# Is it possible for tsunamis to occur on deep water planets?

Assume a water planet with an ocean depth of about 200-500 km.

Would a very strong ground-quake happening at the very bottom of the planet's ocean floor be able to cause a large tsunami to rise up to the surface of the ocean? If it could, how strong would the ground-quake need to be on the Richter scale in order to do that?

• No land no shore, only with deep water layer (ocean) 200-500km. And ok, I will be keep an eye on it.
– ZanMoon-chan
Nov 10, 2022 at 17:55
• From a simplified - shallow-water theory standpoint, tsunamis depend on the conservation of fluid height h and bottom height b. If you don't change b, then h cannot change, hence no tsunami. Nov 10, 2022 at 18:15
• So tsunami most likely can not occur in deep water planet. But what about earthquake? If the earthquake be ridiculously strong, let say about 13.0 magnitude richter scale, it might cause some waves on the ocean surface, right? Nov 10, 2022 at 18:23
• You have to compare b+h before and after the event. If the total delta is negligible, then so are the surface waves. Nov 11, 2022 at 0:44
• Tsunami occurs when oceanic tectonic plates uplift water, at sea it is a wave and will dissipate, however upon reaching land, volume it has to expand decreases exponentially and has but one output, to overlap on land. Deep water planets can have tides, geologic plate upheavels..upheavals... but no land, no tsunami 🌊 Nov 19, 2022 at 9:42

Tsunami are caused by sudden changes in the sea floor. An earthquake that causes a slump or surge in the seabed can cause a tsunami.

Tsunami travel as shallow water waves; they have a wavelength comparable to, or longer than the depth of the water. A large earthquake can cause rupturing on a scale of hundreds of km, (the boxing day quake caused a rupture of 200km) and so can create a tsunami even in 200km deep water. Such a tsunami would have a wavelength of hundreds of km and a wave height of perhaps a few cm. So this is not the great wave off Kanagawa (which was not a tsunami) but it is a true tsunami, and would be undetectable except with high accuracy gauges.

If there is no land, the tsunami will circle the planet several times gradually dissipating due to friction. With no land, it will remain a few cm in amplitude and be of no practical interest.

A planet with significant earthquakes will probably have plate tectonics, so the seabed won't be flat, which will have consequences for its velocity. It will move faster over deeper water.

• For those curious about what the "boxing day quake" is, it appears to refer to the 2004 Indian Ocean earthquake Nov 11, 2022 at 8:55
• With regard to Tsunami travel as shallow water waves: That's true on Earth because tsunamis have a wavelength in the multiple 100s of kilometers. The depth of the oceans on Earth is less than 1/20 of that long wavelength, which is the borderline of shallow water waves and non-shallow water waves. Deep water waves have a wavelength that is at most half the depth. The OP is positing a planet with a very deep ocean, so the tsunami on that planet would traverse more as deep water waves rather than shallow water waves. Nov 11, 2022 at 12:30
• To make sure for Zan... what happens when this few cm tall wave (generated in 200 km water) travels into a zone where the seafloor was only 20 km? Little negligible change because it's a shallow water wave? Nov 11, 2022 at 23:14

Tsunamis occur in the wake of large earthquakes. They grow in amplitude because the speed of a water gravity wave is influenced by water depth, causing the wave to 'pile up'. In particular, if the wavelength is (much) greater than the depth of the water, the speed at which a wave will travel is roughly:

$$v \sim \sqrt{gd}$$

where v is the speed, g is the gravity, and d is the depth. So by keeping the dimensions consistent as a wave enter shallower water the wave length (L) is squished proportionally to the change in speed and amplitude (A) grows accordingly:

$$\dfrac{A_1}{A_2} \sim \dfrac{L_1}{L_2} \sim \dfrac{v_2}{v_1}$$

such that

$$A_2 \sim A_1 \sqrt{\dfrac{d_1}{d_2}}.$$

### Plugging some numbers in:

So even for an earthquake that displaces water over say 500km (e.g. $$M_w9.1$$ Tohoku earthquake). If the water changes in depth from 500km to 200km (as OP asks), the change in height will only be a factor of:

$$A_2 \sim A_1 \sqrt{\dfrac{500km}{200km}} \sim 1.5A_1.$$

JeopardyTempest asks, what if the water got as shallow as 20km then:

$$A_2 \sim A_1 \sqrt{\dfrac{500km}{20km}} \sim 5A_1.$$

### How big?

Now $$A_1$$ (the original wave height) is generally comparable to the average vertical displacement from the earthquake. On earth, the plate boundary configuration is such that the largest earthquakes occur on shallow-dipping subduction zones such that the corresponding average vertical deformation would rarely exceed 1 meter even in the largest earthquakes. Tohoku for instance had a maximum sea floor displacement of tens of meters (at a dip of approximately 10 degrees). Should your configuration be better suited on your mystery planet (steeper dip), and should there be a potential for larger earthquakes than on earth (say $$M_w9.5+$$), only then would there be a potential for a sizable tsunami with an amplitude at least as large as the sea floor displacement and potentially larger should the bathymetric change. Certainly a $$M_w10+$$ would generate a lot (10s of meters) of seafloor displacement, but then, of course, the wavelength would be much bigger than the amplitude.