9
$\begingroup$

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!

|cite|improve this question
$\endgroup$
  • 4
    $\begingroup$ Note, that "water planet" here means that only the surface of this planet is covered in water (and not particularly deep, some few metres). So it shouldn't make any difference regarding density (which seems the only thing relevant for escape velocity given that the gravitational acceleration is fixed, which is missing in the question though). $\endgroup$ – Chris says Reinstate Monica Nov 23 '14 at 20:56
  • $\begingroup$ @ChristianRau: I'm assuming any respondents will be familiar with the film and thus take that into account. $\endgroup$ – Andrew Martin Nov 23 '14 at 21:00
  • 3
    $\begingroup$ Which unfortunately highly limits the possible answerers, though. It's not that the boundary conditions of this whole problem are very complicated or depend on the whole movie, so summarizing them to make for a self-contained question might not be a bad idea. The problem arises when someone rushes to answer ignoring the boundary conditions you didn't list. All you can do then is say "wrong answer, watch the movie to see why", which isn't a good idea on the self-contained and movie-independent site Physics, I guess. $\endgroup$ – Chris says Reinstate Monica Nov 23 '14 at 21:03
  • $\begingroup$ I've edited question but to be honest I do want answers from people who have seen the film. There are too many variables at play for me to summarise movie, so I'm hoping a Physics boffin with more knowledge than me who has seen the film can answer. Although that restricts possible answers, it's far more likely to produce a correct one. $\endgroup$ – Andrew Martin Nov 23 '14 at 21:39
  • 4
    $\begingroup$ In Kip Thorne's The Science of Interstellar, there's a section at the end titled "Some technical notes", and in the notes for Ch. 6 he says that he assumed the planet's density was about that of compressed rock, or $\rho$ = 10,000 kg/m^3. So, he definitely wasn't assuming the entire planet was made up of water. $\endgroup$ – Hypnosifl Nov 23 '14 at 21:41
10
$\begingroup$

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 $\approx 11.2 ~\mathrm{kg/ m^3}\;.$

|cite|improve this answer
$\endgroup$
  • 2
    $\begingroup$ The composition of the planet is only known to the level of "entirely water, roughly knee deep" which begs the question about what it below that... Clearly it would have to be quite dense if the escape velocity is to be lower. Nice start to solving the problem. $\endgroup$ – Floris Nov 23 '14 at 21:44
  • $\begingroup$ @Floris That is unfortunately entirely unknown, so we can't reall completely answer this. Maybe the fact that it doesn't get ripped apart by the black hole's tidal gravity speaks for it being very dense (or not very dense?). But Kip Thorne didn't make any mention in this regard. $\endgroup$ – Chris says Reinstate Monica Nov 23 '14 at 21:47
  • $\begingroup$ With your latest edit you have answered this question as well as it can be answered given the lack of information about its size. I'm +1 on this. $\endgroup$ – Floris Nov 23 '14 at 21:47
  • $\begingroup$ I think this question is largely answered. I'll delete my answer on the other site. Given the amount of work you've put in here, I'd strongly encourage you to reference it/link to it in your answer over there. Well done and thanks for the information! $\endgroup$ – Andrew Martin Nov 23 '14 at 21:51
  • $\begingroup$ @AndrewMartin Wait wait, it seems the escape velocity is lower, I completely missed that Thorne indeed gave a rough estimate for the planet's density. $\endgroup$ – Chris says Reinstate Monica Nov 23 '14 at 21:55
1
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ The problem here, is the fact that the gravitational acceleration of the planet is fixed (to be precise 120% that of earth's), which isn't mentioned in the question, though. This in turn puts additional constraints on the relation between the planets' masses, densities and radii, I think. Actually, it should result in the fact that the planet's density would have to be more than 1.44 times that of earth. Ideas? $\endgroup$ – Chris says Reinstate Monica Nov 23 '14 at 21:13
  • $\begingroup$ @ChristianRau Are you saying that the gravitational acceleration would change? Why? $\endgroup$ – HDE 226868 Nov 23 '14 at 21:14
  • $\begingroup$ No, I'm saying the fact that the gravitational acceleration is fixed puts constraints on the relations between the two planets's densities, masses and radii. $\endgroup$ – Chris says Reinstate Monica Nov 23 '14 at 21:15
  • 1
    $\begingroup$ I know, but the gravitational acceleration of the planet, i.e. small $g$, is fixed to be 1.2 times that of earth. That is an information from the movie the question has unfortunately hidden, but which is largely relevant to solving the whole matter. $\endgroup$ – Chris says Reinstate Monica Nov 23 '14 at 21:18
  • 1
    $\begingroup$ My best guess would be that someone has downvoted following the comments below the question which clarify that the planet is only thinly covered in water, being compressed rock beneath that. This is a major change to the question after you answered... $\endgroup$ – trichoplax Nov 24 '14 at 1:58
0
$\begingroup$

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?

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Though, this answer says the black hole does not have any influence on the escape velocity for the planet alone, since objects in orbit (as the planet around the black hole) are pretty much "weightless". As to the tidal waves, they result rather from a slight "back and forth rocking" the planet experiences due to the black hole's tidal gravity (though I don't know if that's plausible, you'd have to ask Kip Thorne for that, who thinks it is). $\endgroup$ – Chris says Reinstate Monica Dec 4 '14 at 20:35
  • $\begingroup$ So the tides on earth are caused by the same thing then? The moon rocking earth back and forth? $\endgroup$ – Viktor Dec 8 '14 at 11:11
  • $\begingroup$ No, they aren't, the tidal waves on Miller's planet are different from the tides on earth afterall. $\endgroup$ – Chris says Reinstate Monica Dec 8 '14 at 11:30
0
$\begingroup$

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.

|cite|improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.