Lunar orbit distance Can anyone calculate the distance the moon travels in a single year? I can calculate it as it is close to 2.4 million km, but I need adjustments as the earth is moving and so is the Sun. So, I need the absolute distance rather than distance from earth's perspective.
It depends on where we observe it from. There is no absolute velocity reference.

For example, the most distant quasars are moving away from us at nearly the speed of the light. They "see" our Moon going away from them at nearly light speed. So, for them, the Moon travels around a light year each year (its other movements are negligible, compared to that).



If we observe it from the Earth, then we see a Moon on a circular orbit around us. Consider the following well-known facts:


Its orbital radius is $\approx$ 385000 km,
Its orbital time is $\approx$ 29 days,
A year is $\approx$ 365.25 days,


, then we can easily calculate the distance: $385000km \cdot \pi \cdot \frac{365.25}{29} \approx \underline{\underline{3\cdot 10^7 km}}$.



If we observe it from the point of view of the Sun: as we can see detailed in this answer, the orbit of the Moon around the Sun is practically circular: the 400000km orbital radius around the Earth is just a minor perturbation on its $\approx$ 150million km orbit around the Sun. Thus, we can calculate the length of the orbit of the Earth, which is $2 \cdot \pi \cdot 150 000 000 km \approx \underline{\underline{9.4245\cdot 10^8 km}}$.



Well... it is not entirely true that there is no absolute velocity reference: the WMAP probe, measuring the minor perturbations of the cosmic microwave background radiation, found also a relativistic Doppler effect, from which you can read more here. In essence, this is what it measured:



However, the results actually measured were these:



The deviation is due to our galaxy's movement of around 160km/s in the direction of the constellation Hercules, in an reference frame of the Cosmic Microwave Background, that is the closest thing there is to an absolute reference frame.

The answer of your question in this frame: all other movements (of the Sun, the orbit of the Earth around the Sun, and the orbit of the Moon around the Earth) are negligible or cancel themselves out. The result can be gotten by multiplying:


the speed in the frame of the CMB (160km/s)
the seconds of a year ($\approx$ 31.5million)


It is $\approx 160\frac{km}{s} * 31.5\cdot 10^7 s = \underline{\underline{5\cdot 10^{10} km}}$.