I'm building a world that orbits close in to an M-Dwarf sun. I've figured out a bunch of the physical parameters of the world, I just can't quite get a solid handle on my tides.

I know my tides will be substantial, but exactly how substantial is something I need some help on.

I have three main sub-questions:

  1. Am I calculating the tidal force correctly?
  2. Does tidal force correspond 1:1 with ocean tide height?
  3. All else being equal, changing the diameter of the planet should change the tide height. But hot much? How do you calculate this?

Below is the supplementary info, and a basic outline of my thinking on the matter.


  • Elliptical orbit with a semi-major axis of 0.1307 AU, apoapsis of 0.1622 AU, and periapsis of 0.0991 AU.
  • Planet radius 5268km.
  • No moons
  • Solar mass of 0.395 earth suns.
  • Year of 27.44 earth days, with a rotational period of exactly half (13.72 earth days).
  • The planet is in 2:1 spin/rotation resonance, so a solar day (sunrise to sunrise) equals one 27.44 earth-day year (visual). So the tides should ebb/flow based on this number, not the actual planet rotation rate of 13.72.

Basic Calculation

The relative tidal force on the planet is easy to compute. It should be:

T = M/(d^3)

Where M = mass of sun & d = distance.

So that would give us T = 0.395/(0.1307^3). Which comes out to a tidal force of 177.1 times sol's force on the Earth. Likewise the tidal force for apoapsis is 92.6 and for periapsis 405.9.

Now, the tricky parts come in: Changing planet diameter

First, this is the tidal force as measured from the center of the planet. But based on my understanding of tides, its the differential in tidal force between sides of the planet that causes the tides, not the absolute tidal force. The number I computed above would be an accurate number for relative tidal force on an exactly Earth size planet in that location, but not a smaller or larger planet. And since my planet is smaller, the tidal force must logically be smaller than this calculation.

(Example: If you get a theoretical planet of zero diameter the tidal force exerted would be...zero. So the force MUST necessarily scale by some formula from zero upward).

I don't know how to calculate that force for my smaller world and can't find any direct calculations online.

Now the second wrench into the works: Does Tidal Force = Ocean Tide Height

Does a mean tidal force 177 greater than that of sol's on Earth necessarily imply tide heights 177x greater? My intuition says no, but I don't know exactly why. I suspect the tide heights would be ameliorated to an extent by the friction between water and ocean bottom, and perhaps other factors. But...I don't know. It's far too much of a guess for me to be comfortable with. I want to have more concrete numbers based on a more concrete understanding.

Why I care

On of the reasons this is necessary to figure out to a reasonable accuracy is that tides of this size will have a MASSIVE impact on the ecology of my world. And the difference between tides of 40x sols, 100x sols, and 400x sols is huge. At the upper end of my current calculations (405.9x sols) I get max tide heights of about 101.5 meters (based on a solar theoretical tide height of 0.25meters) or maybe 72.7meters (based on the solar semi-diurnal tidal constituent of 0.179meters listed here: https://en.wikipedia.org/wiki/Earth_tide). Those numbers just seem...catastrophic to work with. Even the smaller calculation of 72.7 meters would mean that the tide rises and falls that height every 13.72 earth-days, or 5.3meters per 24 hours. Constantly. So the coasts & coastal plains of my world would be zones of perpetual flooding and draining, potentially moving many kilometers inland over just a dozen or two dozen hours. Heck, if Earth had tides of 72.7 meters a good portion of Florida would be covered/uncovered by seawater regularly. That's some interesting fodder to work with, but perhaps too much of a good thing.


  • 1
    $\begingroup$ I have a feeling that if you could remove all the extra fluff that I didn't get rid of, this would be a good fit on Physics.SE $\endgroup$
    – Aify
    Mar 4, 2018 at 18:40
  • $\begingroup$ Force is measured in Newtons, which are kg m/s^2. Your T is kg/m^3 $\endgroup$
    – L.Dutch
    Mar 4, 2018 at 18:50
  • $\begingroup$ @Aify Thanks for starting the edits. I've made my own more substantial edits for clarity. If there is "fluff" you recommend removing, let me know. I have put most of the supplementary info down below the intro for those interested in my thought process on the topic. $\endgroup$
    – n_bandit
    Mar 4, 2018 at 19:18
  • $\begingroup$ @L.Dutch I believe tidal force rises with the cube of the distance. So twice as close = 8x tidal force. I've seen this equation on numerous sources so I believe it's correct. $\endgroup$
    – n_bandit
    Mar 4, 2018 at 19:20
  • $\begingroup$ @Aify Also, if you think it would be a good question for Physics, I could condense the questions and post it there. I would modify it to ask about shrinking/expanding the radius of the earth and how to calculate the effect on tides. What do you think? $\endgroup$
    – n_bandit
    Mar 4, 2018 at 19:23

1 Answer 1


Why do you expect tides to be substantial?

There are no moons, the distance to the nearest body of significant mass is something like five orders of magnitude greater than the diameter of your world.

Tidal forces in a context where you can disregard general relativity usually has to do with two bodies of mass at different distances resulting in varying gravity across your world.

For Earth, it's the fact that, while the moon is a lot less massive than the sun, its also really closer that as it goes around the Earth, the effects of gravitation on the side of the planet facing the moon is significantly greater than the opposite.

e.g. Let's approximate, for calculation's sake, that:

  • The moon is a $10^{23}\ \mathrm{kg}$ point mass $10^6\ \mathrm{km}$ away from the center of the Earth
  • The Sun is a $10^{30}\ \mathrm{kg}$ point mass $10^8\ \mathrm{km}$ away from the center of the Earth
  • The Earth has a radius of $10^5\ \mathrm{km}$
  • And the water in the ocean has a mass of $m$

And consider the case where the three celestial bodies are in a line.

The distance between the Sun and the two sides of the Earth in line with the Sun and the Moon is $10^{30} \pm10^5\ \mathrm{km}$, which is basically $10^{30}\ \mathrm{km}$. We can basically ignore the Sun's gravity when considering tides, because it affects the entire Earth the same.

When we consider the gravity due to the moon, the differences in the distances matter. $10^6+10^5\neq10^6-10^5$

The gravitational force on the water on one side is $$F_1=\frac{G\centerdot10^{23}\centerdot m}{\left(10^6+10^5\right)^2}\approx8.3\times10^9\centerdot m \centerdot G$$

The gravitational force on the water on the opposite side is $$F_2=\frac{G\centerdot10^{23}\centerdot m}{\left(10^6-10^5\right)^2}\approx1.2\times10^{12}\centerdot m \centerdot G$$

For comparison, the gravitation due to the Sun would be $$F_S=\frac{G\centerdot10^{30}\centerdot m}{\left(10^8\right)^2}=10^{14}\centerdot m \centerdot G$$

The math is rubbish. It does show you how the Sun's gravitation dominates, but the moon's proximity to the Earth causes the difference in gravity that is significant enough that it has observable effects as in the phenomenon of sea tides.

Tidal forces with a single source of gravity would be something like a very long objects very close to (and probably falling into) a very dense body of mass, like a black hole. I think this is the "kind" of tidal forces you are thinking of when you mention differentials in forces. In this case the sole source of tidal forces is the difference in the distance between one end of the long object and the other.

e.g. imagine a very long stick falling into a tiny black hole; let's say the stick is a two-dimensional 2000m long with a density 1kg/m 1 meter away from a 10000kg mass of 1mm (1E-3 m) diameter. (I have no idea if these numbers work out, or even make sense, but it's just for illustration)

If we treat each 1m section of the stick as a point mass (and butcher physics) and calculate the gravitational force on the two sections closest and furthest away from the tiny black hole, we get (gravitational force given by formula $F_G=\frac{GMm}{r^2}$):


for the part of the stick closest, and


If you pull on one end of a wooden stick with $10000G$ of force, but only $0.001G$ on the other, the difference causes the type of tidal force that disintegrates stuff that fall into black holes.

Hopefully this gives you a better starting point in thinking about tides on your world.

But to answer your listed questions for completeness:

  1. Yes. But the Sun's tidal force on the Earth is so small that 177.1 times that is still insignificant. This is the tidal force as in the force that's ripping Earth apart because of the difference in the distance between the two sides of the planet and the Sun. It's negligible.

  2. Uh, no. I guess the easiest way to see this experimentally would be to see how high you can pull up a pile of iron powder (or whatever else ferromagnetic powder/small shavings you might have lying around, if you have access to a workshop, stuff piles up around lathes, mills, etc.) with one magnet, and then try the same with two. The height doesn't double. Chances are this isn't something you can try, but you can consider the fact that the water's weight (mass) is a product of its volume, with units of $\mathrm{m^3}$, while height is a length with units of $\mathrm{m}$.

  3. How much depends on the distance between the nearest significant body of mass vs. the diameter of the planet. As I mentioned, with the numbers you have now, they are about five orders of magnitude apart. I'd imagine you would have to add at least two zeroes to the diameter of your planet before seeing any appreciable change. You can approximate without calculus using the formula for classical gravitational force, use the difference between $F_{G_1}=\frac{G\centerdot M\centerdot m}{(D+r)^2}$ and $F_{G_2}=\frac{G\centerdot M\centerdot m}{(D-r)^2}$, where $D$ is the distance between the star and the planet, and $r$ is the radius of the planet.

  • $\begingroup$ $G$ is a very small number, by the way. Gravity is the weakest force by far. $\endgroup$ Mar 18, 2018 at 17:49
  • 1
    $\begingroup$ "We can basically ignore the Sun's gravity when considering tides, because it affects the entire Earth the same." The Sun's tidal effect on the Earth is roughly one-half that of the Moon. That's why the height of tides varies so much over the course of a month: near quarter moons the two effects partially cancel and near new or full moon they add. Add you need to check some of the calculation here-in as they are not all consistent. $\endgroup$ Mar 19, 2018 at 0:41
  • $\begingroup$ @dmckee You are completely right, we can't ignore the Sun's effect. The Sun also isn't a $10^{30}\ \mathrm{kg}$ point mass, it isn't $10^8\ \mathrm{km}$ away from the Earth (for most of Earth's orbit around it), and generally, not very much of the example I gave has its basis in reality. I was just trying to show how and why the Moon is the biggest factor that causes tides on Earth. I don't know if the Sun accounts for half of the phenomenon we observe as ocean tides (it very well may), but what I meant was that the tidal forces on the Earth due to the Sun are negligible. $\endgroup$ Mar 19, 2018 at 14:53
  • $\begingroup$ @dmckee "Add you need to check some of the calculation here-in as they are not all consistent." No, I should really throw them out the window, they not consistent at all and doesn't describe what really happens. The calculations in my answer were meant to give a very general, crude (if incorrect) idea of the magnitude of gravitation we are dealing with here without needing to resort to calculus. Again, the examples I gave don't model (or attempt to model) what happens in reality, they were only meant to send the OP back to the drawing board and try to think about tides again. $\endgroup$ Mar 19, 2018 at 15:02

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