Escape velocity to leave the water planet in the movie Interstellar 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!
 A: 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 $\approx 11.2 ~\mathrm{kg/ m^3}\;.$
A: Yes, it would be easier.
. . . But only if the planet was similar in size to Earth.
The escape velocity depends on the mass of the body. For a sphere, it's
$$v=\sqrt{\frac{2GM}{r}}=\sqrt{\frac{2G}{r}} \sqrt{M}$$
Earth has a mean density of roughly 5.514 grams per cubic centimeter; liquid water has a density of roughly 1 gram per cubic centimeter, or 1,000,000 grams per cubic meter. This means that a planet made largely of water will be much less massive than another planet of the same size. If this water planet is the same size as Earth, it will be about 2/11 times as massive as Earth; its escape velocity will thus be about $\sqrt{\frac{2}{11}}$ times that of Earth, or 0.426 times that of Earth.

Edit
Okay, so on this planet, $g$ is 1.2 times that of Earth, or 11.76 meters per second squared. It is defined as
$$g=\frac{MG}{r^2}$$
This means that
$$M=\frac{1.2 r^2}{G}$$
Putting this back into the original equation, we have
$$v=\sqrt{\frac{2GM}{r}}=\sqrt{\frac{2G\frac{1.2r^2}{G}}r}=\sqrt{2.4r}$$
If $r$ is the same as Earth, we get an escape velocity 1.55 times that of Earth.
Note: This was posted before the question was changed to explain that much of the planet's composition was unknown. As others have said, this means that there isn't really a great answer to be had. We cannot calculate the density or the mass of the planet; it will be nearly impossible to solve this accurately without making a host of assumptions.
A: If the black hole creates those huge tidal waves, then the black hole's gravity must have an impact on the planet and therefor it would be easier to escape the planet, if you were on the side of the planet that faces the black hole, wich the tidal wave indicates. If the people were on the other side of the planet the effect would be reverse, minus the fact that  gravitational pull would shrink with the diametre of the water planet. And would´nt be very hard to walk on the other side of the planet too? The gravity must be waing you down??
Is this correct? Otherwise, why would the tidal waves be created by the black hole but nothing else on the planet is effected? 
A: Escape velocity is only applicable for non-powered projectiles. A powered vehicle can leave another body at any speed, it only needs to provide a greater force than gravity.
To leave a body near to a black hole, you need only sum all the available forces, and make sure the spacecraft can provide a greater force, in which case it can leave the system at any speed it wishes.
