I'm making a simplified animation for college project on how a star reacts with a black hole. I had known that a black hole sucks in everything and thus should be doing the same for stars. But upon my research I found out that recently a discovery was made that a super-massive black hole ripped apart a star. Now I've always been curious about cosmology and this took my attention and also worried me for my assignment. There were two conditions with the above mentioned calamity - one, the black was a super-massive one and second, the victim star was possibly already stripped off it's outer gaseous by the same black hole. My question is that the ripping apart happens in all conditions and the sucking in thing was wrong or it happens only in mentioned circumstances?

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    $\begingroup$ NOTE: I'm not asking for help for the animation. My question is for the science only. $\endgroup$ Apr 29, 2015 at 10:11
  • $\begingroup$ You might want to look at Shoemaker Levy crashing into Jupiter (videos and simulations of that shouldn't be hard to find). The main things to consider are the Roche limit of the black hole and the size ratio, star to black hole. A super-massive black hole would likely be larger than most stars. Once inside the Roche limit, the stars gravity loses and it crumbles like an 8 day old biscuit in a hungry dog's mouth, or, well, more accurately, what happened to shoemaker levy happens to the star. What that looks like - well, it would put on quite a show. The angle of approach matters too. $\endgroup$
    – userLTK
    Apr 30, 2015 at 1:50

1 Answer 1


A black hole (BH) never sucks anything in. 'Sucking' requires gas pressure, but a black hole only acts via its gravity. A non-rotating (=Schwarzschild) black hole attracts all massive objects, very much like the Sun attracts the planets. Yet, the planets are not falling into the Sun. This is because the planets move on near-circular orbits when the centrifugal force balances the gravitational attraction. The planets orbital angular momemtum is conserved (because the Solar force field is purely radial [to good approximation]) and these orbits are stable.

The difference to a black hole is that very close to the event horizon no such stable circular orbits exist anymore. At those distances, nothing can orbit the BH without falling in.

However, an object orbiting a BH on a stable orbit (including passing trajectories) is still subjected to the BH's tidal forces. These distort the object, very much like the Moon's gravity distorts the Earth, generating tides. In case, the trajectory passes sufficiently close to the BH and the object is rather big and fluffy, like (some) stars, the tidal forces may not merely distort the object, but rip it apart.

Such tidal disruption of stars must happen if a star passes close to a supermassive BH. This can hardly be directly observed, but the it is thought that some of the stellar matter forms an accretion disc around the BH and produces a characteristic light curve. An observed light curve in agreement with this model is often interpreted as circumstantial evidence for a stellar disruption event.

  • $\begingroup$ You talked about non-rotating black holes, what happens with the rotating ones? And what happens after a black hole rips apart an object? Does it simply drift in space or get pulled in the black hole. $\endgroup$ Apr 30, 2015 at 4:10
  • $\begingroup$ Rotating BHs will make things a bit more complicated but the explanations given above remain valid. Once the object is ripped apart, as you say, the same phenomena, again, apply for every individual part: stable versus non-stable orbita, tidal disruption etc. $\endgroup$ May 2, 2015 at 10:39
  • $\begingroup$ @SwapnilRastogi As I said in the answer, some of the stellar matter [from the star ripped apart] forms an accretion disc around the BH. This is essentially all matter that is gravitationally bound to the BH. The remainder escapes. If the star was initially on a parabolic orbit (which to good approximation is the typical situation), half of the gas from the star will accrete and the other half escape. $\endgroup$
    – Walter
    May 3, 2015 at 17:25
  • $\begingroup$ Alright, that sums up my doubt, thanks for the explanation! $\endgroup$ May 5, 2015 at 6:57

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