How close can an observer approach the black hole in an unpowered flyby without falling into it? In classical mechanics by choosing the right trajectory you can approach a planet arbitrarily closely, if there is no atmosphere or anything to slow you down, you can approach the surface then fly away without firing your rockets.
Does the same works in general relativity with black holes? Can you do close approach of the event horizon without falling in?
(My guess is that it's the distance of the photon sphere, but it's just a guess.)
To be more specific: 


*

*observer much smaller than the black hole, so you can assume it's a point particle. And won't be torn apart by tidal forces above the horizon.

*Assume a flyby: The observer must be able to escape to the infinity unpowered.

*Assume the black hole doesn't have accretion disk. 

*I don't mind if it's rotating, but if it's simpler to make calculations in terms of a non-rotating black hole I accept it too.

 A: It's the distance of the photon sphere in the limit. The answer is implied by the Wikipedia entry for photon sphere: "Any free fall orbit that crosses [the photon sphere] from the outside spirals into the black hole. Any orbit that crosses [the photon sphere] from the inside escapes to infinity." The Wikipedia entry implies that any orbit that crosses, from the inside, a sphere that is larger than the photon sphere to an arbitrarily small degree, escapes to infinity. Then our unpowered observer can in principle get arbitrarily close to the photon sphere and still escape to infinity. At the shallowest angle of approach at the photon sphere, light gets trapped into an orbit, so that's the limit.
A: I think your guess is correct for mass-less "rockets" (a photon), but for real rockets (with mass m), the pertinent equations are the same one would use to calculate a satellite's  orbit around the earth, at an altitude = to the radius of the earth, and making the velocity a little larger.  Substitute the mass of the rocket and the equivalent mass of the black hole for the satellite and earth, respectively, and you have your answer.  
