Time for two objects to collide due to gravity

  • #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]
 

Answers and Replies

  • #2
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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.
Does that mean the acceleration is decreasing?
Depends on your sign convention, if acceleration is negative then its increase in magnitude gives a negative jerk.
 
  • #3
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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?
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").
 
  • #4
Khashishi
Science Advisor
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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
 
  • #5
rcgldr
Homework Helper
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Last edited:
  • #6
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.

https://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##
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?
 
  • #7
rcgldr
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There was a part where θ was defined as
[tex]r=r_{0}sin^{2}θ[/tex]
Where was this identity derived from?
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).
 
  • #8
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).
Sorry, I meant the definition. I don't understand how the equation was formed. Could you please elaborate? Thanks :)
 
  • #9
rcgldr
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I don't understand how the equation was formed.
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.)
 
  • #10
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).
Oh. I get it now. Thanks so much! :)
 

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