Page:Popular Astronomy - Airy - 1881.djvu/235

Rh however hereafter allude to the effects of both these circumstances. Now the principle of calculation in this and all similar cases will be the following: in Figure 56, let MN be the arc which the moon describes in her course round the earth E, in one hour., or one minute, or one second, or any other short time that we may choose to fix on (we shall, for the present, take one second). If no force had acted on the moon, she would have moved in a straight line Mm. Therefore the force with which the earth has attracted the moon, has drawn her from m to N, or through the space mN. We must therefore compute the length mN we shall then know how far the earth's attraction draws the moon in one second; we also know how far the earth's attraction makes any-thing at the surface of the earth fall in one second; and the proportion of these will give the proportion of the earth's attraction in these two different places. Now, considering the moon as moving in a circle whose semi-diameter is her mean distance, EM is 238,800 miles. Also the whole circumference of the moon's orbit is 1,500,450 miles; and her periodic time is 27 days, 7 hours, 43 minutes; hence the length of the line Mm, which would have been described in one second if no force had acted, is 0.5356 of a mile. With these two lengths of the sides of the right-angled triangle EMm, by the usual rule of squaring the two sides, adding the squares together, and extracting the square root of the sum, the hypotenuse Em is found to be 238800.0000008459 miles. Therefore the line mN is 0.0000008459 of a mile, or 0.0536 of an inch; and this is the space