads, one destination :-)

    Good point, but I only start from that inertial frame. First I find the acceleration of each mass. Then the relative acceleration, which is just the sum of the two individual accelerations. Which brings me to the the same equation that you found. :smile:
  14. Jul 14, 2004 #13

    This problem is NON-Linear as the force changes with their separation , it is easier to simulate using small time steps especially if they form an orbit rather than meeting head on.
    WRT. cuvature , the rubber sheet example is a reasonable analogy ,( that is two masses placed on an infinite sheet depressing it due to their masses) it's clear that the overall curvature does not 'suddenly' start so that in the initial condition there is already a 'force' caused by the sheet curvature in general due to their combined mass.
    In that example the observer is a third person viewing the system from outside it is convenient for that observer to be attached to the mass center of the system , in that view he sees the larger mass accellerated more slowly than the lighter one both heading towards the mass center.
    The mass center is simply the view of someone with a weighing scale
    M.d1 = m.d2 this point has no tendancy to move because it is at the gravitational nul point.
    Last edited: Jul 14, 2004
  15. Jul 14, 2004 #14
    Since r changes, the force between the bodies changes, and so the acceleration changes. This makes the problem much tougher than expected.

    This problem is solved numerically in the book Computational Physics by Rubin Landau.
  16. Jul 14, 2004 #15


    User Avatar
    Science Advisor

    As has been pointed out, not such an easy problem. I'll sketch out one approach. The first step is to eliminate the Center of mass coord, and go to the relative coordinate which I'll call r = r1-r2 for masses 1 and 2 with coords r1 and r2. if they move along a line then there's no angular momentum, and the energy equation becomes(I'm being sloppy with constant factors)

    (dr/dt)**2 = E + g/r where E is the energy(up to some scale factor) and g is the appropriate normalized gravitational coupling constant. Note that if the particles start at rest, then E=-g/r0, r0 is the initial separation. Now the rest is algebra

    Take the square root to get dr/dt= SQRT(E + g/r),

    and then rewrite dr/SQRT(E + g/r) = dt. On the lhs, will have

    SQRT(r) dr/SQRT(Er + g), and the integral of this this can be found in Tables of integrals. Note that the time integral goes from 0 to T, while the spatial integral goes from s0, the initial separation, to 0. And, most likely, being careful with signs will be crucial. That's the gist of it.

    Reilly Atkinson
  17. Jul 15, 2004 #16


    User Avatar
    Science Advisor
    Gold Member

    btw the equation I posted (just in case anyone does want to solve it), though it is a non-linear second order differntial equation, is one that is (relatively) easy to solve :

    better stated as:

    [tex]\renewcommand{\vec}[1]{\mbox{\boldmath $ #1 $}}
    \vec{\ddot{r}} = \frac{-G(M+m)}{r^2}\vec{\hat{r}}[/tex]
    Last edited: Jul 15, 2004
  18. Jul 15, 2004 #17
    The way we were taught to think of this was as the exchange particle being like a boomerang. If two people face in opposite directions, and one throws a boomerang, which loops back round and the other person catches it, they have exchanged this "particle" and will move towards each other.
  19. Jul 15, 2004 #18
    I hope all realize that two bodies meeting like this will accellerate to infinite velocity if they have zero radius and happen to travel 'in line' ------- i often wondered ( idly) why two photons ( which have momentun ( motional mass) ) don't do the same thing.
  20. Jul 16, 2004 #19
    my mistake. I thought the masses equaled eachother.
  21. Jul 16, 2004 #20
    Guys, honestly, I got lost pretty quick. Who has the right answer? And can't you use the Keplerian equation of motion to solve the collision, even though it's trivial? Wouldn't it be at the barycenter?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook