# The Solar System/Chapter 2

II
MERCURY
Nearest to the Sun of all the bodies of the system, excepting only the swarm of particles which give us the Zodiacal Light, is Mercury.
Till very lately, we knew next to nothing about Till lately very little known of it. this planet. Its doings, as represented by its path, were well determined, but its self not at all. Part cause of this was its nearness to the Sun; a part, its being an inferior planet, and thus being but ill seen when most observable; for when at its greatest apparent distance from the Sun,—at one of its elongations, as it is called,—half of it alone is illuminated, and that half but poorly. Secondly, when it appears to the naked eye, and when in consequence it is generally looked for with the telescope, it is deep sunk in the vapors of the horizon, and the air through which it is seen is so tremulous that its disk, in consequence, is ill-defined. As this was supposed the best time for observation, the disk was deemed inscrutable.
But the obvious is to be avoided. Acting uponMarkings detected in 1889. this principle, Schiaparelli, in 1889, took a new departure by systematically observing Mercury by ​day. He was before long rewarded. Markings began to show themselves upon the little disk, difficult of detection, indeed, but still visible enough to enable him to be satisfied of their permanency ; and then the markings disclosed of themselves a very singular fact. From the stability of their positions, it became evident that the planet rotated upon its axis in the same time that it revolved about the sun.
Let us consider a moment how it was theRotation and revolution isochronous. markings disclosed this fact. Suppose, for simplicity, a body revolving round its primary in a circle and made visible by the light received from it. Furthermore, suppose the revolving body to have markings upon it, and to rotate once upon its axis as it makes one revolution round its sun. Clearly it will always present the same face to the central attracting and illuminating body, and therefore the markings will maintain an invariable position with regard to the illuminated face. To an outsider, the planet, if inferior, will present the phases of the Moon. Unlike the Moon, however, the illumination will not sweep over an invariable face, but lighting and lighted will rotate together ; for in the case of the Moon, we are the attracting, but not the illuminating, body ; in the case of a planet, the Sun is both. ​
Schiaparelli was the only one to see theseFlagstaff corroborates Schiaparelli. markings till 1896, when the subject was taken up at Flagstaff. The planet was at the time coming out from inferior conjunction, and was at first no easy matter to find ; for in relative visibility Mercury behaves like the Moon. Size of disk does not begin to compensate for phase, as calculation would lead one to expect ; because obliquity of illumination greatly enfeebles its amount. The planet presented so faint a contrast with the sky that on one occasion an assistant, coming to look at it through the telescope, could not see it until its exact position was pointed out to him ; and I always picked it up myself by trailing it across the field, an object in motion being much more evident than one at rest, as every hunter knows. Nor could I at first make much out of it ; it was only a pretty little moon nearly lost in the vast blue sky. To my surprise, however, as it left elongation to return to the Sun, it grew brighter and brighter, and distinct dark markings came out upon its disk. The best views occurred when popular almanacs inform their readers : "Mercury invisible during the month." In the clear sky and steady air of Arizona and Mexico the markings were not especially difficult objects, though more difficult than the canals on Mars. They ​were narrow, irregular lines and very dark. They were not in the least like the markings on Mars. There were no large patches of shade on the one hand, nor fine, regular pencilings on the other. Its lines were fairly straight, but broken and of varying width. "Cracks" best explains their appearance, and probably their nature.
Their positions were unmoved, even after as much as five hours' interval.
As I continued to map them, I marked thatMarkings reveal libration in longitude. while their relation to the terminator was unchanged by the hours, it was slowly shifting with the days. The lines were gradually passing over its edge, and it dawned on me what I was witnessing: the swaying, or libration, of the planet in longitude due to the eccentricity of the planet's orbit. ​
Libration in longitude is a necessaryCause of libration in longitude. consequence of the planet's moving in a focal conic. The moment of rotation of a body of Mercury's mass is so great that it would take more than the Sun's might to suddenly alter it. The planet turns upon its axis, therefore, with a uniform spin. But its angular speed in its orbit is not uniform. Since the radius vector sweeps out equal areas in equal times, the angular velocity near perihelion exceeds that near aphelion. The revolution gains on the rotation here, and at the end of a certain time reaches its maximum ; after which the rotation gains on the revolution, and the deficiency is made up again at aphelion.
$\pi ab$
$\scriptstyle {\frac {\pi ab}{T}}\,\!$
$\scriptstyle {\sqrt {ab}}\,\!$
To find, therefore, the point on the ellipse where the radius has the value corresponding to the mean angular velocity, we must take the expression for r of the ellipse referred to its focus as a pole, ​
$\scriptstyle r={\frac {a\;(1-e^{2})}{1+e\cos v}}\,\!$,
$\scriptstyle {\sqrt {ab}}\,\!$
Consequently for the point sought whence, since
$\scriptstyle {\sqrt {ab}}={\frac {a\;(1-e^{2})}{1+e\cos v}}\,\!$,
$\scriptstyle b=a{\sqrt {1-e^{2}}}\,\!$
$\scriptstyle (1-e^{2})^{\frac {1}{4}}={\frac {(1-e^{2})}{1+e\cos v}}\,\!$,
$\scriptstyle \cos v={\frac {(1-e^{2})^{\frac {3}{4}}-1}{e}}\,\!$
In the case of Mercury, e = .205605; v, the true anomaly of the point of maximum libration, is therefore 98° 55'.13.
But $\scriptstyle {\frac {a-r}{a\;e}}=\cos E\,\!$,
$\scriptstyle E-e\sin E=M\,\!$
$\scriptstyle v-M=\zeta \,\!$
The gain or loss of the rotation over the revolution is the same thing as the equation of the centre.
We have, then, in the libration, a most conclusive and interesting proof of the isochronism of rotation and revolution.
The next point to consider is what caused this ​isochronism. This question raises a wholly newNew branch of celestial mechanics. set of problems in celestial mechanics from those in which celestial mechanicians were wont to engage. Until recently, mathematical astronomy dealt almost entirely with solids,—entirely so outside the consideration of the Earth. But no solid is absolutely rigid, and the action of one body upon another must cause mutual deformation of figure and give rise to tides in the two masses. Darwin has shown that this tidal action is an important cosmic factor, one which has played as constructive a part in the evolution of things as gravitation itself.
Not only were the planets not rigid in the past ; they are not rigid to-day. So far as we can judge, all the planets behave as plastic bodies at the present moment. So great are the masses that, even in the case of the denser and cooler ones, deformation of figure seems to be what fluidity and rotary conditions would require. They are, therefore, fit subjects for tidal action.
Owing to the great importance of the subject,Tides—how caused. and to the fact that the explanation given of it in almost all the text-books is erroneous, I shall present it to you with some pains, the more so that the action may, I think, be outlined quite ​simply. The prestige of Sir Isaac Newton's name is responsible for the inertia which still carries the usual explanation rolling down the ages. He attempted to explain the tides statically, and the account he gave has been blindly copied and perpetuated. But the problem is not a static, but a kinematic one ; the body acted on is in motion at the time of the action, and this entirely changes the result. Let me give you an analogous instance of the impossibility of treating a problem of motion as if it were one of rest. The precession of the equinoxes is a case in point, and may be seen in a gyroscope. If a weight be hung on the axis of the wheel while the latter is at rest, the wheel instantly turns into the horizontal plane and stays there. This is a case of statics. If now the wheel be set in motion, however slightly, the wheel, instead of lying down in the plane of the pull once and for all, simply rotates in space without any change of inclination whatever. This is a case of kinematics. Kinematic questions always thus differ from static ones.
Nor can the motion be tacked on afterward, as simultaneity is of the essence of the problem. If the effect of the Earth's rotation was merely to carry forward the crest of the tide through friction, it is the deep-water tides,—those in water ​over 12¾ miles deep, not the shallow, that would be nearest under the Moon.
Disturbing force. Consider a body revolving freely around another in a circle, and disturbed in this motion by a third. This is the case with any particle of the ocean when we neglect pressure and friction. Connect the three bodies by lines, and, keeping their directions, increase their lengths inversely
as their squares, and join the ends. The disturbing force will be represented by the connecting line, on the principle of the composition of forces.
$\textstyle {\frac {CM^{2}}{PM^{2}}}={\frac {PM}{NM}}$;
If M be far away compared with CP,
$\textstyle BN=2CB=2p$
$\textstyle PM=BM=D$
$\textstyle CB=p$
$\textstyle {\frac {(D+P)^{2}}{D^{2}}}={\frac {D}{NM}}$
$\textstyle NM=D-2p$
$\textstyle BM-NM=BN=2p$
The tide-raising force PN may be resolved into a normal disturbing force PL and a tangential disturbing force LN. From the fact that BN is always twice CB, we find for the vanishing points of the normal force a and b, those where the angle BCP = 54° 44’. The whole disturbing force is there tangential.
Now consider the action of the two components; first, that of the tangential factor. At F, the whole force is normal and acting inward. From its minimum here the tangential force rises to a maximum at a, where it comprises the whole force. It then subsides to zero at A. During this quadrant it has been urging the particle onward in its own direction of movement FA. At A, it changes sign and becomes a retarding force, which attains its maximum at b, and then sinks to zero again at E.
In consequence, the velocity of the particle due ​to the disturbing force is a maximum at A — because the force has been adding increments to it up to this point — and a minimum at F and E. The particle, by traveling fast, lessens the curvature of its path about C, since the pull from C has less time to act; and reversely by traveling slowly it increases this curvature. In consequence, then, of this component, the path is flattened at A and bulged at E.
The normal component acts inward at F and is proportional to CP. It helps the central force at F, and curves the path the more. At a it vanishes, and is then reversed, acting outward or against the gravity of C. It thus lessens the curvature from a to b. It thus conspires completely with the tangential component; and the two together squeeze the orbit into an ellipse with its longer diameter at right angles to the line joining C to M.
The tidal action on a particle of the ocean is Tide analogus to moon's variation. thus precisely the same, neglecting pressure and friction, as that of the Sun upon the Moon's orbit. This deformation of the Moon's orbit was detected, probably by Aboul Wefa, nine centuries ago. It is called the Moon's variation. Thus the tidal wave and the variation are analogous exhibitions of the same force.
​Friction now comes in to modify the result. At Effect of frictionF, in consequence of the tide-raising force, the particle is traveling less rapidly than the rest of the Earth. Friction, therefore, urges it on and increases its tangential velocity up to some point P’ , where its speed becomes equal to the mean speed of the earth. After this, its speed being greater than the Earth's, friction retards it, until it again becomes the mean at P. Then friction begins again to accelerate it.
In consequence, the particle is accelerated from Q to P’', retarded from P’ to P, and then accelerated again. From A on, friction thus helps the retarding tangential force, and the Earth causes the particle to turn the corner of the ellipse at E sooner than it otherwise would. The tangential force thus reaches its maximum earlier, and the crest of the tide is thus shifted from E backward to some point P.
On the Earth, in the case of the ocean, we are dealing with superficial tides. In celestial mechanics, it is the substantial tides, or tides of the whole body, with which we are concerned. The latter are immensely the more potent. As the tidal crest lies ahead of the line joining the two bodies, the Sun or the Moon is constantly trying to pull it back into this line, while the Earth is ​striving by friction to set it at right angles to the line. The bulge, therefore, acts as a brake upon the Earth's rotation, and must continue so to act until the Earth's rotation and revolution coincide.
Tide-generating force.Now let us determine the tide-generating force:—
Let $M$= mass of the Earth;
$m$ =mass of the Moon;
$x$,$y$,$z$ = be coördinates of the Moon referred to the Earth's centre;
$r$=its distance;
$\xi ,\eta ,\zeta$ = the coördinates of the particle referred to the Earth's centre;
$\rho$ = its distance.
Then the Earth describes an ellipse round the centre of inertia of the Earth and Moon, and its acceleration is ${\frac {m}{r^{2}}}$ toward this centre.
To bring it to rest, we must apply to it an acceleration, $-{\frac {m}{r^{2}}}$, of which the accelerations along the coördinates are,—
$\textstyle -{\frac {m}{r^{2}}}.{\frac {x}{r}},-{\frac {m}{r^{2}}}.{\frac {y}{r}},-{\frac {m}{r^{2}}}.{\frac {z}{r}}$
Now
$\textstyle cos{z}={\frac {x}{r}}.{\frac {\xi }{\rho }}+{\frac {y}{r}}.{\frac {\eta }{\rho }}+{\frac {z}{r}}.{\frac {\zeta }{\rho }}$
and
$\textstyle r\rho \cos {z}=x\xi +y\eta +z\zeta ;$
$-{\frac {mx}{r^{3}}}$
$-{\frac {mx\zeta }{r^{3}}}$
$\zeta$
$-{\frac {m\rho }{r^{2}}}cos{z}$
The potential necessary to bring the Earth to rest is then $-{\frac {m\rho }{r^{2}}}cos{z}$
The potential of M with regard to the particle is ${\frac {M}{\rho }}$, while the potential of m upon the particle is ${\frac {m}{\sqrt {r^{2}+\rho ^{2}-2r\rho \cos {z}}}}$ plus a constant. This constant we determine by the condition that the potential at the planet's centre shall be zero, since we are seeking the motion of the particle relative to this centre, and it becomes ${\frac {m}{\sqrt {r^{2}+\rho ^{2}-2r\rho \cos {z}}}}-{\frac {M}{r}}$
Since $r$ is very large compared with $p$, we may advantageously expand the last in powers of ${\frac {p}{r}}$, which gives:—
${\frac {m}{r}}\lbrack {\frac {\rho }{r}}cos{z}+{\frac {\rho ^{2}}{r^{2}}}({{\frac {3}{2}}cos^{2}{z}-{\frac {1}{2}}})+$
${\frac {\rho ^{3}}{r^{3}}}({\frac {5}{2}}\cos ^{3}{z}-{\frac {3}{2}}\cos {z})+$ etc.$\rbrack .$.
The first term cancels with the potential for bringing the Earth to rest, and we have for the whole potential urging the particle,—
${\frac {M}{\rho }}+{\frac {m\rho ^{2}}{r^{3}}}({\frac {3}{2}}\cos ^{2}{z}-{\frac {1}{2}})+$ ${\frac {m\rho ^{3}}{r^{4}}}({\frac {5}{2}}\cos ^{3}{z}-{\frac {3}{2}}\cos {z})$.
To get the forces, we must differentiate this expression with regard to the position of the particle.
Tide-rising force for different planets.
In order to compare the tide-raising forces on different different bodies, we will assume $z=0$ whence the tide-raising force at its maximum may be expressed in a rapidly converging series, of which the first two terms are $\textstyle {\frac {2m\rho }{r^{3}}}+{\frac {3m\rho ^{2}}{r^{4}}}.$
If the affected body be distant compared with its size, the first term is enough, and we see that then the tide-raising force is directly as the radius of the second body, and inversely as the cube of its distance from the first, while also directly as the latter's mass.
But the work done by a force is the product of the force into the space through which it acts,—as, for instance, the lifting a weight a certain distance,—and in a given time the space is itself proportional to the force, whence the work in that time is as the square of the force.
$fdt=dv={\frac {ds}{dt}},$
whence
${\frac {1}{2}}ft^{2}=s$
Whence if the time remain constant the force must vary as the space. For the proportionate work ​done in a given time by tide-raising forces, we have, then, $\textstyle ({\frac {2m\rho }{r}}+{\frac {3m\rho ^{2}}{r}})^{2}$ or for most cases sufficiently well, taking only the first term, $\textstyle {\frac {4m^{2}\rho ^{2}}{r^{6}}}$ That is, it is as the square of the attracting mass and the square of the radius of the affected body directly and inversely as the sixth power of the latter's distance.
Work done by Tide-raising force in Unity of time in Ratio to Sun's Action on the Earth taken as Unity.
| By Sun on :- | By Sun on :- | By Sun on :- | | | | | | | $\scriptstyle {\frac {4m^{2}\rho ^{2}}{r^{6}}}$ (approx) | $\scriptstyle {\frac {4m^{2}\rho ^{2}}{r^{6}}}$ (approx) |
| Mercury | Mercury | . | . | . | . | . | . | . | . | 43-26 |
| Venus | Venus | . | . | . | . | . | . | . | . | 6.60 |
| Earth | Earth | . | . | . | . | . | . | . | . | 1.00 |
| Mars | . | . | . | . | . | . | . | . | . | 0.023 |
| Jupiter | Jupiter | . | . | . | . | . | . | . | . | 0.006 |
| On Earth by :- | On Earth by :- | On Earth by :- | | | | | $\scriptstyle \lbrack {\frac {2m\rho }{r^{3}}}+{\frac {3m\rho ^{2}}{r^{4}}}\rbrack ^{2}$ | $\scriptstyle \lbrack {\frac {2m\rho }{r^{3}}}+{\frac {3m\rho ^{2}}{r^{4}}}\rbrack ^{2}$ | $\scriptstyle \lbrack {\frac {2m\rho }{r^{3}}}+{\frac {3m\rho ^{2}}{r^{4}}}\rbrack ^{2}$ |
| Sun | . | . | . | . | . | . | . | . | . | 1.00 |
| Moon | . | . | . | . | . | . | . | . | 4.97 |
| On Satellites by their Primaries :- | On Satellites by their Primaries :- | On Satellites by their Primaries :- | On Satellites by their Primaries :- | On Satellites by their Primaries :- | On Satellites by their Primaries :- | $\scriptstyle \lbrack {\frac {2m\rho }{r^{3}}}+{\frac {3m\rho ^{2}}{r^{4}}}\rbrack ^{2}$ | $\scriptstyle \lbrack {\frac {2m\rho }{r^{3}}}+{\frac {3m\rho ^{2}}{r^{4}}}\rbrack ^{2}$ | $\scriptstyle \lbrack {\frac {2m\rho }{r^{3}}}+{\frac {3m\rho ^{2}}{r^{4}}}\rbrack ^{2}$ |
| Iapetus | . | . | . | . | . | . | . | 27.6 |
| Callisto | . | . | . | . | . | . | 32,549.0 | 32,549.0 |
| Ganymede | . | . | . | . | . | . | 1,385,600.0 | 1,385,600.0 |
| Moon | . | . | . | . | . | . | . | 2,374.4 |
​Professor G. H. Darwin has calculated the relative effect of tidal retardation by the Sun on each of the several planets, that upon the Earth being taken as unity, with the accompanying result:—
| Planet. | Number to which Tidal Retardation is Proportional. | Number to which Tidal Retardation is Proportional. |
| Mercury | 1000 | .(?) |
| Venus | 11 | .(?) |
| Earth | 1 | . |
| Mars | | .89 |
| Jupiter | | .00005 |
| Saturn | $\scriptstyle {\left\{{\begin{matrix}\ \\\ \end{matrix}}\right.}$ | .000020 to .000066 |
Supposing, then, all the bodies to have started in the race for rotary retardation at the same time, the isochronism of rotation and revolution of Mercury is what was to have been expected. For the previously known facts were: first, that the Moon showed this state of things. Now the relative tide-raising effect in a given time of the Earth on the Moon is 2374; that of the Sun on the Earth being unity. Second, that Iapetus did the same; for this satellite is always much brighter on the western side of Saturn than on the eastern. Such a periodic change of brilliancy would be accounted for by isochronism of rotation and revolution. Now the relative tide-raising effect of Saturn on this satellite is 28.
​On the other hand, the Earth's rotation and revolution do not coincide; and the relative effects of Sun and Moon on it are:—
| Sun | 1.00 |
| Moon | 4.97 |
| | 5.97 |
Assuming, therefore, that the retardation began synchronously for all, Mercury, upon whom the effect was 43, should have reached the isochronous condition.
We may note incidentally that Venus on this assumption falls in the debatable ground, since the effect on it is 6.60.
But we do not know either the time of the birth of the Moon nor the relative age of the Earth and Venus. It is quite possible, for aught we know, that Venus may have been subjected to the process practically much longer than the Earth.
It is certainly significant that isochronism ceases just where a first approximation would put it.
Since the date of the detection of Mercury's isochronism by Schiaparelli, the third and fourth satellites of Jupiter, Ganymede and Callisto, have been added to the isochronous list by Mr. Douglass at Flagstaff. These, then, agree with theory. We may safely predict that all the other satellites ​of Jupiter and Saturn will be found to behave similarly.
Consummation of tidal effect marks the last stage in the planetary career. So soon as identity of rotation and revolution is effected, the planet is placed in a changeless, or largely changeless, state, which, so far as we can conceive, means as a world its death. It now turns the same face, except for libration, in perpetuity to the Sun, Day and night, summer and winter, have ceased to exist. One half of it is forever being baked, the other half forever frozen; and from this condition there is no escape. The planet must remain so until the Sun itself goes out.
Mercury, therefore, represents planetary decrepitude; and the symptoms of this old age are: loss of air; isochronous rotation and revolution; rotundity.
* ↑ Pro. Roy. Soc. 1878-81.
* ↑ 1 After G. H. Darwin. Article in the Encyclopedia Britannica on "Tides.")