Total energy of a circular orbit 
A particle of mass $M$ moves in a circular orbit of radius $r$ around a fixed point under the influence of an attractive force $F=K/r^3$, where $K$ is a constant. If the potential energy of the particle is zero at an infinite distance from the force center, the total energy of the particle in the circular orbit is 
  
  
*
  
*$-k/r^2$
  
*$-k/2r^2$
  
*0
  
*$k/2r^2$
  
*$k/r^2$

This is a physics GRE problem and the solution can be found here
What they have said that:

consider conservation of energy. Coming in from far away, the particle
  has E=V=0 the total energy equal to the potential energy equal to 0

But we know  the potential is always considered as zero at the infinite distance from the force center. Then according to them the total energy of the circular orbit will be always zero and that would not depend on the force $F=K/r^3$. But what I know is the total energy zero implies the orbit has to be parabolic. I am bit confused about this problem.  
If the problem would have different force field, let assume $F=K/r^2$    then how would we deal with it?  I mean what would be the shape of thew orbit and the total energy of the orbit. 
 A: Let's take the general case and suppose the force is:
$$ F = \frac{k}{r^n} $$
Then integrating to get the potential gives:
$$ U = -\frac{1}{n-1} \frac{k}{r^{n-1}} $$
If the orbit is circular the acceleration of the object is $v^2/r$ so the force is $mv^2/r$, and equating this to the force law we're given we get:
$$ \frac{k}{r^n} = \frac{mv^2}{r} $$
or with a quick rearrangement:
$$ \frac{k}{2r^{n-1}} = \tfrac{1}{2}mv^2 = T $$
The total energy is then:
$$\begin{align}
 E &= T + U \\
   &= \frac{k}{2r^{n-1}} - \frac{1}{n-1} \frac{k}{r^{n-1}} \\
   &= \frac{k}{r^{n-1}}\left(\frac{1}{2} - \frac{1}{n-1}\right) 
\end{align}$$
This is only zero when $n = 3$. So it isn't obvious to me how vague statements of conservation of energy can be used to make this argument.
A: Considering the total energy of an orbit of mass $m$ under the influence of an attractive central force, $$\vec{F}=-F(r)\hat{r}$$
we can show that the angular momentum, $\vec{L}$, is conserved because the torque produced by the force about the force center is zero:
$$\vec{\Gamma} = \vec{r}\times\vec{F} = rF(r)(\hat{r}\times\hat{r}) = 0.$$
So we can say $|\vec{L}|=L = mr^2\dot{\theta}$ is constant, and the orbit will be planar. We can describe the orbit in terms of the variables $r$ and $\theta.$
The kinetic energy is given by $$K=\frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}^2)$$ which can be re-written using $L$ as
$$K=\frac{1}{2}m(\dot{r}^2)+\frac{L^2}{2mr^2}.$$
Using @JohnRennie 's potential energy (see his answer) for the general $r^{-n}$ central force, we can write that the energy is
$$E=\frac{1}{2}m(\dot{r}^2)+\frac{L^2}{2mr^2}-\frac{K}{(n-1)r^{n-1}}.$$
If we have a circular orbit, $\dot{r} = 0$.
If n=2 (a Newtonian gravitational force), the energy of a circular orbit becomes
$$E=\frac{L^2}{2mr^2}-\frac{K}{r}.$$
With $\frac{K}{r^2}$=$mr\dot{\theta}^2$=$\frac{L^2}{mr^3}$ by a centripetal force relationship,$\frac{K}{r}=\frac{L^2}{mr^2}.$  From this we get the total energy 
$$E=-\frac{L^2}{2mr^2},$$
a negative value.
If we further examine the part of the energy as a function of $r$, known as the ``centrifugal potential'', there is a local minimum of this potential when
$$\frac{K}{r^n}=\frac{L^2}{mr^3}.$$
For $n=2$, this yields precisely the the radius for a circular orbit, and the equilibrium is stable.
On the other hand, for n=3, the $r$ functionality of this minimum potential disappears. This means that it is possible for the orbit to be circular, but the orbit is not stable.
The $E<0$ condition for bound orbits pertains specifically to an $n=2$ central force, and a minimum E corresponds to the circular orbit.  For $n=3$, the energy will be zero, a circular orbit is possible, but it will not be a stable equilibrium.
