Also, does it always take the same amount of time, or does it fractionally differ on each revolution?
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asked Nov 20, 2013 at 6:37
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2 Answers
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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.
answered Nov 20, 2013 at 7:08
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$\begingroup$ What if you want to know the motion of the moon around the sun? How would you compute that? $\endgroup$– ArneNov 20, 2013 at 12:47
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$\begingroup$ @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. ;) $\endgroup$ Nov 20, 2013 at 12:55
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$\begingroup$ 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...? $\endgroup$– ArneNov 20, 2013 at 13:04
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2$\begingroup$ @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. $\endgroup$ Nov 21, 2013 at 12:10
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1$\begingroup$ @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 $\endgroup$ Nov 22, 2013 at 6:03
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Concerning your first question, a simple estimation can be done assuming the distance Earth-Moon ≅ 4·10⁵km, and the orbit circular. So you can calculate the distance as a circumference (C=2πr) like that:
2π·4·10⁵km =8π·10⁵km ≅ 2.4 millions of kilometers
Of course you can do more precise calculations, but sometimes is good to have at first an idea of the orders of magnitude.
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$\begingroup$ Ohh really that's an other way of calculating distance thanks would have voted up if i could.. anyways thanks.. $\endgroup$ Nov 20, 2013 at 7:54
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1$\begingroup$ you're welcome, and don't care about the upvote ahha. I suggest you to use this kind of technique (raw approximations) to have a first clue of what an answer of a problem could be. This will help you furthermore to find errors while using a software or a calculator. $\endgroup$ Nov 20, 2013 at 8:00
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1$\begingroup$ Amazing, how close this answer is to the other one, although you did quite a few approximations! $\endgroup$– ArneNov 20, 2013 at 12:26