homework and exercises - 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

1. 
   
   

   $-k/r^2$

   
   


2. 
   

   $-k/2r^2$

   
   


3. 
   

   0

   
   


4. 
   

   $k/2r^2$

   
   
   


5. 
   

   $k/r^2$

   
   

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.

Answer

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.