I Is there a negative sign in front of the centrifugal potential energy?

[phantomvommand]

TL;DR Summary

I am not sure if there is a negative sign in front of the formula for the centrifugal potential energy.

In orbital mechanics, the effective potential is given by $\frac {1} {2} m r^2 w^2$, which can be expressed in terms of angular momentum $L$ which is conserved.

Yet, https://web.njit.edu/~gary/321/Lecture17.html apparently shows the centrifugal potential as the negative of the above. Deriving the paraboloid shape of some spinning water's surface also involves $- \frac {1} {2} m r^2 w^2$, instead of $\frac {1} {2} m r^2 w^2$.

So which is it? What is the difference?

 

[PeroK]

Does potential energy increase or decrease as $r$ increases?

 

[phantomvommand]

Does potential energy increase or decrease as $r$ increases?

I am not sure if centrifugal potential energy is really a potential energy. To me, it is a kinetic energy that just behaves like a potential energy (since v is only dependent on r). In that sense, I would expect 'potential energy' to increase as r increases, given that v increases. As for the gravitational potential energy, it increases with r too. So potential energy should increase?

Then how would this agree with the $- \frac {1} {2} m r^2 w^2$ formula, which shows it decreasing?

 

[PeroK]

I am not sure if centrifugal potential energy is really a potential energy. To me, it is a kinetic energy that just behaves like a potential energy (since v is only dependent on r). In that sense, I would expect 'potential energy' to increase as r increases, given that v increases. As for the gravitational potential energy, it increases with r too. So potential energy should increase?

Then how would this agree with the $- \frac {1} {2} m r^2 w^2$ formula, which shows it decreasing?

How could PE be KE? What normally happens ti PE as KE increases?

 

[phantomvommand]

How could PE be KE? What normally happens ti PE as KE increases?

In the orbital 'Effective potential energy' case, the tangential KE is treated as a potential energy? Hence PE can be KE?
from conservation of energy point of view, PE should decrease as KE increases.

 

[PeroK]

In the orbital 'Effective potential energy' case, the tangential KE is treated as a potential energy? Hence PE can be KE?

I've no idea what you mean by that.

from conservation of energy point of view, PE should decrease as KE increases.

Precisely!

 

[hutchphd]

I do not understand the Lecture notes. The effective centrifugal potential assumes the angular momentum to be constant. Thus $$U_{eff}(r)=\frac {L^2}{2mr ^2}-\frac {GmM} r $$ and that takes care of the signs for the derivative...

 

[phantomvommand]

I do not understand the Lecture notes. The effective centrifugal potential assumes the angular momentum to be constant. Thus $$U_{eff}(r)=\frac {L^2}{2mr ^2}-\frac {GmM} r $$ and that takes care of the signs for the derivative...

I think the lecture notes is referring to a circular orbit.

I guess I am asking about the derivation of the centrifugal potential energies. In the case of elliptical orbits, the effective potential energy (ignore gravitational potential energy) given by $\frac {L^2} {2mr^2}$ can be obtained from $\int mrw^2 \, dr$, while in the case of water spinning/circular orbit, we use $- \int mrw^2 \, dr$, essentially $- \int F \, dr$.

@PeroK thanks for the replies. But in both forms $- \int mrw^2 \, dr = - \frac {1} {2} mr^2w^2$ and $\frac {L^2} {2mr^2}$, the potential energy decreases with distance. I meant that $\frac {L^2} {2mr^2}$ is equal to $\frac {1} {2} mv^2$, hence my 'KE is a PE' comment.

What is the difference between these 2 systems, that resulted in 2 different methods of potential energy derivation?

For reference to the lecture notes in question: https://web.njit.edu/~gary/321/Lecture17.html

Please do point out any conceptual errors I have.

 

[ergospherical]

The effective centrifugal potential assumes the angular momentum to be constant.

This isn't correct. In a rotating frame (with angular velocity $\boldsymbol{\Omega}$ relative to an inertial frame) one may introduce a centrifugal potential energy term regardless of the dynamics of the system, i.e. whether its angular momentum is constant or not.

For simplicity, consider a unit mass moving in a potential $U(\mathbf{r})$ but otherwise arbitrarily (that is, arbitrary initial conditions). Recall the equation of motion of a particle in a uniformly rotating frame of reference $K$ in which $\boldsymbol{v} = \dfrac{d\mathbf{r}}{dt} \bigg{|}_K$ is\begin{align*}
\dfrac{d\boldsymbol{v}}{dt} &= -\dfrac{\partial U}{\partial \mathbf{r}} + 2\boldsymbol{v} \times \boldsymbol{\Omega} + \boldsymbol{\Omega} \times (\mathbf{r} \times \boldsymbol{\Omega}) \\

\boldsymbol{v} \cdot \dfrac{d\boldsymbol{v}}{dt} &= - \boldsymbol{v} \cdot \dfrac{\partial U}{\partial \mathbf{r}} + 2 \underbrace{\boldsymbol{v} \cdot \boldsymbol{v} \times \boldsymbol{\Omega}}_{=0} + \boldsymbol{v} \cdot \boldsymbol{\Omega} \times (\mathbf{r} \times \boldsymbol{\Omega}) \\

\dfrac{d}{dt} \left( \frac{1}{2}v^2 \right) &= - \frac{d}{dt} \left( U - \frac{1}{2} (\boldsymbol{\Omega} \times \mathbf{r})^2 \right)
\end{align*}therefore $E \equiv \frac{1}{2}v^2 + U - \dfrac{1}{2} (\boldsymbol{\Omega} \times \mathbf{r})^2$ is a constant and the term $- \dfrac{1}{2} (\boldsymbol{\Omega} \times \mathbf{r})^2$ is identified as a rotational contribution.

 

[A.T.]

I do not understand the Lecture notes. The effective centrifugal potential assumes the angular momentum to be constant. Thus $$U_{eff}(r)=\frac {L^2}{2mr ^2}-\frac {GmM} r $$ and that takes care of the signs for the derivative...

This isn't correct. In a rotating frame (with angular velocity $\boldsymbol{\Omega}$ relative to an inertial frame) one may introduce a centrifugal potential energy term regardless of the dynamics of the system, i.e. whether its angular momentum is constant or not.

You are talking about different potentials.

@hutchphd means this one with constant $L$ & $E$:
https://en.wikipedia.org/wiki/Effective_potential#Gravitational_potential

While @ergospherical assumes constant $\boldsymbol{\Omega}$

 

[ergospherical]

Yes the link posted by @phantomvommand analyses the motion of a test mass $m$ in the background potential $U(\mathbf{r})$ using a co-rotating reference system rigidly attached to the binary star system (with [approximately] constant angular velocity $\mathbf{w}$).