# How far does the moon travel in a year?

Asked by GeorgeGee (4914) November 12th, 2010

Unlike the circular orbit of the earth, the moon is traveling in a cycloidal path, so I’m wondering if anyone has computed the distance actually traveled by it.

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• 1.4 million miles = distance the moon travels in 1 trip around the earth
• 27⅓ days = time it takes to make one trip
• 1 year = 365 days

So in a year it would make ~13.35 trips, and travel 18,690,000 miles in year. I got all this info by googling your question and finding this site.

Aqua (2546)

But you’re missing the distance around the Sun.
Maybe we just add the two?

GeorgeGee (4914)

But then you’d be missing the distance around the center of the Milky Way too.

It’s sort of a meaningless question without asking relative to what, though since it says “in a year”, one might assume it means relative to the thing it moves around each year, which would be the sun.

Zaku (19335)

If you wanted to compute it exactly, you’d have to take into account the elliptical path the moon takes around the earth each lunar month, plus the elliptical path taken by the moon around the sun each year as it’s dragged along by the earth-moon system, whose center of mass is closer to earth’s surface than center & whose orbital plane is not precisely the ecliptic (which is why every full moon does not produce an eclipse).

It’s been a long time since I set up and calculated integration paths along parametrized epicycloid curves in 3 dimensions! Even assuming circular, coplanar orbits, I suppose that the calculation is now the job of astrophysical software. Not to mention Google

So I’m sure you could look up the exact distance traveled by the moon (they use Apollo-era corner reflectors to return a laser beam to its source for continuous & precise ranging of Earth-Moon distance). We can do a “back of the envelope” calculation instead…
——————
Simplify by treating the two motions of the moon (orbiting earth, orbiting sun) separately—(A) Calculate how far the moon travels orbiting the earth for one lunar cycle, multiplied by number of lunar cycles per year. This assumes the Earth is standing still. (B) Add the orbital length of the Earth around the sun each year & assume it’s the same for the moon.

The moon is about a quarter-million miles away, more like 240,000 miles mean. Orbital circumference is thus (240,000)*2*pi = 1.51 million miles. If a lunar cycle is 27.3 days, then there are 365 / 27.3 = 13.4 cycles. (1.51)(13.4) = 20 million miles a year orbiting Earth.

At a distance of 93 million miles from the sun, the earth (and its moon) travels about
(93)2*pi = 584 million miles a year. Compare: (20 / 584) = 3%. We thus estimate *the moon’s orbit around the earth adds only 3% to its total distance traveled through the solar system.

What about speeds? Based on our numbers, the earth’s speed is (584 million / 365) / 24 = 67,000 mph, while the moon’s speed relative to earth is (1.51 million / 27.3) / 24 = 2,300 mph. Compare: (2300 / 67,000) = 3%. Thus the moon orbits the earth at only about 3% of the speed with which they both orbit the sun, assuming the solar system is fixed. This is confirmed by this image of the moon’s path around sun here, showing the path as convex.

Although the moon travels “backward” during half of each lunar cycle, this only lowers its forward motion by those 3%. In geometrical parlance the path is approximated by a highly curtate epicycloid.

Here’s a nice quote from the link above:

Imagine you’re driving on a circular race track. You overtake a car on the right, and immediately slow down and go into the left lane. When the other car passes you, you speed up and overtake on the right again. You will then be making circles around the other car, but when seen from above, both of you are driving forward all the time and your path will be convex.”

gasman (11264)

Thanks @gasman, next we’ll calculate the distance that Ginger Rogers danced while circling Fred Astaire, backwards, in heels. :D

GeorgeGee (4914)

Except a lunar cycle is actually 29.53059 days; not 27.3.

@tragiclikebowie: Except it depends on what type of lunar month you’re talking about.
A synodic month takes 29.5 days and a sidereal month takes 27.3 days.

Aqua (2546)

@Aqua Right—depending on viewpoint. Note that (29.5 – 27.3) = 2.2 days—the difference between synodic and sidereal month lengths—times 13.4 cycles (see above) equals 2.2 * 13.4 = 29.5 days—exactly one extra cycle as the Earth-moon system completes one orbit.

@GeorgeGee lol. “Epicycles upon epicycles” in the Ptolemaic system. And the Disneyland teacups make me barf.

gasman (11264)

With respect to what?

Ivan (13434)

For the purpose here, we’ll assume relative to the Sun.

GeorgeGee (4914)

Well, its displacement is zero.

Ivan (13434)

It would travel the same distance as the earth if we are measuring relative to the sun. Since when it passes in one half revolution it is moving forward with the earth, and then moves backwards on the second half of its journey.

Afos22 (3980)

@Afos22 I’m not sure it’s that simple. Suppose the Moon revolved around the Earth many times per hour (such can happen with exotic objects like neutron stars, etc). Then it might very well travel much farther in all its local orbits compared to its annual circuit around the sun. You have to plug in actual numbers to be sure, it would seem.

gasman (11264)

well since space time bends with gravity and speed… you can’t really calculate until you know the time dilation from the speed of the moon and the sun and galaxy… ect

talljasperman (21734)

or

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