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Black Hole's "Point of No Return" Found 130

dsinc writes "Using a continent-spanning telescope, an international team of astronomers has peered to the edge of a black hole at the center of a distant galaxy. For the first time, they have measured the black hole's 'point of no return' — the closest distance that matter can approach before being irretrievably pulled into the black hole. According to Einstein's theory of general relativity, a black hole's mass and spin determine how close material can orbit before becoming unstable and falling in toward the event horizon. The team was able to measure this innermost stable orbit and found that it's only 5.5 times the size of the black hole's event horizon. This size suggests that the accretion disk is spinning in the same direction as the black hole. The observations were made by linking together radio telescopes in Hawaii, Arizona, and California to create a virtual telescope called the Event Horizon Telescope, or EHT. The EHT is capable of seeing details 2,000 times finer than the Hubble Space Telescope."
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Black Hole's "Point of No Return" Found

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  • by mbone ( 558574 ) on Sunday October 14, 2012 @12:34PM (#41649999)

    The event horizon and the innermost stable orbit have a band of space between them. What happens if you go there?

    First, these regions near a black hole tend to be very nasty for our kind of life. Lot's of radiation, and the tidal stresses will kill you for a solar mass black hole. So, suppose you have a multi-billion solar mass black hole to play with, lots of shielding, and a super rocket as well. The ISCO orbit will be about 2 days in that case.

    Could you make a regular orbit inside the ISCO? Yes, in principle, down to the Innermost _Unstable_ circular orbit, AKA the "photon orbit," as this is (at 1.5 Schwarzchild radii) where photons would orbit. It's unstable, so you will need to maneuver frequently to not fall into the black hole.

    Below the IUCO, you have to fire your rockets constantly to avoid being sucked in. Better not run out of fuel !

    A movie is worth a lot of words, so here are some movies of orbiting a back hole ISCO [colorado.edu].

  • Re:Unstable? (Score:5, Interesting)

    by Anonymous Coward on Sunday October 14, 2012 @12:59PM (#41650147)

    Comment posting limits (and time...) won't let me respond to many individual comments, so I will see if I can address a few things at the same time here.

    For a given angular momentum of something going around a black hole, you can work out what potential energy it would have at different radii. In a normal Newtonian case, you can think of having some satellite orbiting at some speed. If you try to push that satellite further in, while still maintaining its angular speed, it will try to pop back out since it is essentially going too fast to orbit at a smaller radius. There is a minimum in the potential energy of the satellite where it would have a circular orbit for that given angular momentum, as it would just stay at that radius. The potential energy about this radius would be like a bowl, if you push the satellite inward, it would roll back down toward the radius corresponding to a circular orbit. Momentum would of course carry it beyond that point, so it would oscillate in radius between some place closer and some place further from the circular orbit. This would give you an elliptical orbit where the radius goes between two values. The potential energy for over radius for a given angular momentum would look roughly like the red curve in the image here [oregonstate.edu].

    Now, for a black hole, GR gives some differences from Newtonian gravity when you get closer. The potential energy curve now looks more like this [amazonaws.com]. There is still a stable orbit, as you can see it could oscillate around the minimum there like a marble in a bowl. In other words, small pushes on a perfectly circular orbit will turn it into a slightly elliptical orbit that is still pretty close to the circular one. However, if you push it far enough inward to get over that bump, the orbital radius would be like a marble just rolling down that hill toward the black hole. Now, the size of that bump changes depending on what angular momentum you are talking about. As you increase the angular momentum, which in Newtonian gravity would just give you a smaller radius for a circular orbit, that bump gets smaller. There is a point where the bump goes away, such that you just now have a curve that decreases with decreasing radius. Hence, a particle in such an orbit would continue to move closer to the black hole, as there is lower potential energy the closer it gets.

    This is all due to the geometry of space around a black hole. Weird stuff like the circumference of a circle not being 2 pi r depending on how you measure the r from the black hole, which is why orbits no longer have the same stability they have in Newtonian gravity. This is not an effect due to gravitational waves. The orbiting particle can be something like a proton where the gravitational waves would be too small to matter. However, if you are talking about the orbit of a massive object, like a star or second black hole, then the gravitational waves become significant. In that case, the orbit at any radius would slowly decay due to emitting gravitational waves. Once the decay orbit hits the radius of the innermost stable orbit, the decay would greatly accelerate.

    This is also not an effect of rotation or frame dragging, as it happens with a non-spinning black hole solution too. However, spinning black holes and frame dragging do factor into it, such that for a spinning black hole, the inner most stable orbit is smaller if you are going in the right direction around the black hole. Although there are other effects that the frame dragging causes. You get things like the ergosphere, a region where due to frame dragging, you would have to go faster than light to look stationary from an outside viewer, so all matter within that region is spinning around the black hole.

    This is also quite distinct from the event horizo

  • by PNutts ( 199112 ) on Sunday October 14, 2012 @04:05PM (#41651433)

    Thanks for that. Not everyone can view /. via a home computer screen with mouse. I was going to try to meticulously "select text" myself on my 2 1/2" smartphone screen and post the link, an excercise in futility at times.

    First world problems.

    Besides not being funny any more, your statement demonstrates a lack of knowledge of mobile devices in developing countries.

In seeking the unattainable, simplicity only gets in the way. -- Epigrams in Programming, ACM SIGPLAN Sept. 1982