What is the distance that the Moon travels during one orbit around the Earth?

  • Also, does it always take the same amount of time, or does it fractionally differ on each revolution?


  • TildalWave

    TildalWave Correct answer

    9 years ago

    The Moon has an orbital eccentricity of 0.0549, so its path around the Earth is not perfectly circular and the distance between the Earth and the Moon will vary from the Earth's frame of reference (Perigee at 363,295 km and apogee at 405,503 km), see for example second animation explaining Lunar librations in this answer.



    But its orbit can be said, in an oversimplified manner, to be periodic, with no significant apsidal precession (not really true, but somewhat irrelevant for my following musings here to be still close enough), so we can calculate its orbital length based on its quoted average orbital speed of 1.022 km/s and orbital period of 27.321582 days.



    So, plugging our numbers in a calculator, $l = v * t$, we get the Moon's orbital length of 2,412,517.5 km (or 1,499,070 miles). Should be close enough. Source of all orbital elements of the Moon is Wikipedia on Moon.


    What if you want to know the motion of the moon around the sun? How would you compute that?

    @Arne Heh, I'll take your question as a well meaning brain teaser. :) There is always this question of what's your frame of reference, of course, but one relatively easy way would be calculating the length of one Earth's orbit around the Sun and the Moon's path as a helix with radius the Moon's Semi-major axis, and height of one rotation 365.25 / 27.321582 days. Should be close enough. ;)

    Yes, I thought of a similar thing. Wikipedia states that the orbit of the Moon around the Sun is convex however, since the Sun's influence is much greater than the Earth's influence. So I don't know if a helix would be a good approximation... Maybe for the Sun/Jupiter/Ganymede system...?

    @Arne Ah yes, the Moon's path around the Sun is convex, but that is irrelevant for calculating the length of the path it makes. You see, you can bend any spring into a convex curvature, but the wire it's made of will still be of equal length. ;)

    Good point! Though I still don't quite know how to calculate it. :)

    @Arne Well the equation for the length of a circular helix is given by Wikipedia on Helix, and the remaining required orbital parameters should be also listed in Wikipedia on Earth, but if you're after a fast approximation, then calculate the length of the Earth's orbit the same way I calculated the Moon's around the Earth, and then simply add to it my Moon around the Earth multiplied by 365.25 / 27.321582. It won't be much off from the real helix length (not more than ±5.5% off).

    @Arne, in ~27.3 days the Moon complete an orbit around the Earth. So in 365 days it completes 13.4 orbits. If one orbit is ≅ **2.4** millions of kilometers (look at my answer), in **13.4** orbits it completes 32.2 millions of kilometers around the Earth. Then, we add the path of the Earth around the Sun: 2π·150 millions of Km (≅ **942** millions of km), and we obtain **974 millions of km around the Sun**.

    @leonardvertighel ok, so orbits are linear and/or additive? I was wondering that. I thought it might be a bit more complicated, since orbits are a function of time, and the orbits are not simple, linear functions.

    @Arne No, orbits are nearly as complicated as you want them to be. For example, we didn't even consider perturbations, anomalies, precession, even radiation pressure and space weather. But here's the catch, you _have to_ decide at which point you stop appreciating any effects as meaningful for your needs, otherwise it becomes impossible to calculate, while you're only moving your object a few millimeters, perhaps. Periodic corrections are your best friend with orbits, otherwise it gets insanely complicated, even with relatively simple Keplerian ones in classical mechanics.

    @Arne, as TildalWave stated, my approach was just a simplification to have a raw estimation. You did not know before if it was 1 million km, 10 millions, 100 millions, or more? Now you have a starting point

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Content dated before 7/24/2021 11:53 AM