ame 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

This fraction is, however, too small, as the ablatitious action of the
sun diminishes the attraction of the earth on the moon, in the ratio of
178 29/40 to 177 29/40. So that we must increase the fall of the moon
in the ratio of 711 to 715, and hence the true fall of the moon from the
tangent of her orbit becomes 0.00451 feet per second.

We have found the fall of a body at the surface of the earth, considered
as a sphere, 16.1067 feet per second, and the force of gravity
diminishes as the squares of the distances increases. The polar diameter
of the earth is set down as 7899.170 miles, and the equatorial diameter
7925.648 miles; therefore, the mean diameter is 7916.189 miles.[36] So
that, reckoning in mean radii of the earth, the moon's distance is
59.787925, which squared, is equal to 3574.595975805625. At one mean
radius distance, that is, at the surface, the force of gravity, or fall
per second, is as above, 16.1067 feet. Divide this by the square of the
distance, it is 16.1067/3574.595975805625 = 0.0045058 feet for the force
of gravity at the moon. But, from the preceding calculation, it appears,
that the moon only falls 0.0044510 feet in a second, showing a
deficiency of 1/82d part of the principal force that retains the moon in
her orbit, being more than double