14
$\begingroup$

The movie Interstellar shows people on a water planet where time is dilated so much that 1 hour is equal to 7 years back on Earth. Even though they lift off from Earth using a Saturn-V two stage rocket, they leave this water planet in a shuttle craft.

Can someone provide the escape velocity for a place where time is dilated by 60,000 times? The stated mass of the black hole creating the gravity well is 100 million solar masses, but I'm not sure that would work into the calculation.

|cite|improve this question
$\endgroup$
  • 2
    $\begingroup$ First of all, the movie is so full of sciencemagicwoo that it's hard to apply actual physics for explanations. But for your specific question, remember that the time dilation is not "felt" by the local observer, so they only need to calculate the gravitaional field (and its change with altitude) to calculate the required escape velocity. $\endgroup$ – Carl Witthoft Nov 18 '14 at 13:24
  • 4
    $\begingroup$ @CarlWitthoft His question is perfectly fine, and the physics here is perfectly well-defined. I don't understand your objection. $\endgroup$ – Jold Nov 18 '14 at 14:14
  • $\begingroup$ @jld I'm only objecting to the movie, not his question. $\endgroup$ – Carl Witthoft Nov 18 '14 at 14:42
  • 3
    $\begingroup$ @Carl Witthoft - Most of the stuff in the movie was calculated to obey general relativity by Kip Thorne, see his book *The Science of Interstellar. Even the most speculative stuff was loosely based on real theories--the idea of a higher-dimensional space was inspired by brane models, and the idea that beings in this dimension could alter the local gravitational constant in our own 3D brane was based on the idea that the strength of gravity depends on the curvature of a large extra dimension. $\endgroup$ – Hypnosifl Nov 18 '14 at 17:05
  • 2
    $\begingroup$ (cont.) Of course the most speculative part is probably the idea that the beings in this higher dimension had the ability to send gravitational signals backwards in time, but Thorne said he was assuming that the inside of the "tesseract" functioned similarly to a traversable wormhole, a GR solution that he had discovered could theoretically be used for backwards time travel (see this paper he coauthored). $\endgroup$ – Hypnosifl Nov 18 '14 at 17:30
7
$\begingroup$

The gravity from the black hole (BH) will have no effect on their ability to take off from the water planet itself. Objects in orbit feel weightless (think astronauts in the ISS). If you're only worried about getting off the water planet then there should be no problem.

However, if they were to try to put some distance between themselves and the BH then they'd find it a much more difficult task.

Assuming the BH isn't rotating very fast, the time dilation factor for a body in orbit relative to a stationary observer at infinity is:

$$\frac{d \tau}{dt} = \sqrt{1-\frac{3GM}{c^2 r}}$$

The escape velocity of a BH looks the same as it does in Newtonian mechanics:

$$v_0^2 = \frac{2GM}{r}$$

So if the mass of the BH is ~100 million solar masses, the radius at which the time dilation is 60,000 times normal is at $r \approx$ 275 million miles $\approx$ 3x average Earth-Sun distance. The event horizon itself is at $r \approx$ 2x the average Earth-Sun distance. Meanwhile the escape velocity at the planet radius is about 82% the speed of light, or $v_0 \approx$ 250 million meters/second.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z♦ Nov 19 '14 at 11:53
  • 1
    $\begingroup$ Hello fellow SEers. On an answer to a related question over on Movies & TV there is currently a bit of confusion about your answer and the fact if you implied anything about density with the term "water planet" (which I think you didn't) and how the escape velocity of that planet could actually be lower than that of earth, given the 1.2 ratio of their respective gravities (which I think is only possible if the planet has a higher density than earth). Maybe you could clarify this up somehow. $\endgroup$ – Christian Rau Nov 23 '14 at 20:50
  • 1
    $\begingroup$ @ChristianRau I replied to the post :). $\endgroup$ – Jold Nov 24 '14 at 5:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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