0
$\begingroup$

We know from research that the Solar System travels around the Galactic plane, and sometimes (periodically) dips below the Galactic equator, only to return and dip back above the Galactic equator later.

Does anything happen to the planets on our solar system as the system changes from over-to-below, and from below-to-over? Does rotation change? Or the axial tilt of rotation? Does the planet’s equator displace one direction or the other? Is there an alteration in the orientation of the spin axis or the entire planet? Does it “flip upside down”? Or, simply, is there no change whatever?

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ Nothing major will happen to the planet itself, but some scientist do think that the change in Earth's geological period has something to do with Earth crossing the Galactic plane. (just a guess) $\endgroup$
    – Polaris5744
    Commented Jun 10 at 22:38
  • 2
    $\begingroup$ I believe that the planet must endure the line-crossing ceremony at least for the first time. $\endgroup$
    – uhoh
    Commented Jun 13 at 0:11
  • 1
    $\begingroup$ What exactly happens to the Earth when it crosses the solar equator? Right - nothing. $\endgroup$
    – Zac67
    Commented Jun 14 at 18:11

2 Answers 2

Reset to default
9
$\begingroup$

Nothing whatsoever.

To be honest I find it a little hard to imagine why you think anything could happen.

The galactic equator is a plane that is roughly in the middle of the disc of stars that compose the milky way. But out by us, this disc is 3000 lightyears thick. There is nothing special in the middle of this disc, and, in fact, it's not easy to even determinine where the "middle" actually is. It is just more empty space.

To cause a planet to flip or change direction would require a massive force. But there is nothing to cause such a force.

Improve this answer
$\endgroup$
9
  • 1
    $\begingroup$ There is a reasonably well-defined mid plane and the Sun is above it and being accelerated back towards it. The thickness isn't really relevant (other than to make SHM a better approximation). $\endgroup$
    – ProfRob
    Commented Jun 10 at 19:10
  • 2
    $\begingroup$ Yes "reasonably" well defined. But still the uncertainty in the location of the mid plane is a lot more than a few AU. And there is nothing particularly on or around the mid plane. "Above" is of course conventional, not in any sense absolute. If we choose to view the galaxy from the other size, it would be rotating in the oppose sense and the sun would be below the plane. $\endgroup$
    – James K
    Commented Jun 10 at 19:37
  • 1
    $\begingroup$ @ProfRob SHM is not a good approximation at all, except perhaps for orbits that don't move more than a few pc out from the mid plane. $\endgroup$
    – Walter
    Commented Jun 12 at 1:44
  • 1
    $\begingroup$ @Walter How so? The epicyclic approximation for near circular orbits is a totally standard bit of galactic dynamics isn't it? The SHM approx is good so long as you can assume the density is constant. Hence as long as the vertical amplitude is less than the vertical scale on which the density varies. galaxiesbook.org/chapters/II-03.-Orbits-in-Disks.html (Sec 10.3.1). Which is what I said. $\endgroup$
    – ProfRob
    Commented Jun 12 at 7:10
  • 1
    $\begingroup$ @ProfRob At one scale height you are already making significant errors with SH. For a vertically exponential density $\rho(z)=(\Sigma/2h)\exp(-\zeta)$ with $\zeta=|z|/h$, the potential is $\Phi=2\pi G\Sigma h[\zeta-1+\exp(-\zeta)] = \tfrac12\omega^2z^2[1-\tfrac13\zeta+O(\zeta^2)]$. Thus, the deviation from SH is first not second order and becomes significant early. The force $F=-\omega^2z[1-\tfrac12\zeta+O(\zeta^2)]$ deviates by ~50% from SH by $|z|=h$. At low $|z|$, the ISM layer becomes important, which has a scale height of $\sim 40\,$pc. So, the SH is perhaps okay for $|z|<20\,$pc. $\endgroup$
    – Walter
    Commented Jun 19 at 10:33
5
$\begingroup$

The Sun and planets are both pulled together by the gravitational field of the Galaxy. But the differential tidal effects could be felt by planets (or other objects) on orbits of size $\sim 10^5$ au (like putative planet X perhaps, or the outer Oort cloud). These effects could radically alter the orbits of such objects, but not their spins because the tidal forces are tiny on planetary (as opposed to orbital) scales. There would be no effect on the known planets.

A further, non-gravitational effect is that the Solar System is somewhat more likely to be within the "danger zone" of a nearby supernova as it crosses the Galactic midplane. This also is unlikely to have any influence on the rotation and spins of the known planets.

Details

We can use Gauss's law for gravitation to get a rough idea of the gravitational acceleration drawing the Solar System (and everything in it) back towards the Galactic plane. $$g \simeq -4\pi G\rho |z|\ ,$$ where $|z|$ is the distance from the Galactic plane and $\rho$ is the mass density in the plane (derivation here).

However, because this affects the whole Solar System, what really matters for your question is a comparison between the gravitational acceleration on a planet exerted by the Sun and the tidal acceleration caused by the Galactic plane. The latter is given by the gradient of the acceleration given above, multiplied by the separation $r$ between the Sun and a planet. i.e. What matters is the ratio $$\frac{g_{\rm tide}}{g_{\odot}} \simeq \frac{4\pi \rho r^3}{M_{\odot}}\ . $$

Quantitatively, we can express $r$ in au and use $\rho \sim 0.1 M_{\odot}$/pc$^{3}$, so $$\frac{g_{\rm tide}}{g_{\odot}}\simeq 10^{-16}\left(\frac{r}{\rm au}\right)^3 . $$ Thus, for star-planet separations of au to tens of au, the tidal influence of the Galactic plane is negligible. Only planets (or comets, or binary companions, or anything else, since the ratio does not depend on the mass of the orbiting object) with orbits of $\sim 10^5$ au might be affected in any significant way.

This is why, for example, the outer Oort cloud should be pseudo-spherical and any planet X on a very wide orbit may not be in the same plane as the rest of the planets. i.e. The effect is sufficient to change the orbits of such objects, but the torque exerted on small length scales would not be anywhere near enough to affect spin axes or rotation rates.

High mass stars are generally born and live their short lives within a few tens of pc from the Galactic midplane. Thus this is also the location of most type II (core collapse) supernovae. A nearby supernova (less than a few parsecs) is likely to be quite damaging to the Solar System planets (to their atmospheres, though possibly not to their orbits and certainly not to their rotation and spins). The vertical motion of the Sun is an oscillation with a period of around 70 Myr and a semi-amplitude of around 100 pc. That means that every 35 Myr or so there is a possibility for the Solar System to be in the vicinity of a core-collapse supernova (though you have to factor in spiral arm location too).

Improve this answer
$\endgroup$
1
  • $\begingroup$ ProfRob how certain is the 35myr semi-period of oscillation? I read the initial calculation was 33myr. How certain is it that the periodicity is not much shorter, closer to 24ka in total, with a semi-period of 12ka? $\endgroup$
    – Jim
    Commented Jun 28 at 4:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .