Gravity well and constant total energy in circular orbit According to this question, the total energy of an object in a circular orbit in a $1/r^2$ potential does not depend on the radius of the orbit.
However, if you look at "circular orbits" in a gravity well, this is not obvious:
https://www.youtube.com/watch?v=4ffI0l8vUK8
In this situation, the particle is constantly losing energy due to friction, but it always stays in a (more or less) circular orbit.
If the total energy of circular orbits stays constant, why does such a particle come closer to the attractor?
 A: You have got your forces and potentials mixed up. If the force is given by:
$$ F(r) = \frac{GMm}{r^n} $$
then the potential is the integral of this:
$$ V(r) = -\frac{GMm}{(n-1)r^{n-1}} $$
In the question you link the force has an inverse cubic dependance, $n=3$, while the gravitational force has an inverse square dependance, $n=2$.
We get a circular orbit when the centripetal acceleration is equal to the gravitational acceleration i.e.
$$ \frac{v^2}{r} = \frac{GM}{r^n} $$
giving:
$$ v = \sqrt{\frac{GM}{r^{n-1}}} $$
So the kinetic energy is:
$$ T = \frac{GMm}{2r^{n-1}} $$
To get the total energy we add the potential energy:
$$ E = \frac{GMm}{2r^{n-1}} - \frac{GMm}{(n-1)r^{n-1}}$$
For the inverse cube force described in the linked question $n = 3$ and we get:
$$ E = \frac{GMm}{2r^{3-1}} - \frac{GMm}{(3-1)r^{3-1}} = 0 $$
so we have the slightly surprising result that the total energy is independent of $r$ and in fact is always zero. For gravity $n=2$ and we get:
$$ E = \frac{GMm}{2r^{2-1}} - \frac{GMm}{(2-1)r^{2-1}} = -\frac{GMm}{2r} $$
so the energy gets lower (more negative) as $r$ decreases. That's why the balls in your video move inwards as they lose energy.
This is an example of the virial theorem that tells us for a potential $V(r) = ar^{-n}$ the kinetic and potential energy are linked by:
$$ 2T = -nV $$
For the cubic force the potential has an inverse square dependance, $n = 2$, so we get:
$$ T = -V $$
and we can see immediately that the sum of the kinetic and potential energy is always zero. For gravity the potential depends on $r^{-1}$ so the virial theorem tells us:
$$ 2T = -V $$
and the sum of the kinetic and potential energy is always $-T$.
