TITLE: How does adding a inverse cube force field to a inverse square force field change the orbit?
QUESTION [5 upvotes]: Let us say that we modify the law of gravitation to the following force field
$$
\vec{F}(r) = \frac{-GMm}{r^2}\hat{r} + \alpha\frac{|L|^2}{m^2 r^3}\hat{r}
$$
where, 
$L = m_2(\vec{r} \times \dot{\vec{r}} )$ which is the angular momentum of $m$ in the reference frame of $M$.
$M$ is mass of parent body which is stationery in our reference frame.
$m$ is mass of the moving body.
$\vec{r}$ is position vector of $m$ with $M$ as the reference (origin).
$\dot{\vec{r}}$ is the velocity of $m$ with M being the reference or origin again.

First Question
Find the equation of orbit of a planet around the Sun and draw an approximate orbit for $\alpha \ll  1.$

Second Question
For the Sun-Mercury system find the value of $\alpha$. Here $M$ is the Sun.

I could not solve this question. However I feel that Newton's Theorem of revolving Orbits is somehow needed for the solution. I would like to see a proof of the theorem as well as a complete solution for the problem.
I did not post this question for Physics SE mainly because (i) this question mainly needs knowledge of vector calculus (or so I feel) and (ii) My experience is that Math SE is a much better community so I prefer posting a borderline Maths-Physics question on Mathematics SE rather than Physics SE.

Edit 1
I know how to derive the two body problem by using the Rung Lenz vector method. So any answers may skip over details such as how to prove $\vec{L}$ is constant etc.

Edit 2
An article by Baez on the inverse cube force law.

Major Edit/Typo
I had made a mistake while writing the title. I had mistakenly written inverse cube potential instead of inverse cube force field.

REPLY [3 votes]: Question 1:
This does not change qualitatively anything in the sense that your two-body problem accepts, formally speaking, the same kind of solutions than the Kepler's two body problem, i.e. conics but now paramterized by $\alpha$.
The two-body problem is best solved in the frame of the center of mass of the system. In that reference frame it appears as central force field 
problem applied to a particle of reduced mass $\mu = \frac{Mm}{m+M}$. 
Since in our case $M\gg m$, by asumption, the motion of the mass M around the center of mass can be safely ignored (the center of mass coincides with the star). Under this approximation the reduced mass $\mu\sim m$ and the total angular momentum of the two body system 
$\vec{L}=\mu \vec{r}\times \dot{\vec{r}}$ with that of the planet $\vec{L}\sim m \vec{r} \times \dot{\vec{r}}$.
The central problem in the fixed star reference frame has the force field given by, 
$$\vec{F} = \frac{-GMm}{r^2} \hat{r} +\alpha\frac{L^2}{m^2r^3} \hat{r} = -\frac{dU}{dr} \hat r$$ 
deriving from potential energy, 
$$U(r) = \frac{GMm}{r}-\frac{1}{2}{\alpha}\frac{L^2}{m^2r^2}$$
It is a well known result that in a central force field the angular moment $L$ is a first integral of the motion, i.e. stay constant,
$$\dot{\vec{L}}=0.$$ This implies that the motion at any time keeps being planar, in a plane orthogonal to the angular momentum. Making use of polar coordinates in this plane of normal unit vector $\hat{e}_z$, 
one has, ${L} = L \hat{e}_z = mr^2\dot \phi \hat{e}_z$, yielding,
$$L=mr^2\dot \phi = \mbox{Cte}.$$
Another first integral of the motion is the total energy $E$ of the system, defined as the sum  of the kinetic energy and the potential energy of the planet,
$$E=\frac{1}{2} m(\dot r^2 + r^2\dot \phi^2)+U(r)=\frac{1}{2} m {\dot r^2} + \frac{L^2}{2mr^2}+U(r)$$.
This equation of energy, shows that the motion in a central field is formally similar to a "linear or translational" motion (i.e. with a single degree of freedom, the radius $r$) but in a field of "effective" potential"energy,
$$U_{\rm eff}(r)=\frac{L^2}{2mr^2}+U(r) = (1-\alpha)\frac{L^2}{2mr^2}+\frac{GMm}{r}$$
Now setting $L_\alpha = L\sqrt{1-\alpha}$, one gets,
$$U_{\rm eff}(r)=\frac{L^2}{2mr^2}+U(r) = \frac{L_\alpha^2}{2mr^2}+\frac{GMm}{r}$$. 
and a total energy 
$$E=\frac{1}{2} m {\dot r^2} + U_{\rm eff}(r) = \frac{1}{2} m {\dot r^2} +
\frac{L_\alpha^2}{2mr^2}+ \frac{GmM}{r}$$
Both equations (effective potential or total energy) describe the motion of a classical two-body Keplerian system that would have $L_\alpha$ instead of $L_0=L$ as total angular momentum. Hence we are led to conclude that a substitution of $L$ by $L_\alpha$ in all the results of the classical Keplerian two body system theory yields the correct solution to your problem. One obtains the same kind of motion than for a classical two-body Keplerian system (the case $\alpha =0$) but the trajectories instead of being entirely governed by the pair $(E,L_0)$ are now governed by $(E, L_\alpha)$. Solutions for $\alpha\neq 0$ are deduced from the Keplerian case $\alpha=0$ homothetically by the ratio $\sqrt{1-\alpha}$ (possibly pure imaginary complex when $\alpha>1$).  
With a gravitational force field (of potential $\frac{GMm}{r}$) ---and more generally speaking in every Coulombian force field (of potential $-k/r$) 
one can easily integrate the trajectory as a conics of equation, 
$$p/r=1+e\cos\phi$$ 
where,
$$p=\frac{L^2}{km}=-\frac{L^2}{GMm^2}$$ is known as the parameter and $e$ the excentiricty of the conics is given by, $$e=\sqrt{1+\frac{2EL^2}{mk^2}}=\sqrt{1+\frac{2EL^2}{G^2M^2m^3}}$$.
Substitute $L \rightarrow L_\alpha$ in this classical result and you demonstrate that in your case, you obtain a (family of) conics (indexed by $\alpha$) of corresponding parameter(s) $p_\alpha$ and excentricitie(s) $e_\alpha$.
When $\alpha\leq 1$ the centrifugal term $\frac{L_\alpha^2}{2mr^2}$ in the effective potential is positive. (i) if $E<0$, the motion is bounded, $e_\alpha$ < 1 and you get an ellipse; (ii) if $E=0$ you get a round circle. (iii) if $E>0$, the motion is unbounded, $e_\alpha$ > 1 and you get an hyperbola. These results still hold for $\alpha=1$, simply it is the Keplerian case of null angular momentum giving a degenerated ellipse, i.e. a motion accelerated along a straight line towards the central star. 
When $\alpha>1$ the centrifugal term $\frac{L_\alpha^2}{2mr^2}$ in the effective potential becomes negative. I do not see any substantial changes to the preceding results except that now, one get bounded motion along ellipses when $E>0$ and unbounded motion along hyperbolas when $E<0$.
Question 2: 
I bet that $\alpha=0$ since the inclusion of your extra force term is not really necessary. Suppose you introduce a force term with let say some $L$ and $\alpha=0.1$. Your model is equivalent in a sense to one without extra force term but with planet momentum $L$ decreased to an adequate value i.e. multiplied by $\sqrt{1-\alpha}$. As a physicist, I would prefer this simpler formal solution.