When is the effective potential equal to the total energy?

Question

I have a question about the energy of a particle in orbit due to a gravitational attraction. The effective potential given by the gravitational force is defined to be $$ U_{\text{eff}} = \frac{L^2}{2mr^2}- \frac{GmM}{r} $$ On the other hand, using conservation of energy and writing $v^2 = \vert{\dot{\vec{r}}}\rvert^2$ in polar coordinates we see that $$ \frac{1}{2}m\dot{r}^2 = E - U_{\text{eff}}\tag{1} $$ The above expression got me thinking, and I wanted to ask if I correctly understood what the equation implied.

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If $ E = U_{\text{eff}}$ then $(1)$ tells us that $\dot{r} = 0 \iff r = \text{constant}$, but since $r = \text{constant}$ describes a circular orbit, is the statement

The effective potential is equal to the total energy of a particle under a gravitation force if and only if the orbit of the particle is circular.

correct? Or am I misunderstanding? Thank you in advance!

Answer 1

The effective potential is obtained as follows. The kinetic energy in polar coordinates is $T = { 1\over 2}m v^2={ 1\over 2}m( \dot r^2 + r^2\dot \theta^2)$. (Your expression for $v^2$ is incorrect.)

Using conservation of energy $E = T + V$ is constant where $E$ is total energy, $T$ is kinetic energy, and $V$ is the gravitational potential energy, $V = -GmM/r$. So ${ 1\over 2}m( \dot r^2 + r^2\dot \theta^2) - GmM/r = E$, a constant. Consider the terms ${ 1\over 2}mr^2\dot \theta^2 - GmM/r$. The angular momentum $L = mr^2 \dot \theta$ is constant. So these terms are ${ {L^2} \over {2mr^2}} - {{GmM} \over {r }}$. We define the effective potential energy $U_{eff} = { {L^2} \over {2mr^2}} - {{GmM} \over {r }}$. Therefore, ${ 1\over 2}m\dot r^2 = E - U_{eff}$.

If $\dot r = 0$ the orbit is circular and $E = U_{eff}$ as you say. Note, your conclusion is correct for the effective potential energy.

In general, an effective potential energy includes terms that are dependent on position only; that is the case here since the angular momentum is constant. This approach is used in evaluations using the Lagrangian and the Hamiltonian. See for example the text Mechanics by Symon.