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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.


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!

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  • $\begingroup$ Your equations are incorrect. In the kinetic energy terms, m should be the reduced mass. $\endgroup$
    – Nick
    Jun 22, 2021 at 21:09
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    $\begingroup$ I think @Robert Lee assumes M is very much greater than m in which case the reduced mass is essentially m. This assumption should be explicitly stated. $\endgroup$
    – John Darby
    Jun 22, 2021 at 21:55
  • $\begingroup$ Best to use equations that are correct, not ones that are almost correct under certain assumptions. This causes a lot of confusion. $\endgroup$
    – Nick
    Jun 22, 2021 at 22:03

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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.

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  • $\begingroup$ In the kinetic energy and angular momentum terms "m" should be the reduced mass. $\endgroup$
    – Nick
    Jun 22, 2021 at 21:18
  • $\begingroup$ I assumed M is much greater than m; I think this is the intent of the question. In this case the reduced mass is essentially m. If not, as you say, the reduced mass should be used. Thanks for the clarification. $\endgroup$
    – John Darby
    Jun 22, 2021 at 21:52

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