# When is the effective potential equal to the total energy?

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!

• Your equations are incorrect. In the kinetic energy terms, m should be the reduced mass.
– Nick
Commented Jun 22, 2021 at 21:09
• 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. Commented Jun 22, 2021 at 21:55
• Best to use equations that are correct, not ones that are almost correct under certain assumptions. This causes a lot of confusion.
– Nick
Commented Jun 22, 2021 at 22:03

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.