Barbicane, however, lost not one moment amid all the enthusiasm of which he had become the object. His first care was to reassemble his colleagues in the board-room of the Gun Club.There, after some discussion, it was agreed to consult the astronomers regarding the astronomical part of the enterprise.Their reply once ascertained, they could then discuss the mechanical means, and nothing should be wanting to ensure the success of this great experiment.

A note couched in precise terms, containing special interrogatories, was then drawn up and addressed to the Observatory of Cambridge in Massachusetts. This city, where the first university of the United States was founded, is justly celebrated for its astronomical staff.There are to be found assembled all the most eminent men of science.Here is to be seen at work that powerful telescope which enabled Bond to resolve the nebula of Andromeda, and Clarke to discover the satellite of Sirius.This celebrated institution fully justified on all points the confidence reposed in it by the Gun Club.

So, after two days, the reply so impatiently awaited was placed in the hands of President Barbicane. It was couched in the following terms：

The Director of the Cambridge Observatory to the President of the Gun Club at Baltimore.

CAMBRIDGE, October 7.

On the receipt of your favor of the 6th instant, addressed to the Observatory of Cambridge in the name of the members of the Baltimore Gun Club, our staff was immediately called together, and it was judged expedient to reply as follows：

The questions which have been proposed to it are these—

1.Is it possible to transmit a projectile up to the moon?

2.What is the exact distance which separates the earth from its satellite?

3.What will be the period of transit of the projectile when endowed with sufficient initial velocity?And, consequently, at what moment ought it to be discharged in order that it may touch the moon at a particular point?

4.At what precise moment will the moon present herself in the most favorable position to be reached by the projectile?

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.?”

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 form 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°42'20"，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—

1st. The cannon ought to be planted in a country situated between 0°and 28°of N.or S.lat.

2nd. It ought to be pointed directly toward the zenith of the place.

3rd. The projectile ought to be propelled with an initial velocity of 12，000 yards per second.

4th. It ought to be discharged at 10hrs.46m.40sec.of the 1st of December of the ensuing year.

5th. It will meet the moon four days after its discharge, precisely at midnight on the 4th of December, at the moment of its transit across the zenith.

The members of the Gun Club ought, therefore, without delay, to commence the works necessary for such an experiment, and to be prepared to set to work at the moment determined upon；for, if they should suffer this 4th of December to go by, they will not find the moon again under the same conditions of perigee and of zenith until eighteen years and eleven days afterward.

The staff of the Cambridge Observatory place themselves entirely at their disposal in respect of all questions of theoretical astronomy；and herewith add their congratulations to those of all the rest of America.

J. M.BELFAST，

Director of the Observatory of Cambridge.

不过，巴比凯恩并未因受到众人的欢呼而忘乎所以。他首先要做的，是把他的同事们召集到大炮俱乐部的办公室里来。在那儿，经过一番讨论，大家同意就方案的天文学部分请教一下天文学家。等天文学家的回音一到，大家就将着手讨论机械装备的问题；而且，为保证这一伟大试验的成功，任何细节都不可疏忽。

于是，一份包括一些专业问题的十分明确的纪要便拟好了，寄给了位于马萨诸塞州的剑桥天文台[15]。剑桥城是美国第一所大学的诞生地，而且也正是因为它的天文台而享誉世界。那儿聚集着一些顶尖的科学家；那里的一台高性能望远镜使邦德[16]解析了仙女座星云，使克拉克发现了天狼星。这座著名的天文台完全值得大炮俱乐部信赖。

两天后，大家焦急不安等待着的回信寄到了巴比凯恩主席的手中。内容如下：

剑桥天文台台长致巴尔的摩大炮俱乐部主席：

您本月六日以巴尔的摩大炮俱乐部全体会员的名义，寄给剑桥天文台的信函，我台业已收悉。我们立即开了会，并做出如下我们认为较为合适的答复。

您提出的问题归纳如下：

1.可不可能向月球发射一颗炮弹？

2.地球与它的这颗卫星的精确距离是多少？

3.在给炮弹以足够的初速度的情况下，它能飞行多长时间？而为了让它落在月球上的某一个特定地点，应该何时发射为好？

4.炮弹落在月球上的最佳位置应该是在什么时候？

5.发射炮弹的大炮应该对准天空中的哪一个点？

6.炮弹射出时，月球将在天空中的什么位置？

就第一个问题：可不可能向月球发射一颗炮弹？现回答如下。

可以。如果能使炮弹的初速度达到每秒一万二千码的话，就可以向月球发射。经过计算，这一速度足够了。随着物体离开地球，地心引力的作用与距离的平方成反比，因而引力在逐渐减小，也就是说，如果距离变为原来的三倍，这一引力就将减小到原来的九分之一。因此，炮弹的重量在迅速减小，到月球的引力与地球的引力持平的时候，也就是说，在射程达到五十二分之四十七的时候，炮弹的重量就会减少到零。这时，炮弹就不再有重量了。如果它超越了这个点，它就将在唯一的月球的引力之下落在月球上。试验的理论性完全得到了验证。至于成功与否，那就只取决于发射装置的功率了。

就第二个问题：地球与它的这颗卫星的精确距离是多少？现回答如下。

月球围绕地球运转的轨迹并不是圆形的，而是椭圆形的，我们的地球占据着这个椭圆中心中的一个，因此，月球有时离地球较近，有时又较远；用天文学术语来说就是，时而在远地点，时而在近地点。可是，最大距离与最小距离之间的差距是很大的，大到不容我们忽略。事实上，在远地点时，月球距离地球二十四万七千五百五十二英里；而在近地点时，距离则只有二十一万八千六百五十七英里，相差两万八千八百九十五英里，超过总射程的九分之一。因此，近地点的距离应该作为考虑的基础。

就第三个问题：在给炮弹以足够的初速度的情况下，它能飞行多长时间？而为了让它落在月球上的某一个特定地点，应该何时发射为好？现回答如下。

如果炮弹始终保持发射时的初速度——每秒一万二千码的话，它大约只需九个多小时便可到达目的地。但是，由于这个初速度将逐渐减小，经过推算，炮弹将要花费三十万秒，亦即八十三小时二十分，才能到达地球引力和月球引力相抵消的点。然后，从这个点起，它将在五万秒之后，亦即十三小时五十三分二十秒之后落到月球上。因此，应该在炮弹将到达月球上的那个瞄准点之前的九十七小时十三分二十秒之前发射它。

就第四个问题：炮弹落在月球的最佳位置应该是在什么时候？现回答如下。

根据刚才上面所说的，首先必须选择月球在近地点的时刻，同时也要是它通过天顶[17]的时刻，这将能够减少相当于一个地球半径的距离，亦即三千九百一十九英里；这样的话，最终射程则为二十一万四千九百七十六英里。不过，如果说月球每月都经过近地点的话，那它并不是在这一时刻总是处在天顶。而这两个条件同时具备的话，必须有一个很长的间隔。因此，必须等待月球到达近地点同时又在天顶的时刻的到来。不过，巧得很，明年十二月四日，月球将正好具备这两个条件：午夜时分，它将到达它的近地点，也就是说离地球最近的距离；与此同时，它又经过天顶。

就第五个问题：发射炮弹的大炮应该对准天空中的哪一个点？现回答如下。

根据上述看法，大炮应该瞄准天顶，这样，炮弹飞出时与地平线呈垂直状，它因此也就能尽快地摆脱地球引力。不过，要使这种情况出现，即月球到达天空最高点的话，就必须让这个地方的纬度不高于地球的赤纬[18]，也就是说，它必须位于北纬或南纬的0°至28°之间。在其他任何地点，就必须倾斜发射，而这就可能影响试验的成功。

就第六个问题：炮弹射出时，月球将在天空中的什么位置？现回答如下。

当炮弹将向天空发射时，每天以十三度十分三十五秒向前运行的月球，应该出现在离天顶是这个度数四倍距离的地方，亦即五十二度四十二分二十秒的地方，这一空间正好符合炮弹轨迹中月球的运行路线。不过，由于必须同时考虑到地球自转所引发的炮弹的偏差，而且还由于炮弹只是经过一个相当于十六个地球半径的偏差距离之后才到达月球（从月球轨道来看，这个偏差大约为十一度），因此，我们必须把这十一度加到所提及的月球离天空最高点的距离中去，变为六十四度整。这样一来，在发射炮弹时，月球视线方位将与发射点的垂直线构成一个六十四度的夹角。

这就是剑桥天文台对大炮俱乐部的会员们提出的问题的回复。

概括起来，就是：

1.大炮必须安放在一个北纬或南纬0°至28°之间的地方。

2.大炮必须瞄准天空最高点。

3.炮弹的初速度必须达到每秒一万二千码。

4.炮弹应该在明年十二月一日晚上十点四十六分四十秒发射。

5.炮弹将于发射后的第四天，即精确时间十二月四日午夜时分，在通过天空最高点时到达月球。

因此，大炮俱乐部的会员们应该立即着手进行这样的一次试验所必需的工作，做好在规定时刻发射的准备；因为如果错过了十二月四日这个日期的话，那就必须等到十八年零十一天以后，才能遇上月球同时符合既位于近地点又处在天空最高点的条件。

剑桥天文台愿意毫无保留地回答所有有关天文学理论方面的问题，并与全国人民一起恭祝诸位马到成功。

剑桥天文台台长

J. M.贝尔法斯特