I have been interested in black holes for some time, and am still trying to wrap my head around some of their more obscure properties.
Now I know that the Schwarzschild radius is $r= \frac{2GM}{c^2}$, and my knowledge of orbital mechanics tells me that the orbital velocity at the Schwarzschild radius is $v= \sqrt\frac{GM}{r} = \frac{1}{\sqrt 2}c$ .
This would also suggest that the time dilation and compression of space at the Schwarzschild radius is $\frac{1}{\sqrt 2}$ compared to flat spacetime, using $t'=t(\sqrt{1-\frac{v^2}{c^2}}$
This compression of space would then suggest that the surface area at the Schwarzschild radius is actually $A_s=2\pi r^2$, or half of the surface of a sphere for the same radius without a singularity in the centre.
Playing with this idea of compression of space, I wanted to know if there is a way to work out the "surface area" of spacetime at a given distance from the singularity. The best I could come up with was $A_s=(4-\frac{2r_s^2}{r^2})\pi r^2$ .
What I found interesting about this is it made 2 "predictions":
At the radius $\frac{r_s}{\sqrt2}$, the surface area is 0.
As you approach the singularity, it predicts a negative surface area, which approaches the inverse of the Schwarzchild surface area as the radius tends towards 0.
Am I just clutching at straws trying to work on this or is there something to my equation?
It seems to suggest a black hole is actually an inverse sphere contained in a sphere, and that the "singularity" is acually a sphere with no surface located at the radius $\frac{r_s}{\sqrt2}$.
Can someone show me where I went wrong in my understanding?