# Energy diagram of a planet with fixed mechanical energy

Consider the following energy diagram for the motion of a planet about a star.

The centrifugal potential curve can be represented once I fixed the angular momentum $\vec{L}$ of the planet. To fix the angular momentum vector of a planet in a orbit I have to fix not only its velocity vector $\vec{v}$ but I must fix this vector at a particular distance $\vec{r}$ from the point $O$. Is this correct?

Now suppose to fix the mechanical energy of the planet $E$ instead (i.e. the horizontal line cannot change). Suppose that I can change the angular momentum of the planet, for istance I can move approximately up or down the centrifugal potential curve. For fixed $E>0$ values, if $L$ grows, the minimum distance approached by the planet from the star increases too. But consider the ellipse situation, i.e. $E<0$. If I move up the centrifugal potential curve the perihelion gets further (and that seems right) but the aphelion gets closer to the star. I don’t see the reason of this last fact, if the planet rotates faster ($L$ increases) shouldn’t it get further both in the min and max distance in its orbit?

Is there something I am missing here?

For your second case, you can change the angular momentum, but remember that you have fixed your total energy. You can't make the planet revolve arbitrarily fast or it will have more energy than allowed. By increasing the angular momentum without adding energy, you are circularizing the orbit.

To add, you might take a look at the Specific Orbital Energy equation

$$\epsilon = -\frac{1}{2}\frac{\mu ^2}{h^2}(1-e^2)$$

Considering the case where energy and masses are constant, we can rewrite this as

$$k = \frac{(1-e^2)}{h^2}$$

$$h^2 = k(1-e^2)$$ $$h^2 = k - ke^2$$

So for a given energy, angular momentum is at a maximum when eccentricity is 0 (or circular).

• Thanks a lot for the answer, I'm aware of the fact that conservation of energy cannot be violated, my doubt was about the "circularizing": the orbit looks more and more as a circle as angular momentum increases, until when the orbit is a perfect circle $U_{eff}=E$ and that's the minum of mechanical energy. But why is this caused by an increase of angular momentum (at constant energy)? To increase angular momentum means increase $v_{\theta}$ but it seems also that $v_{r}$ decreases (at limit orbit is a circle so $v_{r}=0$) – Sørën Apr 17 '16 at 10:08