# Tide without moons

I’m creating a planet which is relatively similar in size towards earth being slightly smaller. It’s covered in oceans and about only 17% of it is covered in land. The planet has no moons.

I don’t know if any of this would effect the question but the planet is much warmer than earth with a thin atmosphere. The planet has a small number of very large tectonic plates and are less active than those on earth.

So my question is would this planet have any sort of tides or waves or anything of the sort?

• Waves on earth are associated with seismic action (tsunamis) and storms. So it seems like your planet should have waves at least. Commented Mar 31, 2022 at 18:40
• I was hoping someone would say something about waves. Commented Mar 31, 2022 at 19:39
• Even on Earth, the moon is only partially responsible for tides. The sun is an equally strong component. Checking out the tides forecast for my location. You can clearly see it's the sum of two waves (moon and sun). Commented Mar 31, 2022 at 22:09
• Waves can come not just from storms, but any wind. Commented Apr 1, 2022 at 0:07
• Thinner atmosphere probably means lower temperatures, relative to a fixed orbital radius. At least at night. In the day, you may get irradiated to a crisp of course. If the Sun is about the same size as ours, then the oceans being liquid implies a similar orbital radius to ours. Any further and your ice caps would be bodacious. Commented Apr 1, 2022 at 9:34

Tidal forces just require nearby things that are big enough to have a decent gravitational pull on a planet (or other body).

The strength of the tidal force on the planet is proportional to the planet's radius, the mass of the star it orbits and inversely proportional with the cube of its distance from the star: $$F_T \propto {rM \over d^3}$$

Without trying to calculate the actual strength and direction of the tidal force (which is hard), you can use the figures for the Earth, the Moon and the Sun to see that the Moon has the biggest tidal impact on Earth but the Sun also has quite a large effect... about half as much as the moon.

Tides on your planet would be at the same solar time each day. You can tailor the size of your planet, the size of your star and the separation between them to tweak the strength of the tidal force and hence the heights of the tides, though it is quite a lot harder to go from these simple numbers to the actual height of a tide. All else being equal, a smaller world without a moon will have much smaller tidal ranges than Earth.

• Why would it get one tide a day? At each rotation, a given point will pass through two maxima and two minima of the tidal force. (The tidal force is maximum both at the closest point and at the farthest point from the star.) (And anyway, if the tides due to the star occur only once a day, why would those due to the Moon occur more than once?) Commented Mar 30, 2022 at 18:55
• Yep, should be two tides a day, but always at the same solar times. Commented Mar 30, 2022 at 19:01
• Tide would also be very regular in height. On earth, it is a resonance with sun and moon together, every tide is different, you can have extreme low and extreme high tides. On a moonless planet, it would be very regular and moderate. Commented Mar 30, 2022 at 22:15
• @Goodies - excellent point. I was curious if other planets could have a discernable effect on tides when there's no moon. Using Jupiter, Earth and our sun as a benchmark, I got that the moon's effect is more than double that of the sun (2.18 ratio for moon/sun), but Jupiter, at its closest to Earth, has a negligible effect (0.000016 ratio for Jupiter/sun) - so I guess the answer to that is no. Commented Mar 31, 2022 at 11:58
• Jupiter also has very small moons. The earth's moon is a giant.. Commented Mar 31, 2022 at 14:45

First of all, if the atmosphere is too thin, there will be no liquid water at all, in particular if the temperature is higher than on Earth: with lower atmospheric pressure, water evaporates more easily.

That said, there will still be other bodies which will exert a tidal force on the water body, the most important would surely be the central star, in the same our Sun gives a small contribution to the tides.

Being alone it won't create tides as impressive as the one caused by the Moon, but still the effect will be measurable.

• "Our Sun gives a small contribution to the tides": That small contribution is about one third. Moon's tidal force is only two times as large as the tidal force of the Sun. Commented Mar 30, 2022 at 18:21
• It would be possible for a moonless planet close to a large star to experience tidal forces bigger than those experienced on earth. Tidal forces result from the gradient of a body's gravitational field, which is steeper closer you are. I don't think we can compare to earth tides without knowing anything about the central star or the planet's orbital distance. Commented Mar 30, 2022 at 19:18
• Regarding the oceans evaporating, a less volatile liquid could be considered, for example oil oceans. If the world had a less important magnetic field maybe it would also reduce the volatility of of the liquids, as of sciencedirect.com/science/article/pii/S2211379717317230 (I don't know how such a world would handle solar flares for instance) Commented Mar 31, 2022 at 7:55

The sun also produces tides, though on earth it's somewhat smaller compared to that of the moon. So yes, if your planet is not a rogue planet, it will probably have tides.

For any given tide, local coastal features can exaggerate the effects. I'm not sure but I would have thought that a steep underwater slope with a flat exposed coast would amplify the effects. You could look up surf beaches that are strongly affected by tides to see how they work.

TL;DR: You need a planet with a big, red sun.

This is intended only a complement to Starfish Prime's excellent answer. A complement too long to be posted as a comment.

The formula $$F_T \propto {rM \over d^3}$$ can be rearranged by noting that:

• the mass of the star is proportional to its density multiplied by the cube of its diameter
• the ratio of that diameter to the distance is almost exactly the star's apparent size, as seen from the planet.

This gives the formula $$F_T \propto r \rho \delta^3$$, where ρ is the density of the star and δ its apparent size.

An interesting thing we can deduce from this formula is that, our Sun and our Moon having the same apparent size, the ratio of their tidal forces is just their density ratio: the Moon is about twice as dense as the Sun.

Back to your planet. If you want it to have tides similar to our own, this planet would need a Sun that is bigger (as seen from the surface) than our Sun is to us. For example, in order to match the tides we would have with the Moon alone (a rough average between high tides and low tides), their Sun should be about 30% bigger, assuming it has the same density as ours.

You want a hot planet, but beware it doesn't get too hot. Ignoring greenhouse effects and the like, the average surface temperature of a planet should be roughly proportional to $$T \sqrt{\delta}$$, where T is the star's effective temperature. If we make the Sun 30% larger with the same temperature, we could expect an average Earth temperature around 60 °C (more if you account for the greenhouse effect of water vapor). To avoid that, make your star cooler, i.e. redder. Don't go read giant though: these stars have a very small density, and you would need to orbit unreasonably close in order to get significant tides.

What if your planet is a moon? Moons around gas giants have plenty of tidal activity, and can have atmospheres, liquid water, habitable temperatures, and the like. Europa and Io come to mind, in my home solar system.

• I failed to mention this in my answer but it is tidal forces from Jupiter that keep Io molten and highly tectonically active, and keep Europa Commented Apr 1, 2022 at 14:28

Eccentricity of the planet's orbit around its Sun will determine the height of the tide throughout the year. Think annual, rather than monthly, Spring tides.

As others have stated, the lack of moons means no M2 component, so the S2 component will dominate.

Looking at what happens on Earth, when you get into enclosed waters (eg, Gulf of Mexico, which has a basically diurnal tide), things can change.

Also, there will still be overtides, ie, the tide will be slowed in shallows and start to be forced out while it is still coming in, and vice versa.

All in all, there should be a number of tidal harmonic constituents, which when combined, should result in complex behaviour in some areas closer to land, and probably fairly simple ones in deep oceanic waters, varying over the course of a year due to the eccentricity of the planet's orbit.

The inclination of the planet's orbit will also probably manifest itself in additional constituents.

The tide, as others have indicated, is not caused by the Sun sucking the water up on one side. There will be two tides a day in deep oceans, not one.

If your hero wishes to analyse tides, they will need to record the tidal heights at a reasonable frequency for a reasonable period of time (longer than the periods of the constituents). They would then determine precise orbital periods and tidal harmonic periods. These will then allow determination of corresponding phases. Prediction of tide times, critical to marine operations throughout the planet, would then be possible.

Having a central authority sample tide heights and predict tides in advance allows great scope for mindless bureaucracy, and a bloated public sector rife with bribery and corruption.

Resisting privatisation will allow for a 1984-style tidal dictatorship, while full private ownership will allow your shipping operations to be held for ransom by shadowy private tidal contractors.

As others pointed out the planet could still have solar tides, but they would be small. But on the other hands you can have currents due to the planet rotation. Since your planet has a small land mass the currents would have few obstacles and could get quite fast. This could create some nasty maelstroms close to the islands.