How does adding a inverse cube force field to a inverse square force field change the orbit? 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.

The key fact here is a very nice one that I learned from a mathematical physics book somewhere, but deserves to be better known: If $F(r)$ is a rotationally symmmetric force (with positive $F$ meaning towards the origin), and we have an almost circular orbit of radius $r_0$, then the angle between perihelions of the orbit will be
$$\frac{2 \pi}{\sqrt{3+\tfrac{F'(r_0) r_0}{F(r_0)}}}.$$
When $F$ is the inverse square, we have $r F'/F = -2$, so perihelions are precisely $2 \pi$ apart, in other words, orbits close up. To solve your problem, you can look up how fast the orbit of Mercury precesses and match that with the proposed inverse cube correction term.
Derivation Let's suppose that our orbit has angular momentum $L$. That means its angular velocity is $L/(m r^2)$, so the centripetal force is $m r \left( \tfrac{L}{m r^2} \right)^2 = \tfrac{L^2}{m r^3}$. In other words, we have
$$m r'' = -F(r) + \frac{L^2}{m r^3}.$$
If we had a perfect circular orbit at radius $r_0$, we would have $F(r_0) = \tfrac{L^2}{m r_0^3}$.
Now, suppose that we perturb our perfect circular orbit a little, to $r_0+\epsilon(t)$. If $\epsilon$ is small enough that we can approximate formulas with their Taylor series up to linear order, then
$$m \epsilon'' = -F(r_0) - F'(r_0) \epsilon + \frac{L^2}{m r_0^3} - \frac{3 L^2}{m r_0^4} \epsilon = - \left(F'(r_0)+ \frac{3 L^2}{m r_0^4} \right) \epsilon.$$
So, by the standard analysis of a mass on a spring, $\epsilon$ oscillates with period
$$2 \pi \sqrt{\frac{m}{F'(r_0) + \tfrac{3 L^2}{m r_0^4}}}.$$
To get the angle traveled between oscillations, multiply the period by the angular velocity $L/(m r_0^2)$ to get
$$\theta = 2 \pi \sqrt{\frac{m}{F'(r_0) + \tfrac{3 L^2}{m r_0^4}}} \frac{L}{m r_0^2}=\frac{2 \pi}{\sqrt{3+\tfrac{m F'(r_0) r_0^4}{L^2}}}.$$
Finally, remember that $F(r_0) = \tfrac{L^2}{m r_0^3}$ to rewrite this as 
$$\frac{2 \pi}{\sqrt{3+\tfrac{ F'(r_0) r_0}{F(r_0)}}}.$$
