It's not clear to me, what you mean by "centrifugal" energy. So maybe, I misunderstand you by just guessing what you mean.
The usual treatment of planetary orbits (aka "the Kepler problem") is to consider the Sun and a single planet, ignoring the interaction of these with any other objects around. This has the advantage to be not too bad an approximation (because the Sun has much larger mass than all other objects around her) and as well being treatable analytically.
To make the discussion even simpler we can also assume that the Sun's mass is so much larger than the considered planet's mass that we can as well make the Sun fixed at the origin of a Cartesian coordinate system. Then the planet moves simply according to
$$m \ddot{\vec{r}}=-\gamma m M \frac{r}{r^2}.$$
Now first we see that taking the cross product with ##\vec{r}## gives 0 on the right-hand side, which gives
$$m \vec{r} \times \ddot{\vec{r}}=\mathrm{d}_t (m \vec{r} \times \dot{\vec{r}})=0$$
or
$$\vec{L}=m \vec{r} \times \dot{\vec{r}}=\text{const}.$$
That's angular-momentum conservation. Now we make ##\vec{L}## point along the 3-axis. The motion is then in the (12)-plane, and we can introduce polar coordinates
$$\vec{r}=(r \cos \varphi,r \sin \varphi,0).$$
From this
$$\dot{\vec{r}}=\dot{r} (\cos \varphi,\sin \varphi_0)+r \dot{\varphi} (-\sin \varphi,\cos \varphi,0).$$
$$\vec{L}=m \vec{r} \dot{\vec{r}}=m r^2 \dot{\varphi} (0,0,1).$$
Further taking the scalar product of the equation of motion with ##\dot{\vec{r}}##, we find the energy-conservation law
$$\frac{m}{2} \dot{\vec{x}}^2-\frac{\gamma m M}{r}=E=\text{const}.$$
Now plugging in our polar coordinates this gives
$$\frac{m}{2} (\dot{r}^2+r^2 \dot{\varphi}^2)-\frac{\gamma mM}{r}=E.$$
Now we use
$$\vec{L}^2=m^2 r^4 \dot{\varphi}^2,$$
to eliminate ##\dot{\varphi}## from the energy-conservation equation:
$$\frac{m}{2} \dot{r}^2 + \frac{L^2}{2 m r^2} - \frac{\gamma m M}{r}=E.$$
Now you can reinterpret this as the equation of motion of a mass point moving along the ##r## axis moving in an effective potential
$$V_{\text{eff}}=\frac{L^2}{2m r^2} - \frac{\gamma m M}{r}.$$
The ##r## coordinate can be taken as the distance of an observer sitting at the planet to the Sun, which is a rotating reference frame, and seen from this frame ##L^2/(2m r^2)## is the potential of the corresponding centrifugal force.
Taking the time-derivative of the energy-conserving equation indeed gives (after deviding by ##\dot{r}##)
$$m \ddot{r}=\frac{L^2}{m r^3}-\frac{\gamma m M}{r^2}.$$
The first term can be rewritten as
$$F_Z=\frac{L^2}{m r^3}=m r \dot{\varphi}^2,$$
and this is indeed the well-known expression of the centrifugal force. That's why the first term in ##V_{\text{eff}}## is called the "potential of the centrifugal force".
The 2nd term is of course the gravitational attraction of the Sun. So from the point of view of the "corotating" observer in addition to the "true" force (gravitation) you have the centrifugal force as an "inertial force".