How can I interpret the equation of the orbit in Schwarzschild metric?

Given the standard geodesic equation:
$$frac{d^2 x^mu}{dlambda^2}+Gamma ^mu _{sigma rho}frac{d x^sigma}{d lambda}frac{d x^rho}{d lambda}=0$$
we want to apply it to the Schwarzschild metric; conceptually this simply means using the proper $Gamma$ coefficients, but in practice this leads to a gargantuan calculation. But fortunately we can apply some simplification to get the following equation:
$$frac{1}{2}left(frac{dr}{dlambda}right)^2+V(r)=mathcal{E} (1)$$
where:
$$V(r)=frac{1}{2}epsilon-epsilonfrac{GM}{r}+frac{L^2}{2r^2}-frac{GML^2}{r^3}$$
$$mathcal{E}=frac{1}{2}E^2$$
we should also specify that:

1. This equations should hold for massive and massless particles alike, so $lambda$ is in one case the proper time and in the other an affine parameter of the proper time.
2. $r equiv r(lambda)$ which is the radius of the orbit
3. $epsilon := -g_{munu}frac{d x^mu}{dlambda}frac{d x^nu}{dlambda}$
4. $E=left(1-frac{2GM}{r}right)frac{dt}{dlambda}$
5. $L=r^2frac{d phi}{dlambda}$ (where $phi$ is the fourth coordinate in our Schwarzschild’s spherical coordinates)

End of the setup; my question is: I followed the proof of this statement, involving killing vectors ecc., and I get the mathematical gist of it, but I have problems interpreting the result. Is there an intuitive way to read statement (1)?

(For example: One problem is that it is said that $V(lambda)$ should represent the potential and $mathcal{E}$ should represent the energy, but I don’t see how.)