frame-dragginggeneral-relativitygravitysatellitesvelocity
We all know about the spin of black holes. That spin and of course it's gravity determines the minimum distance a star can orbit the black hole with a min orbital velocity.
Do we see similar result? Does spin of earth also effect the orbital velocity of its satellites?
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)
The transit time across your field of view would depend on your position on the surface of the earth as well as the speed and altitude of the satellite. That gets really complicated. I'm thinking that to estimate the altitude, you should observe it passing overhead several times and use your best estimate of the period of revolution to calculate the altitude. Don't forget to allow for your rotation with the earth.
Best Answer
The effect of black holes on prograde and retrograde orbits is due to frame dragging, which is a general relativistic effect. Hence the effect is microscopic on Earth satellites.
The LAGEOS satellites tried to detect frame dragging around Earth, but did not succeeded directly. NASA has claimed success with GRACE, though. A more recent paper used LAGEOS and other satellites to find the effect. The drift due to the effect is about 30.68 milli-second of arc per year, so it is totally minuscule.
The big source of noise in these measurements is tidal effects due to the lumpiness of Earth, which is also changing over time due to other bodies. That brings up a bigger, classical effect on satellites that does matter in practice. A rotating planet with a tidal bulge will pull and push on the orbiting satellite depending on how the orbital period fits with the planetary rotation. There is also a difference in how easy it is to capture natural satellites in prograde and retrograde orbits (classic paper). Retrograde orbits average the bulginess more, and are somewhat more stable, while prograde orbits tend to change more.
For big enough satellites (or rather moons), the bulges they induce themselves matter. If they are above prograde geosynchronous orbit the bulge will tend to be ahead of them (Earth rotates faster than the moon orbits) and pull them ahead, moving them into a higher orbit and slowing Earth's rotation. The opposite would happen in a retrograde orbit. But this is all classical physics.