Slingshot around a black hole, dipping into event horizon

Question

Suppose we have an object swinging by a black hole. My understanding is that if it remains outside the black hole, it just swings by and doesn't lose any energy (except maybe gravitational waves?). But if it enters the event horizon, it's gone forever. i.e. the blue trajectory in the picture is possible, but the red one isn't.

How does this work? How does the object on the red trajectory lose its kinetic energy? Is it an "all or nothing" thing where the object will retain all its energy unless it dips into the event horizon and then lose it all? Or does the object gradually lose more energy as it gets closer to the event horizon?

Edit:

Thank you to the respondents. Summary (if I've understood correctly - please correct me if not) is:

• In Newtonian physics, a small object orbiting a big object without losing energy will follow either an ellipse or hyperbola, depending on the speed. What goes down, must come up. In Relativity, this is not True - objects follow Schwarzchild geodesics - and the difference from Newtonian orbits becomes larger as we get very close to very massive objects. Neither the "red" nor the "blue" object in the below image loses its energy - in relativity, objects can spiral in without losing energy.

• Re: The all-or-nothing thing. There exists a sphere (the Photon Sphere - dotted line on below image) outside the event-horizon, where any photon entering it will eventually spiral into the black hole. So at least for massless objects, if the object dips into the photon sphere, it's in for good.
  
  Image of Schwartzchild geodesics near a black hole from General-Relativistic Visualization.

• What happens to the red object after falling past the event horizon depends on who's watching. For an outside observer, the red object will just park at the event horizon and hang out there forever. For the observer aboard the red object, it will just cross the event horizon and hit the singularity in a finite (and small) time. After that, the rules don't say what happens.

• There appears to be some controversy on that last point (at least in the answers below). Some seem to argue that an observer on the red object ceases to exist once it hits the event horizon, rather than the singularity (i.e. the left image above is also true for the observer on the red trajectory). It may be a moot point since there's no way to verify this.

Answer 1

In a plain old Newtonian context, there are some things in your question that show some incorrect understanding. Kinetic energy isn't conserved, it's total energy that's conserved. It's not true in general, for motion under the influence of a central force, that conservation of energy forbids a collision with the origin. That depends on how the force varies with distance.

Re general relativity, Wikipedia has a nice article on the orbits of test particles in the Schwarzschild spacetime: https://en.wikipedia.org/wiki/Schwarzschild_geodesics . Another good reference is the book Exploring Black Holes by Taylor and Wheeler.

A test particle in the Schwarzschild spacetime has a conserved energy and a conserved angular momentum. These are conserved both inside and outside the horizon. The energy doesn't have a nice interpretation in terms of a split into kinetic and potential terms.

As a particle infalls past the event horizon and approaches the singularity, its energy and angular momentum remain constant. The singularity represents the end of time in this spacetime, so the particle's energy and angular momentum are literally conserved until the end of time. If these remarks about the end of time don't make sense to you, I would suggest you learn how to interpret Penrose diagrams. I have a simple nonmathematical presentation of Penrose diagrams in my book Relativity for Poets, which is free online: http://www.lightandmatter.com/poets/

Answer 2

As long as the object is outside of the event horizon it will not lose kinetic energy (disregarding any gravitational wave phenomena). You can think of this case as an elastic scattering of two bodies.

If the object falls inside the event horizon, it will become a part of black hole and will not go out. Let's assume that the object has no charge and falls straight into the black hole so that it doesn't add any angular momentum to the black hole. The falling object will increase the size of the black hole according to the first law of black hole thermodynamics:

$$ d E = \frac{1}{32 \pi M} d A $$

where $d E$ is the change in energy and $dA$ is the change in the horizon area of the black hole. Hence, in both cases energy will be conserved. The extra kinetic energy of the object will be compensated by an increase in the gravitational potential of the black hole.