- #1

- 31

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## Homework Statement

Let $$\frac{1}{2}\dot{r}^2=e+\frac{m}{r}-\frac{L^2}{2r^2}$$ where L is angular moment, and e is energy (so I guess I'll take as constants for now...)

## Homework Equations

Not sure for now.

## The Attempt at a Solution

So, if I let $$u=\frac{1}{r}$$ then my equation becomes $$\frac{1}{2}\dot{r}^2=e+mu-\frac{L^2u^2}{2}$$ However, I'm not sure if I should also get my differential in terms of u as well. IE $$r=\frac{1}{u} \rightarrow dr=-\frac{1}{u^2} du$$ or just continue down the first path.

If I don't do the differential change, then I'd get $$\frac{1}{2}\dot{r}^2=e+mu-\frac{L^2u^2}{2} \rightarrow \dot{r}^2 = 2e+2mu-\frac{2L^2u^2}{2}$$ finally giving us... $$ \dot{r} =\sqrt{2e+2mu-L^2u^2}$$ which is just... yuck.

If I do the substitution, i'll get... $$\frac{1}{2}\dot{r}^2=(\frac{dr}{2dt})^2 \rightarrow (\frac{-\frac{1}{u^2} du}{2dt})^2 \rightarrow (\frac{-du}{2u^2dt})^2 $$ From here, we set it equal to the righthand side, and then see that $$\frac{-du}{2u^2dt} = \sqrt{e+mu-\frac{L^2u^2}{2}} \rightarrow \frac{du}{u^2} =-2\sqrt{e+mu-\frac{L^2u^2}{2}} dt$$ which is also, bad looking. So before I go ahead and solve either, I was looking to maybe see if there is an easier way to look at these since the next one is even *worse* (cubic in r..)

Could it also be that I'm not seeing the simplicity of it all? I'm not sure if r *depends* on t really, since this comes from our metric that ##ds^2=dr^2+r^2d\phi^2##(working in equatorial plane thus, ##\sin\theta = 1## and ##d\theta = 0##) Can go more into this if needed. So, it could be that the right hand side on both is simple because none of the terms depend on t.