The circumference of the Moon's orbit around the sun? 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.
 A: 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
