For practical purposes, I'm considering a non-rotating black hole or neutron star, because I find Kerr black holes a little confusing.
The ratio between orbital velocity and escape velocity in Newtonian physics is
$$ \frac{v_\text{escape}}{v_\text{orbit}} = \sqrt{2}$$
I've read several times that the escape velocity at the event horizon is exactly the speed of light $c$, mentioned here, among other places.
At the photon sphere at 1.5 times the Schwarzschild radius, the orbital speed is $c$, so a photon can orbit the black hole. But this is weird, because it seems to be the case that near a black hole orbital speed is greater than the escape velocity.
Can this be explained in layman terms? Is there an error in my train of thought?
If somebody wants to do the math, that's cool, but I'm not sure I'd understand it.
Assuming the premise of the first question isn't flawed, does this happen around a dense neutron star as well? Dense neutron stars may also have photon-spheres. Wiki says it's possible, Andrew says it's not. Whether neutron stars can or can't make photons fly in circles isn't strictly relevant to this question, but they are dense enough to not be far off and perhaps get relativistic effects and, perhaps, that funky relation where orbital velocity approaches and even surpasses escape velocity (which makes absolutely no sense).
Is there a neat and tidy mathematical relation between orbital speed and escape velocity for these ultra-high gravity situations?
Best Answer
The local escape velocity is
$$v_{esc} = \sqrt{\text{2 G M}/\text{r}}$$
At infinty you observe that velocity slower by a factor of
$$1-\text{r}_s/\text{r}$$
so at infinity you observe
$$\text{v}_{esc} = \sqrt{\text{2 G M}/\text{r}} \cdot (1-\text{r}_s/\text{r})$$
because of gravitational length contraction radial to the mass and gravitational time dilation in all directions.
The local orbital velocity is
$$v_{orb} = \sqrt{\text{G M}/\text{r}}/\sqrt{1-\text{r}_s/\text{r}}$$
which is at infinity observed to be slower by a factor of
$$\sqrt{1-\text{r}_s/\text{r}}$$
At infinity you observe simply
$$\text{v}_{orb} = \sqrt{\text{G M}/\text{r}}$$
So the observer at infity should observe an orbiting particle orbit with it's newtonian velocity, while locally this velocity is higher.
The Plots show the velocities in terms of v/c and the Schwarzschild r coordinate in units of GM/c². Left is the system of a local observer, and right like an observer at infinity would see the Shapiro-delayed velocities.
Locally the escape velocity equals the orbital velocity at r=4GM/c², while at infinity their equality is observed at r=2·(2+√2)=6.8284GM/c².
At the photon sphere a local observer would observe the orbiting particle with c, while an observer at infinity would measure it slower by a factor of √(1-2/3), so with 0.577c.
An escaping particle near the event horizon would need a local radial velocity of c, and seem to have zero velocity for the observer at infinity.
The different factors for the radial and the transversal components is due the graviational length contraction which is only in radial direction (there is more radius inside the circle than the circumference divided by 2π).
To sum it up:
Locally, the radial escape velocity for Einstein is the same as for Newton. At infinity it is observed slower than that.
Also locally, the angular orbital velocity for Einstein is higher than for Newton, but at infinity it is observed to be the same as it would be under Newton.
The escape velocity can be lower than the orbital velocity, therefore orbits near the photon sphere are unstable in a sense that if your velocity is not only transversal but splits into a transversal and a radial component you will escape to infinity. With Newton you would just get an elliptical orbit with everything else staying the same.
Radial:
Transversal:
Comparison with Newton:
The index is in german, but I'm sure you'll find the local $v$ and the external $\text{v}$ (the latter with Shapiro delay).
Further reading: Equation of Motion and Geodesics, page 4, eq(9)
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