Spin of Mercury

Introduction

The planet Mercury has an unusual spin-orbit relation. It is similar to tidal locking seen in circular orbits, such as the Earth-moon system. However, where the moon spins once for every orbit, Mercury spins three times for every two orbits.

Two explanations for this peculiar resonance exist in Colombo [1]. One, a common theory, has a high rate of spin being reduced by tidal friction into the resonance we see today. The other has Mercury locked into a retrograde orbit with Venus. For a history on the subject of mechanisms, see [2].

The Spin of Mercury (or tidal locking) is one the last arguments in a series explaining solar system formation (the origin of rings and craters are the others). As explained earlier, solar systems are believed to have a liquid phase early in development. Demonstrating Mercury will tidally lock to the Sun as a liquid, will reinforce this theme. In fact, all moons tidally lock as a liquid.

Place a liquid (molten) Mercury in an elliptical orbit around the Sun. There is a problem with the present theory. If Mercury was captured from the outside, how do they know which direction it spins when placed in orbit. We start with a rule for liquid moons that eliminates this problem:

  • A molten (or liquid) moon placed in an orbit begins with zero spin.

Then tidal forces spin the moon up (not down) in the correct direction.

Since liquids deform easily, we can model the planet/moon as a dumbbell to calculate the torque and spin. For a circular orbit, the spin of the moon is the spin of the orbit. For an elliptical orbit, the spin of the planet/moon lies between the spins produced at the apse points.

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[1] Colombo, G. and Shapiro, I. I. 1965. {\it The Rotation of the Planet Mercury},American Astronomical Society.

[2] Noyelles, B., Frouard, J., Makarov, V. V. and Efroimsky, M. {\it Spin-orbit evolution of Mercury revisited}, Icarus. 241 (2014): 26–44

Tidal Shape of Liquids

We generate ellipsoids [use Roche ellipsoid?] to examine the behavior of liquid moons in orbit around a planet. For a liquid there is little to no tensile strength. If the change in the radius is simply proportional to the forces at the surface, particularly at the axis, then the equations of the major (a) and minor (b) axis are as follows:

\begin{align}
a1 = r_{moon}*(1 + F1/F_{moon})\\
b1 = r_{moon}*(1 - F2/F_{moon})
\end{align}

where

\begin{align}
F_{moon} &= Gm/r_{moon}^2 \\
F1 &= GM/(d-r_{moon})^2-(d-r_{moon})w^2 \\
F2 &= (r_{moon}/R)*GM/R^2,\textrm{ where  } R= \sqrt{d^2+r_{moon}^2}
\end{align}

Then adjusting for constant volume

\begin{align*}
\textrm{ratio} &= a1/b1\\
b &= r_{moon}/\textrm{ratio}^{1/3} \\
a &= b*\textrm{ratio}
\end{align*}

The only comparison for a homogeneous liquid was an approximation (e << 1) between the Earth and the moon [3], where we both calculated a tidal distortion of 50 meters — between a and b. Calculations for some moons of Saturn are listed in Table 1 along with the measured values.

Most moons match well. We conclude the method we use to generate ellipsoids is adequate for what we wish to demonstrate.

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[3] Fitzpatrick,R. (2016-01-22), {\it Roche Ellipsoids}, \ (https://farside.ph.utexas.edu/teaching/336L/Fluid/node37.html), 10/9/2019.

Model the Liquid

With little to no tensile strength, a liquid planet or moon should take the shape of an ellipsoid when tidally distorted. We wish to model this ellipsoid as a dumbbell of half length, \ell, as a possible convenience to evaluate torque. A diagram of the ellipsoid is shown in Fig. 1. The scheme is to subtract a sphere of radius b from the ellipsoid and find the center of mass of the remainder.

Figure 1: he moment, \ell, of the liquid remaining after subtracting a sphere of radius b from the ellipsoid.

The equation for the ellipse is x^2/a^2 + y^2/b^2 = 1, a>b, and the equation of a sphere is x^2 + y^2 = b^2, of radius b. We find the centroid, \overline{x}, for half the volume as follows:

\begin{align*}
\int{x dV} &= \pi \int{x y^{2}dx} = \int_{0}^{b}  [\frac{b^{2}}{a^{2}}  (xa^{2} - x^{3}) - (xb^{2} - x^{3})]dx \\
  &\qquad \qquad \qquad+ \int_{b}^{a}\frac{b^{2}}{a^{2}}(xa^{2} - x^{3})dx
\end{align*}
\begin{align}
&\qquad \qquad\qquad\quad \quad= \pi \frac{b^{2}}{4}(a^{2}- b^{2}) \\
\int{dV} &= \frac{\pi}{2}(\frac{4}{3}ab^{2} - \frac{4}{3}b^{3}) \;=  \pi \frac{2}{3}b^{2}(a - b) \\
\overline{x} &= \int{x dV} / \int{dV} \:\:=  \frac{3}{8} (a + b) = \it{\ell}
\end{align}

Model the Torque

We are interested in how the spin of the planet evolves in an orbit. This is governed by torque applied to the distortion of the planet. The spin of our moon is synchronous with its orbit (tidally locked), or \Omega = \omega_{orbit}, where \Omega is the spin of the moon. The planet Mercury is in an elliptical orbit. We claim the spin will be between the apse points of the orbit, or \omega_a < \Omega < \omega_p .

A model of the planet/sun system showing forces on the distortions is shown in Fig. 2. Here, the torque equals the change in angular momentum (spin) or, \tau = I\dot{\Omega}, where the following apply:

\qquad\tau = \textrm{Torque} = \textrm{sin}(\theta)\, \it{\ell}\textrm{\,F,\quad where}
\qquad\qquad\textrm{F} = \textrm{GmM}_{2}\textrm{/R}^{2}
\qquad\qquad\textrm{M}_{2} = \textrm{mass of the Sun}
\qquad\qquad\textrm{m} = \textrm{mass of dumbbell}
\qquad\qquad\theta = \textrm{angular position of dumbbell}

\qquad\textrm{I} = \textrm{ Inertia} = \frac{2}{5}\textrm{M}_{b}b^{2} +2\textrm{m}\it{\ell}^{2}, \quad\textrm{where}
\qquad\qquad\textrm{m \;\;= }\;(\textrm{M}_{1} - \textrm{M}_{b})/2
\qquad\qquad  \textrm{M}_{1} =  \textrm{mass of the planet/moon}
\qquad\qquad  \textrm{M}_{b} =  \textrm{mass of sphere with radius b}

\qquad\Omega  =  \textrm{spin of the planet}
Figure 2: Dumbbell representation of distorted planet (Mercury), M_{1}, acted on by the Sun, M_{2}.

Outline of Computation

The tidally distorted planet is modeled as a dumbbell and the torque is calculated by the following equations (see Fig. 2):

\begin{align}
R_{1}&= \sqrt{(r-\ell cos(\phi))^{2} + (\ell sin(\phi))^{2})} \\R_{2}&= \sqrt{(r+\ell cos(\phi))^{2} + (\ell sin(\phi))^{2})}  \\s_{1}&= r sin(\phi)/R_{1} \\s_{2}&= r sin(\phi)/R_{2} \\F_{1}&= s_{1} \ell m M_{2} G/R_{1}^{2} \\F_{2}&= s_{2} \ell m M_{2} G/R_{2}^{2} \\
\end{align}

where,

\begin{align}
\phi&= (\omega - \Omega)/\omega
\end{align}

Then the Torque, \tau = \tau_{1} – \tau_{2}. The value, \phi, is the lag angle between the distortion and the sun. A positive angle (1^{o} when \Omega = 0) adds spin. A negative angle subtracts. When \Omega > \omega_{a}, the contributions begin to decrease. If the spin of one orbit is positive, it adds to the total. If the spin of one orbit is < 0, the program ends.

The spin of one orbit is the sum of spin time intervals, as follows:

\begin{align}
\dot{\Omega}&= -\tau/I \\
Omega&= \Omega + \dot{\Omega} \Delta t \\
\end{align}

Calculation of Spin

We start by dividing the orbit into intervals of \Delta \it{t}. Calculating the torque, we sum the spins at each interval using equation Eq. (17) for one orbit. The orbits are summed until the spin of one orbit is negative. This is the orbit of resonance, \omega_{r}.

The contribution of spin depends on the lag angle, \phi (see Fig. 2). When \Omega < \omega, the lag angle, \phi, and the torque, \tau, are positive. Each calculated spin adds to the total. When \Omega > \omega, \phi and the torque, \tau, are negative. Each calculated spin subtracts from the total. As the spin increases, there is some point on the orbit where the additions equal the subtractions, at some \omega_{r} in Fig. 3. This is the maximum (tidally locked) spin.

Figure 3: Points, \omega‘s, slower than \omega_{r}, decrease spin. Points, \omega‘s, faster than \omega_{r}, increase spin. Equivalently, \phi < 0 => the torque, \tau < 0 and \phi > 0 => the torque, \tau > 0 (see Fig. 2). At \omega_{r}, the contributions from both cancel for one orbit.

The calculated spin is\,\omega_{r} = 1.04×10^{-6}rads/sec, or 2.54 m/sec denoted on Fig. 4, with a calculated spin up time of 2.8×10^{6} years. Fig. 4 shows the spin velocity is between the maximum and minimum orbital velocities — an indication of this type of tidal locking for elliptical orbits.

Figure 4: Calculated spin velocity of Mercury is between the angular velocities at the apse points of the orbit. Calculated spin velocity is less than the observed value.

The calculated spin is below the measured (actual) spin velocity of 3.026 m/sec. If this method is correct, this would indicate an initial orbit closer to the Sun, i.e. the spin developed closer to the Sun. We can choose a more eccentric orbit a little closer to the Sun by trail and error. Then r_{p} = 4.2×10^{10} m (about 4×10^{9} meters closer to the Sun) and r_{a} = 8.2×10^{10} m (about 1.2×10^{10} meters farther from the Sun). The calculated spin is \omega_{r} = 1.24×10^{-6}rads/sec, or 3.026 m/sec denoted on Fig. 5, with a calculated spin up time from rest of 2.3×10^{6} years. The spin up as a function of the number of orbits (or time) is shown in Fig. 6.

Figure 5: Calculated spin velocity matches the observed spin velocity for an eccentric orbit closer to the Sun.
Figure 6: Spin up versus \# of orbits (or time, \sim 2 Myrs) for an eccentric orbit closer to the Sun

Conclusion

As a liquid in an elliptical orbit, Mercury is tidally spun up to an orbital value, \omega_{r}. This is a steady state value between apse points (\omega_{p}, \omega_{a}), where the torque contributions cancel. This is an example of tidal locking for an elliptical orbit. The calculated spin value did not match the observed value, so we suggest Mercury may have formed closer to the Sun.

A case can be made defining a moon, as those objects tidally locked around a primary. Mercury fits this definition.