Submitted by Sol33t303 t3_11g1vky in askscience
My basic understanding of orbital mechanics would suggest that this should be possible right? I assume there's nothing to get in your way to slow your orbit in the event horizon. In a normal orbit you gain speed as you approach periapsis, then that speed flings you back out to apoapsis. I don't see any reason this wouldn't apply in the event horizon of a black hole.
> My basic understanding of orbital mechanics would suggest that this should be possible right?
No, because it would require the velocity to exceed the speed of light when you're approaching the periapsis. This is where the "classic" calculations fail. In "regular" orbital mechanics problem the closer to periapsis you are, the faster you're moving, and this way you "climb up" from the gravity well. However in reality there is a limit of how fast you can move, and in your example you'd have to exceed the speed of light in order to climb up from the gravity well after crossing the event horizon, and this is not possible.
> it would require the velocity to exceed the speed of light when you're approaching the periapsis
No! Once you pass the event horizon your worldline will terminate at the singularity. There is no way you would be able to exit the black hole, even with a hypothetical engine that provides infinite thrust.
Exiting the black hole is equivalent with going back in time. Approaching the singularity is equivalent with going forward in time. How do you accelerate away from tomorrow?
An object traveling faster than the speed of light is going backwards in time in some frames, so there's not actually any disagreement between "going back in time" and "it would require the velocity to exceed the speed of light".
>An object traveling faster than the speed of light is going backwards in time in some frames
Where are you taking this information from? This "follows" from special relativity, where massive objects travel at strictly less than c.
Such statement don't have any physical meaning. You're braking the assumptions that were used to derive the equations you're relying on.
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Consider Newtonian gravity. If an object falls directly into the gravitating body with no sideways motion, it will simply collide. It's orbital angular momentum that causes the object to be ejected back outward.
How does this work? If you write down the equation of motion for the orbital distance r, two forces emerge. One is the gravitational attraction, which scales as 1/r^(2). The other is a centrifugal repulsion term, which scales as L^(2)/r^(3), where L is the angular momentum. As the distance r becomes small, the centrifugal repulsion eventually dominates, ejecting the object back out to apocenter, as you say.
This works because the attractive force scales as 1/r^(2) at distance r. If the attractive force scaled as 1/r^(3) or steeper, then the centrifugal repulsion would be no longer guaranteed to overpower the gravitational attraction at sufficiently small radii, so there would be nothing to prevent the orbiting object from eventually colliding with the central gravitating body.
While gravity in general relativity can't be exactly described by a radial force law, the same basic idea applies. See for example how a number of 1/r^(3), 1/r^(4), etc. terms arise in the post-Newtonian expansion (scroll to equation 203).
That's true for nonrotating black holes, anyway. In the idealized rotating black hole solution, it is actually possible for the centrifugal repulsion to overpower the gravitational attraction! This is what leads to the crazy conformal diagram for a rotating black hole that suggests you can fall in and emerge back out in a different universe. However, there are many good reasons to expect that this idea does not work for realistic black holes and is just an artifact of the idealized construction.
Found the answer online.
A Kerr black hole (one that rotates around its axis), has an ergosphere in which you can enter and get out because of the Lense-Thiring effect. Basically, the rotation of the black hole drags space-time and changes the frame of reference in which you move. Therefore if an object passes into the ergosphere it can still be ejected by gaining energy from the rotation.
But! If anything passes the event horizon, the surface limit where the escape velocity is the speed of light, then no material thing can escape.
It's on daviddarling.info.
Right, the infalling object can't escape into our universe after it crosses the event horizon. In the idealized black hole construction, the infalling object will, however, escape into another universe (via a white hole) if it avoids hitting the singularity.
It wouldn't be an event horizon if you could get out.
Newtonian mechanics doesn't have event horizons so it's important to consider general relativity here. If you are inside the photon sphere (1.5 times the radius of the event horizon for a non-rotating black hole) then tangential motion makes you fall in faster. Every attempt to orbit there makes you spiral into the black hole.
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> any object that crosses the event horizon is moving at or faster than the speed of light relative to an observer outside the black hole [..] In summary, while it is possible in theory to cross the event horizon of a black hole and orbit it
But hasn't the first bit just explained exactly why that's not possible, not even in theory?
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I'm pretty sure that this is when you get "spaghettified."
Basically, the rate at which gravity changes (as a function of distance from the black hole) is so huge, that the gravitational force on your feet would be so much greater than that force on your head, that you would be stretched like taffy...or, um, spaghetti.
If the black hole is large enough (like the 4 million solar mass at our galaxy center), the point where tidal forces become unbearable lies way inside the event horizon. You could fall through the event horizon and not even notice unless paying attention to your navigation instruments. You'd continue to fall for the next 30 seconds blissfully unaware that there is now no chance to escape. You WILL get spaghettified in the last second before impact with the singularity though.
No. Just...no.
You're completely ignoring the fact that once you cross the event horizon, you are effectively pulled to the "surface" of the "body". This is an effect which is inescapable, which is why a black hole is not a planet or star and should not and cannot be thought of as such.
cygx t1_jany3wu wrote
No. The solutions to the relativistic version of the Kepler problem are different from the Newtonian ones:
You still have hyperbolic-like, parabolic-like and elliptic-like solutions. However, the parabolic ones can loop around the black hole a couple of times before going off to infinity, and the elliptic ones will have their perihelion precess around it. Additionally, you get solutions that cross the horizon and never come back out, eventually hitting the singularity, and trajectories that asymptotically approach a circular orbit.