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.?"

_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