Tides and Dumbbell Dynamics

Abstract

We discuss a model describing the effects of tidal dissipation on the satellite’s orbit in the two-body problem. Tidal bulges are described in terms of a dumbbell, coupled to the rotation by a dissipative interaction. The assumptions on this dissipative coupling turn out to be crucial in the evolution of the system.

APPENDIX A. COMPUTATION OF $$\varepsilon(t)$$

In this appendix we want to show the detailed computation of $\varepsilon(t)$ starting from Eq. (2.17), which we rewrite in this form:

$$\alpha\dot{\varepsilon}(t)-\frac{\partial\mathcal{L}}{\partial\varepsilon}-\alpha\left(\dot{\varphi}-\dot{\vartheta}\right)=0.$$

(A.1)

First of all, we compute $\frac{\partial L}{\partial\varepsilon}$ from Eq. (2.4):

$$\frac{\partial\mathcal{L}}{\partial\varepsilon}=-\frac{gm}{\rho}\frac{\mu}{M_{E}}\frac{r^{2}}{{\rho}^{2}}3\cos{\varepsilon}\sin{\varepsilon}\simeq-3\frac{km\mu}{\rho}\frac{r^{2}}{{\rho}^{2}}\varepsilon,$$

(A.2)

where we have used $g=kM_{E}$ after developing $\cos{\varepsilon}\sin{\varepsilon}$ in power series of $\varepsilon$ and keeping only the linear terms in $\varepsilon$.

Recall that $\mu$ represents the mass of the ocean’s bulges. These bulges can be imagined as an ellipsoid of radii $R_{E}$, $R_{E}$ and $R_{E}+h$ deprived of a sphere of radius $R_{E}$ concentric to it, $h$ being the tidal height.

If we replace $h$ with the Newton formula for tidal height:

$$h=\frac{3}{2}\frac{m}{M_{E}}\left(\frac{R_{E}}{\rho}\right)^{3}R_{E},$$

(A.3)

then $\mu$ becomes

$$\mu=\delta_{W}\frac{4}{3}\pi R_{E}^{2}h=\delta_{W}\frac{4}{3}\pi R_{E}^{2}\frac{3}{2}\frac{m}{M_{E}}\left(\frac{R_{E}}{\rho}\right)^{3}R_{E}=\frac{3}{2}\frac{\delta_{W}}{\delta_{E}}m\left(\frac{R_{E}}{\rho}\right)^{3}w,$$

(A.4)

where $\delta_{W}$ and $\delta_{E}$ represent the densities of the liquid part and the solid part of the planet. Finally, in (A.2), $r$ is the distance between the center of the Earth and the center of mass of each bulge, that is, the center of mass of the aforementioned ellipsoid deprived of a sphere. It is a standard calculation to show that $r=\frac{3}{4}R_{E}$ up to terms that go to zero as $\frac{h}{R_{E}}$.

Therefore,

$$\frac{\partial\mathcal{L}}{\partial\varepsilon}=-\frac{81}{32}\frac{\delta_{W}}{\delta_{E}}km^{2}\frac{R_{E}^{5}}{{\rho}^{6}}\varepsilon=-Dkm^{2}\frac{R_{E}^{5}}{{\rho}^{6}}\varepsilon,$$

(A.5)

with $D=\frac{81}{32}\frac{\delta_{W}}{\delta_{E}}$ a dimensionless constant.

Equation (A.1) thus becomes

$$\alpha\dot{\varepsilon}+Dkm^{2}\frac{R_{E}^{5}}{{\rho}^{6}}\varepsilon-\alpha\left(\dot{\varphi}-\dot{\vartheta}\right)=0.$$

(A.6)

Finally, we have to consider the time dependence of $\rho$ and $\vartheta$. We also assume that $\Omega$, $\omega$ and $a$ remain constant during each revolution and change their values only at the end of each revolution. This assumption yields

$$\vartheta\simeq\lambda+2e\sin{\lambda}=\omega t+2e\sin{(\omega t)}\implies\dot{\vartheta}\simeq\omega+2\omega e\cos{(\omega t)}$$

(A.7)

and

$$\rho(t)=\frac{p}{1+e\cos{(\omega t)}}=\frac{a(1-e^{2})}{1+e\cos{(\omega t)}}\simeq a[1-e\cos{(\omega t)}]\implies\frac{1}{\rho^{6}}\simeq\frac{1}{a^{6}}[1+6e\cos{(\omega t)}].$$

(A.8)

So, Eq. (A.6) becomes

$$\dot{\varepsilon}+\frac{\gamma_{c}}{\alpha}[1+6e\cos{(\omega t)}]\varepsilon-\Omega+\omega+2\omega e\cos(\omega t)=0,$$

(A.9)

where $\gamma_{c}=Dkm^{2}\frac{R_{E}^{5}}{{a}^{6}}$.

A trivial solution of this equation is

$$\varepsilon(t)=\frac{\alpha}{\gamma_{c}}(\Omega-\omega)-e\frac{6\Omega-4\omega}{\omega}\cos{\delta}\sin(\omega t+\delta)=A+eB\sin(\omega t+\delta),$$

(A.10)

with $A=\frac{\alpha}{\gamma_{c}}(\Omega-\omega),B=\frac{6\Omega-4\omega}{\omega}\cos{\delta}\text{ and}\tan\delta=\frac{\gamma_{c}}{\alpha\omega}$.

Remark 1

The ratio $\frac{\alpha}{\gamma_{c}}$ contains, as a whole, the information about the quantities appearing in the system. This suggests, for instance, that the friction coefficient $\alpha$ has to decay very fast for increasing $a$, the semimajor axis of the orbit, in order to balance the dependence of $\gamma_{c}$ on $a$ (see the comments immediately following (2.5)). This sounds reasonable, since the friction should depend on the total amount of liquid involved in the motion of the bulges, and this clearly depends on $a$. However, we do not have a detailed model of the tidal currents inside the oceans (in the Earth – Moon case) or in the inner mantle (Io – Jupiter case). We think that one of the virtues of the simple model we presented in this work is the fact that it suggests the correct relations among the various elements of the orbits, while their quantitative evaluation has to be based on empirical observations.

APPENDIX B. CHARACTERISTIC TIMESCALES OF THE SYSTEM

In this appendix we want to compare the characteristic timescales of the system in order to show that the assumptions we made in our model are justified.

In particular, as we explained above, we are interested in the comparison between the timescale of elongation of the semimajor axis, which we call $\tau_{a}$, and the timescale of circularization of the orbit, that is, $\tau_{M}$ or $\tau_{I}$ in Eqs. (2.29) and (3.12), respectively.

From Eq. (2.32) we see that $\tau_{a}\simeq\tau_{\Omega}$, with $\tau_{\Omega}$ the characteristic timescale of evolution of $\Omega$. Moreover, we can rewrite Eq. (2.16) as

$$I\dot{\Omega}\simeq-\alpha\Omega,$$

(B.1)

hence we have $\tau_{\Omega}\simeq\frac{I}{\alpha}$.

On the other hand,

$$\tau_{M}=\frac{L^{4}}{G\dot{G}L^{2}+LG^{2}C}\simeq\frac{L}{\dot{G}+C}\simeq\frac{L}{\alpha}\frac{1}{18\Omega\frac{\Omega}{\omega}}.$$

(B.2)

Therefore, in the case of the Earth – Moon system, $\tau_{M}\simeq\frac{L}{500\alpha\Omega}$ and hence

$$\frac{\tau_{a}}{\tau_{M}}\simeq\frac{500I\Omega}{L}\simeq 500\frac{\Omega}{\omega}\frac{\frac{2}{5}M_{E}R^{2}_{E}}{mR^{2}_{EM}}\simeq 500\cdot 28\cdot\frac{2}{5}\frac{M_{E}}{m}\left(\frac{R_{E}}{R_{EM}}\right)^{2}\simeq 120>1.$$

(B.3)

Conversely, in the case of the Jupiter – Io system, the analogue of Eq. (B.2) is $\tau_{I}\simeq\frac{L}{80\alpha\Omega}$, yielding

$$\frac{\tau_{a}}{\tau_{I}}\simeq\frac{80I\Omega}{L}\simeq 80\frac{\Omega}{\omega}\frac{\frac{2}{5}M_{J}R^{2}_{J}}{mR^{2}_{JI}}=32\cdot 1\cdot\frac{2}{5}\frac{M_{J}}{m}\left(\frac{R_{J}}{R_{JI}}\right)^{2}\simeq 1.9\times 10^{4}\gg 1.$$

(B.4)

These calculations show that the assumption of a constant $\mu$ appears to be very reasonable already in the case of the Earth – Moon system. In the case of Jupiter and Io the approximation is much better and the difference between a variable and a constant $\mu$ is really negligible.