_Answer._—After what has been said above, it will be necessary, first of all, to choose the period when the moon will be in perigee, and _also_ the moment when she will be crossing the zenith, which latter event will further diminish the entire distance by a length equal to the radius of the earth, _i. e._ 3,919 miles; the result of which will be that the final passage remaining to be accomplished will be 214,976 miles. But although the moon passes her perigee every month, she does not reach the zenith always _at exactly the same moment_. She does not appear under these two conditions simultaneously, except at long intervals of time. It will be necessary, therefore, to wait for the moment when her passage in perigee shall coincide with that in the zenith. Now, by a fortunate circumstance, on the 4th of December in the ensuing year the moon _will_ present these two conditions. At midnight she will be in perigee, that is, at her shortest distance from the earth, and at the same moment she will be crossing the zenith. On the _fifth_ question, “At what point in the heavens ought the cannon to be aimed?” _Answer._—The preceding remarks being admitted, the cannon ought to be pointed to the zenith of the place. Its fire, therefore, will be perpendicular to the plane of the horizon; and the projectile will soonest pass beyond the range of the terrestrial attraction. But, in order that the moon should reach the zenith of a given place, it is necessary that the place should not exceed in latitude the declination of the luminary; in other words, it must be comprised within the degrees 0° and 28° of lat. N. or S. In every other spot the fire must necessarily be oblique, which would seriously militate against the success of the experiment. As to the _sixth_ question, “What place will the moon occupy in the heavens at the moment of the projectile’s departure?” _Answer._—At the moment when the projectile shall be discharged into space, the moon, which travels daily forward 13° 10′ 35″, will be distant from the zenith point by four times that quantity, _i. e._ by 52° 41′ 20″, a space which corresponds to the path which she will describe during the entire journey of the projectile. But, inasmuch as it is equally necessary to take into account the deviation which the rotary motion of the earth will impart to the shot, and as the shot cannot reach the moon until after a deviation equal to 16 radii of the earth, which, calculated upon the moon’s orbit, are equal to about eleven degrees, it becomes necessary to add these eleven degrees to those which express the retardation of the moon just mentioned: that is to say, in round numbers, about sixty-four degrees. Consequently, at the moment of firing the visual radius applied to the moon will describe, with the vertical line of the place, an angle of sixty-four degrees. These are our answers to the questions proposed to the Observatory of Cambridge by the members of the Gun Club:— To sum up—