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.?”
_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