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arildno

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What's your argument?

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- #26

arildno

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What's your argument?

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arildno said:What's your argument?

Hummm, this is where I am stuck. I don't know how to proceed with the integral 3 posts above.

Integration of that integral would allow me to solve for R in terms of theta, which will in turn tell me the shape of the orbit of any test particle about the origin, in terms of its energy and angular momentum.

But the thing is, I don't know how to use the term [tex]V(u)[/tex] in the integral, where [tex] u = R(\Theta)[/tex]

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arildno

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I meant why you think the sign should be positive.

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Is it true that [tex] V = -\int F[/tex]? =\

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Wait. If that's true then it should be negative. AHHHH!! ><

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Odyssey said:[tex]t-t_{0}=\int_{R_{0}}^{R(t)}\frac{du}{\sqrt{2mEL^{-2}u^4-u^2-2mL^{-2}u^4V(u)}}[/tex]

where, u = R of theta. So should I plug the V = -(1/2)Amr^-2 into V(u)?

Now I have the integral. plug the V = -(1/2)Amr^-2 into V(u)? How should I proceed from here?

Thank you again for the help!

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