Black Hole surface area at Schwarzschild radius is half? 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?
 A: The Schwarzschild radius $r$ is not simply the distance to the centre of the black hole. If you measured that distance by letting down a tape measure you'd find the distance was substantially greater than $r$. See for example my answer to How much extra distance to an event horizon?, where I do this calculation.
We actually define $r$ to be the circumference of a circle centred on the black hole divided by $2\pi$ (i.e. the circumference is $2\pi r$). That means the area of a sphere of radius $r$ centred on the black hole is $4\pi r^2$ by definition. So the surface area doesn't go to zero then go negative. Sorry :-)
The time dilation at the distance $r$, compared to an observer at infinity is:
$$ \frac{t_r}{t_\infty} = \sqrt {1 - \frac{2GM}{r c^2}} $$
Note that this goes to zero at the event horizon i.e. time slows to a stop there. You'll find lots of Q/As on this site that discuss this further.
Finally, your analysis of the orbital period is flawed because you're using the Newtonian expression for the orbital velocity:
$$ v_{Newt} = \sqrt{\frac{GM}{r}} $$
and this doesn't apply to the curved spacetime around a black hole. Calculating the orbits in curved spacetime is a rather more involved procedure. For massive objects there are no stable orbits closer than three times the Schwarzschild radius i.e.
$$ r \lt \frac{6GM}{c^2} $$
Even light cannot maintain a stable orbit for $r \lt 3GM/c^2$.
