In order to explain the cause of the tides, I have since I saw you last prepared a little drawing for you, (Plate IV. fig. 3.) where S represents the sun, M the moon at change, E the center of the earth, and A B C D its surface, covered with water. It is obvious, from the principles of gravitation, that if the earth were at rest the water in the ocean would be truly spherical, if its figure were not altered by the action of some other power. But, daily experience proves that it is continually agitated.

Pupil. What is the cause of this agitation?

Tutor. The attraction of the sun and moon, particularly the latter: for, as she is so much nearer the earth than the sun, she attracts with a much greater force than he does, and consequently raises the water much higher, which, being a fluid, loses as it were its gravitating power, and yields to their superior force.

Pupil. What proportion does the attractive power of the sun bear to that of the moon?

Tutor. As three to ten. So when the moon is at change, the sun and moon being in conjunction, or on the same side of the earth, the action of both bodies is on the surface of the water, the moon raising it ten parts,^[17] and the sun three, the sum of which is thirteen parts, represented by B b. Now it is evident, that if thirteen parts be added by the attractive power of those bodies, the same number of parts must be drawn off from some other part, as A a, C c. It will now be high-water under the moon at b, and its opposite side d, and low-water at a and c.

Pupil. That the attraction of the sun and moon must occasion a swelling of the waters on the side next them, I can readily conceive, and that this swell must cause a falling off at the sides: but that the tide should rise as high on the side opposite to the sun and moon, in a direction contrary to their attraction, is what I am not able to account for.

Tutor. This difficulty will be removed when you consider that all bodies moving in circles have a constant tendency to fly off from their centers. Now, as the earth and moon move round their center of gravity, that part of the earth which is at any time opposite to the moon will have a greater centrifugal force than the side next her, and at the earth’s center the centrifugal force exactly balances the attractive force: therefore, as much water is thrown off by the centrifugal force on the side opposite to the moon, as is raised on the side next her by her attraction. Hence, it is plain, that at D, fig. 3, the centrifugal force must be greater than at the center E, and at E than B, because the part D is farther from the center of motion than the part B. On the contrary, the part B being nearer the moon than the center E, the attracting power must there be strongest, and weakest at D. And, as the two opposing powers balance each other at the earth’s center, the tides will rise as high on that side from the moon, by the excess of the centrifugal force, as they rise on the side next her by the excess of her attraction.

Pupil. In this explanation you have mentioned nothing of the sun.

Tutor. From what I have already said it must be plain to you that if there were no moon the sun by his attraction would raise a small tide on the side next him; and, it is as evident that the tides opposite would be raised as high by the centrifugal force: for the sun and earth, as well as the earth and moon, move round their center of gravity. This may be exemplified by an easy experiment. Take a flexible hoop, suppose of thin brass, tie a string to it and whirl it round your head, and it will assume an elliptical shape; the tightness of the string drawing out the side next to your hand, and the centrifugal force throwing off the other.

Pupil. This I clearly comprehend.

Tutor. I shall now refer you to the next figure, (fig. 4.) where F represents the moon at full: the sun and moon are in opposition, and yet the tide is as high on each side as in the former case. I wish you to shew me the cause.

Pupil. I will use my endeavour to do it, Sir.

Tutor. Then I doubt not you will accomplish it.

Pupil. When the moon is at full, ten parts of water are raised from that side of the earth next her, by her attraction; and, as the side which is next her is opposite to the sun, three parts must be thrown off by his centrifugal force, the sum of which will be thirteen parts next the moon.—From the side opposite to the moon, and under the sun, ten parts are thrown off by her centrifugal force, and three raised by his attraction, making thirteen, the same as before.

Tutor. I could not have done it better. These are called Spring Tides. But when the moon is in her quarters, the action of the sun and moon are in opposition to each other; that is, they act in contrary directions (see fig. 5.) The moon of herself would raise the water ten parts under her, and throw off ten parts by her centrifugal force on the opposite side; but, the sun being then in a line with the low-water, his action keeps the tides from falling so low there, and consequently from rising so high under and opposite to her. His power, therefore, on the low-water being three parts, leaves only seven parts for the high water, under and oppose the moon. These are called Neap Tides.

Pupil. This is very plain.

Tutor. You would naturally suppose that the tides ought to be highest directly under and opposite to the moon: that is, when the moon is due north and south. But we find, that in open seas, where the water flows freely, the moon is generally past the north and south meridian when it is high-water. For, if the moon’s attraction were to cease when she was past the meridian, the motion of ascent communicated to the water before that time would make it continue to rise for some time after: as the heat of the day is greater at three o’clock in the afternoon than it is at twelve; and it is hotter in July and August than in June, when the sun is highest and the days are longest.

Pupil. These are convincing reasons. And, pray what time after the moon has passed the meridian, is it high-water?

Tutor. If the earth were entirely covered with water, so that the tides might regularly follow the moon, she would always be three hours past the meridian of any given place when the tide was at the highest at that place. But, as the earth is not covered with water, the tides do not always answer to the same distance of the moon from the meridian at the same places, because the regular course of the tides is much interrupted by the different capes and corners of the land running out into the oceans and seas in different directions, and also by their running through shoals and channels. But, at whatever distance the moon is from the meridian on any given day, at any place, when the tide is at its height there, it will be so again the next day, much about the time when the moon is at the like distance from the meridian again.

Pupil. Are not the tides later every day than they were the preceding day?

Tutor. Yes; and the reason is obvious: for, whilst the earth is revolving on its axis in twenty-four hours, the moon will be advancing in her orbit; therefore the earth must turn as much more than round its axis before the same place which was under her can come to the same place again with respect to her, as she has advanced in her orbit during that interval of time, which is 50 minutes. This being divided by 4, gives 12-1/2 minutes; so that it will be 6 hours 12-1/2 minutes from high to low-water, and the same time from low to high-water: or 12 hours 25 minutes from high-water to high-water again.

Pupil. This I understand perfectly well.

Tutor. I have now finished my description of the tides, and having a little time to spare, if you wish to know how to find the proportionate magnitude of the planets with that of the earth, and to calculate their distances from the sun, I will employ it that way.

Pupil. At our first conference I remember you shewed me the proportion that the other planets bear to the earth, with their periods and distances from the sun; but to have it in my power to make the calculations myself, will certainly give me great pleasure.

Tutor. To find what proportion any planet bears to the earth; or, that one globe bears to another, you must observe that, all spheres or globes are in proportion to one another as the cubes of their diameters. So that you have nothing more to do than to cube the diameter of each, and divide the greatest by the least number, and the quotient will shew you the proportion that one bears to the other.

Pupil. The operation appears very simple; but, as I do not know what a cube number is, I cannot perform it.

Tutor. You cannot forget what a square number is.

Pupil. The product of any number multiplied into itself is a square number, as 4 is the square of 2.

Tutor. Any square number multiplied by its root, or first power, will be a cube number. Thus 4 multiplied by 2 will be 8, which is the cube of 2; 9 is the square or second power, and 27 the cube or third power of 3, &c. This you will perhaps better understand by

Pupil. I do, Sir; and am now prepared for an example.

Tutor. The diameter of the sun is 893552 miles, of the earth 7920 miles; how much does the sun exceed the earth in magnitude?

Pupil. The cube of 893522, the sun’s diameter, is 713371492260872648; and of 7920, the earth’s, 496793088000. And 713371492260872648 divided by 496793088000 is equal to 1435952, and so many times is the bulk of the sun greater than that of the earth.

Tutor. This one example may suffice, as I intend by and by to give you a table of diameters, &c.; you may then calculate the rest at your leisure.

Pupil. I shall now, Sir, be glad to have the other explained.

Tutor. The periods of the planets, or the times they take to complete their revolutions in their orbits, are exactly known; and the mean distance of the earth from the sun has been also ascertained. Here, then, we have the periods of all, and the mean distance of one, to find the distances of the rest; which may be found by attending to the following proportion:

The cube root of this quotient will be the distance sought.

Pupil. Here again I find myself at a loss, as I have not learnt to extract the cube root.

Tutor. I will give you ^[18]Doctor Turner’s rule, which I think will answer your purpose.

“First, having set down the given number, or resolvend, make a dot over the unit figure, and so on over every third figure (towards the left hand in whole numbers, but towards the right hand in decimals); and so many dots as there are, so many figures will be in the root.

Next, seek the nearest cube to the first period; place its root in the quotient, and its cube set under the first period. Subtract it therefrom; and to the remainder bring down one figure only of the next period, which will be a dividend.

Then, square the figure put in the quotient, and multiply it by 3, for a divisor. Seek how often this divisor may be had in the dividend, and set the figure in the quotient, which will be the second place in the root.

Now, cube the figures in the root, and subtract it from the two first periods of the resolvend; and to the remainder bring down the first figure of the next period, for a new dividend. Square the figures in the quotient, and multiply it by 3, for a new divisor; then proceed in all respects as before, till the whole is finished.”

The following example will, I trust, make it clear to you.

It is required to find the cube root of 15625.

Point every third figure, and the first period will be 15; the nearest cube to which, in the table I gave you just now, you will find to be 8, and its root 2; the 8 you must place under the 15, and the 2 in the quotient: take 8 from 15 and 7 will remain, to which bring down 6, the first figure of the next period, and you have 76 for a dividend. The figure put in the quotient is 2, the square of which is 4, which multiplied by 3 is 12, for a divisor. Now 12 in 76 will be 5 times; cube 25, and you will have 15625, which, subtract from the resolvend, and nothing will remain; which shews that the resolvend is a cube number, and 25 its root.

Pupil. You say 12 in 76 is 5 times; I should have said 6 times.

Tutor. In common division it would be so; but as the cube of 26 would be greater than the resolvend from which you are to subtract it, it can go but 5 times.

Pupil. Now, Sir, I think I have a sufficient knowledge of the rule to solve a problem.

Tutor. The earth’s period is 365 days, and its mean distance from the sun 95 millions of miles; the period of Mercury is 88 days—what is his mean distance?

Pupil. As the distance of the earth is given, I must make the square of 365 the first term, the cube of 95 the second, and the square of 88 the third term of the proportion.

Tutor. Certainly.—Take your slate, or a piece of paper, prepare your numbers, and make your proportion.

Pupil. I find the square of 365 = 133225; of 88 = 7744; and the cube of 95 = 857375.

Then 133225 : 857375 :: 7744 to a fourth term.

I now multiply the second and third terms together, and divide the product by the first, the quotient 49836 is the cube of the mean distance of Mercury from the sun in millions of miles, and the fourth term sought.

Tutor. So far you are right. Now extract the root.

Pupil. The root I find to be 36, which is the mean distance of Mercury from the sun, in millions of miles.

Tutor. You now see, that although 27 in 228 will go 8 times, yet here it will go but 6 times; and, as there is a remainder, it shews you that the resolvend is not a cube number.

Pupil. I see it clearly.

Tutor. You now seem perfect in the rule; I shall therefore not trouble you with any more examples, but shall give you the table I promised you.

Pupil. I shall take the first opportunity of calculating the rest, in which I am certain I shall have great satisfaction.

Tutor. I have now conducted you through the elementary parts of astronomy, have given you a general view of the system of the world, and prepared you to pursue the study with profit and pleasure.—In your future researches, the more accurate you are, the more you will discover of regularity, symmetry, and order in the constitution of the frame of nature.

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