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circles; and since circles are to one another as the squares of their diameters (Euclid, Book XII., Prop. 2,) or, which is the same thing, since spherical surfaces are to each other as the squares of their radii,-therefore, from twice the logarithm of the earth's diameter, subtract twice the logarithm of that of the moon; and the remainder will be the logarithm of the number of times that the earth appears larger to the inhabitants of the moon than the moon does to the inhabitants of the earth.

Example.

The diameter of the earth is 7917.5 miles, and that of the moon 2164.2 miles; required how much larger the earth appears from the moon than the moon does from the earth?

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Number of times the earth is larger than the D= 13.38 Log.=1.1265814

PROBLEM XXII.

To find the Rate at which the Moon revolves round her Orbit. Note. This is performed by Problem XVIII., page 617; as thus:

Since the moon's mean distance from the earth is 236692.35 miles, the diameter of her orbit must be twice that distance, or 473384.70 miles : hence its circumference is 1487182 miles. And since the moon goes through this circuit, or orbit, in 27:7:435, her hourly motion may be determined in the following manner; viz.,

=

As one lunation=277435, in secs.=2360585 Log.ar.co.-3.6269804 ·
Is to the circumference of the moon's orbit= 1487182 Log. 6. 1723641
So is one hour, in seconds =
3600 Log.-3.5563025.

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To the moon's hourly motion in her orbit 2268 miles Log. 3. 3556470

PROBLEM XXIII.

To find the mean Distance of a Planet from the Sun.

RULE.

It has been demonstrated, by the celebrated Kepler, that if two or more bodies move round another body as their common centre of motion, the

squares of their periodical times will be to each other in the same proportion as the cubes of their mean distances from the central body; and hence the following rule :—

As the square of the earth's periodical or annual motion round the sun, is to the cube of its mean distance from that luminary; so is the square of any other planet's periodical revolution round the sun, to the cube of its mean distance therefrom; the root of which will be the distance sought.

Example.

The earth's periodical or annual motion round the sun is completed in 365 days, 5 hours, 48 minutes, 48 seconds, and that of Venus in 224 days, 16 hours, 49 minutes, 11 seconds.. Now, if the earth's mean distance from the sun be 94546196 miles, what is Venus' distance from that luminary?

Earth's periodical revolution

365 5 48 48, in secs. 31556928 Ar. co. of twice its log. 5.0018106 Earth's mean dist. from sun, in miles,=94546196 Thrice its log. 23.9269323 Venus' per.rev.224 16:49 11, in secs. 19414151 Twice log. 14.5762368 Reject 20 from the index; and, to extract the root, divide by 3) 23. 5049797 Venus' mean distance from sun, in miles, 68390098 Log.=7.8349932}

PROBLEM XXIV.

To find how much more Heat and Light the Planets adjacent to the Sun receive from that Luminary than those which are more remote.

RULE.

Since the effects of heat and light are reciprocally proportional to the squares of the distances from the centre whence they are generated,— therefore, from twice the logarithm of the remote planet's distance from the sun, subtract twice the logarithm of the adjacent planet's distance therefrom; and the remainder will be the logarithm of the number of times that the planet adjacent to the sun is hotter and more luminous than that which is more remote.

Example.

If the earth's distance from the sun be 94546196 miles, and that of Venus 68390098 miles, required how much more heat and light the latter planet receives from the sun than the former?

Earth's mean dist. from sun=94546196 miles Twice its log.=15.9512882 Venus' =68390098 miles Twice its log.=15.8699865

ditto

Heat and light Venus receives more than the earth-1.205 Log. 0.0813017

PROBLEM XXV.

Given the apparent Diameter of a Planet; to find the Measure of its true Diameter.

RULE.

Find the difference between the earth's and the planet's mean distances from the sun, and it will show the planet's mean distance from the earth; with which and the planet's apparent semi-diameter, compute her true diameter, by Problem XIX., page 617.

Example.

Let the earth's distance from the sun be 94546196 English miles, that of Venus 68390098 such miles, and her apparent diameter 58".79; required the true measure of her diameter, in English miles?

Earth's distance from the sun = 94546196 miles

Venus'

ditto

= 68390098 miles

Venus' mean dist. from earth = 26156098 miles Log.
Venus' apparent semi-diameter = 29".395

= . 7.4175729 Log. tang. 6. 1537885

=

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True measure of Venus' diam.= 7454 English miles.

Note. If the ratio of the magnitudes of the earth and a planet be required, it may be determined by Problem XIV., page 615; thus in the case of Venus :

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Ratio of the magnitudes of the earth and Venus=1. 198 Log.=0.0785961

The velocity or rate at which a planet moves round its orbit may be determined by Problem XVIII., page 617.

PROBLEM XXVI.

To find the Time that the Sun takes to turn round its Axis.

RULE.

If the bright face of the sun be carefully observed through a good telescope, large black spots will be found to make their appearance on its eastern limb: from this they gradually advance to the middle of the disk, and thence to the western limb, where they disappear. After being absent for nearly the same period of time that they were visible, they will be observed to appear again on the eastern limb as at first; thus finishing their career in 27 days, 12 hours, and 20 minutes. Call this time the observed interval.

Find the number of degrees and parts of a degree that the earth has moved eastward in the ecliptic during that interval, in the following manner; viz.,

As the earth's annual motion

=

round the sun=36554848, in secs.31556928 Log. ar. co.2. 5009053 Is to eclip., or great circle of 360°, in secs. 1296000 Log. = 6.1126050 So is the obs. interv. 27:12:20:, in secs. 2377200 Log. = 6.3760657 Earth's advance in ecliptic during obs. interv. 97628" Log.= 4.9895760 Ditto, in degrees and parts of a degree, =

27:7:8"

Now, as 360 degrees, augmented by the earth's advance in the ecliptic during the observed interval, thus found, is to the observed interval; so is the great circle of 360 degrees, to the absolute time of the sun's rotatory motion on its axis; thus :

=

As 360: +27:7'8"=387:7:8", in secs.=1393628 Log. ar. co.3. 8558532 Is to the obs. int. = 271220", in secs. =2377200 Log. 6.3760657 So is the great circle of 360: in secs. 1296000 Log. 6.1126050

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To the time of the sun's rotatory motion=2210670: Log. = 6.3445239 Ditto, in days and parts of a day = 25:14:430; which, therefore, is the true time that the sun takes to turn round once upon its axis, as required.

PROBLEM XXVII.

To find the Length of a Pendulum for Vibrating Seconds in the Latitude of London. RULE.

It has been found by actual experiment that a heavy body let fall in the latitude of London, will descend, by the force of gravity, 16 feet in 1 second of time ;-now, since the circumference of a circle, whose diameter is unity or 1, is found by computation to be 3.14159265; and since the pendulum vibrates in the arc of a circle, or cycloid, the radius of which is equal to the length of the pendulum from the centre of oscillation; therefore if twice the space passed through by a falling body in one second of time, be divided by the square of the computed circumference of a circle, as above, the quotient will be the length of the pendulum for vibrating seconds in the parallel of London.

Thus. 16 feet 193 inches x 2 = 386 inches, Log. = 2.5865873 Circumf. of circle to diam. 13. 14159265, twice its Log. 0.9942998

Length of the pend. in inches 39.11 Logarithm =1.5922875 Note. By actual experiment the length of a pendulum for vibrating seconds in London is 39 inches, or 39. 125.

PROBLEM XXVIII.

To find the Length of a Pendulum for vibrating Half-Seconds.

RULE.

To the arithmetical complement of twice the logarithm of 120 (the number of vibrations in a minute for the half-seconds' pendulum), add twice the logarithm of 60 (the number of vibrations in a minute for the seconds' pendulum), and the logarithm of the length of the latter pendulum: the sum of these three logarithms (abating 10 in the index,) will be the logarithm of the length of the pendulum for vibrating half-seconds.

Example.

Let the length of a pendulum for vibrating seconds be 39. 125 inches; required the length of a pendulum that will vibrate half-seconds? Vibrations for 4-secs.' pendulum = 120 Ar. co. of twice its log.=5.841638. Ditto for the seconds' pendulum = 60 Twice its log. = . Length of the pendulum for secs. 39. 125 inches Log. =

Length of half-seconds' pendulum = 9.781 inches Log. =

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3.556302 1.592454

0.990394

Hence the length of a pendulum for vibrating half-seconds, is 94 inches.

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