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Consider a moon orbiting its planet on a circular orbit. The moon is tidal locked to its planet, and has two permanent bulges, not exactly aligned with the planet. According to Newton's theory of gravitation, the moon should oscillates around its equilibrium alignment with the planet. I'm considering a simple moon represented by two small parts, with gravitationnal force acting on each part. The line joining both parts is confined in the orbital plane only, for simplicity. The moon should behaves like a pendulum. For very small angular displacements $\vartheta$, I've found this differential equation : \begin{equation}\tag{1} \ddot{\vartheta} + \frac{3 G M}{r_{\text{cm}}^3} \, \vartheta = 0, \end{equation} where $M$ is the planet's mass, and $r_{\text{cm}}$ is the distance from the planet to the center of mass of the moon. This is the equation of harmonic oscillations, and the oscillations angular frequency is thus \begin{equation}\tag{2} \Omega = \sqrt{\frac{3 G M}{r_{\text{cm}}^3}}. \end{equation} For our Moon, this gives a period of 15.8 days. On the surface of the Earth, holding a long rod by its center of mass, gives a period of 48.7 minutes (this is too long to be measurable, because of the friction that will stabilize the rod in its equilibrium vertical position. Also, the imprecision in the support would tilt the rod much faster in a direction or another).

Now, I never saw this anywhere, and I need a confirmation that it is right. Searching with Google about moon's oscillations gives me nothing.

I'm very surprised that the moment of inertia doesn't show in the angular frequency formula (2).


EDIT : Here are some details. The moon and its bulges are modeled as a light rod with a spherical mass on each end (dumbell-like moon). The spin angular momentum is defined relative to the center of mass of the moon : \begin{equation}\tag{3} \vec{S} = m_1 \, \vec{\tilde{r}}_1 \times \vec{\tilde{v}}_1 + m_2 \, \vec{\tilde{r}}_2 \times \vec{\tilde{v}}_2, \end{equation} where $m_1 = m_2 = \tfrac{1}{2} \, m_{\text{moon}}$, and vectors with a tilde are defined relative to the center of mass frame. We have $\vec{\tilde{r}}_2 = -\, \vec{\tilde{r}}_1$ (see the picture below).

enter image description here

Using the right hand rule for the cross product, I find this : \begin{equation}\tag{4} S = (m_1 \, \tilde{r}_1^2 + m_2 \, \tilde{r}_2^2) (\omega_{\text{rev}} - \dot{\vartheta}) \equiv I \, \omega_{\text{rot}}. \end{equation} The time derivative of the spin vector is equal to the torque applied on the moon : \begin{align} \frac{d\vec{S}}{dt} &= \vec{\tilde{r}}_1 \times \vec{F}_1 + \vec{\tilde{r}}_2 \times \vec{F}_2 \\[12pt] &= -\, \vec{\tilde{r}}_1 \times \frac{G M m_1}{r_1^3} \, \vec{r}_1 - \vec{\tilde{r}}_2 \times \frac{G M m_2}{r_2^3} \, \vec{r}_2 \\[12pt] &= -\, \frac{G M m}{2} \Big( \frac{1}{r_1^3} \, \vec{\tilde{r}}_1 \times (\vec{r}_{\text{cm}} + \vec{\tilde{r}}_1) + \frac{1}{r_2^3} \, \vec{\tilde{r}}_2 \times (\vec{r}_{\text{cm}} + \vec{\tilde{r}}_2) \Big) \\[12pt] &= -\, \frac{G M m}{2} \Big( \frac{1}{r_1^3} - \frac{1}{r_2^3} \Big) \, \vec{\tilde{r}}_1 \times \vec{r}_{\text{cm}} \tag{5} \end{align} Expanding the last parenthesis to lowest order gives \begin{equation} \frac{1}{r_1^3} - \frac{1}{r_2^3} \approx \frac{6 \, \tilde{r}_1}{r_{\text{cm}}^4} \, \cos{\vartheta}. \end{equation} Substituting this and (4) into equ. (5), using the small angle approximation : $2 \sin{\vartheta} \, \cos{\vartheta} \equiv \sin{2\vartheta} \approx 2 \vartheta$, and simplifying, gives equ. (1).

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  • $\begingroup$ what is the configuration of the bulges? Are they permanent, and in the plane of rotation? A diagram and your derivation would be helpful - but I don't understand the motion, and why the moment of inertia wouldn't be in your expression. $\endgroup$ – Floris Jun 12 '17 at 13:35
  • $\begingroup$ @Floris, the moment of inertia cancels out from the equation. The bulges are permanent, and in the plane of revolution, yes. $\endgroup$ – Cham Jun 12 '17 at 13:48
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    $\begingroup$ This was a problem on my Ph.D. qualifying exam. I dug out the solutions (since apparently I never throw anything out), and they agree with your result. Note that $\sqrt{GM/r_\text{cm}^3}$ is better known as the angular velocity of the orbit. Also note that this general sort of motion is called libration, which might give you something else to search on. $\endgroup$ – Michael Seifert Jun 12 '17 at 16:11
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    $\begingroup$ ... Thus, the total torque and the total moment of inertia of the body should be proportional, resulting in an oscillation frequency independent of $I$. The factor of 3 presumably can be written in terms of an angular integral. (I could try writing this up as a full answer if you're interested.) $\endgroup$ – Michael Seifert Jun 12 '17 at 16:51
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    $\begingroup$ @Cham sorry, default settings makes the picture as big as the full width. I made it "medium" sized now (just add a letter "m" at the end of the imgur link, in case you want to do this yourself in future; see edit history). $\endgroup$ – Floris Jun 12 '17 at 19:44
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Now, I never saw this anywhere, and I need a confirmation that it is right. Searching with Google about moon's oscillations gives me nothing.

I'm very surprised that the moment of inertia doesn't show in the angular frequency formula (2).

While your model is correct for your simple dumbbell model, it does not describe the Moon's librations. The Moon's librations are the result of the Moon's elliptical orbit rather than gravity gradient torque.

The reason moments of inertia don't show up in your simple model is that they cancel in that simple model. As is well-known (well-known in the artificial satellite community), the gravity gradient torque on an orbiting body expressed in the body-fixed frame of the orbiting object is approximately $$\tau_{gg} = 3\frac{GM}{r^3}\hat r \times (\mathrm I \hat r)\tag{1}$$ where

  • $\tau$ is the gravity gradient torque,
  • $GM$ is the gravitational parameter of the central body,
  • $r$ is the magnitude of the distance between the central body and orbiting body,
  • $\hat r$ is the unit vector from the central body to the orbiting body (or from the orbiting body to the central body; changing the sign has no effect), expressed in the body-fixed coordinates of the orbiting body, and
  • $\mathrm I$ is the orbiting body's moment of inertia tensor.

Suppose the orbiting body is rotating about the orbital angular momentum axis, that this rotation axis is a principal axis, that its rotation rate is, on average, equal to the orbital rate, and that the principal axis with the smaller of the remaining two principal moments of inertia is nearly co-aligned with the displacement vector connecting the two bodies.

Denoting $\hat x$ and $\hat y$ as the in-plane principal axes, with the $\hat x$ axis pointing more or less toward the central body, the unit vector $\hat r$ is $\cos\theta\,\hat x - \sin\theta\,\hat y$, where $\theta$ is small by assumption. The orbiting body's moment of inertia tensor is $I = \begin{bmatrix} A&0&0 \\ 0&B&0 \\ 0&0&C \end{bmatrix}$ where $A<B$, by assumption.

Applying equation (1) yields a gravity gradient torque of $$\tau_{gg} = -3 \frac{GM}{r^3} (B-A)\sin\theta\cos\theta \hat z = -3 \frac{GM}{r^3} (B-A)\frac12\sin(2\theta) \hat z$$ Premultiplying the above by the inverse of the inertia tensor yields the angular acceleration: $$\ddot\theta = \mathrm I^{-1}\tau_{gg} = -3\frac{GM}{r^3} \frac{B-A}C\frac12\sin(2\theta) \approx -3\frac{GM}{r^3} \frac{B-A}C \theta$$

This is a simple harmonic oscillator with angular frequency $$\Omega = \sqrt{\frac{3GM}{r^3} \frac{B-A}C}$$ In the special case $A=0, B=C$ (which is the case for your dumbbell model, or for a slender rod), this simplifies to $$\Omega = \sqrt{\frac{3GM}{r^3}}$$ In the case of the Moon, the principal moments of inertia are very close to one another, making the term $\sqrt{(B-A)/C}$ very small, which in turn makes the oscillation period much larger than the Moon's orbital period. This does not explain the Moon's libration.

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  • $\begingroup$ Ah, so you confirm the factor of 3, in the dumbull case ? $\endgroup$ – Cham Jun 16 '17 at 9:28
  • $\begingroup$ @Cham -- Yes. See the second link (the one beneath "well-known"). That factor of three should have been there all along. Thank's for catching the error. $\endgroup$ – David Hammen Jun 16 '17 at 12:05
  • $\begingroup$ I'm usrprised that in the torque formula that you cited, it isn't the quadrupole tensor, instead of the inertia tensor. Physically, it would make more sense if it was the quarupole tensor. $\endgroup$ – Cham Jun 16 '17 at 12:38

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