1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Time for two objects to collide due to gravity

  1. May 15, 2016 #1
    When two objects, there is a change in the force they exert on each other due to gravity because of the change in distance between them. Therefore, there's also changing acceleration. How do I find the time it takes for the two to collide? Their final velocities?
    I've already looked at quite a few other threads on this forum but I still haven't been able to find one that answers this question nor gives me any clues on how to proceed.
    First Thread
    Second Thread
    These 2 links proved to be particularly helpful. There's just a part I don't quite understand because of arildno's answer on the second thread which has messed up LaTeX formatting which was referred to in rcqldr's answer on the first thread.
    I don't quite understand where u came from nor what it represents.

    This is what I've come up with so far on my own:

    I originally tackled this as an object approaching a planet so I only looked at it from the perspective of the object.

    Finding the time taken:
    [tex]v-v_{0} = a_{0}t + \frac{1}{2}jt^{2}[/tex]
    [tex]\frac{1}{2}jt^{2}+a_{0}t-(v-v_{0})=0[/tex]
    [tex]t=\frac{-a_{0}\pm \sqrt{a_{0}^{2}+2j(v-v_{0})}}{j}[/tex]

    Finding acceleration:
    [tex]a=\frac{GM}{r^{2}}[/tex]

    Finding jerk. I read on another thread that the derivative of gravitational force is jerk. Is this right? What's making me wonder is the fact that it's negative. Does that mean the acceleration is decreasing?
    [tex]j=\frac{\partial F}{\partial r}=\frac{\partial }{\partial r}\left(\frac{GmM}{r^{2}}\right)=-\frac{2GMm}{r^{3}}[/tex]

    Finding final velocity:
    [tex]KE_{i} + PE_{i} = KE_{f} + PE_{f}[/tex]
    [tex]\frac{1}{2}mv_{i}^{2}-\frac{GMm}{r_{i}}=\frac{1}{2}mv_{f}^{2}-\frac{GMm}{r_{f}}[/tex]
    Assuming that the other object has no initial velocity, the starting KE can be cancelled out.
    [tex]\frac{1}{2}v_{f}^{2}=\frac{GM}{r_{f}}-\frac{GM}{r_{i}}[/tex]
    [tex]v=\sqrt{\frac{2GM(r_{i}-r_{f})}{r_{i}r_{f}}}[/tex]
     
  2. jcsd
  3. May 15, 2016 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    All the quantities (position, velocity, acceleration, jerk, ...) will vary in a nonlinear way. You cannot use equations that require them to be linear, you'll need a differential equation to solve.
    Depends on your sign convention, if acceleration is negative then its increase in magnitude gives a negative jerk.
     
  4. May 15, 2016 #3
    This is actually not a trivial problem. I believe that the maths you need is here. This first example is worked for Newtonian gravity, but as it happens the solution translates identically to GR if you use proper time as the time variable (this is how we can work out the proper time to "hit the singularity").
     
  5. May 16, 2016 #4
    This is a classical central force problem.
    (cf https://en.wikipedia.org/wiki/Classical_central-force_problem)
    A two body problem can easily be turned into a one-body problem using the reduced mass.
    If they are traveling in a straight line toward each other, it's simpler still.
    You can use the equation given in the wiki, changing m to mu, and taking r to mean the relative distance.
    ##\mu = \frac{mM}{m+M}##
    ##|t-t_0| = \sqrt{\frac{\mu}{2}} \int \frac{|dr|}{\sqrt{E_\text{tot} - U(r)}}##
    ##U(r) = -\frac{mMG}{r}##
    ##E_\text{tot}=U(r_0) + \frac{1}{2}\mu v_0^2##
    Try to take it from there
     
  6. May 16, 2016 #5

    rcgldr

    User Avatar
    Homework Helper

    u was used for a substitution to help solve that last integral. There is a more recent thread, using an alternate substitution (θ) for the last integral.

    http://www.physicsforums.com/threads/rectilinear-motion-of-two-attracting-masses.849750

    You'll need to need to determine θ0 and θ1 based on r. If these are point masses, then

    ##θ_0 = \frac{\pi}{2}##
    ##θ_1 = 0##
     
    Last edited: May 16, 2016
  7. May 18, 2016 #6
    Thanks so much for the link. I understand the solution now. I just have one question. There was a part where θ was defined as
    [tex]r=r_{0}sin^{2}θ[/tex]
    Where was this identity derived from?
     
  8. May 18, 2016 #7

    rcgldr

    User Avatar
    Homework Helper

    It's not an identity, it's a definition of θ based on observation designed to simplify the integral, so that after substitution, the square root expression becomes tan(θ).
     
  9. May 18, 2016 #8
    Sorry, I meant the definition. I don't understand how the equation was formed. Could you please elaborate? Thanks :)
     
  10. May 18, 2016 #9

    rcgldr

    User Avatar
    Homework Helper

    Either by observation or trial and error. Starting off with r / (r0 - r), you choose some substitution r = r0 f(θ), in which case r / (r0 - r) becomes r0 f(θ) / (r0 - r0 f(θ)) = r0 (f(θ) / (1 - f(θ))), and at this point some insight leads to choosing f(θ) = sin2(θ). (f(θ) = cos2(θ) could have also been chosen.)
     
  11. May 18, 2016 #10
    Oh. I get it now. Thanks so much! :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted