What is a nongeodesic orbit? I have read that in the Schwarzschild spacetime for a nongeodesic circular orbit the radial acceleration becomes positive for $r<3r_S$. Intuitively, the acceleration should be negative, pulling the object in. 
My questions:


*

*What does a nongeodesic orbit even mean? Is it a solution to the general equation 
$$f^\mu=ma^\mu$$
where $a^\mu=u^\nu\nabla_\nu u^\mu$ is the proper acceleration? What is $f^\mu$ in this case?

*Does this result make sense? If so, how? Why does the central mass seem to repel satellites? 

*If the equation in 1. is correct, then why does this behave differently than the geodesic case? I don't see how a term on the other side could change the behavior of the orbit this much.
 A: Let's consider a circular orbit in Schwarzschild coordinates, taken to be in the equatorial plane for simplicity. The test particle's position has components $x^\mu = (t, r, \pi/2, \phi)$, where $t$ and $\phi$ vary linearly with time/proper time and $r$ is constant. Then the $4$-velocity is $u^\mu = (\dot{t}, 0, 0, \dot{\phi})$, where dots denote derivatives with respect to the particle's proper time.
The proper acceleration has radial component
$$ a^r = \left(1 - \frac{2M}{r}\right) \left(\frac{M}{r^2} \dot{t}^2 - r \dot{\phi}^2\right), $$
as you are encouraged to check by crunching connection coefficients, either in the above equation or in the inhomogeneous geodesic equation (that is, where the right-hand-side is $a^\mu$ rather than $0$ but where $\ddot{r} = 0$). This holds for any circular (and equatorial) orbit, geodesic or not. The orbit will be geodesic if and only if $a^r$ vanishes (we can ignore the other components because we are already assuming uniform circular motion). A nongeodesic orbit will be going at the wrong velocity for its radius, and this situation will be maintained by an applied external acceleration $a^r$.
Now let's look at allowed angular velocities. Any massive particle should have a velocity (and therefore angular velocity) bound by a photon moving in the same (i.e. purely azimuthal) spatial direction. A circular, equatorial, null orbit will obey
$$ 0 = g_{\mu\nu} u_\mathrm{null}^\mu u_\mathrm{null}^\nu = -\left(1 - \frac{2M}{r}\right) \dot{t}_\mathrm{null}^2 + r^2 \dot{\phi}_\mathrm{null}^2, $$
where here dots refer to derivatives with respect to some affine parameter. This gives the bound on angular frequencies:
$$ \omega_\mathrm{max}^2 \equiv \left(\frac{\mathrm{d}\phi_\mathrm{null}}{\mathrm{d}t_\mathrm{null}}\right)^2 = \frac{1}{r^2} \left(1 - \frac{2M}{r}\right). $$
For massive objects, angular frequency $\omega \equiv \mathrm{d}\phi/\mathrm{d}t$ is bounded by $0 \leq \omega^2 < \omega_\mathrm{max}^2$. Rewriting our formula for radial acceleration, we have
$$ a^r = \left(1 - \frac{2M}{r}\right) \left(\frac{M}{r^2} - r\omega^2\right) \dot{t}^2. $$
Taking into account the positivity of the first term (since we are outside the horizon) and the particle's motion through time, we have
$$ a^r > \left(1 - \frac{2M}{r}\right) \left(\frac{M}{r^2} - r\omega_\mathrm{max}^2\right) \dot{t}^2 = \frac{1}{r^2} \left(1 - \frac{2M}{r}\right) (3M - r) \dot{t}^2. $$
For $2M < r < 3M$, $a^r$ must be strictly positive in order to maintain a circular orbit.
Note it isn't the central mass that is providing the repelling force. The idea is that within $3/2$ Schwarzschild radii of an object, there are no geodesic circular orbits. You can move in a circle, but you need a continuous source of outward acceleration to do so. Outside $r = 3M$, you can have geodesic orbits ($a^r = 0$), too-slow orbits ($a^r > 0$), or too-fast orbits ($a^r < 0$).
