PROBLEM XXX. To find the Rate at which the Earth moves in the Ecliptic during the Time of its Annual or Periodical Revolution round the Sun. RULE. Since the earth's mean distance from the sun is 94546196 miles (Problem XXIV., page 738), the diameter of the orbit in which it moves round that great luminary is 189092392 miles; and since the diameter of a circle is to its circumference in the ratio of unity or 1, to 3.14159265, the circumference of the earth's orbit is 594051320 miles. Now, as the earth describes this circumference in 365:5'48"48: (last Problem), or 8766 hours nearly, we have the following computation by logarithms :— As the length of the year, in hours=8766 Log. ar. comp,=6.0571985 Log.=0.0000000 To the earth's hourly motion in its orbit=67768 miles Log.=4.8310224; which is about 141 times swifter than the ordinary flight of a cannonball. PROBLEM XXXI. Given the Moon's Mean Distance from the Earth, and her Apparent Semidiameter, at a Mean Rate; to find the True Measure of her Diameter, in English Miles. RULE. It is shown in page 9, under the head "Augmentation of the Moon's Semidiameter," that the moon's mean distance from the earth is 236692.35 miles. Now, since her apparent semidiameter is 15:43′′ at a mean rate,-therefore, to the logarithm of her mean distance from the earth, add the logarithmic tangent of her apparent semidiameter; and the sum (abating 10 in the index) will be the logarithm of the moon's semidiameter in English miles: the double of which will be the measure of her whole diameter. Example. Let the moon's distance from the earth be 236692.35 miles, and her semidiameter 15:43; required the true measure of her diameter in English miles? Moon's mean distance from the earth-236692.35 miles Log.=5.3741842 Moon's apparent semidiameter = 15:43 Log. tang, = 7.6600896 True measure of the moon's diameter=2164. 2 English miles. PROBLEM XXXII. Given the Diameters of the Earth and the Moon; to find the Ratio of their Magnitudes. Note. This is performed by Problem XXVI., page 739. Example. If the earth's diameter be 7917. 5 English miles, and that of the moon 2164. 2 such miles, required the ratio of their magnitudes? Ratio of the magnitudes of the earth and moon=48.96 Log.=1,6898721 PROBLEM XXXIII. To find how much larger the Earth appears to the Lunar Inhabitants than the Moon appears to the Terrestrial Inhabitants. RULE. Since the distance between the earth and the moon is such as to cause their opposing hemispheres to appear, reciprocally from each other, like flat 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? = Number of times the earth is larger than the ▷ 13.38 Log.=1.1265814 PROBLEM XXXIV. To find the Height of a Mountain in the Moon. RULE. When the moon is viewed through a good telescope, her face appears to be diversified with extensive ranges of remarkably high mountains: -now, although the present Problem may appear as merely chimerical to the uninformed in astronomy, yet, there is not a more rational and demonstrable Problem, or proposition, in the whole range of the mathematics, as may be seen in the following operations, viz. :— In the annexed diagram let ADBE be the disc or face of the moon at the time of the quadratures; for, it is at this time, or when the moon is 90 degrees from the sun, that the altitudes of her mountains are most easily determined. D T B Let the diameter A CB be the boundary. of light and darkness, and RT the mountain called St. Catherine, the summit of which, at the point T, is just beginning to be enlightened by the solar ray SA T. Now, by means of a micrometer fitted in a suitable telescope, the distance between the points T and A may be easily measured; and thus their ratio will be known:-this, according to Hevelius and other astronomers, is the part of the diameter A B, or the part of the semidiameter A C. Now, we have seen in the preceding Problem that the value of the semidiameter C A, or CR, is 1082.1 miles; therefore, agreeably to the above ratio, 1082. 1÷÷13= 83.24 miles, is the value of TA: then, in the right angled triangle TAC, the legs are given; viz. C A=1082.1 miles, and TA=83.24 miles; to find the hypothenuse TC; from which take away the semidiameter CR (CA) and the remainder, or RT, will be the height of the mountain. Hence, by right angled plane trigonometry, Problem IV., page 175, The remainder, or RT = 3.2-hence, the height of the lunar mountain called St. Catherine is 3 miles, or 3 miles, and 352 yards; and thus it is manifest that the apparently paradoxical Problem for determining the height of a mountain in the moon is simply the resolution of a plane right angled triangle. PROBLEM XXXV. To find the Rate at which the Moon revolves round her Orbit. Note. This is performed by Problem XXX., page 742, 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., Asone lunation=27743" 5,in secs.=2360585 Log.ar.co.=3.6269804 To the moon's hourly motion in her orbit=2268 miles Log.=3.3556470 PROBLEM XXXVI. 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's distance from that luminary ? Earth's periodical revolution 36554848, in secs. 31556928 Ar.co. of twice its log.-5.0018106 Earth's mn.dist. from sun,in miles=94546196 Thrice its log. 23. 9269323 Venus's per.rev.22416'49" 11,in secs. 19414151Twice log. 14. 5762368 Reject 20 from index; and, to extract the root, divide by 3) 23.5049797 Venus's mean distance from sun, in miles=68390098 Log.=7.8349932} PROBLEM XXXVII. 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 |