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OK, here's another that might be simpler than it looks but I am about ready to give up. Given the following:

$$U_\mathrm{eff}(r) = - \frac{k}{r}+ \frac{L^2}{2mr^2}$$

$$\frac{U_\mathrm{eff}}{U_0} = -\frac{2}{r / r_0} + \frac {1}{(r/r_0)^2}$$

$$r_0= \frac{L^2}{mk}, \qquad U_0= \frac{k}{2r_0}$$

EDIT - L is angular monentum and k is a constant. m is mass r is radius.

I am supposed to make $U$ dimensionless. But I can't seem to get rid of the $r$ in the denominator. Maybe that isn't what I am supposed to do. The question is "Convert the effective potential to a dimensionless function by scaling to the radius of a circular orbit. " Which makes zero sense to me right now.

One suggestion was to plug in the $r_0$ expression in $U_0$ but I can't seem to see how this is supposed to work. I end up with $U_\mathrm{eff} = \frac{-2L^2}{kmr} + \frac{L^4}{k^2mr^2}$. Maybe that's what is supposed to happen, I don't know.

This isn't even calculus I bet, This should be simple but for some reason I am losing it.

ANyhow, any help is appreciated. thanks.

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  • $\begingroup$ I'm guessing $r$ is radius and maybe $m$ is mass. What are $k$ and $L$, and why haven't you told us what they are? $\endgroup$
    – Michael Hardy
    Oct 16, 2013 at 21:10
  • $\begingroup$ In your equation $\frac{U_\mathrm{eff}}{U_0} = -\frac{2}{r / r_0} + \frac {1}{(r/r_0)^2}$ the quantity $\frac{r}{r_0}$ is dimensionless, so I think you are done. $\endgroup$
    – Nicky Hekster
    Oct 16, 2013 at 21:29
  • $\begingroup$ @Nicky Heksetr - but if that s the case there's no problem to do. That was one of the givens. $\endgroup$
    – Jesse
    Oct 16, 2013 at 22:41
  • $\begingroup$ I figured it out! he wanted us to show the intervening STEPS. Arrrgh! That was really stupid of me. $\endgroup$
    – Jesse
    Oct 17, 2013 at 1:30
  • $\begingroup$ @Jesse not stupid and it learned you to really think over these equations! Keep posting, good luck! $\endgroup$
    – Nicky Hekster
    Oct 19, 2013 at 17:40

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