I saw this question and this question on the site a few days ago. It asks about escape velocity from the water-based planet in Interstellar and whether the black hole had any effect. Now, one question is unanswered whilst the other has an answer focussing on the effect of the black hole (it said the effect was non existent.
My question is: If the black hole had no effect, then does the fact it was a water based planet mean it is easier to achieve escape velocity, or harder?
I'm aware leaving the water planet is one of the contentious parts of the movie. If anybody has any further comments on its possibility, I'd love to hear them.
Edit: for anyone unfamiliar with movie, gravity on water planet is 1.2 times that of earth. We've no idea what the planet is composed of, other than it is entirely water, roughly thigh deep.
On a final note, I'll add that I'm an active member of the Movies & TV Stack Exchange. I'm asking this question here as we've had a plethora of questions there about issues like this and frankly none of us are physicists. Therefore, I'll cheekily request that answers be kept on the simple side!
Best Answer
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 $\approx 11.2 ~\mathrm{kg/ m^3}\;.$