Separate the variables by using kinetic energy and potential energy

  • Thread starter Tomsk
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  • #1
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Is it possible to solve [tex]x^2\ddot{x}=\frac{q_{1}q_{2}}{4\pi\epsilon_{0}m}[/tex] to get x(t)? I can't see how!! Maybe I'm just missing something...
 

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  • #2
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Tomsk said:
Is it possible to solve [tex]x^2\ddot{x}=\frac{q_{1}q_{2}}{4\pi\epsilon_{0}m}[/tex] to get x(t)? I can't see how!! Maybe I'm just missing something...

You can separate the variables by using kinetic energy and potential energy, and conservation of angular momentum. I did that. I put
it all into polar coordinates - and I came with an integral of a function
of r (radius) that there's probably a formula for somewhere. It's not
an easy integral though.

It's simpler if you assume the particles don't have any angular momentum relative to each other. That's how I got curious about it - I did an exercise about the enormous acceleration a proton would have, jetting out of the nucleus, if there weren't any strong nuclear forces holding it
in. So I wondered, what's its equation of motion?

You could figure out the proton's final velocity without doing any
complicated integrals - that would be (sort of) interesting too.

Laura
 
  • #3
arildno
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Certainly.
Let the squared position stand in the denominator on the right-hand side, and multiply your diff. eq. with the velocity.
You now will get a first integral (take note of the sign of the square root used!), this can be integrated one more time.
 
  • #4
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Spooky....

Well, I got
[tex]x^3=-\frac{9}{2}\frac{q_{1}q_{2}}{4\pi\epsilon_{0}m}t^2[/tex]

The sign threw me though, so I'm not sure of it.

Thanks! I might try it relativistically, to stop my brain from rotting before I go back to uni.
 
  • #5
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One thing, there's only one m taken into account, which must be the mass of the particle which moves a distance x, or is x the distance between the two particles? I'm assuming they're both free to move, so wouldn't you need to take both masses into account? Hmmm
 
  • #6
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in two dimension this is known as the one body problem (using a equilvalent equation in 2D)... you might wanna try to solve that (it is impossible to get r(t) explicitly, but you can find out the shape of the orbit) its a lot of "fun". after that go to two body problem... more fun awaits...
 

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