- #1
cyborg6060
- 10
- 0
When classically deriving the Schwarzschild radius of a black hole, the kinetic energy of an outgoing particle (moving at the impossible-to-achieve maximum of the speed of light) is equated with the gravitational potential of the black hole at that point.
[itex]\frac{1}{2}mc^2 = \frac{GMm}{r}[/itex]
[itex]r=\frac{2GM}{c^2}[/itex]
This seems reasonable for justification that light cannot escape, but what about objects that can do work or exert forces?
The gravitational force of the black hole at the event horizon is:
[itex] F = \frac{GMm}{r^2} [/itex]
One could imagine a spaceship throwing off fuel in the opposite direction so that
[itex] \frac{Δp}{Δt} = \frac{GMm}{r^2} [/itex].
I'm well aware that this is not the case, so I'm curious as to what the reasoning is. Thanks for the help!
[itex]\frac{1}{2}mc^2 = \frac{GMm}{r}[/itex]
[itex]r=\frac{2GM}{c^2}[/itex]
This seems reasonable for justification that light cannot escape, but what about objects that can do work or exert forces?
The gravitational force of the black hole at the event horizon is:
[itex] F = \frac{GMm}{r^2} [/itex]
One could imagine a spaceship throwing off fuel in the opposite direction so that
[itex] \frac{Δp}{Δt} = \frac{GMm}{r^2} [/itex].
I'm well aware that this is not the case, so I'm curious as to what the reasoning is. Thanks for the help!