# Safe distance from a black hole?

Okay so we all know that one planet from the movie "Interstellar" that orbited the black hole Gargantua. How do I NOT get that? What's the minimum safe distance for a planet to be from a black hole ("safe" as in no funky relativistic time dilation and no megatsunamis)?

• "Okay so we all know that one planet...". Actually: no... I have not seen that film. I am most likely not the only one. Kindly edit your question to not assume people know what you are talking about when you refer a specific element to a specific film. – MichaelK Jun 23 '16 at 8:13
• Ah, well, this sounded like a question I might be able to answer but unfortunately I can't understand the question, having not seen the movie that the question is contained in. – Devsman Jun 23 '16 at 15:42
• Interstellar - Another film that was pretending to be scientifically acurate while in reality it features severe pseudoscientific facts further hindered by severe lack of logic. At least it was not so bad as 2012. – Victor Stafusa Jun 23 '16 at 18:11
• @VictorStafusa The astrophysicists consulting for Interstellar published several papers based on a groundbreaking new method of visualizing a black hole that was developed for use in the film (the most accurate visualization ever). So, while there were some handwavium moments for the magic of film, to take the tone that it was nothing more than pseudo science is ignorant, at best. – TylerH Jun 23 '16 at 19:34
• @TylerH I didn't say that it is "nothing more than pseudo science". I said that it "features severe pseudoscientific facts". Those two claims are very different. There is a lot of good science there indeed, but unfortunately, it also got mixed with a large chunk of pseudoscience. – Victor Stafusa Jun 23 '16 at 19:43

A black hole does not have any magic properties, it does not "radiate" time dilation or any other nonsense like that. Noticable time dilation happens when one observer moves at relativistic speeds with reference to another, or when one observer is under much higher gravitational acceleration than another.

In the movie "Interstellar", the crew of a spaceship landed on a planet that orbited a supermassive black hole. Being on the surface of that planet is said to cause a time dilation of 7 years on Earth for each hour on the planet surface, in other words a Lorentz factor of > 61,000.

The scientific problem with that is that the numbers don't nearly add up. The planet itself has 1.3g surface gravity. That causes a negligible time dilation compared to Earth.

That only leaves relativistic speed compared to Earth as the source of time dilation. For a Lorentz factor of 61,000 and ignoring the effect of the planet's surface gravity, you would need a relative speed of 99.999999987% of c, the speed of light. That is how fast the planet would have to orbit around its black hole in order to cause the time dilation presented in the story, and is obviously impossible. Even if you had a planet rotating that quickly, in order to land on it, you would have to match its orbit, i.e. you would at least have to accelerate your spaceship to the same speed.

The movie got a lot of things right, and even caused some scientific papers to be written, but that bit about time dilation, while necessary for the story, was the worst kind of nonsense: Scientifically sound, but the numbers were way, way off.

• I thought it was the closeness to the black hole that caused the time dilation, not the gravity of the planet itself... – colmde Jun 23 '16 at 10:27
• @Selenog My whole point is that the astronauts don't experience the gravity of the black hole and are not affected by it. If they did, time dilation would be the least of their worries. The dominant gravitational force, as depicted in the movie, came from the planet and was 1.3g, which is not nearly enough to cause that much time dilation. And yes, the spacecraft should have experienced pretty much the same time dilation, because it was at most a few hundred kilometers away from the astronauts. Another reason why the science failed on that part. – Hackworth Jun 23 '16 at 12:54
• I believe the astronauts are affected by the black hole's gravity, just as much as the planet is. The only reason the "dominant gravitational force" they experience is the planet's is because their frame of reference (the planet) is accelerated by the same gravitational force as the astronauts on it. So yes, the black hole's gravity should cause time dilation relative to an observer that isn't as close to the black hole as the planet. How much farther away the Endurance was and how much less affected it would be, I can't say though. – Martin Ender Jun 23 '16 at 14:44
• @Hackworth The example with the sun doesn't matter. This central body's force on the planet's surface can easily be larger than the planet's own force. The sun's gravitational acceleration on the Earth's surface could be 10g and you still wouldn't feel it or fall off the planet as long as you're orbiting together with the Earth. As long as the gravitational force form the SMBH doesn't vary significantly on the length scale of the planet (which would tear it apart), it can be arbitrarily strong without affecting anything but the orbital speed and of course the time dilation. – Martin Ender Jun 23 '16 at 15:26
• This answer completely misrepresents how gravitational time dilation works. Gravitational time dilation is due to gravitational potential; it doesn't matter whether you "experience" the gravity as a normal force from whatever surface you're standing on, or if you're in orbit, or even if you're standing between two masses and their pull cancels out. – user2357112 Jun 23 '16 at 17:25

Orbiting a black hole and you don't want extreme tidal forces? This depends on several things, including:

• The mass of the black hole; the Interstellar one looks like it's in the largest category, a supermassive black hole, because you can actually see the the black hole is much larger than a planet in front of it. This puts in anywhere from $10^{5}$ to $10^{10}$ solar masses. (If it was about the same size as the planet, it would still weigh 1000 ($10^3$) times as heavy as our sun!)
• The mass of the planet in question is pretty unknown, but I'm assuming earth-like mass. ($~5.7 * 10^{24} kg$)

Now, let's talk about tidal forces. Wikipedia goes through some derivations of the tidal force, but eventually ends up with

$$\vec{a}_g \approx \hat{r}*2 \Delta r* G*\frac{M_{bh}}{R_{bh}^3}$$

with:

• $a$ being the acceleration of the tidal force
• $G$ being the universal gravitational constant
• $\Delta r$ being the distance from the planet's center to the the surface
• $M_{bh}$ is the mass of the black hole
• $R_{bh}$ is the orbital radius of the planet to the black hole
• $\hat{r}$ just lets us know the force is in the direction of the black hole.

but wait, this gets easier! We only want the orbital radius, and Earth-like tidal forces on an Earth-sized planet. That means we want to solve the following for $R_{bh}$

$$\frac{M_{bh}}{R_{bh}^3} = \frac{M_{sun}}{R_{earth}^3}$$

Just looking at this, I can tell you that $R_{bh}^3$ needs to scale with the size of your black hole, producing an extra $10^5$ times $R_{earth}^3$. $\sqrt[3]{10^5} \approx 46$, so the planet is about 46 AU out from the middle of the black hole.

• Thanks mate! That's really helpful. How would those numbers change for a planet slightly smaller than Earth orbiting a stellar mass (315 sol) black hole? Or rather, a planet slightly smaller than Earth orbiting a star that orbits a stellar mass black hole? :) – Z.Schroeder Jun 23 '16 at 13:31
• @Z.Schroeder Yep, the numbers would change; the equations hold for all two-body systems. You could assume a Star+Small black hole acts like a really massive thing at the center of mass for the star+black hole system, reducing that three body problem to a two body. This only works if the sun/black hole are really far away. – PipperChip Jun 23 '16 at 16:31
• @Z.Schroeder, if I've done the math correctly, a 315 sol black hole would have Earth-like tidal forces at an orbital radius of 6.8 AU. – Mark Jun 23 '16 at 20:53

Orbiting a black hole is no more or less dangerous than orbiting a small sun: You get too close, bad things happen. If your orbit decays without correction, you are going to be obliterated.

As long as you properly insert yourself into an orbit at a safe distance at the appropriate velocity, there is no particular danger from a black hole.

And leaving is no special challenge either. By being in orbit you are already at escape velocity, so you don't need any particularly special 'oomph' to get back home. [EDIT: Sorry, I was wrong here. Your speed is about 0.7 of what you would need for escape, but it's in the ball-park!]

Now a planet close to a black hole might be a more complex story.

But don't believe the hype of 80's SciFi - You won't inevitably have to hold the rails of your starship as your head gets stretched. Unless of course, you get too close...

There was an episode of Battlestar Galactica where the Cylons were orbiting a black hole and my brief reaction was "How ridiculous!", but then I thought about it, and concluded "Why not!?!".

I will trust @PipperChip on the math!

• "By being in orbit you are already at escape velocity" - no you're not. If you were at escape velocity, you would be on an escape trajectory, not orbit. – user2357112 Jun 23 '16 at 17:20
• @user2357112 - I am no orbital mechanics expert, but I can read Wikipedia: "this object's speed at any point in the orbit will be equal to the escape velocity at that point". There is a difference between "at escape velocity" and "above escape velocity". In stable orbit, you are precisely "at". In mathematical terms - 0.001 joule less energy and you are decaying, 0.001 joule more energy and you are escaping (I believe...) That is what is special about a stable orbit - You are sitting on the knife-edge of escape velocity. Neither decaying, nor escaping. – spechter Jul 25 '16 at 3:33
• No, you've completely misread that section of the article. That section describes a particular type of escape trajectory sometimes referred to as an escape orbit, which occurs in an ideal 2-body system when there is exactly enough energy in the system for escape. With less than escape velocity, you are not decaying; orbital mechanics do not work that way. Less than escape velocity produces an elliptical or circular orbit (or a crash if your path goes through the central body). – user2357112 Jul 25 '16 at 4:02
• (I don't know whether using the term "escape orbit" to refer to such parabolic trajectories is actually common; I could only find one source using such terminology, which the Wikipedia article seems to have plagiarized.) – user2357112 Jul 25 '16 at 4:05
• OK, I concede. But is it true that in a stable circular orbit, your "speed" is equivalent to the "speed of escape velocity at that same altitude" - Just in the wrong direction? (90 degrees off) That's what I read from that Wikipedia section. My main point still stands though - There is nothing magical about orbiting a mass after it tips over to being a black hole. – spechter Jul 27 '16 at 0:18

Yes, orbiting a two solar-mass black home should be no different to orbiting a two solar-mass star. Stay at a sensible planetary distance and you will be OK ...

... unless or until something falls into the back hole. The process of its being chewed up and eaten will convert a good fraction of its mass into energy, much of it gamma rays. Most of that energy will be tightly beamed more or less perpendicular to the plane of the accretion disk into which that object will be transformed. So an orbit at right angles to the disk would be especially unhealthy.

If the black hole has "hoovered" its stellar neighbourhood many aeons ago and there's no active accretion disk when you arrive, you are probably safe.

Edit ... should have said, something large falls into it. The sun converts 5 million tons of mass into energy per second(!), most of which comes out as heat and light rather than gamma rays. But at a planetary distance I don't think a few tons of rock being converted into gamma rays will be lethal.