I mean, we experience tides here on earth because of combined effect of gravitational field of moon and sun. So I thought that tides could also occur on other planets.

Let us consider Neptune where the gas transits into a slushy ice and water layer. The water-ammonia ocean serves as the planet's mantle, and contains more than ten times the mass of Earth. Also it has 14 moons. What I was thinking was that more moons would mean more tidal effect would be seen even if distance from sun has increased.

But then I thought that the moons of Neptune would all have different orbits and different time periods of rotation. So this will definitely impact the extent of tides.

-To what extent will this affect tides?

  • And how will tides formed on Neptune be different than that formed on earth? I mean, how will they vary in properties?
  • $\begingroup$ Every extended body experiences tides to one degree or the other. $\endgroup$ – AtmosphericPrisonEscape Feb 1 '19 at 16:07
  • $\begingroup$ Tides on Earth do not occur only to the water, they influence also the solid crust and the gaseous atmosphere of Earth. The effect of the tides to water is visible to the naked eye, but is measurable also for the solid and gaseous parts of Earth. $\endgroup$ – Uwe Feb 1 '19 at 22:58
  • $\begingroup$ You might want to check out farside.ph.utexas.edu/teaching/336L/Fluid/node161.html $\endgroup$ – irchans Feb 3 '19 at 3:07
  • $\begingroup$ Tidal forces happen whenever there's a body orbiting a rotating body. (or vice versa). The magnitude of observed motion is a separate question. $\endgroup$ – Carl Witthoft Feb 5 '19 at 15:03

There will be tides, but they will be not very large. It's pretty easy to get a good estimate of their size.

There are two things that control tides on Earth and both will be present anywhere else: The tidal forces from celestial bodies, and the size of the body of liquid in which the tides are raised along with resonances int he body of liquid. Taking them in turn.

The tidal force is a bit funny, since the size of the tides raised on a planet by a celestial body is

(1) proportional to the mass of the celestial body raising the tide,

(2) inversely proportion to the cube of the distance away,

(3) inversely proportional to the surface gravity of the planet,

(4) and (roughly) proportional to the diameter of the body of liquid in which the tides are being raised.

That inverse cube of the distance means that the ability of a body to raise a tide drops off very quickly with distance.

Comparing Earth and Neptune: The Sun is 30 times further from Neptune than Earth, so Solar tides will be roughly 27,000 times smaller on Neptune. They can be completely ignored.

Triton (Neptune's largest moon) is the jackpot, since it's a third as massive as Earth's Moon, and about the same distance away. Neptune is 17 times more massive than Earth and four times the diameter. So Triton tides of Neptune will be about fifteen times smaller than lunar tides on Earth.

All the other moons are much smaller or further away or both and will raise completely negligible tides.

So from gravitational effects alone, Triton will raise small tides and nothing else will be measurable without good instruments and a lot of care.

(Note, I'm talking about tides in oceans comparable in size to the planet -- for smaller bodies of liquid, the tides will scale down.)

On Earth, tides vary all over the place, and this is due to the configuration of the body of water and resonance effects. Basically, if a body of water has a resonance -- a natural period for sloshing back and forth -- that matches the period of the tidal force, the tide can build up like repeated pushes to a kid in a swing.

Additionally, there can be a funnel effect with two arms of land coming together, or shallowing water that can amplify the tides. (E.g., the Bay of Fundy.) Since as far as we know, Neptune has no land, it's effectively an planetary ocean and will get the full effect of the gravitational tides, but no resonances.

So, bottom line: Small Triton tides and nothing else.

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    $\begingroup$ Thanks for this Mr. Mark. This helped. But one more doubt emerged in my mind while reading your answer. Do neighbouring planets affect tides ? $\endgroup$ – Garima Feb 2 '19 at 16:19
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    $\begingroup$ Sure, but they are very, very small effects because of the inverse cube of the distance. $\endgroup$ – Mark Olson Feb 2 '19 at 16:25
  • $\begingroup$ Minor nit - if a putative Neptune-sized planet had "oceans" of some low-density fluid, the physical effect would be greater than for, say liquid water. Otherwise, nice clean explanation. $\endgroup$ – Carl Witthoft Feb 5 '19 at 15:04

Neptune's largest moon, Triton, is about a third the mass of Earth's moon, and roughly the same distance away from its primary, so it will indeed have noticeable tidal effect on Neptune's atmosphere and oceans.

Neptune's other moons are all tiny (all of them together being about 0.5% of Triton's mass), so while they do exert tidal forces, those effects would be dwarfed by Triton even when all the small moons are on the same side of Neptune.

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Neptune’s moon Triton will cause tidal potential changes that are less than Earths, but still good fractions of a meter.

But whether there are visible tides is a harder question. On Earth, we have visible tides because the land doesn’t respond as much as the water does. A planet completely covered by deep water would have no detectable ocean tides: everything would just move together.

If Neptune’s surface is slush & liquid, it may move together enough that Neptunians don’t notice any tides.

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As Russell Borogove and Bob Jacobsen write, indeed Triton wil raise tides on Neptune. But they won't be like the tides people envision here on Earth.

Complex modeling of the vertical structure of Neptune (behind a paywall, but the abstract says enough) by researchers like Sushil Atreya suggests there is no liquid water ocean at Neptune, but instead an "ionic ocean" of water with dissolved ammonia very deep in the atmosphere, deep enough to have pressures in the thousands of bars. At those depths the temperatures are so high (2,000-5,000 K) they are well above the 647 K critical temperature of water, so liquid water doesn't exist. Instead you have a supercritical fluid that is dense like a liquid but compressible like a gas, with other characteristics (like chemical reactivity) that you don't find in the liquid or gaseous state.

Notably, at Neptune (and at other giant planets) there is no discontinuous surface where above it is one state and below it is another. Here on Earth we're quite accustomed to such a discontinuous (at least on human scales) surface, like the surface of our oceans: above that surface is a gas, our atmosphere, and below it is a liquid. Instead, at Neptune the water density and content of other species grades gradually as you go down. So you can't see a discontinuous surface rising and falling with respect to its surroundings, like an ocean level at a shoreline.

Here on Earth we see the tides at the boundaries of liquids and solids—shorelines of continents and islands. The response of the liquid in our oceans to tidal forces is larger than the response of the solid part of Earth. Seeing that difference with human eyes is only possible when you have solid Earth extending above the ocean's level. Were Earth's lithosphere (the solid part) completely smooth, i.e. no topography, and the oceans a uniformly deep layer above that, someone floating on that ocean wouldn't detect the tidal rise and fall without instruments that go far, far beyond human senses.

Similarly, you couldn't detect the tidal response at Neptune. It's made even more difficult to detect by the absence of a discontinuous surface as a reference.

By the way, this reference (that I found while searching for Sushil's) attributes the same paper to the wrong Atreya!

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  • $\begingroup$ One way you might detect tides in the "smooth Earth" scenario would be to fix a buoy on a rigid (no detectable stretching) tether to the ocean bottom, such that the tether is just taut at low tide. At high tide, more of the buoy would be submerged. $\endgroup$ – Tom Spilker Feb 3 '19 at 3:04
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    $\begingroup$ @uhoh OK...when people talk about tidal forces existing without a 2nd (or 3rd, etc.) body involved, what they really mean is tidal accelerations, and that translates to "the change in gravitational acceleration with position". Those of us who speak vector calculus know that this is just the spatial gradient of the gravity field. To get a force from this requires a mass: F=ma. The fellow who talked about a dark object distorting a circular ring of space stations into an oval ring got it right, but the explanation might be a bit unclear, and lacking some math. $\endgroup$ – Tom Spilker Feb 3 '19 at 6:02

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