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Is there any specific trajectory that an object can take in a black hole without it getting spaghettified? I am aware that the intance gravity of a black hole would rip apart any object if it falls into one, but would there be any points or paths in the black hole's gravitational field that does not affect the object and the object continues to move towards the singularity, undisturbed.

(Like in the movie Interstellar, the lead character played by Matthew McConaughey plummets into a black hole without getting spheghettified. I am quoting this example because of the fact that Kip Thorne was the scientific advisor for the movie and the entire modelling of the black hole scene was done by him.)

Related article:here

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  • $\begingroup$ Would an electron (indeed an object) get ripped apart? Otherwise, I'm thinking that 'safe descent' and 'black hole' just don't go together... $\endgroup$ – Jon Custer Oct 20 '16 at 14:09
  • $\begingroup$ @JonCuster I'm not sure about that, I know that atoms rip into elementary particles. But if we talk about an electron and a BH, either we are to see it as a classical particle (which it's not) into a classical BH or a quantum particle into a quantum BH. But I'm not sure how tha latter might work as I have no knowledge about quantum gravity. What do you think? $\endgroup$ – Naveen Balaji Oct 20 '16 at 14:34
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    $\begingroup$ Naveen, google on Interstellar pseudoscience. It was touted as being solid science, but it isn't. $\endgroup$ – John Duffield Oct 20 '16 at 16:05
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    $\begingroup$ The tidal forces need not be strong at the event horizon. The larger the black hole the lower the tidal forces at the horizon, so for a large enough black hole you can pass through the horizon unharmed. However the tidal forces always tend to infinity as you approach the singularity. $\endgroup$ – John Rennie Oct 20 '16 at 16:25
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    $\begingroup$ This is an interesting question. It's presumably possible to calculate the tidal force in the rest frame of the infalling observer, though offhand I don't know how to do it. But how will that tidal force depend on the geodesic they follow towards the singularity? It's far from obvious that the tidal force would be independent of the angular momentum of the infalling observer. Neither of the two (at the time of writing) answers have addressed this. $\endgroup$ – John Rennie Oct 20 '16 at 17:01
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No, there is no such trajectory for a (classical, Schwarzschild) black hole. Once you have passed the horizon then you will reach the singularity in finite proper time, and tidal forces increase without bound as you approach it.

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  • $\begingroup$ what if it's a Quantum mechanical black hole? $\endgroup$ – Naveen Balaji Oct 22 '16 at 17:52
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    $\begingroup$ @NaveenBalaji: well we have no model for such a thing that is even close to working, so I think all bets are off. $\endgroup$ – tfb Oct 22 '16 at 18:02
  • $\begingroup$ so if a classical object, like a human dives into a BH, he would get spaghettified and eventually broken down into atoms which would further break down into nucleons and electrons; now why wouldn't they break down further into quarks and leptons and so on? $\endgroup$ – Naveen Balaji Oct 22 '16 at 18:17
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    $\begingroup$ @NaveenBalaji I presume they would. But, to my knowledge (I'm not a quantum gravity person) there is no working theory which describes what happens when gravity is strong compared to the other forces. $\endgroup$ – tfb Oct 23 '16 at 10:24
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    $\begingroup$ @NaveenBalaji Don't know, sorry: that might itself be a good question here. $\endgroup$ – tfb Oct 23 '16 at 15:49
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I was also rather intrigued by that scene in INTERSTELLAR, and honestly, i am not knowledgeable enough to comment on such a theoretical situation, but let me share some on the information i gathered.

Tidal forces due to a massive body, can break apart a rigid body, due to the varied forces acting on various parts of the body, causing it to ultimately disintegrate. Suppose $M$ is the mass of the black hole, and $R$ is the radial length of the line joining its center to the center of a body of mass $m$ near it. Then , The tidal acceleration $a_t$ = 2$\Delta r$$.\frac{GM}{R^3}\hat r$ ( in the radial direction). Here $\Delta r$ denotes a the distance of a particle of the mass m measured from its center. The flesh of humans is unusually elastic, and in generally incompressible, but compared to a gravitational field that large, it can get stretched for some length before getting torn apart. The real fun starts, when $R$ gets really small(about a 1000m or so from the center of the black hole suppose). Thats when everything begins to get ripped to shreds. Reading upto this part you may think that your question has a negative answer, but thats not entirely true. Analysing further, i happened to recall a crucial statement in the movie 'If you are travelling fast enough, you may avoid the singularity and survive'. Someone in his team said this to Matthew McConaughey in the movie. See where that comes in ? Remember that the real elongation and ripping to shreds all starts, when $R^3$ is small enough to make even something as small as $\Delta r$ significant. So, if the object $m$ is aimed with an impact parameter(w.r.t the center of M) $b$ which is much larger, then $R^3$ will never have to be small enough. But for that,$m$ would need to move insanely fast, or it will never be able to miss the center(due to the huge gravitational pull). The picture looks something like this: enter image description here

Hope this helped somehow, although even i am not very satisfied with it.

PS: What i could not find out at all , is what would happen if m missed the center once. Once inside the event horizon, it cannot really escape the black hole technically(or can it?). What will happen then? Again insane speeds might save it, but that can only happen a few times, and eventually it will end up at the singularity, will it not ? Any clarifications are welcome.

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  • $\begingroup$ Thanks a lot for answering. So the impact parameter of the spaceship he was in was large so that $R^{3}$ did not have to be small. But if you recall, the ship was dead and he detached from the main ship in the movie, so he did not have high speeds as he was moving into the BH. So that doesn't make sense right? $\endgroup$ – Naveen Balaji Oct 20 '16 at 17:02
  • $\begingroup$ That part, i cannot explain. I distinctly remember the advice to 'go fast' though. Maybe when he dropped from the spaceship, it had a large enough velocity whih was imparted to him during detachment, causing the impact parameter to be non zero. $\endgroup$ – Lelouch Oct 20 '16 at 17:09
  • $\begingroup$ Oh ok. Any idea about how to calculate the speed with which they have to pass to be 'undisturbed and unaffected' by the BH? $\endgroup$ – Naveen Balaji Oct 20 '16 at 17:12
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    $\begingroup$ I can suggest a classical, non relativistic approach. Simply put, it will be a very rough estimate. For simplicity assume a schwarschild black hole. Let b denote the minimum safe distance from the singularity. Then mv^2/b = GM/(b^2) should give you the velocity at distance b. His launch velocity can br calculated from this info and some others like the lauch angle, distance of lauch , using conservation of angular momentum in central force motion. $\endgroup$ – Lelouch Oct 20 '16 at 17:30

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