The Study of Astronomy, adapted to the capacities of youth: In twelve familiar dialogues, between a tutor and his pupil: explaining the general phænomena of the heavenly bodies, the theory of the tides, &c.Stedman, John, teacher of astronomy
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The Study of Astronomy, adapted to the capacities of youth: In twelve familiar dialogues, between a tutor and his pupil: explaining the general phænomena of the heavenly bodies, the theory of the tides, &c.
Stedman, John, teacher of astronomy
Astronomy -- Juvenile literature -- Early works to 1800
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.
Public-domain text, read in full here on John Shaqi.
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