Geodesic equation for Newtonian gravity The usual force due to gravity is given by 
$$ \mathbf{F} = -\frac{GM}{r^2}\mathbf{\hat{r}},$$
where $r=|\mathbf{r}|$ and $\mathbf{\hat{r}} = \mathbf{r}/|\mathbf{r}| $. To derive the standard second order ODE that describes orbital motion given by 
$$ \frac{d^2 u }{d \varphi} + u = C,$$
is quite easy and is found throughout undergrad mechanics books. 
I'm interested in deriving the ODE using the geodesic equation given by
$$ \frac{d^2 x^i}{dt^2} + \Gamma^i_{jk} \frac{dx^j}{dt}\frac{dx^k}{dt} = 0, $$
where $x^i=(r,\theta,\phi)$. My question is:  Does it make sense to set the geodesic equation up in such a way that 
$$ \frac{d^2 x^i}{dt^2} + \Gamma^i_{jk} \frac{dx^j}{dt}\frac{dx^k}{dt} = F_i, $$
where $F_i=(-GM/r^2,0,0)$. With this method the ODE comes straight out but I'm not sure if the move is legal.
 A: Yes this is a perfectly valid approach to take. We can construct any coordinate system $\mathbf x$ and the acceleration $\mathbf a$ in this system will be given by:
$$ a^i = \frac{d^2 x^i}{dt^2} + \Gamma^i_{jk} \frac{dx^j}{dt}\frac{dx^k}{dt} $$
So your equation is just Newton's second law.
In practice we'd rarely bother with calculating all the Christoffel symbols then using them to calculate the acceleration. We'd just to the coordinate transformation directly. But conceptually this nicely illustrates what is going on.
The obvious example of where we do generally calculate the Christoffel symbols is in general relativity where the equation above gives us the four-acceleration.
A: Let me throw in my two cents to this topic. In general in Newtonian mechanics, one gets the complete information on a particle's path by knowing 
$$
(t,x^{\alpha}(t))
$$
where the $x^{\alpha}$ are the the 3 spatial coordinates given as functions of time. One can make the smart observation, that the time $t$ can be interpreted as an affine parameter. What comes next is just a reformulation of the second law of Newton. First let's identify $t$ with $x^0$, subsequently re-write newtons equations $\bf{F}=m\bf{a}$ component wise , by observing that $\frac{dx^0}{dt}=1$
$$
\frac{d^2x^0}{dt^2}=0\\
\frac{d^2x^{\alpha}}{dt^2}=f^{\alpha}({x(t)})\frac{dx^0}{dt}\frac{dx^0}{dt}
$$ 
The equations given above take exactly the form of the Einstein geodesic equations
$$
\frac{d^2x^{\mu}}{ds^2}=-\Gamma^{\alpha}_{\mu\nu} \frac{dx^{\mu}}{ds}\frac{dx^{\nu}}{ds},
$$
where the forces $f^{\alpha}({x(t)})=$ play the role of the Christoffel symbols $-\Gamma^{\alpha}_{00}$ (note the minus sign). Moreover, one can deduce some conditions imposed by the above system. For instance from the first equation it follows $\Gamma^0_{\mu\nu}=0$. Where, $\mu$ and $\nu$ run from 0 to 3. Knowing this one can compute the Riemann tensor and the corresponding metric. There is an excellent review of this topic in the book "Gravity" by K. Thorne and J.A. Wheeler.     
