Maximizing the time near a black hole I just learned about the Schwarzchild solution, following Carrol's book. Now, there is a question that I want to answer:
Consider an observer that starts at infinity with some velocity, comes close to the black hole (always free-falling) and then goes off to infinity. Looking at Carrol's book, it's easy to see that the closest that the observer can get to the black hole is $3GM$ (actually, not $3GM$ itself: the value of the minimum $r$ goes to $3GM$ in the limit where the angular momentum $L$ goes to $\infty$).  
Now, I want to figure out how to control how much time the observer is "near the black hole". For example: how can he maximize the time that he spends at $r<4GM$?
A first naive answer would be: "Just start with the biggest $L$ possible", but just because you go closer to the black hole it does not mean that you spend more time in its neighborhood: you may also be traveling faster.
Note: from Carrol's book, the differential equation for $r$ is
$$
     \frac{1}{2} (\frac{dr}{d\lambda})^2 + V(r) = \mathcal{E}
$$
 A: An observer can spend infinite time at $r=4GM$ if he falls from infinity with a precise angular momentum $\ell = \pm 4GM$. (Units $c=1$). 
Consider launching an observer from infinity such that 1) he has infinitesimal velocity 2) he has a finite amount $\ell$ of angular momentum. We claim that we can choose $\ell$ so that the observer falls into circular orbit. 
From $$V_{eff}(r) = - \frac{GM}{r} + \frac{\ell^2}{2r^2}- \frac{GM\ell^2}{r^3}$$
and the energy equation you mentioned
$$\frac12 \left(\frac{dr}{d\tau}\right)^2+ V_{eff}(r) = \mathcal{E}= \mbox{const}$$
we see that $\mathcal{E} = 0$ because, initially, we start at spatial infinity with infinitesimal radial velocity. 
A circular orbit at radius $r_c$ will have
$$V'_{eff}(r_c)=0$$
$$\left.\frac{dr}{d\tau}\right|_{r_c} = 0$$
and the latter equation together with the energy equation tells us
$$V_{eff}(r_c) = 0.$$
The pair of equations is
$$\frac{3 G \ell^2 M}{r^4}+\frac{G M}{r^2}-\frac{\ell^2}{r^3}=0$$
$$-\frac{G \ell^2 M}{r^3}-\frac{G M}{r}+\frac{\ell^2}{2 r^2}=0$$
which amounts to solving a pair of quadratics. You can check that the only solutions are 
$$\ell = \pm 4GM,\quad r=4GM.$$
Not only is it possible, $r=4GM$ is the only possible orbit with the initial conditions we specified. Now maybe you can generalize to some finite amount of initial radial velocity to see if a closer orbit is possible. 
A: Generalizing the answer by Dwagg: For any radius $3M<r\leq4M$ (using units with $G=c=1$), there is an orbit that comes in from infinity an asymptotes to a circular orbit. These orbits are known as homoclinic orbits. These orbits have the same energy and angular momentum as the (unstable) circular orbit that they asymptote to, which can be found by solving the equations
$$V'_{eff}(r)=0$$
and
$$V_{eff}(r)=\mathcal{E}.$$
This gives
$$ \mathcal{E} = \frac{(r-2M)}{\sqrt{r(r-3M)}}$$
and
$$ \ell = \pm\frac{Mr}{\sqrt{M(r-3M)}}.$$
These orbits spend an infinite amount of time below $r=4M$. If, as you specify, you want an orbit that dips below $r=4M$ and returns to infinity, then you can take the energy as in the formula, and an angular momentum that is just a bit higher. As you let the angular momentum approach the homoclinic limit value, the time spent below $r=4M$ will approach infinity.
