- #1

Tomsk

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- Thread starter Tomsk
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- #1

Tomsk

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- #2

lark

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Tomsk said:

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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Dearly Missed

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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

Tomsk

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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

Tomsk

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- #6

tim_lou

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