According to Fishbone, 1973, ApJ, 185, 43 the minimum density defining the Roche limit for something in the innermost stable, equatorial, circular orbit around a Kerr black hole is the same as that for an orbiting object in Newtonian physics$^1$.
Using Fig.2 in that paper, for a maximally spinning Kerr black hole, the limiting Roche density is $\sim 3 \times 10^{18} (M/M_\odot)^{-2}$ g/cm$^3$. The black hole that Miller's planet orbits has $M \sim 10^8M_\odot$ and so Miller's planet needs to have a density of $>300 $ g/cm$^3$ in order to survive.
Double-checking these figures, a simple use of the Newtonian Roche limit is that
$$ \rho > (2.44)^3 \left(\frac{3}{4\pi}\right)\left(\frac{M}{r^3}\right)\ , $$
where $M$ is the black hole mass, $r \simeq 0.5 r_s$ is the orbital "radius" and the $(2.44)^3$ coefficient accounts for the planet being a deformable fluid ellipsoid (see here). For $M = 10^8M_\odot$, then $r = 1.48\times 10^{11}$ m, and this gives gives $\rho > 200$ g/cm$^3$, in fair agreement.
Theis density looks 1-2 orders of magnitude too high for any kind of sensible planet.
I'm quite unhappy with this answer, since Kip Thorne designed the world/black hole system so that it would survive tidal forces...??
If you make the planet a rigid, undeformable sphere, then the $(2.44)^3$ gets replaced by $(1.26)^3$, which reduces the threshold density to $\sim 30$ g/cm$^3$. This is still too high for a terrestrial-type rocky planet.
You really need the black hole to have had a mass $>2\times 10^8 M_\odot$ to give a sensibly dense planet a chance of surviving break-up. Apparently Kip Thorne didn't do that because he wanted to make the black hole mass a nice round number. In a footnote in Chapter 6 of his book The Science of Interstellar, where the black hole mass is discussed, Thorne says
A more reasonable value might be 200 million times the Sun’s mass, but I want to keep the numbers simple and there’s a lot of
slop in this one, so I chose 100 million.
I guess that isn't the only (or even nearly the biggest) scientific inaccuracy in the film (see also Wouldn't Miller's planet be fried by blueshifted radiation?).
$^1$ Note that is by no means obvious that this should be the case. In general, the tidal force on an orbiting object is not given by the simple Newtonian formula. For example, the expression for the tidal forces in the radial direction for an object in a stable circular orbit around a Schwarzschild black hole are 1.5 times larger than the Newtonian value at the ISCO at $r=3r_s$ and would blow up to infinity at the unstable circular orbit at $r=3r_s/2$ (e.g. see chapter 9 of Exploring Black Holes by Taylor, Wheeler & Bertschinger).
Nevertheless if you consult chapter 19 of the same book, it gives general expressions for the tides for orbits around a Kerr black hole. If you work through these expressions, then when $a$ is maximal, and the ISCO is at $r = r_s/2$ then the tidal acceleration in the radial direction is given by
$$ g_{\rm tidal} \simeq \frac{2GM}{r^3}\Delta r\ ,$$
as in Newtonian physics. I think this is because at this radius, the orbiting object is essentially co-rotating with the dragged spacetime.
Best Answer
That would not be possible, since the required local orbital velocity around a Schwarzschild black hole is higher by the same amount the time slows down due to gravity, so that cancels out and the observed angular velocity is the same as with Newton:
So what remains is that the radius by which we multiply gets smaller and the angular velocity by which we divide larger, so the period gets smaller with smaller radius as it would under Newton.
If we plot the observed time period for one revolution (in units of GM/c³) by the radial coordinate (in GM/c²) we see that the period gets shorter with smaller radius, and is shortest at the photon sphere at r=3GM/c² where the required velocity is 1c and the proper time therefore goes to 0:
The red curve is the orbital period t in the frame of the coordinate bookeeper far away from the black hole, and the blue one in terms of the proper time τ of the orbiting clock.