The Moon that went up a Hill but came down a planet

By Phil Plait | September 29, 2008 4:25 am

Edited to add: I have taken the topic of this post and logically continued it in a post about Pluto.

Every now and again I get an email from someone who tries to tell me that the moon orbits the Sun more than it orbits the Earth.

On the face of it, their claim appears to hold water. For example, they’ll say that if you compare the orbital velocity of the Moon around the Earth (about 1 km/sec) to the orbital velocity of the Moon around the Sun (the same as the Earth’s velocity around the Sun, about 30 km/sec), you’ll see that the Moon’s orbit is always convex to the Sun; in other words, it doesn’t make a loopy pattern around the Sun, as you might expect:

Shape of the Moon’s orbit

That diagram is exaggerated; on that scale the Moon’s combined path around the Earth and Sun would look pretty much like a circle (and to be clear, the link above doesn’t make the claim that the Moon orbits the Sun more than it does the Earth, it’s just pointing out the patterns).

Moreover, if you calculate the force of the Sun’s gravity on the Moon, you find it’s more than twice the force of Earth’s gravity on the Moon!

Whoa. So does the Moon orbit the Earth, or the Sun?

Turns out, it orbits the Earth, despite these claims. The above claims are true, but are not important in this argument. Instead, you have to look at something called the Hill sphere. Basically, it’s the volume of space around an object where the gravity of that object dominates over the gravity of a more massive but distant object around which the first object orbits.

OK, in English — and more pertinent to this issue — it’s the volume of space around the Earth where the Earth’s gravity is more important than the Sun’s. If something is orbiting the Earth inside Earth’s Hill’s sphere, it’ll be a satellite of the Earth and not the Sun.

The derivation of the math isn’t terribly important here (and it’s on the Wikipedia page if you’re curious), but when you plug in the numbers, you find the Earth’s Hill sphere has a radius of about 1.5 million kilometers. The Moon’s orbital radius of 400,000 km keeps it well within the Earth’s Hill sphere, so there you have it. The Moon orbits the Earth more than it orbits the Sun. In reality it does both, and saying it orbits one and not the other is silly anyway.

But what about those other claims? Well, the orbital shape is pretty meaningless for this argument. You’ll get that always-convex pattern for any moon that orbits a planet far enough out compared to the planet’s distance from the Sun. But as long as that moon is inside the planet’s Hill sphere, it orbits the planet more than the Sun.

And the fact that the Sun’s force effect of gravity on the Moon is stronger than the Earth’s is interesting, but not relevant. Why not? Because the Sun is pulling on both the Earth and the Moon! Since both the Earth and Moon are roughly the same distance from the Sun (the Sun is 400 times farther away from us than the Moon is), the force acceleration due to the Sun’s gravity on the Moon and the Earth is about the same. That means the Sun’s gravity is important in keeping the Earth and Moon together orbiting the Sun, but doesn’t affect the Earth and Moon as a system. The Moon is still gravitationally bound to the Earth.

Think of it this way: if the Sun were to disappear, the Moon would still orbit the Earth pretty much as it does now. That means the Sun’s gravity doesn’t affect the Earth/Moon system much. On the flip side, if the Earth disappeared, the Moon would continue to orbit the Sun as well. That means the Earth’s gravity doesn’t much affect the Moon’s orbit around the Sun, either.

Now, if we moved the Earth/Moon system closer to the Sun, then this all starts to matter. That’s because eventually you get to a point where the size of the Moon’s orbit gets to be significant compared to the size of the Earth’s orbit around the Sun. In that case, as the Moon orbits the Earth, when its closest to the Sun, the gravity of the Sun gets significant compared to the Sun’s gravity on the Earth (remember, right now that force effect is about the same). But that’s exactly how the Hill sphere idea works!

It turns out that if you moved the Earth and Moon to about 40 million km from the Sun — closer to the Sun than Mercury! — then the Hill sphere of the Earth would be 400,000 km, equal to the moon’s distance from the Earth now. That means the Moon could be stripped from the Earth by the Sun, and so then you could say the Moon orbits the Sun and not the Earth.

But out here, 150 million kilometers from the Sun, the Moon orbits the Earth quite nicely, thank you very much. You can keep your convex orbits and your forces and everything else, and examine them as matters of interest. But the Moon will keep pacing out its orbit just the same, and it’ll keep doing it that way for a long, long time.

The orbital diagrams in this post are from Helmer Aslaksen’s webpage at the Department of Mathematics, National University of Singapore. I modified them a bit to compare them. You should read that page, as it has an excellent description of the Moon’s orbital shape. Oh, and stay tuned, because I have another thing to say about all this in a later post

Edited to add: in a couple of places I used the word "force" when I mean "acceleration". I fixed that, or used the word "effect". This may seem like a nitpick, but in fact they mean different things physically, and therefore their magnitudes, the sizes of their effects, changes. That’s what I get for writing this at 4 in the morning.

CATEGORIZED UNDER: Astronomy, Cool stuff

Comments (78)

  1. Todd W.

    Thanks for the explanation, Phil. But, what about Hugh Grant?

  2. Blaidd Drwg

    OT, but amusing. I copied this from a comment on HuffPo:

    Experiment. Load McCain and Palin into the Large Hadron Collider. When they collide do they create a tiny black hole that sucks out logic and reason from the universe? Or provide proof that particles with so much negative spin increase the heavy elements of anger and ignorance needed to inflate the alternative Bushverse?

  3. It’s not “convex”, it’s a “star-shaped region”. Convex regions contain the line segment between any two points inside them (which your diagram on the right clearly doesn’t. Star-shaped regions have (at least) one point so that the region contains the segment from that point to any other point in the region.

    Come on. Mathematicians have put in all this effort to create a precise vocabulary, the least you astro/physics guys could do would be to use it.

  4. Chris

    What’s with the sharp changes in direction on the “actual” pattern, wouldn’t it be a smooth sinusoidal pattern?

  5. andy

    You’ve illustrated the difference between a non-self-intersecting curve and a self-intersecting one. The second diagram is NOT a convex one – you copied the wrong diagram from your reference.

  6. John Armstrong, the moon’s orbit is convex – it’s the diagram that’s misleading (exaggerated, as he says). Go to the link Phil provided.

    I did a quick sum to work out the at what distance you would start to get looping. Turns out it’s about 500km from the centre of the earth – and that’s assuming that the entire mass of the earth is inside that radius, which it isn’t.

    I think Io and Europa manage it, though.

  7. The diagrams are not supposed to be perfectly accurate; they are illustrative of the idea. I said they are exaggerated in the post.

  8. Phil: the problem is the diagram is NOT illustrative of the idea. If you’re going to illustrate the idea of a convex orbit, you shouldn’t do it with a picture of an orbit that isn’t convex.

  9. 01101001

    Bad second diagram

  10. Mark

    What is it with people named “Hill” and their obsession with spheroids?
    (that bad one-liner’s not going to work for people unacquainted with the works of the late British comedian Benny Hill …) ;)

  11. Metre

    Wouldn’t it be better to just say that the earth and moon orbit around their common center of gravity, and the common center of gravity orbits around the sun? Trying to decide whether the moon orbits the earth or the sun smacks of trying to compute how many angels can fit on the head of a pin.

  12. Lovely article. It tickled my brain in exactly the right places. (But the math geeks are probably right about the second diagram – myself, I think you should have smoothed out those inner spikesy thingies in the orbit … because they are “not good to behold”. ;)

  13. Cheyenne

    Awhile ago I was playing around with a website that had a physics engine were you could manipulate orbits (you could change mass, velocity, etc. of different objects). One of their default ones was just a simple moon, earth, and sun. It had the moon trace out a path like the one on the left in BA’s diagram. Which did seem, at the time, to make sense.

    I have no doubt that BA is correct here, but it looks a little funky to me (looks like the moon “stops” and then starts again in a new direction on the diagram on the right). I completely don’t understand the math or the “convex” bit.

    Here’s my question – does anybody have any other websites and/or diagrams that might show a little more detail on this?

    Also, can somebody please merge quantum mechanics with the general theory of relativity and pass that along? Something short and not too complicated would be great.

  14. In the lower right corner of the Wikipedia article is an illustration showing a fragment of the earth/moon system’s orbit around the sun.

  15. @Cheyenne, regarding your second question: I think a bloke named Douglas Adams did just that, quite a few years ago.

  16. Digital Cosmos

    Love the blog! Be the blog!

    How would this idea play if one uses these ideas with the sun, earth (or perhaps Jupiter instead to make a closer mass ratio to the sun), and center of the galaxy?

  17. @Metre, yes, you are correct. Considering that the center of rotation is about 2900km from the center of the earth, placing it still inside the earth; one can still say that the moon orbits us.

  18. Ivan M.

    Minor quibble: the force of the Sun’s gravity on the Moon and the Earth is NOT about the same, but rather the acceleration that that force induces.

  19. That was fascinating. Thank you. :)

  20. Kristin

    Phil, you said “Since both the Earth and Moon are roughly the same distance from the Sun (the Sun is 400 times farther away from us than the Moon is), the force of the Sun’s gravity on the Moon and the Earth is about the same.”

    But the force of gravity between two objects depends on the masses of both of those objects. So the force of gravity between the Sun and the Moon is much smaller than the force of gravity between the Sun and Earth, since Earth is 81 times more massive than the Moon.

  21. Torbjörn Larsson, OM

    Trying to decide whether the moon orbits the earth or the sun smacks of trying to compute how many angels can fit on the head of a pin.

    Not if you want to sort out dominating gravitational influence and, I presume, orbits – which seems is what the Hill sphere does.

    can somebody please merge quantum mechanics with the general theory of relativity and pass that along?

    That would be quantum field theory, at the simplest – merging QM and special relativity. I haven’t studied it, but I doubt the subject is “short and not too complicated”.

    You are welcome. ;-)

    But along that line you can AFAIU merge QM and general relativity as the quantized effective (low energy) theory of gravitation Jacques Distler describes:

    It’s often said that it is difficult to reconcile quantum mechanics (quantum field theory) and general relativity. That is wrong. We have what is, for many purposes, a perfectly good effective field theory description of quantum gravity.

    It is governed by a Lagrangian …

    This is a theory with an infinite number of coupling constants … Nonetheless, at low energies, … , we have a controllable expansion in powers of ε. To any finite order in that expansion, only a finite number of couplings contribute to the amplitude for some physical process. We have a finite number of experiments to do, to measure the values of those couplings. After that, everything else is a prediction.

    In other words, as an effective field theory, gravity is no worse, nor better, than any other of the effective field theories we know and love.

    Next step, to cover all energies and/or avoid measurements of parameters, is perhaps string theory. But there you have it, according to Distler it isn’t “difficult” to make a minor merge into something “we know and love”.

  22. James

    Phil, does this concept help explain why Mercury and Venus have no moons, or are they not close enough?
    Also, here is a link to the University of Colorado at Boulders orbital simulation. Lots of fun to play with!http://phet.colorado.edu/simulations/sims.php?sim=My_Solar_System

  23. Andy Beaton

    The pointiness in the 2nd diagram seems to imply the moon stops and changes direction, at least from the suns pint of view. It didn’t make sense until I imagined twirling a lighted yo-yo on a skateboard, and the faster the skateboard goes, the pointier the curve gets. Very cool, Phil.

  24. Kristin

    Just a thought – I’m wondering if explaining this whole topic in a different way would help (or just confuse things more, as I’m really good at making things more complicated than they need to be :-) )

    Right now, as the Moon orbits Earth, it gets closer to and farther from the Sun each orbit (I’m simplifying things by assuming circular orbits). However, this change in distance is very small compared to the average distance between the Moon and the Sun, so it’s essentially negligible, so the Moon will continue to orbit Earth while the two orbit the Sun.

    Now, put the Earth and Moon real close to the Sun, like Phil did, 40 million km. Now the Moon is considerably closer to the Sun at new moon than it is at full moon, so the gravitational force between the Moon and the Sun is much greater at new moon than at full moon, and as a result the Moon will be pulled out of it’s Earth-orbit.

    Does that make any sense? To me it does (in fact it seems very similar to tidal forces), but I know I think differently from most people :-)

  25. James, good point. Venus’s Hill radius is about a million km, which seems large enough to hold on to a moon. The debris from which our Moon formed was much closer to the Earth than the Moon is now, so a similar event would probably have worked with Venus. However, the impact that formed our Moon was relatively rare.

    Mercury’s Hill radius is much smaller, about 200,000 km (Mercury is less massive and much closer to the Sun than Earth). So yeah, I would guess that’s why it has no Moon.

  26. It turns out that if you moved the Earth and Moon to about 40 million km from the Sun — closer to the Sun than Mercury! — then the Hill sphere of the Earth would be 400,000 km, equal to the moon’s distance from the Earth now. That means the Moon could be stripped from the Earth by the Sun, and so then you could say the Moon orbits the Sun and not the Earth.

    Yes and no, Phil. It’s actually more interesting than that. It turns out that prograde (he says, hoping no one complains about that neologism) orbits are only stable out to about 40% of the Hill radius. After they, they get stripped in fairly short order. Retrograde orbits are far more stable, pretty much right out to a full Hill radius. (I’ve done some work on this and I know that Doug Hamilton at UMD has, too.)

    Now… moving the system closer to the Sun is even more interesting! It turns out that if you move the system adiabatically (“slowly” compared to the orbital timescales), prograde orbits also move away from the planet, while retrograde orbits evolve inward. So prograde moons will actually move outward to meet the dropping Hill sphere and be lost even sooner than you’d think. (Retrograde orbits will persist longer, conversely.)

    Inclined orbits make things more interesting, as do eccentric orbits.

    Sorry if this is more info than anyone wants, but I did the research on this. (Never got the paper published… long story. Still love to have it published, but I’d have to re-run the models. Presented it at a few meetings, though.)

  27. Terry Heatlie

    Oh, and I suppose I should believe you astronomers more than I should believe Asimov, a biochemist? (Darn, I really enjoyed Asimov’s essay on this, and I’ve been going around saying the moon orbits the sun ever since. Guess I have to stop now. Pesky facts.)

  28. A friend of mine created this animation to show the motion of the Moon relative to the Earth and Sun: http://www.mogi-vice.com/Scaricamento/Luna%20eliogeocentrica.zip (more of his educational astronomy animations: http://www.mogi-vice.com/Pagine/Downloads.html )

  29. Phil, thanks for answering my question on episode 83 of Skepticality about pointing the Hubble at the moon. I try my best to convince my friend that the moon landing wasn’t faked but I just don’t know enough about it to get into specifics. I posted a link to the podcast and to your website on my blog, and I thought you might be interested in hearing his rebuttal, which I’ve pasted below. If you want to address any factual errors on his part I’d be elated to pass them along. Thanks! (I will also email this to the Skepticality crew)

    “When the idea of photographing objects left behind by astronauts was first presented (to dispel hoax claims), Nasa said that the Hubble telescope could not take pictures of the moon because it was “too close” (as Hubble was designed to take pictures of far away galaxies). This was disproved. Then Nasa said that Hubble could not take pictures of the moon because it was “too bright”. This was also disproved. Hubble has since taken pictures of the moon in 1999 and 2005.

    Now Nasa says that Hubble has a maximum resolution of 0.014 arc seconds and can only resolve objects 88.5 feet or larger on the moon. The largest object reportedly left behind on the moon is the lunar module (16.5 feet). Nasa claims that there is no chance Hubble can see such a small object.

    I am no conspiracy theorist, but I would be remiss in my curiosity to overlook how many times Nasa changed their story, first claiming that something is “impossible” (in regards to Hubble observing the moon), having their official statements disproved, and then returning with “new facts” to trump their “old facts”.

    Now Brazilian astronomers have calculated that Hubble has a maximum resolution of 0.1 arc seconds and can actually resolve an object the size of a car. Not only could such a resolution observe the lunar module, but the smaller lunar rovers as well.

    Can’t see it; too close.
    Disproved.
    Correction. Yes we can.

    Can’t see it; too bright.
    Disproved.
    Oh wait. Yes we can.

    Can’t see it; too small.
    Are you sure about that?
    No rush, take your time.
    History is patient…

  30. Cheyenne

    James – Thanks for the link! I think that is what I saw before. Perfect illustration. Think I understand it now. It’s a little easier to grasp when you set it in motion.

  31. Cheyenne

    Come to think of it (after playing with that website a bit) – the long term stability of our solar system is pretty remarkable. Hope it stays that way….

  32. James

    Cheyenne, I was using that site in my astronomy class last year, and the students had to click on four objects and set up the proper masses and velocities in order to simulate the paths of Mercury, Venus and Earth. It was NOT easy to do, and took them a lot of calculation time! It is pretty amazing how stable our system is, if you just tweak one thing…

  33. The point that was being made about the Moon’s orbit is convex, i.e. that it does not have any of those bits that point inwards towards the Sun. In other words, the Moon’s orbit curves towards the Sun at all times (as it says at the reference, “the curvature never changes sign”). When Phil Plait says “That diagram is exaggerated”, he is actually using a diagram that is completely different to the argument at hand. In Phil Plait’s “exaggerated” diagram, there are points where the Moon is curving away from the Sun, which is not the case in reality. If he wanted an exaggerated version for clarity (at this scale, a rounded dodecagon – which is the closest rounded polygon to the Moon’s path – looks suspiciously like a circle), he could have used a rounded polygon of fewer sides, e.g. a rounded hexagon. Instead he misrepresents the original argument and the true situation (which is a lot more interesting than what he has in the diagram) in the name of exaggeration. I don’t know how bad Phil Plait is at astronomy, but his geometry sure leaves a lot to be desired.

  34. JM Gallant

    @Paolo:

    Thank you for that video. It does a nice job giving visuals to all of this.

  35. I thought I was the only person who still had a Spirograph in the attic.

  36. Dr. Phil Plait, according to Alex Alemi and David Stevenson of the California Institute of Technology, their recent study of models of the early solar system shows that it is very likely that, billions of years ago, Venus had at least one moon, created by a huge impact event. About 10 million years later, according to Alemi and Stevenson, another impact reversed the planet’s spin direction. The reversed spin direction caused the Venusian moon to gradually spiral inward until it collided and merged with Venus.*

    *Source: Wikipedia — Click on my name for the link.

  37. James

    From Websters:
    Tact (takt) n. delicate perception of the right thing to say or do without offending.

    A trait sorely lacking in a large portion of blog posters, as well as some scientists, which leads to things like not being able to explain what their work does, which in turn leads to things like getting their budgets cut (i.e. the Superconducting Supercollider, 1993).

  38. Zippy the Pinhead

    Asimov wrote a book “The Double Planet” (1960) where he promoted this canard that the Moon orbited the Sun. Glad you knocked the old coot down.

  39. Murff

    Seems so simple. The Moon orbits the Earth, together they orbit the Sun, together with the other planets, they orbit the galactic center…

  40. As others have said, I don’t think Dr. Plait’s diagrams quite have it right, even if they are exaggerated. Your right-hand diagram shouldn’t have cusps.

    I created some simulations on Gravity Simulator to illustrate some concepts discussed here.
    Here’s an image similar to the one Lars points out in the Wikipedia article. It shows no cusps.

    http://www.orbitsimulator.com/BA/earthmoonorbit.GIF

    Here’s an image of Jupiter and its Galilean moons as they orbit the sun. Only Io and Europa form the loops shown in Dr. Plait’s first image, although it would be more obvious if I zoomed in more. Ganymede and Callisto form cusps, like Dr. Plait’s 2nd image.

    http://www.orbitsimulator.com/BA/jupitersystemorbit.GIF

    As John Weiss points out, the Hill Sphere is not the edge of stability for prograde satellites. Here’s a screen shot of a simulation of the the moon orbiting the Earth, orbiting the sun (sun not visible). The Earth is 0.55 AU from the Sun. The moon starts out with a semi-major axis of 384,000 km, same as the real moon. In this simulation, the moon makes a few complete orbits before being stripped away by the sun, despite being less than halfway to the edge of Earth’s Hill Sphere for that distance.
    http://www.orbitsimulator.com/BA/earthmoonp55.GIF

    Retrograde moons perform better, but still, a retrograde moon at the edge of the Hill Sphere will be short-lived as well.

    Regarding the comment that if the Earth disappeared, the moon would still orbit the sun, that is correct for Earth’s moon, but not necessarily correct for Jupiter’s moons. Io and Europa both have orbital velocities around Jupiter that exceed solar escape velocity at that distance from the sun. If Jupiter disappeared, depending on where they were in their orbits at the time, Io and Europa could escape to interstellar space. So for parts of their orbits, they orbit the sun, and for parts of their orbits, they don’t. In fact, depending on how you define a “solar system object”, you might argue that Jupiter continuously ejects and recaptures Io and Europa from the solar system.

  41. Chris A.

    @Toren:
    As a former employee of the Space Telescope Science Institute (Hubble’s “home”), I must respond to some of your accusations regarding NASA’s official comments on Hubble pointing at the Moon.

    1) I was there when the first Hubble Moon images (of Copernicus crater) were under consideration. I know of no one, NO ONE, in an official capacity who ever claimed Hubble couldn’t focus on something as close as the Moon for optical reasons, as you seem to imply. The Earth, no. In fact, we routinely pointed Hubble at the Earth for a nice white (due to out-of-focus clouds) flat field for camera calibration (we called them “Earth flats”). More likely they may have been referring to the fact that early on in the mission, Hubble’s pointing control software lacked the ability to track moving (with respect to the celestial sphere) targets. In fact, early on, shots of solar system targets were done in “ambush” mode, by pointing Hubble where the planet/asteroid/comet was going to be, and taking the image when it moved into the field of view. Needless to say, the Moon has a rather large rate of apparent motion with respect to the stars, so I’m guessing the comment you refer to might have had something to do with that. Later on, the software was upgraded to include moving target capability.

    2) While some of Hubble’s instruments would indeed have been overwhelmed by the Moon’s brightness and would not produce a non-saturated image, I believe the “Moon is too bright” claim was likely referring to Hubble’s need to acquire guide stars. Since the Moon is so large compared to Hubble’s field of view, it would have been essentially impossible to use the Fine Guidance Sensors (which lock onto guide stars for pointing) to image the Moon, which fills both the imaging instrument (near the center of Hubble’s FOV) and the FGS apertures which comprise the outer ring of the FOV. The solution, later arrived upon, was to tell Hubble to lock onto “guide craters” on the Moon’s surface, effectively exploiting the FGS’s need for something round and bright to fix on. In other words, a careful choice of small, bright craters in just the right location around the selected target area “fooled” the FGSs into treating them as guide stars. This solution wasn’t immediately obvious, which may be why, after someone figured it out, the “story changed.”

    3) Resolution can be defined in a number of ways when one is forming an image made of pixels. Sometimes, you’ll see resolution quoted as equivalent to the apparent angular diameter of one pixel (i.e. if you have a binary with one dark pixel in between the components, you can claim to have resolved the binary). Needless to say, by this definition, finding one pixel in an image of the Moon that looks different than those around it would be insufficient to convince someone of what the “oddball” pixel was (in this case, the base of the LEM). Other considerations, like the filling factor of the pixels (i.e. how much gap is there between adjacent pixels), come into play as well. And don’t forget that over the 18 years Hubble has been in orbit, instruments have been swapped out, causing Hubble’s maximum resolution to change with each upgrade. When I was there (1988-94), the highest resolution imaging instrument on board was the Faint Object Camera, and we once calculated that even its best resolution was about 2x-3x too low to resolve the base of the LEM.

    In the end, it just doesn’t matter, because in the coming years they’re going to build some gigantic ground based instruments like the OWL (OverWhelmingly Large) Telescope, which will be able to sensibly image lunar hardware. They’ll release the images to the public. And the Moon hoax wing nuts will dismiss them with a single word: “Photoshopped.”

    I’m convinced that you could strap these woo-woos into a rocket, fly them there in person, show them the hardware first hand, and they’d tell you that the whole experience was simulated in a virtual reality computer plugged directly into their brains using super-secret CIA/NSA/DARPA mind-control technology, and that they had never actually left the Earth, because the radiation environment in the Van Allen belts would have killed them, and well, they’re still alive! QED!

  42. Andrew

    Phil,

    Two points, (and please correct me if I’m wrong):

    1) One important reason why Venus and Mercury have no moons is that they rotate so slowly; the Earth’s angular speed is much higher than the Moon’s so tidal effects transfer momemtum to the Moon, leading to the Moon’s slowly increasing distance from the Earth. Because Mercury and Venus rotate so slowly, any moon at a reasonable distance from one of those planets would lose momentum due to tidal effects and “rapidly” (relative to the age of the solar system) end up colliding with the planets

    2) The question of whether the Moon orbits the Earth or the Sun seems to me to be a semantic issue. If you’re trying to send a rocket to the Moon from Earth, you’re much better off considering the Earth unmoving and the Moon orbiting it (and dealing with the Sun as a perturbation);on the other hand, first order calculations about a trip from the Moon to Mars could be done by considering the Moon to be orbiting around the Sun and treating the Earth’s gravity as a perturbation.

    Andrew

  43. Osric

    So the question that hits me is …

    How big the our Moon’s Hill radius, is the Earth within it by any chance?

  44. Pouria

    Stupid question… But isn’t a good explanation to the moon orbiting earth be that if the moon orbited the sun, it would essentially have to decellerate, and accellerate once every 15 days roughly? So that earth can pass it, it passes earth etc, wouldn’t that continuous shift in accelleration put too much force on the moon?

  45. Michael

    Good article, but at which point would the Moon and the Earth be considered a binary planet system? I would probably define it as if the barycenter of the system lies outside that of the larger body, would that be correct? How far off the center of the Earth is the Barycenter of the Earth system? and as the moon travels further from the Earth does the pull the Barycenter further from the Earth, and if it does will the barycenter ever become outside of the Earth before the Moon leaves the hill sphere?
    Also if my assumed definition is correct where is the Barycenter of Pluto and Charon, would that make them technically a binary dwarf-planet system rather than a dwarf-planet with a moon?
    Sorry for swamping the comments with questions.

  46. h-bar

    “Moreover, if you calculate the force of the Sun’s gravity on the Moon, you find it’s more than twice the force of Earth’s gravity on the Moon!”

    This is false. In fact the gravitational force between the earth and moon is around 6 orders of magnitude larger than between the sun and moon. If what you said were true, every time a solar eclipse occurs we would lose the moon to the sun ( if it were fictitiously in a the original orbit to begin with ).

    Also, you should probably mention that it is more correct to say that the earth and moon orbit around their common center of mass, which is located inside the earth.

  47. confusion man

    I think the first time the word “force” is struck out the strikeout is unnecessary. The forces compared are on the same object, so comparisons of force and acceleration are equivalent. This caused me to try to think of another reason for the strikeout. I was afraid it was some pedantic general-relativity thing (i.e. there are no gravitational “forces,” just space expanding toward massive objects blah blah blah).

  48. h-bar: check your math. The Sun is 400 times farther away from the Moon than the Earth is, and has a mass about 330,000 times the Earth’s. 400^2/330,000 = 0.5, so the Earth’s force on the Moon is about 0.5 times the Sun’s.

  49. Osric: No, the Earth is not in the Moon’s Hill Sphere. You can’t *really* just compute the Moon’s Hill Sphere with respect to the Sun (it doesn’t make sense, what with the Moon orbiting Earth), but if you do the math without regards to this, the Earth is a bit outside the Moon’s sphere. (Which makes sense: at about 1/81 the mass of Earth, the Moon’s Hill Sphere is about 1/3 the radius.) On the other hand, if you do the somewhat more reasonable thing and compute the Moon’s Hill Sphere with respect to its orbit about the Earth, we’re well outside of it, although it is also something like 40 times the radius of the Moon. (Which is good, since the Moon is spherical.)

  50. h-bar

    Oops you’re right Phil. I quick google search for ‘sun earth distance’ reports a value that misses the decimal point after the first digit. I mistakenly add 3 more orders of magnitude to this distance.

  51. Abbinormal

    I remember as a kid reading an article in a national geographic from the 60s (I collected these as a pre-teen, I was a geek) that discussed this issue. I wish I still had that issue, there was a diagram in it that lead me to believe the moon orbited the sun because of the path of orbit they showed the moon never coming back around the earth.

    Glad you finally put this to rest for me.

  52. I think the problem here is that we’re splitting hairs, and talking like we’re splitting hares.

    We can all agree that the Moon’s orbit around the Sun is everywhere concave. That’s nice, and might mean something if the Moon were orbiting all by itself. (We’d have to explain the eccentricity somehow – maybe epicycles.)

    But it isn’t. It’s bound to the Earth. In a reference frame fixed to the Earth, the Moon goes round and round with a rotational speed of about 28 or so Earth-days.

    Now if the Moon were going round the Sun in the opposite direction from the Earth, we might have a point; or even if it were going in a synchronous solar orbit not connected with ours.

    Another idea is that we can come up with a farily simple equation to describe the Earth’s orbit. To describe the Moon’s orbit around the Sun would probably take a lot more terms than the equation describing its orbit around the Earth.

    I like Andy Beaton’s “yo-yo on a skateboard” analogy. Another one is the track of a bicycle pedal, when the bicycle is traveling at right angles to your line of sight. To the rider, the pedal spends half its time going backward, but to the observer, it’s always going forward.

    So maybe the Moon’s orbit question is really one about the observer. To us, no question. To an observer looking down into the plane of the Solar System, it’s orbiting the Sun, but bound to the Earth in a complex pattern.

  53. Buzz Parsec

    I’m pretty sure that if the Moon’s path around the Sun (or the Earth’s, for that matter) was accurately drawn to scale in the diagrams, it would look just like a perfect circle. If both paths were drawn in contrasting colors, maybe you could tell they didn’t coincide exactly, but it would be very hard to tell what was going on. On a 100 dpi monitor, with a 4″ radius, the maximum displacement of the Moon’s position from the Earth’s would be 1 pixel. On my 15″ laptop, the orbits are about 2.5-3″ diameter, so the maximum displacement would be less than 1/2 a pixel. However, if you try to exaggerate the wobbliness of the path enough to be easily visible, you probably end up making the curve non-convex.

    Is there any way to illustrate this accurately with a picture that will fit in the blog? Maybe one of those super high res multi-megapixel drawings with instructions to anti-embiggen* it in your browser to see the whole thing? (*Cloverleaf-minus in Safari)

  54. @Andrew: “Because Mercury and Venus rotate so slowly, any moon at a reasonable distance from one of those planets would lose momentum due to tidal effects and “rapidly” (relative to the age of the solar system) end up colliding with the planets”

    Not true at all. Consider the earth and the moon. Tidal effects are causing the earth’s rotation period to increase by about 20 seconds every million years while at the same time the moon’s orbit gets farther and farther away from the earth. One way to think of it is this. The moon causes a tidal bulge in both the earth’s oceans and also solid mantle/crust facing towards the moon and also on the opposite side of the earth. The earth’s rotation tends to push this bulge to the east of the sublunar point. The moon is gravitationally attracted towards the bulge which has been pushed to the east of the sublunar point. This causes the moon to accelerate in the direction of it’s orbital motion which causes its radius from the earth to increase.

    So it’s all caused by the rapid rotation of the earth pushing the tidal bulge forward. Now consider Venus with a rotation period of 243 days retrograde or Mercury with a rotation period of 59 days prograde. A satellite orbiting in the same direction as the planet’s rotation will increase its distance over time and a satellite orbiting in the opposite direction will decrease its distance over time. But with such slow rotations as Venus and Mercury the tidal bulge will be pushed out of line with the sub-satellite point only slightly. We thus expect that the rate at which any satellite increases or decreases its distance to be much less than for our moon. The moon has spiraled out from about 20,000 km from the earth to 384,000 km but it took it 4.5 billion years to do so. If it had been orbiting Venus or Mercury then its orbital shift over 4.5 billion years would have been probably two orders of magnitude less.

  55. Not to nitpick, but I’m nitpicking anyway. I know it’s a disproportionate example and all in your diagram… but you have 12 “loops” (or non-loops) there.

    Supposed to be 13.

  56. Andrew

    @Tom,

    If the Moon were 250,000 miles from Venus, it would take about a month to orbit. Since Venus is rotating so much slower than Earth, the moon would be always be ahead of the tidal bulge, instead of behind it. Wouldn’t this cause the moon to lose angular momentum and Venus to gain it, thus lowering the moon (and thus making the orbital period even shorter, and the tidal effect even greater).

    This article http://en.wikipedia.org/wiki/Tidal_acceleration#Tidal_deceleration also describes what I’m saying – for prograde satellites, the important threshold is the synchronous orbit – satellites higher than that will slowly recede from their primary, while satellites below that will approach their primary. Because Venus rotates so slowly, synchronous orbit is very far out, so a satellite of Venus is very likely to be inside the synchronous orbit, and thus destined to collide with Venus.

  57. “Since Venus is rotating so much slower than Earth, the moon would be always be ahead of the tidal bulge, instead of behind it.”

    You have it backwards. It is the moon that causes the tidal bulge. If the planet had zero viscosity then the tidal bulge will be exactly at the sub-moon point. But planet’s don’t have zero viscosity. Therefore the planetary rotation pushes the bulge away from the sub-moon point and it takes time for the bulge to re-form at the sub-moon point. I’m not denying the effect – just it’s magnitude. For a slowing rotating planet the tidal bulge will be pushed away from the sub-moon point only slightly and therefore the moon will be affected only slightly. It may not have much cumulative effect even after 4.5 billion years.

  58. Andrew

    Yes, the Moon causes the bulge, but when the planet rotates more slowly than the moon orbits, the bulge can’t keep up with the motion of the moon(as opposed to what happens in case of the Earth, where the rotation of the earth is moving the bulge “ahead” of the sublunar point). This is why Mars’s nearer satellite is expected to crash into Mars at some point, isn’t it?

  59. “This is why Mars’s nearer satellite is expected to crash into Mars at some point, isn’t it?”

    I believe your original statement was that all satellites of slowly rotating planets such as Venus or Mercury must inevitably crash into their planet over the age of the solar system which is still not true. It all depends upon what the offset is of the tidal bulge from the sub-satellite point which depends on many factors such as rotation period of the planet, revolution period of the satellite, viscosity of the planet, etc. Imagine a slowly rotating planet and a distant moon. The tidal bulge lags the sub-moon point due to friction for a stationary planet but the rotation of the planet is just enough to compensate by pushing the bulge forward so that it is always at the sub-moon point. Then the moon’s orbit will neither decrease nor increase over time. So it depends on many, many factors.

  60. Mang

    Interesting. So with the Moon distancing itself from us at the blistering rate of 3.8 cm/year it will pass the hill sphere limit in about 29 billion years (max). But if I got it right, the orbit isn’t stable past 40% of the sphere which is about 5 billion years – which is just about when the Sun turns into a red giant and messing up the whole process.

    Wow, I want to see the CGI annimation of that!

    (hopefully I didn’t slip a digit).

  61. Mang

    nuts … nix the last and

  62. Andrew

    Tom wrote:

    “I believe your original statement was that all satellites of slowly rotating planets such as Venus or Mercury must inevitably crash into their planet over the age of the solar system which is still not true. It all depends upon what the offset is of the tidal bulge from the sub-satellite point which depends on many factors such as rotation period of the planet, revolution period of the satellite, viscosity of the planet, etc. Imagine a slowly rotating planet and a distant moon. The tidal bulge lags the sub-moon point due to friction for a stationary planet but the rotation of the planet is just enough to compensate by pushing the bulge forward so that it is always at the sub-moon point. Then the moon’s orbit will neither decrease nor increase over time. So it depends on many, many factors.”

    I think we are now in basic agreement – the situation you describe above is the geosynchronous case where the speed of the planet’s rotation matches the orbital speed of the satellite. Satellites further out than the geosynch orbit will tend to grow further away, satellites closer will tend to approach the planet (the viscosity of the planet only affects (I think) the magnitude of the tidal force not the sign). For Venus, “geo”-synch altitude is about a million miles (if I’ve done the back of the envelope calculation correctly) which is outside Venus’s Hill Sphere. So it seems that any satellite that Venus could “own” (ie. hold within its Hill Sphere) would have to orbit in the range that would make its orbit gradually contract – but as you said, the rate of that decrease would depend significantly on the composition of Venus, so my flat statement that the satellite would definitely crash was an overstatement.

    Thanks, Andrew

  63. There’s a very simple example you can use to utterly demolish the “orbit doesn’t look loopy” argument.

    Take a wheel. Paint a bright red spot somewhere on its side. Roll the wheel on a flat surface. Graph the trajectory of the red spot.

    The red spot will move in a series of semicircles above the flat surface (moving fastest at maximum distance from the surface and lingering longest at the points nearest the surface).

    Would you argue with a straight face that the red spot is not revolving around the centre of the wheel because the graph is not loopy?

  64. THE EARTH IS ORBITING THE SUN TO MAKE NIGHT AND DAY ARMSTRONG MUST OF WENNT UP INTO SPACE FOR A LONG TIME TO SEE THAT

  65. sina

    man im doing my project about the moon

  66. Chris

    Fail. You have made everyone who has read this article DUMBER through your efforts.

    Anyone with a high school physics education will know that neither the Earth nor the moon orbits the sun. The Earth-moon barycenter, the center of mass which sits on a line between the moon and the earth 1700 km beneath the earth’s crust, orbits the sun. The moon and the Earth rotate around this barycenter as it orbits.

  67. Corscaria

    Sorry Chris, you are wrong. The Earth and Moon both orbit the sun. The Barycenter is an imaginary point in space, at which the Earth and Moon orbiting each other, seem to be centered. They don’t actually orbit this point, it is an imaginary point cause by the equilibrium of thier masses. They Orbit each other, with Earth dominationg, AND they both seperately orbit the sun.

  68. Does the position of the barycenter have anything to do with the precession of the equinoxes?

  69. Ibrahim

    Kinda ironic how we are debating something that was mentioned 1400 years ago..
    Check this out
    http://www.answering-christianity.com/moon_orbit_miracle.htm

  70. Lawrence D’Oliveiro

    But the Moon and the Earth are both inside the Sun’s Hill sphere, are they not? And the Moon is proportionately deeper inside the Sun’s Hill sphere than it is inside the Earth’s one, is it not?

    So what does all this Hill sphere business really prove?

  71. Tos

    This discussion is fascinating but in the end this article is about promoting your definition of what it means to “orbit more” a body than another.

    In ordinary language we don’t have such a definition. You offer one but I find it premature to state in bold “The Moon orbits the Earth more than it orbits the Sun.”.
    Yes, you make a valid point in stating a dynamic defintion of “orbiting more”. But there is also a static definition, ignoring dynamics, just looking at the most general shape of the orbit in a given frame. In front of the same facts, maybe a painter would use the static defintion. The choice of the meaning of words should involve all users, maybe leading to multiple defintions.

    In his lectures on N-body motion, astronomer André Brahic made the exact same descriptions but took the opposite conclusion: “The Moon orbits the Sun more than it orbits the Earth”.

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