Note N. 136 — SECTION VIII.

Note 136, p. 55. As hardly anything appears more impossible than that man should have been able to weigh the sun as it were in scales and the earth in a balance, the method of doing so may have some interest. The attraction of the sun is to the attraction of the earth as the quantity of matter in the sun to the quantity of matter in the earth; and, as the force of this reciprocal attraction is measured by its effects, the space the earth would fall through in a second by the sun’s attraction is to the space which the sun would fall through by the earth’s attraction as the mass of the sun to the mass of the earth. Hence, as many times as the fall of the earth to the sun in a second exceeds the fall of the sun to the earth in the same time, so many times does the mass of the sun exceed the mass of the earth. Thus the weight of the sun will be known if the length of these two spaces can be found in miles or parts of a mile. Nothing can be easier. A heavy body falls through 16·0697 feet in a second at the surface of the earth by the earth’s attraction; and, as the force of gravity is inversely as the square of the distance, it is clear that 16·0697 feet are to the space a body would fall through at the distance of the sun by the earth’s attraction, as the square of the distance of the sun from the earth to the square of the distance of the centre of the earth from its surface; that is, as the square of 95,000,000 miles to the square of 4000 miles. And thus, by a simple question in the rule of three, the space which the sun would fall through in a second by the attraction of the earth may be found in parts of a mile. The space the earth would fall through in a second, by the attraction of the sun, must now be found in miles also. Suppose m n, fig. 4, to be the arc which the earth describes round the sun in C, in a second of time, by the joint action of the sun and the centrifugal force. By the centrifugal force alone the earth would move from m to T in a second, and by the sun’s attraction alone it would fall through T n in the same time. Hence the length of T n, in miles, is the space the earth would fall through in a second by the sun’s attraction. Now, as the earth’s orbit is very nearly a circle, if 360 degrees be divided by the number of seconds in a sidereal year of 3651⁄4 days, it will give m n, the arc which the earth moves through in a second, and then the tables will give the length of the line C T in numbers corresponding to that angle; but, as the radius C n is assumed to be unity in the tables, if 1 be subtracted from the number representing C T, the length of T n will be obtained; and, when multiplied by 95,000,000, to reduce it to miles, the space which the earth falls through, by the sun’s attraction, will be obtained in miles. By this simple process it is found that, if the sun were placed in one scale of a balance, it would require 354,936 earths to form a counterpoise.