“5. What point in the heavens ought the cannon to be aimed at which is intended to discharge the projectile? “6. What place will the moon occupy in the heavens at the moment of the projectile’s departure?” Regarding the _first_ question, “Is it possible to transmit a projectile up to the moon?” _Answer._—Yes; provided it possess an initial velocity of 12,000 yards per second; calculations prove that to be sufficient. In proportion as we recede from the earth the action of gravitation diminishes in the inverse ratio of the square of the distance; that is to say, _at three times a given distance the action is nine times less._ Consequently, the weight of a shot will decrease, and will become reduced to _zero_ at the instant that the attraction of the moon exactly counterpoises that of the earth; that is to say at 47/52 of its passage. At that instant the projectile will have no weight whatever; and, if it passes that point, it will fall into the moon by the sole effect of the lunar attraction. The _theoretical possibility_ of the experiment is therefore absolutely demonstrated; its _success_ must depend upon the power of the engine employed. As to the _second_ question, “What is the exact distance which separates the earth from its satellite?” _Answer._—The moon does not describe a _circle_ round the earth, but rather an _ellipse_, of which our earth occupies one of the _foci_; the consequence, therefore, is, that at certain times it approaches nearer to, and at others it recedes farther from, the earth; in astronomical language, it is at one time in _apogee_, at another in _perigee_. Now the difference between its greatest and its least distance is too considerable to be left out of consideration. In point of fact, in its apogee the moon is 247,552 miles, and in its perigee, 218,657 miles only distant; a fact which makes a difference of 28,895 miles, or more than one-ninth of the entire distance. The perigee distance, therefore, is that which ought to serve as the basis of all calculations. To the _third_ question:— _Answer._—If the shot should preserve continuously its initial velocity of 12,000 yards per second, it would require little more than nine hours to reach its destination; but, inasmuch as that initial velocity will be continually decreasing, it will occupy 300,000 seconds, that is 83hrs. 20m. in reaching the point where the attraction of the earth and moon will be _in equilibrio_. From this point it will fall into the moon in 50,000 seconds, or 13hrs. 53m. 20sec. It will be desirable, therefore, to discharge it 97hrs. 13m. 20sec. before the arrival of the moon at the point aimed at. Regarding question _four_, “At what precise moment will the moon present herself in the most favorable position, etc.?”