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
A TABLE OF
Roots. 1. 2. 3. 4. 5. 6. 7. 8. 9.
Squares. 1. 4. 9. 16. 25. 36. 49. 64. 81.
Cubes. 1. 8. 27. 64. 125. 216. 343. 512. 729.
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:
As the square of the period of any one planet,
Is to the cube of its mean distance from the sun;
So is the square of the period of any other planet,
To the cube of its mean distance.
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.
[Footnote 18: Young Geometrician’s Companion.]
“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.
EXAMPLE.
It is required to find the cube root of 15625.
. .
15625 (25
8
─────
12) 76
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.
. .
49836 (36 3 36
27 3 36
─── ── ─────
27) 228 Sq. of 3 = 9 216
46656 Mul. by 3 108
───── ── ─────
3180 Divisor 27 1296
═════ ══ 36
─────
7776
3888
─────
Cube of 36 = 46656
═════
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.
┌─────────────────────────────────────────────────────────────────────┐
│ TABLE. │
├──────────┬──────────┬───────────────┬───────────────┬───────────────┤
│ Names │Diameters,│ Magnitude, │ Periods, │ Mean Distance │
│ of the │in English│ compared │ in │ from the Sun, │
│ PLANETS. │ Miles. │with the Earth.│Years and Days.│ in Mil. of │
│ │ │ │ │ Miles. │
├──────────┼──────────┼───────────────┼───────────────┼───────────────┤
│Sun │ [A]893522│ 1435952 │ —— │ —— │
│ │ │ │ │ │
│Mercury │ 3261│ 1/14 │ 0 —— 88 │ 36 │
│ │ │ │ │ │
│Venus │ 7699│ 5/49 │ 0 —— 224 │ 68 │
│ │ │ │ │ │
│Earth │ 7920│ 1 │ 1 or 365 │ 95 │
│ │ │ │ │ │
│Moon │ 2161│ 1/49 │ —— │ —— │
│ │ │ │ │ │
│Mars │ 5312│ 1/3 │ 1 and 322 │ 145 │
│ │ │ │ │ │
│Jupiter │ 90255│ 1479 │ 11 —— 314 │ 494 │
│ │ │ │ │ │
│Saturn │ 80012│ 1031 │ 29 —— 167 │ 906 │
│ │ │ │ │ │
│Georgian │ 34217│ 82 │ 83 —— 121 │ 1812 │
└──────────┴──────────┴───────────────┴───────────────┴───────────────┘
[Footnote A: The Diameters were taken from Adams’s Lectures, Vol. IV.
p. 39.]
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.
“Hail, Sov’reign Goodness! all-productive Mind!
“On all thy works thyself inscrib’d we find;
“How various all, how variously endow’d,
“How great their number, and each part how good!
“How perfect then must the Great Parent shine, ⎫
“Who, with one act of energy divine, ⎬
“Laid the vast plan, and finish’d the design!” ⎭
THE END.
------------------------------------------------------------------------
Directions to the Bookbinder.
Plate I. _to face the_ Title.
———— II. —— _page_ 40.
———— III. —— —— 88.
———— IV. —— —— 131.
------------------------------------------------------------------------
Transcriber’s note:
All instances of ‘disk’ changed to ‘disc’
Errata, instance of ‘disk’ on page 79 added, “—— 79. — 5. ⎭”
Page 11, ‘Years’ changed to ‘years,’ “130 years after Christ”
Page 20, ‘h e’ changed to ‘the,’ “would have as much the appearance”
Page 24, ‘cannon ball’ changed to ‘cannon-ball,’ “the time a cannon-ball
would”
Page 63, comma changed to full stop after ‘TUTOR,’ “TUTOR. Why?”
Page 65, ‘a’ changed to ‘_a_,’ “carry a planet from A to _a_”
Page 74, ‘itaxis’ changed to ‘its axis,’ “if the earth revolve on its
axis every”
Page 78, ‘Mercury’ struck after ‘Sun,’ “Sun, Venus, Mars, and Jupiter
are known to revolve on their axes”
Page 93, ‘cancer’ changed to ‘Cancer,’ “is the _tropic of cancer_; that”
Page 93, ‘capricorn’ changed to ‘Capricorn,’ “the _tropic of capricorn_”
Page 115, ‘othes’ changed to ‘other,’ “and other stars rise to his”
Page 115, ‘bnt’ changed to ‘but,’ “out from; but, by going round”
Page 116, ‘it’s’ changed to ‘its,’ “How can its surface be round”
Page 128, full stop inserted after ‘eclipses,’ “explain the cause of
eclipses.”