Suppose a body is moving about another body under a central force, such that the path is bounded. We can take the example for planets around the sun, and the energy is $E>0$, such that the orbits are bounded.
We can write : $$E=\frac{1}{2}mv^2+U(r)=\frac{1}{2}m\dot{r}^2+U_{eff}(r)$$
Thus, we have managed to convert our $2D$ problem to an equivalent $1D$ problem. We can easily draw a graph for $U_{eff}(r) $ vs $r$, and on this, we can draw the energy as a straight line. Let $r_1$ and $r_2$ be the points where the energy $E$ intercepts the curve of $U_{eff}(r).$ These points would represent the turning points in $1D$ and thus, the aphelion and the perihelion in $2D$.
However, we can do another interesting thing. If the two following conditions are satisfied:
$$\frac{dU_{eff}}{dr}|_{r=r_0}=0$$
$$\frac{d^2U_{eff}}{dr^2}|_{r=r_0}>0$$
Then we can say, about $r_0$, the particle would have a stable circular orbit. In this case, $E=U_{eff}(r_0)$, and there would be no radial velocity, as every point would be a turning point. Hence $\dot{r}=0$.
My question is, what would happen if this object in stable orbit is perturbed? In the $U_{eff}(r)$ graph, if this object is perturbed, it would undergo small oscillations about the mean position in an SHM fashion, where the frequency of oscillation would be given by:
$$\omega = \sqrt{\frac{U^{''}(r_0)}{m}}$$
However, what would the motion in $2D$ become due to this perturbation? What does small oscillations in $1D$ translate into, in case of $2D$?
According to my source, the particle remains in the circular path, but 'wiggles' back and forth, like oscillating, while moving in this circle. However, this doesn't make sense to me. Isn't perturbing the object equivalent to giving it a little bit of energy, so that the energy $E>U_{eff}(r_0)$ ? In that case, instead of 'wiggling' or oscillating while moving in a circular orbit, shouldn't the orbit itself become slightly elliptical? Unless perturbation doesn't transfer any extra energy to this object, it doesn't make sense to me.
Why would perturbation cause radial oscillation instead of making the orbit more elliptical ? If we give energy to this object, the orbit should become more elliptical instead of the object oscillating radially. If perturbation doesn't provide extra energy, what exactly does it do? What exactly is a perturbation for that matter?
Can anyone help me understand this intuitively?
EDIT :
The last picture in the bottom right corner is what I'm talking about. Instead of that, shouldn't the orbit just become slightly more elliptical?
Best Answer
This entire confusion boils down to a misinterpretation of the source material. When you have a circular orbit, in a central force problem, and you perturb the orbit, the object does undergo radial oscillations or 'wiggles' about the circular trajectory.
However, the nature of this orbit, or what it looks like would depend totally on the nature of the central force. Each force would result in a distinct type of oscillation.
For a force of the nature: $$F(r)=kr^\alpha$$ the frequency of these oscillations would be : $$\omega=(\sqrt{3+\alpha})\space\Omega_0=(\sqrt{3+\alpha})\frac{v}{r}$$
Here $\Omega_0$ is the angular velocity of the unperturbed circular orbit.
As you can see, for gravitation, $\alpha=-2$, hence $\omega=\Omega_0$. So, the object would oscillate only once during its entire revolution. This essentially means that the orbit would become an ellipse.
The cases that you see, where are orbits become pentacle or some other shaped, is due to other forces, not for gravitation. Moreover, for the orbit to be closed, $\sqrt{3+\alpha}$ must be an natural number $\mathbb{N}$.
You can find more information, in this following link