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In "Physics of the Earth" by Frank Stacey and Paul Davis, they approximate the height of the tides on page 103.

They assume the Earth is perfectly spherical, covered in a uniform layer of water and that the Earth spins as fast as the Moon orbits around it, so as to present a constant face to the Moon. They use the below diagram to help with the upcoming derivation:

Figure 1

Now, they claim that the potential $W$ (their choice of variable, not mine) at the point $P$ is the sum of the potential due to the Moon's gravitational force at point $P$ and the Earth-Moon system's orbital motion about its barycentre. They state the following expression for $W$: $$W = -\frac{Gm}{R'} \color{red}{-} \frac{1}{2}\omega_L^2 r^2.\tag{1}$$

Could someone explain to me why the potential due to the orbital motion is negative? I did the following working to find it myself:

A point-like test mass at $P$ with mass $M$ would have total mechanical energy $$E = E_K + E_{rot} + E_{moon} = E_K + \frac{1}{2}I\omega_L^2 - \frac{GmM}{R'}.\tag{2}$$

Thus, $$E_K = E - \left(\frac{1}{2}I\omega_L^2 - \frac{GmM}{R'}\right).\tag{3}$$

I can call the right-hand expression the 'effective potential energy' of the system, which means that potential $W$ is $$W = \frac{U_{eff}}{M} = -\frac{Gm}{R'} \color{red}{+} \frac{1}{2}\omega_L^2 r^2.\tag{4}$$

As you can see, my 'rotational potential' is positive, as opposed to their negative value. Could someone please explain what I did wrong (or if their expression is a typo)?

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    $\begingroup$ I'm a bit confused about this. Why isn't there a term for the gravitational potential of the Earth? I assume we're using a co-rotating reference frame, i.e the Earth & Moon are at rest in the frame & hence there are fictitious centrifugal forces. That's what's done when finding Lagrange points. $\endgroup$
    – PM 2Ring
    Commented Sep 21, 2022 at 10:24
  • $\begingroup$ I was similarly confused about that. I ended up justifying it to myself with "it cancels out with the normal force and so doesn't affect tide magnitude". Not sure if that's correct reasoning. And yes, there would be a centrifugal force. $\endgroup$
    – Trisztan
    Commented Sep 21, 2022 at 10:40

3 Answers 3

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The centrifugal force is known to be $$\vec{F}_\text{centrifugal} = M\ \omega_L^2 r\ \vec{e}_r$$ where $r$ is the distance from the axis of rotation and $\vec{e}_r$ is the unit vector pointing away from the axis.

A force (both gravitational, and centrifugal as well) is defined by the negative gradient of a potential. $$\vec{F}=-M\ \vec{\nabla}\ W$$

So with the centrifugal force from above we have $$\vec{F}_\text{centrifugal}=-M\ \vec{\nabla}\ W_\text{centrifugal}$$ or $$M\ \omega_L^2 r\ \vec{e}_r=-M\ \vec{\nabla}\ W_\text{centrifugal}$$

The solution of this equation is $$W_\text{centrifugal} = - \frac{1}{2}\omega_L^2 r^2$$ because $\vec{\nabla}(\frac{1}{2}r^2)=r\ \vec{e}_r$.

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  • $\begingroup$ So, my expression for the particle's energy at $P$ is inherently incorrect? Because a few days after making this post, I myself found this expression for the potential due to the centrifugal force, but still didn't quite understand why my energy expression is wrong. Doesn't the particle have rotational kinetic energy $\frac{1}{2}I\omega_L^2$ at $P$? $\endgroup$
    – Trisztan
    Commented Sep 30, 2022 at 8:05
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    $\begingroup$ @Trisztan You are confusing the rotating and non-rotating frames of reference. For understanding the tides you need the rotating frame of reference. In this frame the water is nearly at rest (i.e. no kinetic energy) and feels the centrifugal potential. Only in the non-rotating frame the earth and water would have large kinetic energy (from velocity $\omega_L r$), and doesn't feel any centrifugal potential. $\endgroup$ Commented Sep 30, 2022 at 9:21
  • $\begingroup$ Ah that's true. Makes sense now, thank you. $\endgroup$
    – Trisztan
    Commented Sep 30, 2022 at 11:37
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lets look at the reduce "two body" equations at the center of mass coordinate

I ) non rotate CM coordinate system

with

$$T=\frac{\mu}{2}\,\dot r^2\\ U=-\frac{G\,m\,M}{r}$$ and the Lagrange $~L=T-U~$ you obtain the EOM

$$ \mu\,\ddot r=\underbrace{-\frac{G\,m\,M}{r^2}}_{F_G}\quad\Rightarrow\\ U=-\int F_G\,dr$$

II ) rotate CM coordinate system

if the CM coordinate system rotate you obtain additional fictitious forces . the EOM is now

$$ \mu\,\ddot r=\underbrace{~-\frac{G\,m\,M}{r^2}~}_{F_G}\underbrace{~+\mu\omega^2\,r~}_{F_Z}\,\quad\Rightarrow\\ U=-\int (F_G+F_Z)\,dr=-\frac{G\,m\,M}{r}-\frac{\mu}{2}\,\omega^2\,r^2$$

so your potential energy is not correct !!


the EOM with EL

$$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot r}\right)=\left(\frac{\partial L}{\partial r}\right)$$

Edit:

$$\vec F_Z=-m\,(\vec\omega\times\,(\vec\omega\times \vec R))$$

with

$$\vec\omega=\begin{bmatrix} 0 \\ 0 \\ \Omega \\ \end{bmatrix}\quad, \vec R=\begin{bmatrix} r \\ 0 \\ 0 \\ \end{bmatrix}\\ \vec F_Z= \begin{bmatrix} m\,\Omega^2\,r \\ 0 \\ 0 \\ \end{bmatrix}$$

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    $\begingroup$ I'm sure your working is correct, but I haven't done Lagrangians yet, so unfortunately I don't know quite understand your answer. But thank you for taking the time to answer regardless. $\endgroup$
    – Trisztan
    Commented Sep 30, 2022 at 8:03
  • $\begingroup$ did you unterstand the answer from @fritsch ? $\endgroup$
    – Eli
    Commented Sep 30, 2022 at 10:26
  • $\begingroup$ Yeah, his answer and the comment made it clear. $\endgroup$
    – Trisztan
    Commented Sep 30, 2022 at 11:37
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I haven't done Lagrangians or Hamiltonians yet, so I didn't actually understand much of the mathematics of the posts, but these two questions answered mine as well:

Lagrangian of an effective potential

Effective potential of a two-body system

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