# Slingshot around a black hole, dipping into event horizon

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/

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.