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I am trying to create a graph of effective potential for the Earth-Moon system in Excel. It should result in a graph that looks like this: Graph of effective potential with radius

But this is not what I end up with. (I've made a Google Docs version of my spreadsheet here.)

  • I start by calculating the gravitational potential at r: $V(r)=-\frac{GM_E}{r}$.
  • Then multiply by the mass of the Moon to get the gravitational term (i.e. $U(r)=-\frac{GM_EM_M}{r}$).
  • Then calculate the gravitational force with $F=G\frac{M_E M_M}{r^2}$ and calculate the orbital velocity with $v=\sqrt{\frac{Fr}{M_M}}$.
  • Calculate angular momentum with $L = M_Mvr$.
  • Calculate the value of $\frac{L^2}{2\mu r^2}$ where $\mu$ is the reduced mass $\mu = \frac{M_M M_E}{M_M+M_E}$.
  • Find the resultant $\frac{L^2}{2\mu r^2} - \frac{GM_E}{r}$.

I'm making an obvious mistake somewhere, as I end up with a negative effective potential at all radii. I can't post an image because I don't have enough reputation for more than two links, but it's in the Google Docs spreadsheet linked above.

Qmechanic
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Alby
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1 Answers1

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First off, a suggestion: don't make people go digging through a complicated spreadsheet so that they have a chance of answering your question.

You must be doing something wrong, because it can be shown that for small enough $r$ and for any fixed value of $L$, the potential becomes positive:

$$\frac{L^2}{2\mu r^2} - \frac{GM_E}{r} > 0 \\ \iff \frac{L^2}{2\mu r^2} > \frac{GM_E}{r} \\ \iff \frac{L^2}{2\mu} > G M_E r \\ \iff \frac{L^2}{2 \mu G M_e} > r$$

That's all I can say, really. Maybe stick your formula into Wolfram Alpha and ask it to graph it for you.

Javier
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