Correia & Laskar, in their recent article, in J. Nature 429, 848-850 (2004), said that the spin-orbit resonance of Mercury is 3/2 (sidereal orbit period / sidereal rotation period). We may understand from their article that Mercury performs 1.5 cycle around its axis during its orbit-cycle (87.969 days). At NASA's web site, David William, mentioned in the "notes on the fact sheets" that, the sidereal rotation period (which is 1407.6 hrs for Mercury) is "the time for one rotation of the body on its axis relative to the fixed stars." As I understand, the frame of reference for this method of observation is the fixed stars and the observer may be standing on earth. Besides, the data of sidereal rotation may be the result of measuring the time that any physical-object on the planet's surface, or a specific face of Mercury, takes to return to its same position in relation to the fixed stars. By this way, the identified spin (3/2) is irrelevant to the daily motion of the sun when observed from the planet's surface; the latter identifies the true length of Mercury's solar day (4222.6 hrs). Contrary to the case Mars and other planets in our solar system excluding Vinous, Mercury's sidereal rotation period does not equal to its solar day; for Mercury the former is 1/3 of the latter. It appears, therefore, in the case of Mercury, we should differentiate between two types of spin that we might call them: the sidereal spin (3/2) and the diurnal spin (1/2). Which of them is the true spin of Mercury and what causes this difference?
During the last three days, I tried to figure out the answer for this question, using the following analysis that I call "3/6 spin-orbit resonance." We may divide Mercury's orbit into thee equal segments, that we may name them anti-clockwise as: AB, BC, and CA. Then, divide the equatorial perimeter of Mercury into 6 equal segments that we may name them clockwise as: ab, bc, cd, de, ef, and fa. If the motion starts when the center of Mercury was at "A" and the point "a" on Mercury's equator was in the same plane that includes "A" and the center of the sun. The projection of this plane is the line that links the three points in the drawing. Then, after 29.323 Earth-days, Mercury will be at "B" and the points: "b", "B", and the center of the sun will be in one plane. When Mercury reaches "C", after another 29.323 Earth-days, the points: "c", "C" and the center of the sun will be in one plane (one line in drawing). At "C", the point "a" on Mercury's equator will be observed as it did return to its initial orientation when Mercury was at "A", indicating that Mercury has completed one cycle around its axis, while, it performed only 1/3 of it. Then, when Mercury returns again to "B" after another 58.646 Earth-days, the point "a" will have similar orientation, as its initial position, indicating that Mercury has completed two cycles around its axis, while, it performed only 2/3 of it. Again, after 58.646 Earth-days, Mercury will be in its third transit at "A". The points "a" , "A" and the center of the sun will be again in one plane. By then, Mercury has consumed the time of two orbital rounds (176 Earth days) and has completed one rotational cycle around its axis, while the observer of the point "a" will think that Mercury has performed 3 rounds around its axis during two orbit periods.
This analysis of 3/6 spin-orbit resonance may show, as well, why the diurnal motion of the sun when observed from Mercury's surface will not be as normal as the case of the Earth. The observed motion of the sun, from the planet's surface, will nearly stop during the planet's orbital motion between "B" and "C".
Sorry, there is no figure for this analysis; I will try to find out how to attach a figure with this text.
