One can find the effective potential a particle experiences for instance in the most frequent case, the Schwarzschild case. Outside of a non-rotating massive object the effective potential can be found
by computing the 4-momentum square in that metric ($r_s$ is the Schwarzschild radius):
$$(mc)^2 = p_\mu p^\mu = m^2 \frac{dx_\mu}{d\tau}\frac{dx^\mu}{d\tau} = m^2 \left[ \left(1-\frac{r_s}{r}\right)c^2\dot{t}^2 -\frac{\dot{r}^2}{1-\frac{r_s}{r}} -r^2 \dot{\varphi}^2\right] \tag{1}$$
The motion is governed by 2 constants of motion $E$ and $L_z$.
EDIT:
This can be seen considering the expression $m^2 \frac{dx_\mu}{d\tau}\frac{dx^\mu}{d\tau}$ as Lagrangian (more precisely said the Lagrangian multiplied by $-2m$):
$$-2mL = m^2 \frac{dx_\mu}{d\tau}\frac{dx^\mu}{d\tau}$$
In order to find the canonical momenta the derivatives with respect to $\dot{t}$, $\dot{r}$ and $\dot{\varphi}$ are taken. Most interesting are $\dot{t}$ and $\dot{\varphi}$ since $t$ and $\varphi$ are cyclic variables ($r$ is not a cyclic variable since the Lagrangian depends explicitly on $r$):
Then we get (where energy $E$ and angular momentum $L_z$ appear as constants of motion scaled by $2m$ as the Lagrangian):
$$const =\frac{\partial (-2mL)}{\partial \dot{t} } =2m^2c^2 \dot{t} \left(1-\frac{r_s}{r}\right) =2mE$$
and
$$const =\frac{\partial (-2m L)}{\partial \dot{\varphi} } =-2m^2c^2 r^2\dot{\varphi} =-2mL_z$$
which turn into:
EDIT END
$$\dot{t} =\frac{E}{mc^2(1-\frac{r_s}{r})} \quad\text{and}\quad \dot{\varphi} =\frac{L_z}{mr^2}$$
After a bit of algebra one gets from (1):
$$ \frac{E^2}{(mc^2)^2} = 1 + \frac{\dot{r}^2}{c^2} +V_{eff}$$
where the effective potential equals:
$$ V_{eff}= -\frac{r_s}{r} + \frac{L_z^2}{m^2c^2}(\frac{1}{r^2} - \frac{r_s}{r^3})$$
The minima and maxima of this potential provide you the requested
information.
If the solution of Einstein's field equations (EFE) is already known as here assumed the task is rather easy. Of course to find the rotational- symmetric EFE's solution outside a non-rotating massive object one has to solve the EFE's in vacuum, i.e.
$$ R_{\mu\nu}=0 \quad\text{with the corresponding symmetry conditions} $$
where $R_{\mu\nu}$ is the Ricci tensor.
