The following interpretations are taken from Thorne [2014].
Chapter 17, entitled Miller's Planet, discusses the issue of the large waves on the water planet in the movie Interstellar. There Kip mentions that the waves are due to tidal bore waves with height of ~1.2 km. In the appendix entitled Some Technical Notes, Kip estimates the density of Miller's planet to be $\sim 10^{4} \ kg \ m^{-3}$. For comparison, Earth's density is $\sim 5.514 \times 10^{3} \ kg \ m^{-3}$. We are also told that the planet itself has ~130% of the gravitational acceleration of Earth. From this we can estimate the mass and radius of Miller's planet (ignoring tidal distortions to make things easy):
$$
\begin{align}
r_{M} & = \frac{3 g_{M}}{4 \pi \ G \ \rho_{M}} \tag{1a} \\
& = \frac{3.9 g_{E}}{4 \pi \ G \ \rho_{M}} \tag{1b} \\
r_{M} & \sim 4546-4572 \ km \tag{1c} \\
M_{M} & = \frac{ 9 g_{M}^{3} }{ \left( 4 \pi \ \rho_{M} \right)^{2} \ G^{3} \ \rho_{M}} \tag{2a} \\
& = \frac{ 19.773 g_{E}^{3} }{ \left( 4 \pi \ \rho_{M} \right)^{2} \ G^{3} \ \rho_{M}} \tag{2b} \\
M_{M} & \sim 3.936 \times 10^{24} - 4.002 \times 10^{24} \ kg \tag{2c}
\end{align}
$$
For reference, the Earth's mean equatorial radius is $\sim 6.3781366 \times 10^{3} \ km$ and the Earth's mass is $\sim 5.9722 \times 10^{24} \ kg$.
The water is very shallow, as shown by the characters walking through it. So how can there be several hundred meter waves?
Unfortunately, the answer is extremely boring. The planet is tidally locked with the nearby black hole and nearly all of the surface water of the planet is locked into two regions on opposite sides of the planet. The planet itself is shaped much like an American football rather than an oblate spheroid.
There is a slight problem with this interpretation, though. In the movie, the Ranger appears to float. Though I do not doubt that the vehicle is well sealed, I am curious if it could displace more water than its weight allowing it to float on the massive waves.
Is this a wave or just an extreme tide?
Just an extreme tide, and according to the wiki on this planet, they do not actually propagate, the planet rotates beneath you due to a slight difference in the planet's rotation rate and its orbital motion (i.e., the planet "rocks" back-and-forth during its orbit about the black hole).
would there not be (extreme?) weather changes near these mounds of water?
I would be very surprised if such large mounds of water were not surrounded by or at least affecting the nearby weather, much the same as mountains on Earth. However, this is starting to split hairs in an already speculative subject I guess.
Updated Thoughts
I updated the following computations for fun merely because I found them more interesting than the tidal bores.
Gravity Waves
If we assume that the wave height were the same as the wavelength and we assume these were gravity waves, then their phase speed would be ~49 m/s.
Shallow Water Waves
If we assume the wavelength is $\sim r_{M} \gg h$ (i.e., from Equation 1c), where we now assume $h$ ~ 1.2 km, then the phase speed would go to ~124 m/s.
References
- Thorne, K. "The Science of Interstellar," W.W. Norton & Company, New York, NY, ISBN:978-0-393-35137-8, 2014.
Typos and/or mistakes in book
I only found a few typos/mistakes in the book, which are listed below:
- Chapter 2
- He confuses the north and south magnetic poles (i.e., the north magnetic pole is located near the south geographic pole, not the north).
- He assigns the source of the aurora to protons. However, the aurora are due to energetic electrons exciting oxygen and nitrogen.
- Chapter 7
- He states that the Cassini spacecraft used "...Saturn's moon Io..." for a gravitational slingshot. However, Io is one of the four Galilean moons of Jupiter and Saturn is the planet to which Cassini was headed.
I consider these fairly minor and honest mistakes, but worth taking note of...
Best Answer
Basically the whole kinetic energy is transfered to pressure, and then this pressure will be transfered to kinetic energy again; this time only the direction is as defined by hydrostatic pressure; perpendicular to surface.
This above gives a following basis;
Kinetic energy of the ball is also it's potential energy (no friction on fall) Ekin = m g H This is then transferred to pressure through the ball surface; A = 4 pi r^2
This pressure then splashes the fluid up;
In optimal case the diameter of the ball is almost zero, and viscosity of the fluid is such, that the ball would stop in a distance of slightly more than r. This would lead to a situation where the vertical velocity of the water is very low, and thus the water would jump almost directly upwards. This doesn't actually matter too much, if the air friction is not considered.
Ok, so an answer, if the density of the ball is same as the density of the fluid. Then the fluid would jump to same height as the ball was dropped, if we also consider that there is no viscous losses. This is never true, and thus the ball drops deeper in the fluid and the losses reduces the available energy.
This all could be calculated. But the interesting thing is that there is a hole in the water when it goes deeper; And this means that the fluid which has maximum pressure has now a surface with no pressure. And therefore the fluid goes with even higher velocity back to fill this hole; as the velocity come from pressure difference, what happens;
It collides in the middle of the hole, but this time there is many velocities reaching the same point at the same time. Again all these velocities are transferred to pressure and the fluid takes new direction.
In 2-dimensional world this new velocity component would be 2-times the original. In 3-D reality it's more, and in true reality it's limited by viscous losses, surface tensions, etc. etc.
So to conclude this all; The splash hight can be anything.
At this video found from comments, there is a golf-ball used to do the splash. And such a Golf ball makes a higher middle-splash than a round ball, because the boundary layer of the ball makes less losses, but also disturbs the fluid less. And therefore the returning middle splash is so big in this video; the collision happens with minimal disturbances; and the velocity vectors really hits against each others.