HOW TO WEIGH THE SUN. 693 



Weighing the Sun. — When we wish to weigh anything we must have scales ; 

 but what scales will weigh the solar mass? How shall we begin ; and when we 

 have told its weight, how shall we know it to be correct ? 



Our balance shall consist, not of instruments of steel, but of motion, whose 

 amount is only 386 inches in one second, or that velocity developed at the close 

 of a second on an insignificant world more than 92,000,000 miles away from the 

 great sun it is proposed to handle ! How absurd such an undertaking seems, 

 but no matter ; men became bold as soon as they learned that gravity varies as 

 distance squared inversely, and attacked problems that would have appalled the 

 human intellect before. 



The distance of the centre of the sun from the centre of the earth is 23,440 

 times greater than the surface of the earth is from its center. We know gravity 

 to be measured by the 386 inches, at a distance of 3,962.72 miles from the centre 

 of our earth ; how much will it be weakened at a distance 23,440 times as great? 

 The law says it will be lessened as the square of the ratio of the distance, in- 

 versely; hence, 23,440 squared equals 549,433,600; that is, gravity will be 

 diminished that many times at the sun's distance. Divide 386 by this number; 

 the quotient will be .00000070254 of an inch velocity, at the end of a second, 

 that the attraction of the earth is able to impart to a falling body that far from 

 its centre. Proceeding as we did in the case of the moon, we multiply the num- 

 ber of inches from the centre of the earth to the centre of the sun by the force of 

 gravity, or the decimal given, resulting in 4,134,635 inches per second, which is 

 the square of the velocity with which a body at that distance must move in order 

 to counteract that much attraction and not fall. Extracting the square root, we 

 have 2,033 iiiches per second as the real orbital motion. Now, by finding the 

 number of inches in the circumference of a circle having the above radius, and 

 dividing by 2,033, '"'^ ^^^ ^^^ time of revolution, 576.3657 years. But to make 

 circuit in one year it must move 576.3657 times as fast, or 1,171,751 inches per 

 second. 



It is clear that if we can find how much centrifugal force this velocity gener- 

 ates, we shall at once know how strong gravity is at that distance, and how much 

 matter is required to exert that attraction. As before, centrifugal force is found 

 by dividing the square of the velocity by the radius of the orbit. Therefore 

 1,171,751 squared equals 1,373,000,406,000, which divided by 5,885,266,907,- 

 520, the inches in 23,440 times the earth's mean equatorial radius, or the mean 

 distance of the sun, gives a quotient of .233 of an inch per second, velocity gen- 

 erated at the end of a second by a force equal to the centrifugal force found act- 

 ing during that time. 



But this is also equal to gravity; now, how many times stronger is it than what 

 we found by dividing 386 by the square of the distance? We find how much by 

 dividing .233 by .00000070254, and find it to be 331,654 times as great. 

 Hence, the earth must become 331,654 times heavier than it is to make a body 

 at the sun's distance revolve around it in a siderial year, and not fall. Since the 



