Escape velocity to leave the water planet in the movie Interstellar

Okay, trying my luck with a physics answer. Let's first look at the boundary conditions given in the movie, since we're particularly talking about that here. The water planet is said to have $130\%$ of earth's gravitational acceleration on the surface. So we have \begin{equation} g_W = 1.3 g_E \end{equation} This is a given and not to be violated. And in fact it poses constraints on the relation between both planets masses, radii and densities. With the fact that the planet's volume (a supposed sphere for the sake of simplicity) is $\frac{4}{3}\pi r^3$ We can thus express the planet's radius as a function of its density and its gravitational acceleration: \begin{equation} r = \frac{3g}{4\pi G\rho} \quad\sim\quad \frac{g}{\rho} \end{equation}

We can then fill this into the formula for the escape velocity (and drop some constants): \begin{equation} v = \sqrt{2gr} = \sqrt{\frac{6g^2}{4\pi G\rho}} = \sqrt{\frac{3}{2\pi G}}\frac{g}{\sqrt{\rho}} \quad\sim\quad\frac{g}{\sqrt{\rho}} \end{equation}

So now lets look at the relation between the escape velocities. We want the planet's escape velocity to be lower than that of earth, so: \begin{align} v_W &< v_E \\ \frac{g_W}{\sqrt{\rho_W}} &< \frac{g_E}{\sqrt{\rho_E}} \\ \sqrt{\rho_W} &> \frac{g_W}{g_E}\sqrt{\rho_E} \\ \rho_W &> 1.69 \rho_E \end{align}

So to have a lower escape velocity than earth, the planet would have to have more than $169\%$ of earth's average density.

But in fact, Kip Thorne actually gives an estimate of the planet's average density (in the Technical Notes of his book The Science of Interstellar), namely $10,000 ~\mathrm{kg/ m^3}$, which is indeed $181\%$ of earth's $5,515 ~\mathrm{kg/ m^3}\;.$ Since this is the only actual information we can rely on (and is totally independent of how much water there is on the surface) we can indeed conclude that the escape velocity of Miller's planet is lower than that of earth.

More exactly, the planet's escape velocity would be $\approx 10.8 ~\mathrm{kg/ m^3}$ compared to earth's $\approx 11.2 ~\mathrm{kg/ m^3}\;.$