I'm not talking only about the kind of waves one sees at a sea shore, but also smaller waves like ripples and chops, and bigger waves like tsunamis.

Assume a similar tidal force to that of Luna.


4 Answers 4


The best way to study water waves of many sizes is to use Airy wave theory (see also here), a mathematical model using several simplifications that nonetheless produces reasonable results. The linear theory works best when the amplitude ($a$) is small in comparison to the water depth ($h$) and the wavelength ($L$). Using the continuity equation for fluids, the Navier-Stokes equations, Bernoulli's principle, Laplace's equation, appropriate boundary conditions, and the assumption of linearity, we reach the surprisingly simple approximation of $$\eta(x,t)=a\cos(\omega t-kx)\tag{1}$$ where $\eta$ is the elevation of a particle above the plane of water at rest, $\omega$ is the angular frequency, and $k$ is the wavenumber. We also get a nice dispersion relation and a somewhat complicated expression for the phase velocity of the wave, $c$: $$\omega^2=gk\tanh(kh)\tag{2}$$ $$c=\sqrt{\frac{gL}{2\pi}\tanh\left(\frac{2\pi h}{L}\right)}\tag{3}$$ where $g$ is of course the acceleration due to gravity; we get this from the equation $k=2\pi/L$. Additionally, as Green pointed out, the mean kinetic energy $\bar{E}$ is given by1 $$\bar{E}=\frac{1}{16}\rho gH^2\tag{4}$$ where $\rho$ is the density of the water. If an energy $E_w$ is imparted to a wave, then we have the relation $$H=\sqrt{\frac{16E_w}{\rho g}}\to H\propto\frac{1}{\sqrt{g}}$$ as JDługosz said. Stronger gravity means smaller waves.

Holding all other variables constant, $$\omega\propto\sqrt{g},\quad c\propto\sqrt{g},\quad H\propto\frac{1}{\sqrt{g}}$$ Therefore, on a planet with higher gravity, you'll see . . .

  • Smaller waves
  • Faster waves

and on a planet with lower gravity, you'll see . . .

  • Larger waves
  • Slower waves

This is all very straightforward, but what happens when waves reach the shore? At this point, linear wave theory breaks down (pun intended), and numerical modeling is often your best shot (see here). The shallow water equations are useful here, especially when discussing tsunamis. Let's look at one analytical case where we do get results. We look at the Carrier-Greenspan criterion for wave-breaking. When approaching a beach where the seafloor slopes up at a rate $s=\left|\frac{dH}{dx}\right|$, a wave will break if $$s^2<\frac{\omega^2a_{\text{shore}}}{g}\tag{5a}$$ where $a_{\text{shore}}$ is the amplitude of the shore. Substituting in the dispersion relation for $\omega^2$ shows that this is entirely independent of $g$! A more rigorous model gives us a better relation (still without influence from $g$): $$s^{5/2}<\sqrt{2\pi}ka\tag{5b}$$ This is elegant. While gravity will, of course, influence how specifically a wave moves near the shoreline, it will not affect whether it breaks or reflects.

Further reading:

  • $\begingroup$ I feel like I am trolling, but since you went to that much trouble... You are missing the effects of gravity on wind speed and wind wave interactions. And no, I have no idea what, if any, those are and if they are significant, but I am curious. $\endgroup$ Mar 7, 2017 at 20:16
  • $\begingroup$ @VilleNiemi The pressure in the atmosphere is affected by $g$, and given that wind is largely due to pressure differences, you'd probably see some different behavior on the whole. However, local pressure gradients probably wouldn't be that much different with different $g$. At any rate, I'm more concerned with how waves would behave after formation. I suppose that the probabilistic distribution of wave energy would have a different peak, but I don't think it would change that much. So I'll bet that the changes are insignificant, but I don't know for sure. $\endgroup$
    – HDE 226868
    Mar 7, 2017 at 21:15

It's a simple harmonic oscillator with a stronger spring.

The same amount of energy (e.g. from wind) makes lower waves in stronger gravity. Just like changing the tension on a string, waves will travel faster.

Since there is a smaller range of wave sizes since the big waves are smaller and 0 is still zero, in stronger gravity I think the dispursion effects will be reduced.

On the other hand smaller waves will hit harder. Same as the larger size of waves on Earth, just less to show.


Intuitively, it takes more energy to create a wave of equal height on a planet with more gravity compared to standard Earth gravity. Inversely, a planet with lower gravity will have lower energy waves. Using equations for wave energy incorporating gravity, we see: $$ E = \rho_{ w} g \langle \zeta^2 \rangle $$

where ρw is water density, g is gravity, and the brackets denote a time or space average.

So, as gravity goes up, the energy in a wave also goes up, if we keep water density the same. There are more equations wrapped up in the above cited equation, such as what $$\zeta^2$$ means. The equations link has a full description.

The actual shape of the waves probably won't be any different though the energy contained in a wave will be radically different. Waves on a higher gravity planet are going to hit a lot harder. If a medium sized seashore wave on Earth will knock a person over, an equal height wave on super-Earth will absolutely flatten them.

  • 1
    $\begingroup$ I'm getting a 'Page Not Found' message on the linked webpage. (Also, I've edited the brackets to use \langle and \rangle, which are commonly used in $\LaTeX$ for averages instead of < and >. Feel free to revert them if I'm mistaken.) $\endgroup$
    – HDE 226868
    Mar 7, 2017 at 21:26
  • $\begingroup$ Here is a live link to the equation of interest: en.wikipedia.org/wiki/…. Incidentally, according to this version of the equation, the time or space average is of the wave height, so you can say that wave height is proportional to the square root of the inverse of the waves. Also, this is for shallow water only, the Wikipedia link runs down the differences in shallow/intermediate/deep water. $\endgroup$
    – kingledion
    Mar 8, 2017 at 13:35
  • $\begingroup$ Please refer to HDE226868's answer to this question. It incorporates all the information I've found plus a bunch more. His is the best answer. $\endgroup$
    – Green
    Mar 8, 2017 at 14:19

A previous answer said that it can be equated to a harmonic oscillator with a stronger/weaker spring. I'm not sure if that's true. Most of the force and movement in a wave is lateral rather than vertical, and so the presence of waves would not be affected much, and neither would their speed, since wind and water currents play a much stronger role in their creation than gravity would.

You could look into the effects gravity change would have on your wind convexion and weather patterns, which would affect the water too, but that's probably going too far.


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