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I know that it orbits the sun in what looks like a 12 sided polygon with rounded corners. But I can't seem to find the radius/circumference anywhere.

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Cute question, but: add the moon's position relative to earth to the earth's position relative to the sun and Bob's yr uncle. –  Carl Witthoft Nov 21 '13 at 14:23
    
We can measure can find the distance that the moon loops around the earth and then since the moon orbits the earth once every 27-28 days then we multiply by 12. I'm not asking for the exact circumference. –  MAyman Nov 21 '13 at 14:28
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While you are at it, add the motion of the sun around the galaxy and the moon goes through a fancy spiral. –  ja72 Nov 21 '13 at 14:31
    
Looking at the wiki picture en.wikipedia.org/wiki/File:Moon_trajectory1.svg I think that the first approximation with the orbit of Earth would be enough. If not, you could integrate it yourself (the page for the picture even has relevant equations). –  user23660 Nov 21 '13 at 14:35

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up vote 3 down vote accepted

To a first approximation distance covered by the Moon is the same as the Earth's, but you can also estimate the correction to first order. Assume both orbits are circular and in the same plane since any deviations will affect only smaller order corrections.

Represent the position in the orbital plane as a complex number

$Z = R e^{2\pi i (t/Y)} + r e^{2\pi i (t/M)}$

where $R$ = radius of Earth orbit, $r$ = radius of Moon orbit, $Y$ = one year and $M$ = a siderial month

The velocity $V$ is given by

$\frac{V}{2\pi i} = \frac{R}{Y} e^{2\pi i (t/Y)} + \frac{r}{M} e^{2\pi i (t/M)}$

The speed $s = |V|$ is given by

$\frac{s^2}{4{\pi}^2} = \frac{R^2}{Y^2} + 2\frac{Rr}{YM}cos(2\pi t(\frac{1}{M}-\frac{1}{Y}))+\frac{r^2}{M^2}$

Take the square root assuming that $\frac{R}{Y} \gg \frac{r}{M}$ and use

$\sqrt{A+\epsilon} = A(1+\frac{\epsilon}{2A}-\frac{\epsilon^2}{8A^2}+O(\frac{\epsilon^3}{A^3}))$

$\frac{s}{2\pi}= \frac{R}{Y}(1+\frac{rY}{RM}cos(2\pi t(\frac{1}{M}-\frac{1}{Y}))+\frac{1}{2}(\frac{rY}{RM})^2 - \frac{1}{2}(\frac{rY}{RM})^2cos^2(2\pi t(\frac{1}{M}-\frac{1}{Y}))+O(\frac{rY}{RM})^3))$

We want the average speed, the cosine averages to zero, but the cosine squared averages to a half, so

$\bar{s}= 2\pi \frac{R}{Y}(1+\frac{1}{4}(\frac{rY}{RM})^2+O(\frac{rY}{RM})^3))$

$\frac{rY}{RM} = 0.03$ so the distance traveled by the Moon is greater than the distance travelled by the Earth by about one part in 4200

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That's hell of a treatment on celestial mech. –  Cheeku Nov 21 '13 at 16:57

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