The vibrations of the pendulum give the force of gravity at the surface of the earth, and it is found to vary in different latitudes. The intensity in any place being as the squares of the number of vibrations in a given time. This inequality depends on the centrifugal force of rotation, and on the spheroidal figure of the earth due to that rotation. At the equator the fall of a heavy body is found to be 16.045223 feet, per second, and in that latitude the squares of whose sine is ⅓, it is 16.0697 feet. The effect in this last-named latitude is the same as if the earth were a perfect sphere. This does not, however, express the whole force of gravity, as the rotation of the earth causes a centrifugal tendency which is a maximum at the equator, and there amounts to 1/289 of the whole gravitating force. In other latitudes it is diminished in the ratio of the squares of the cosines of the latitude; it therefore becomes 1/434 in that latitude the square of whose sine is ⅓. Hence the fall per second becomes 16.1067 feet for the true gravitating force of the earth, or for that force which retains the moon in her orbit. The moon's mean distance is 59.96435 equatorial radii of the earth, which radius is, according to Sir John Herschel, 20.923.713 feet. Her mean distance as derived from the parallax is not to be considered the radius vector of the orbit, inasmuch as the earth also describes a small orbit around the common centre of gravity of the earth and moon; neither is radius vector to be considered as her distance from this common centre; for the attracting power is in the centre of the earth. But the mean distance of the moon moving around a movable centre, is to the same mean distance when the centre of attraction is fixed, as the sum of the masses of the two bodies, to the first of two mean proportionals between this sum and the largest of the two bodies inversely. (Vid. Prin. Prop. 60 Lib. Prim.) The ratio of the masses being as above 80 to 1 the mean proportional sought is 80.666 and in this ratio must the moon's mean distance be diminished to get the force of gravity at the moon. Therefore as 81 is to 80.666, so is 59.96435 to 59.71657 for the moon's distance in equatorial radii of the earth. Multiply this last by 20.923,713 to bring the semi-diameter of the lunar orbit into feet = 1.249.492.373, and this by 6.283185, the ratio of the circumference to the radius, gives 7.850.791.736 feet, for the mean circumference of the lunar orbit. Further, the mean sidereal period of the moon is 2360591 seconds and the 1/2360591th part of 7.850.791.736 is the arc the moon describes in one second = 3325.77381 feet, the square of which divided by the diameter of the orbit, gives the fall of the moon from the tangent or versed size of that arc. 1106771.36876644 = ---------------- = 0.004426106 feet. 2498984746