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Could any forces from the moon, the planets or the sun in orbit hypothetically influence seismic events on earth? And if yes how to approximately calculate and compare the magnitude of the forces?

EDIT: On the How Stuff Works website they actually mention the possible influence of the moon and the sun on the earth's crust. But what about the possible influence of other planets in the solar system and their interactions, with earthquakes?

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    $\begingroup$ There is a joke about it. Tides from the Sun deform the Earth's crust. This flexing can set off earthquakes. The effect is so strong that 50% of all earthquakes happen within 3 hours of sunrise or sunset. Joking aside, I don't know how much this matters. $\endgroup$
    – mmesser314
    Commented Feb 6, 2023 at 16:04
  • $\begingroup$ I get the joke, but i was hoping for some more serious answers. On the how stuff works website, they actually mention the influence of the moon on the earth's crust. It is not clear if this is an effect that is only due to periodic "sloshing/wobbling of the crust " in resonance w.r.t the moon. There are also spring tides in the sea, when the moon lines up with the sun, so at least to me its not that clear that causal effects from other planets are completely irrelevant/negligible. $\endgroup$
    – Hjan
    Commented Feb 6, 2023 at 16:18
  • $\begingroup$ Better over on Earth Science, such as earthscience.stackexchange.com/questions/24740/… $\endgroup$
    – Jon Custer
    Commented Feb 6, 2023 at 19:05
  • $\begingroup$ I really do not understand the close! Also previously the question was formulated in a general way. It is not about the specific theory in the video or link. Yes, the video is not mainstream, but there is nothing non-mainstream about this question. Furthermore i am interested in a physics answer. Not in an answer related to the traditions of geologists in earth science or so. Actually i like the answers given, and i think many people with similar questions could be helped. $\endgroup$
    – Hjan
    Commented Feb 6, 2023 at 20:35
  • $\begingroup$ I suggest deleting everything before "Could any forces from planets …", since that is where the valid Physics or Earth Science question starts. There is a long history of flaky catastrophic planetary alignment theories that have sensitized physicists, earth scientists, and astronomers. See, for example, The Jupiter Effect. $\endgroup$ Commented Feb 6, 2023 at 21:43

2 Answers 2

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The planets do not affect Earth seismicity, but the Sun and Moon do.

Earthquakes are caused by strains in the Earth's crust, and the gravitational strains produced by the planets, Moon, and Sun on the Earth's crust are tidal effects which fall off as the cube of the distance. Only the Sun (because of its large mass) and the Moon (because it is so close) produce significant tidal affects on the Earth. The tidal effect of Venus, the planet that comes closest to the Earth is 4 orders of magnitude less than that of the Sun and Moon. The tidal effect of the most massive planet, Jupiter, is 5 orders of magnitude less. Any claim that planetary positions can be used to predict earthquakes is unfounded.

The tides caused by the Sun and the Moon can affect the likelihood of an earthquake, but the effect is small, e.g. less than 1% according to one study. For earthquakes in or near ocean basins, both solid earth tides and water tides contribute. (The weight of water tides can induce stresses in the ocean bottom that can be an order-of-magnitude larger than the direct solid-Earth tidal effect.)

Tidal stresses are 3–5 orders of magnitude less than the stress relieved in typical earthquakes, so the tide can only trigger an earthquake that was likely going to happen soon anyways. Studies have claimed that high tidal stress can significantly affect the timing of shallow thrust earthquakes or make it more likely that an earthquake will grow to a larger size.

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  • $\begingroup$ Great Explanation and Links. The review article and another paper from 2023 show it's still a relevant topic. $\endgroup$
    – Hjan
    Commented Feb 6, 2023 at 21:16
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The idea is that tidal deformation (gravitational gradients) can increase strain, thereby triggering strain-relief: a quake.

In the extreme case of the Earth passing inside the Roche limit of a larger ($M$) celestial body, the answer is clearly "yes".

For instance, a rogue black hole would produce a gradient large enough to induce The Rapture (people on the line of syzygy would be lifted off the surface on both sides of the planet, those transverse to that would be pulled down into the crumbling crust, for a red hot molten demise...have I read the somewhere before?).

In more mundane circumstances, tidal gradients go as:

$$ \frac{ M }{R^3} $$

The sun competes with $M$, while the moon wins "out" on $R$. From there, look at the planets and see their relative contribution.

Being a tensor alignment, direction doesn't matter, so if Venus and Jupiter line up: they add, even though they are on opposite sides of the planet.

You just have to plug in the numbers.

Edit (based on comments):

With a gravitational potential $U(\vec x)$, the Hessian is:

$$ J_{ij} = \frac{\partial^2 U}{\partial x_i \partial x_j} $$

The tidal tensor is the symmetric trace free (aka "natural rank-2 tensor"):

$$ \Phi_{ij} = J_{ij} - \frac 1 3 {\rm Tr}(J) $$

which has 5 components. These induce quadruple deformations in the Earth's shape that are characterized by the 5 (real) spherical harmonics with $l=2$... e.g. prolate stretching. Because it's traceless, the longitudinal stretch is twice the 2 horizontal squishes, so volume is preserved. (Remember, the field lines converge so they don't point in the same direction on opposite sides of low tide).

One can liken this to the (dipole) polarization of an atom in a (vector) field, where the deformation is characterized by the $Y_{l=1}^{m \in (-1,0,1)}$ spherical harmonics.

It's all very geometric.

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  • $\begingroup$ Very interesting. Why do the gradients scale with M/R^3 and not with R^2 such as the normal gravity forces do? Or with M/(d*R^2) where d is the earths diameter. And i also do not understand why it does not matter if two planets line up behind each other, or are at other sides of the planet. It seems counter intuitive: In one case they would pull the earth's crust to elongate in the same side, while otherwise in 2 opposite directions. To get the picture i compare the earth with a water balloon filled with magnetic liquid covered by a layer of dried yogurt, and 2 magnets around it. $\endgroup$
    – Hjan
    Commented Feb 6, 2023 at 18:57
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    $\begingroup$ @Hjan: That would be true if the Earth was fixed in space, but it's not. For example, everything on Earth accelerates towards the Sun. Stuff on the near side is closer to the sun, so that stuff is pulled away than the center of the Earth. Stuff on the far side is farther from the Sun, so the center of the Earth is pulled away from that stuff. The result is two bulges of "stuff", one on either side. $\endgroup$ Commented Feb 6, 2023 at 19:11
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    $\begingroup$ Moreover, it's not the acceleration of the center of the Earth that mattters (which you correctly note is proportional to $R^{-2}$), but instead the difference in the acceleration from one side of the Earth to the other. This difference is approximately proportional to $d/R^{3}$, where $d$ is the diameter of the Earth, under the assumption that $d \ll R$. $\endgroup$ Commented Feb 6, 2023 at 19:12
  • $\begingroup$ @MichaelSeifert Thanks. could you maybe explain more where the d/R^3 comes from. I would think that the influence of the sun on a particle on the "nearest to the sun" point on the surface of the earth would be $G*(m_{particle}*m_{sun})/(R-d)^2$ and on a similar particle on the "furthest" point $G*(m_{particle}*m_{sun})/(R+d)^2$. And that the gradient is the difference. Probably i am missing something, but what? $\endgroup$
    – Hjan
    Commented Feb 6, 2023 at 20:49
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    $\begingroup$ @Hjan: The tidal effect would be exactly proportional to the difference between those, or $$\frac{1}{(R-d)^2} - \frac{1}{(R+d)^2} = \frac{4 R d}{(R-d)^2 (R + d)^2}.$$But if $R \gg d$ then $(R- d) \approx (R + d) \approx R$ and this becomes approximately $4 R d/R^4 = 4d/R^3.$ $\endgroup$ Commented Feb 6, 2023 at 21:04

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