What could possibly produce the two gigantic water waves, 1.2
kilometers high, that bear down on the Ranger as it rests on Miller's
planet (Figure 17.5)? I searched for a while, did various calculations
with the laws of physics, and found two possible answers for my
science interpretation of the movie. Both answers require that the
planet be not quite locked to Gargantua. Instead it must rock back
and forth relative to Gargantua by a small amount [snip Thorne's
explanation of how Gargantua's tidal gravity will naturally provide a
sort of restoring force back to its preferred orientation, explaining
why the planet would rock this way] ... The result is a simple rocking
of the planet, back and forth, if the tilts are small enough that the
planet's mantle isn't pulverized. When I computed the period of this
rocking, how long it takes to swing from left to right and back again,
I got a joyous answer. About an hour. The same as the observed time
between giant waves, a time chosen by Chris without knowing my science
interpretation.
The first explanation for the giant waves, in my science
interpretation, is a sloshing of the planet's oceans as the planet
rocks under the influence of Gargantua's tidal gravity.
A similar sloshing, called "tidal bores," happens on Earth, on nearly
flat rivers that empty into the sea. When the ocean tide rises, a wall
of water can go rushing up the river; usually a tiny wall, but
occasionally respectably big. ... But the moon's tidal gravity that
drives this tidal bore is tiny—really tiny—compared to Gargantua's
huge tidal gravity!
My second explanation is tsunamis. As Miller's planet rocks,
Gargantua's tidal forces may not pulverize its crust, but they do
deform the crust first this way and then that, once an hour, and those
deformations could easily produce gigantic earthquakes (or
"millerquakes," I suppose we should call them). And those millerquakes
could generate tsunamis on the planet's oceans, far larger than any
tsunami ever seen on Earth
Best Answer
Miller's world would be fried by a strong flux of extreme ultraviolet (EUV) radiation. The cosmic microwave background (CMB) would be blueshifted by gravitational time dilation and then would be very strongly blueshifted and beamed coming from the direction of orbital motion. The overall effect would be a very strong dipolar distribution of temperature that is then distorted by the curved ray paths close to the black hole, whose shadow would fill nearly half the sky.
However, the size of the ultra-blueshifted spot is correspondingly very small. A detailed numerical calculation$^\dagger$ comes up with an equilibrium temperature for Miller's world of 890 $^{\circ}$C (Opatrny et al. 2016), with a flux of about 400 kW/m$^2$ from an EUV blackbody(!) arriving from the CMB "hotspot". I guess you would classify this as "fried"$^{\dagger\dagger}$. It is hotter than Mercury anyway.
$\dagger$ According to Opatrny et al. the peak blueshift in the direction of orbit is $275000$ - i.e. wavelengths are shortened by a factor of $275000+1$. Since temperature goes as redshift, then a tiny spot on the sky is an intensely bright (brightness goes as $T^4$) blackbody source of soft X-rays and EUV radiation. The source size is of order angular radius $1/275000$ radians due to Doppler beaming. Back of the envelope - the source is 130 times hotter than the Sun but covers a $(1200)^2$ times smaller solid angle in the sky. Thus the power per unit area received should be $130^4/1200^2 = 200$ times greater than from the Sun. This is in pretty good agreement with Opatrny et al.'s calculation that also claims to take into account the lensing effects.
$\dagger\dagger$ Apparently, the typical temperature in a frying pan is 150-200$^{\circ}$C