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According to this answer on Astronomy.SE,

The Sun executes oscillations around its mean orbit in the Galaxy, periodically crossing the Galactic plane. I borrowed this illustration (not to scale!) from https://www.nbcnews.com/id/wbna40787453 to show this oscillatory motion. As the Sun is currently above the plane and moving upwards, and each cycle takes about 70 million years with an amplitude of 100pc (Matese et al. 1995), it will be roughly 30 million years before we cross the plane again.

Sun's motion around the Galaxy

What causes this kind of motion? Does Newtonian Physics have an explanation? If yes, what is it? If no, what General Relativity explanation have we got?

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    $\begingroup$ astronomy.stackexchange.com/q/1997 $\endgroup$
    – BowlOfRed
    Commented Aug 30, 2016 at 7:12
  • $\begingroup$ related: physics.stackexchange.com/q/67133/23473 $\endgroup$
    – Jim
    Commented Aug 30, 2016 at 13:20
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    $\begingroup$ I feel like your drawing makes this look way more counterintuitive than it actually is, because the frequency of the oscillation that it shows is higher than you'd expect in practice. It's actually quite intuitive: try to imagine an inhomogeneous distribution of mass approximately as a disk, and imagine some object coming into orbit around it at some random angle close to (but not equal to 0). Do you intuitively expect it to suddenly get locked into an orbit on a single plane magically, or do you expect it to wobble around it? $\endgroup$
    – user541686
    Commented Aug 30, 2016 at 16:52
  • $\begingroup$ Throw a tennis ball into a swimming pool at an angle and watch what it does :) Only a simulation but there are some similar-ish principles involved. The galaxy didn't materialise one day as a single swirling mass all in nice tidy equilibrium; it's the swimming pool that someone's already jumped in. $\endgroup$ Commented Aug 30, 2016 at 23:03

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It's good old Newtonian gravity! The plane of the galaxy can be approximated as a disk made up of stars and gas, with a density $\rho(|z|)$, that decreases with absolute distance $|z|$ from the plane.

If you were to assume that the Sun was close enough to $z=0$ and that the radial variation in $\rho$ was negligible enough to treat the disk as an infinite plane (this is not bad, the amplitude of the Sun's motion is only about 10% of the radial scale length of the disk density), then you could construct a little cylinder through the plane, with one face at $z=0$, where $g=0$, and use Gauss's law for gravity to estimate the gravitational acceleration at height $z$. $$ g(z) \simeq -4\pi G \int_0^{z} \rho(z)\ dz$$

Now $\rho(z)$ approximates to an exponentially decaying function with a scale height of maybe 200-300 pc. If we are closer to $z=0$ than that, then the density can be roughly said to be a constant $\rho_0$. Putting this into the equation above, we see that $$g(z) =-4\pi G\rho_0 z.$$ But this is simple harmonic motion with an angular frequency $\sqrt{4\pi G\rho_0}$.

The density of the disk near the Sun is estimated to be 0.076 solar mass per cubic parsec (Creze et al. 1998). Using this value, we get an approximate predicted oscillation period up and down through the disk plane of 95 million years.

Note added: The previous paragraph is the reverse of what is actually done - the dynamics of stars in the solar vicinity are used to estimate the density in the plane. However, just counting up stars and estimating the contribution of gas does give a similar result - and in the process, illustrates that the contribution of dark matter to the density of the disk is very small.

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  • $\begingroup$ Is this essentially equivalent to saying that the sun both orbits radially (around the galactic center, in the plane of the galactic disk) and also has an orbit perpendicular to the disk - so if the sun suddenly stopped orbiting around the galactic center, then there would be a point on the disk that the sun orbited around, and this oscillation is that orbit described during the period of the radial orbit? $\endgroup$
    – Joe
    Commented Aug 30, 2016 at 14:45
  • $\begingroup$ @Joe - you can look at the motion of the Sun in a frame of reference that orbits around the Galactic centre at a fixed angular velocity. In such a frame the Sun will appear to execute a (relatively) small orbit around a point in the plane of the Galaxy. The up and down motion is just an approximate simple harmonic oscillation. The epicyclic orbital period in the Galactic plane and the simple harmonic period along the z-axis are not the same. $\endgroup$
    – ProfRob
    Commented Aug 30, 2016 at 15:25
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    $\begingroup$ In short, the sun is pretty much bouncing on (and through) the "surface" of the galaxy, which is pretty cool. :) $\endgroup$ Commented Aug 30, 2016 at 23:01
  • $\begingroup$ @Joe I would hesitate to call it an "orbit", but yes, you can look at it that way. In reality there's just one thing going on: gravitational attraction between the sun and all the other mass in the galaxy. But the in-plane and out-of-plane components are only weakly coupled to each other so you can look at them as two separate and superimposed motions: an orbit and an oscillation. $\endgroup$
    – hobbs
    Commented Aug 31, 2016 at 5:21
  • $\begingroup$ Thanks for the clarification! I like the bouncing metaphor... $\endgroup$
    – Joe
    Commented Aug 31, 2016 at 14:49
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The plane where the Sun and most of the Milky Way are comfined to is neither a plane nor a disk: it is a dense, matter filled section with non-zero width. This means this "disk" also generates a gravitational potential that traps the Sun. In a way, we can say that the Sun has its galactic orbit centered both at the galactic centre and its planar oscillations centred at the galactic "disk". If this approximated disk had a constant density and the Sun were never perturbed in non-planar directions, then the orbit would indeed be confined to a plane; but matter is non-uniformly arranged within the galaxy and this anisotropy provides the non-planar perturbation needed to create an harmonic motion around the galactic plane not only for our Sun, but (at least theoretically) for all celestial bodies.

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I feel like the existing answers make the explanation seem too complicated, so I'll add a simple explanation.

Imagine "throwing" a new star into the galaxy, aiming for the orbit to be approximately (but NOT exactly) along the galactic plane. Would you expect the orbit to magically get locked into the galactic plane? Of course not--that would violate conservation of momentum. Would you expect it to forever continue in its own plane? Of course not--that's only expected for a spherical mass, and the galaxy isn't a spherical mass. The closer parts of the galaxy to the star will pull the star toward them too, and hence you get a wobble.

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    $\begingroup$ This simple explanation is incorrect in appealing to conservation of momentum - how does that explain a wobble? The linear momentum of the Sun is continually changing. In general you do expect an orbit to get pulled into the same plane as everything else, but only if there are dissipative processes. $\endgroup$
    – ProfRob
    Commented Aug 31, 2016 at 11:42
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    $\begingroup$ @RobJeffries: Uh, did you read it correctly? Conservation of momentum wasn't the explanation for the wobble, it was the explanation for why you can't expect a newly added star to suddenly get locked onto the galactic plane. And note that "locked onto" is not the same as "pulled into". The former is talking about convergence in a short amount of time, but you're talking about convergence in infinite time. $\endgroup$
    – user541686
    Commented Aug 31, 2016 at 17:53
  • $\begingroup$ @Rob Jeffries .. this should be a different question, but: I'm a little intrigued about the lack of dark matter (proportionally) in the disk. This tells me that disk formation is as much (or more) due to electromagnetic interaction as gravitational. $\endgroup$ Commented Sep 4, 2016 at 16:42

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