(G.)-ON TIDAL FRICTION, by G. H. DARWIN, F.R.S. (a.) The retardation of the earth's rotation, as deduced from the secular acceleration of the Moon's mean motion. of earth's Numerical In my paper on the precession of a viscous spheroid [Phil. Retardation Trans. Pt. II., 1879], all the data are given which are requisite rotation. for making the calculations for Professor Adams' result in § 830, estimates. viz. that if there is an unexplained part in the coefficient of the secular acceleration of the moon's mean motion amounting to 6", and if this be due to tidal friction, then in a century the earth gets 22 seconds behind time, when compared with an ideal clock, going perfectly for a century, and perfectly rated at the beginning of the century. In the paper referred to however the earth is treated as homogeneous, and the tides are supposed to consist in a bodily deformation of the mass. The numerical results there given require some modification on this account. If E, E', E" be the heights of the semidiurnal, diurnal and fortnightly tides, expressed as fractions of the equilibrium tides of the same denominations; and if e, é, é" be the corresponding retardations of phase of these tides due to friction; it is shown on p. 476 and in equation (48), that in consequence of lunar and solar tides, at the end of a century, the earth, as a time-keeper, is behind the time indicated by the ideal perfect clock 1900-27 E sin 2€ + 423·49 E' sin e' seconds of time ......(a), and that if the motion of the moon were unaffected by the tides, an observer, taking the earth as his clock, would note that at the end of the century the moon was in advance of her place in her orbit by 1043"-28 E sin 2e + 232′′-50 E' sin é' ................... ·(b). This is of course merely the expression of the same fact as (a), in Lastly it is shown in equation (60) that from these causes in a 630" 7 E sin 2€ + 108′′-6 E' sin é – 7′′·042 E" sin 2e".................(c). In adapting these results to the hypothesis of oceanic tides on a heterogeneous earth, we observe in the first place that, if the Retardation of earth's rotation. fluid tides are inverted, that is to say if for example it is low water under the moon, then friction advances the fluid tides*, and therefore in that case the e's are to be interpreted as advancements of phase; and secondly that the E's are to be multiplied by, which is the ratio of the density of water to the mean density of the earth. Next the earth's moment of inertia (as we learn from col. vii. of the table in § 824) is about 83 of its amount on the hypothesis of homogeneity, and therefore the results (a) and (b) have both to be multiplied by 1/83 or 1.2; the result (c) remains unaffected except as to the factor Thus subtracting (c) from (b) as amended, we find that to an observer, taking the earth as a true time-keeper, the moon is, at the end of the century, in advance of her place by {(1.2 x 1043"-28-630" 7) E sin 2€ + (1.2 x 232′′·50-108"-6) E' sin e' + 7"-042 E" sin 2e"}, which is equal to (621"-24 Esin 2e + 170"-40 E' sin e'+7"-04 E" sin 2e"}...(d) and from (a) as amended that the earth, as a time-keeper, is behind the time indicated by the ideal clock, perfectly rated at the beginning of the century, by {2280-32 E sin 2e + 508-19 E' sin e'} seconds of time........(e). Now if we suppose that the tides have their equilibrium height, so that the E's are each unity; and that e' is one half of € (which must roughly correspond to the state of the case), and that e” is insensible, and small, (d) becomes The second term, both in the numerator and denominator of (h), depends on the diurnal tide, which only exists when the ecliptic * That this is true may be seen from considerations of energy. If it were approximately low water under the moon, the earth's rotation would be accelerated by tidal friction, if the tides of short period lagged; and this would violate the principles of energy. result. is oblique. Now Adams' result was obtained on the hypothesis Adams Accordingly 6" in the coefficient gives 22 secs. at the end of a results. Thus taking Hansen's 12"-56 with Delaunay's 6"-1, we have the Other It is worthy of notice that this result would be only very (b.) The Determination of the Secular Effects of Tidal Friction by a Graphical Method. (Portion of a paper published in the Proc. Roy. Soc. No. 197, 1879, but with alterations and additions.) problem of friction. Suppose an attractive particle or satellite of mass m to be General moving in a circular orbit, with an angular velocity 2, round a tidal planet of mass M, and suppose the planet to be rotating about an axis perpendicular to the plane of the orbit, with an angular velocity n; suppose, also, the mass of the planet to be partially or wholly imperfectly elastic or viscous, or that there are oceans Special units. on the surface of the planet; then the attraction of the satellite must produce a relative motion in the parts of the planet, and that motion must be subject to friction, or, in other words, there must be frictional tides of some sort or other. The system must accordingly be losing energy by friction, and its configuration must change in such a way that its whole energy diminishes. Such a system does not differ much from those of actual planets and satellites, and, therefore, the results deduced in this hypothetical case must agree pretty closely with the actual course of evolution, provided that time enough has been and will be given for such changes. Let C be the moment of inertia of the planet about its axis of rotation; r the distance of the satellite from the centre of the planet; h the resultant moment of momentum of the whole system; e the whole energy, both kinetic and potential of the system. It will be supposed that the figure of the planet and the distribution of its internal density are such that the attraction of the satellite causes no couple about any axis perpendicular to that of rotation. I shall now adopt a special system of units of mass, length, and time such that the analytical results are reduced to their simplest forms. Let the unit of mass be Mm/(M+m). Let the unit of length y be such a distance, that the moment of inertia of the planet about its axis of rotation may be equal to the moment of inertia of the planet and satellite, treated as particles, about their centre of inertia, when distant y apart from one another. This condition gives Let the unit of time be the time in which the satellite revolves through 57°.3 about the planet, when the satellite's radius vector is equal to y. In this case 1/7 is the satellite's orbital angular where μ is the attraction between unit masses at unit distance. Then by substitution for y This system of units will be found to make the three following functions each equal to unity, viz. μ3Mm (M +m) ̃3, μMm, and C. The units are in fact derived from the consideration that these functions are each to be unity. 1 83 values of earth and In the case of the earth and moon, if we take the moon's mass Numerical asnd of the earth's, and the earth's moment of inertia as the units for Ma2 [see § 824], it may easily be shown that the unit of mass moon. is of the earth's mass, the unit of length is 5.26 earth's radii or 33,506 kilometres, and the unit of time is 2 hrs. 41 minutes. In these units the present angular velocity of the earth's diurnal rotation is expressed by 7044, and the moon's present radius vector by 11.454. momentum The two bodies being supposed to revolve in circles about Moment of their common centre of inertia with an angular velocity 2, the and energy moment of momentum of orbital motion is of system. Then, by the law of periodic times, in a circular orbit, Ñ3μ3 = μ (M + m) whence r2 =μ3 (M+m)a ra. And the moment of momentum of orbital motion =μ3 Mm (M+m)1 post, and in the special units this is equal to 3. The moment of momentum of the planet's rotation is Cn, and |