Reply from the Cambridge Observatory

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.

Despite all the enthusiasm surrounding him, Barbicane wasted no time. His first step was to gather his colleagues in the Gun Club boardroom. After some discussion, they agreed to consult astronomers about the astronomical aspects of the project. Once they had the astronomers’ reply, they could discuss the mechanical means. Nothing would be left undone 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.

A carefully worded note containing specific questions was then sent to the Observatory of Cambridge in Massachusetts. Cambridge, where the first university in the United States was founded, is rightly famous for its astronomers. All the most eminent scientists are gathered there. The Observatory houses the powerful telescope that allowed Bond to resolve the Andromeda nebula and Clarke to discover Sirius’s satellite. The famous institution fully justified the Gun Club’s confidence in every respect. So, after two days, the eagerly awaited reply was placed in President Barbicane’s hands.

It was couched in the following terms:

It read as follows:

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

The Director of the Cambridge Observatory to the President of the Gun Club in 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:

CAMBRIDGE, October 7. We received your letter of October 6, addressed to the Observatory of Cambridge on behalf of the Baltimore Gun Club members. Our staff was immediately called together and decided to reply as follows:

The questions which have been proposed to it are these—

The questions put to it are:

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

“1. Is it possible to send a projectile to the moon?

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

“2. What is the exact distance between Earth and 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?

“3. How long will the projectile take to reach the moon if it is launched fast enough? And when should it be launched so that it reaches a particular point on the moon?

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

“4. At exactly what moment will the moon be in the best position for the projectile to reach it?

“5. What point in the heavens ought the cannon to be aimed at which is intended to discharge the projectile?

“5. What point in the sky should the cannon be aimed at to fire the projectile?

“6. What place will the moon occupy in the heavens at the moment of the projectile’s departure?”

“6. Where will the moon be in the sky when the projectile is launched?”

Regarding the first question, “Is it possible to transmit a projectile up to the moon?”

Regarding the first question, “Is it possible to send a projectile 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.

Answer. Yes, provided it has an initial velocity of 12,000 yards per second. Calculations prove that this speed is sufficient. As we move away from the earth, gravity weakens in inverse proportion to the square of the distance. In other words, at three times a given distance, gravity is nine times weaker. Consequently, the weight of a shot will decrease. It will fall to zero when the moon's attraction exactly balances the earth's attraction. This will happen at 47/52 of the way through its journey. At that moment, the projectile will have no weight at all. If it passes that point, it will fall into the moon solely because of the moon's attraction. The theoretical possibility of the experiment is therefore absolutely demonstrated. Its success will depend on the power of the engine used.

As to the second question, “What is the exact distance which separates the earth from its satellite?”

As for the second question, “What is the exact distance between Earth and 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.

Answer. The moon does not travel in a circle around the earth, but rather in an ellipse. The earth occupies one of the ellipse’s foci. As a result, the moon is sometimes closer to the earth and sometimes farther away. In astronomical terms, it is sometimes at apogee and sometimes at perigee. The difference between its greatest and least distances is too large to ignore. At apogee, the moon is 247,552 miles from the earth. At perigee, it is only 218,657 miles away. That is a difference of 28,895 miles, or more than one-ninth of the entire distance. Therefore, all calculations should be based on the perigee distance.

To the third question:—

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.

Answer. If the shot maintained its initial speed of 12,000 yards per second, it would reach its destination in just over nine hours. But its initial speed will continually decrease. It will take 300,000 seconds, or 83 hours, 20 minutes, to reach the point where the Earth's and Moon's gravitational pulls are in balance. From there, it will fall to the Moon in 50,000 seconds, or 13 hours, 53 minutes, 20 seconds. Therefore, it should be fired 97 hours, 13 minutes, 20 seconds before the Moon reaches the point aimed at.

Regarding question four, “At what precise moment will the moon present herself in the most favorable position, etc.?”

Regarding question four, “Exactly when will the moon be 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.

Answer. First, we must choose a time when the moon is at perigee. We must also choose a moment when it is crossing the zenith. This will reduce the total distance by the radius of the earth, i. e. 3,919 miles. The remaining distance to travel will then be 214,976 miles. But although the moon reaches perigee every month, it does not always cross the zenith at exactly the same moment. These two events occur together only at long intervals. We must therefore wait until the moon reaches perigee and crosses the zenith at the same time. Fortunately, the moon will meet both conditions on December 4 of the coming year. At midnight it will be at perigee, its shortest distance from the earth. At the same moment, it will be crossing the zenith.

On the fifth question, “At what point in the heavens ought the cannon to be aimed?”

On the fifth question, “Where in the sky should the cannon 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.

Answer. If the preceding remarks are accepted, the cannon should be aimed at the point directly overhead. It will then fire perpendicular to the horizon, and the projectile will escape Earth's gravitational pull as quickly as possible. But for the moon to pass directly overhead, the location's latitude must not exceed the moon's declination. In other words, the location must be between 0° and 28° north or south latitude. Anywhere else, the cannon must be fired at an angle, which would seriously reduce the experiment's chances of success.

As to the sixth question, “What place will the moon occupy in the heavens at the moment of the projectile’s departure?”

As for the sixth question, “Where will the moon be in the sky when the projectile is launched?”

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.

Answer. At the moment the projectile is fired into space, the moon will be four times its daily travel distance from the zenith point. The moon travels 13° 10′ 35″ each day, so this distance will be 52° 41′ 20″. That is the distance the moon will travel during the projectile’s entire journey. But we must also account for the deviation the earth’s rotation will impart to the shot. The shot cannot reach the moon until it has deviated by a distance equal to 16 radii of the earth. On the moon’s orbit, this distance equals about eleven degrees. We must therefore add these eleven degrees to the degrees given for the moon’s lag just mentioned. In round numbers, the result is about sixty-four degrees. Consequently, at the moment of firing, the line of sight to the moon will form an angle of sixty-four degrees with the vertical line at the firing location.

These are our answers to the questions proposed to the Observatory of Cambridge by the members of the Gun Club:—

These are our answers to the questions the members of the Gun Club asked the Observatory of Cambridge:

To sum up—

To sum up:

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

1st. The cannon should be placed in a country between 0° and 28° north or south latitude.

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

2nd. It should point straight up from that location.

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

3rd. The projectile should be launched at an initial speed 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.

4th. It should be fired at 10:46:40 on December 1 of the following 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.

5th. It will meet the moon four days after being fired, precisely at midnight on December 4, as the moon passes directly overhead.

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 members of the Gun Club should therefore begin the necessary preparations for the experiment without delay. They should be ready to start work at the appointed time. For if they let December 4 pass, they will have to wait eighteen years and eleven days. Only then will the moon again be at perigee and at the zenith under the same conditions.

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. For the Astronomical Staff, J. M. BELFAST, Director of the Observatory of Cambridge.

The staff of the Cambridge Observatory are entirely at their disposal for all questions about theoretical astronomy. We also add our congratulations to those of the rest of America. For the Astronomical Staff, J. M. BELFAST, Director of the Cambridge Observatory.