98 KANSAS CITY REVIEW OF SCIENCE, 
the radius=333, 426-7-11,950=27.83. Therefore solar gravity is 27.83 times- 
stronger than the Earth's ; that is : a stone that would weigh one pound on the 
surface of the Earth would weigh 27.83 pounds on the Sun. Also, gravity exerted 
by the mass of a sphere at any distance from it, is equal to gravity on its surface 
divided by the square of the distance. 
That being the case, let us learn how much solar gravity is weakened at a 
distance of 92,882,917 miles from its centre The radius of the Sun is 433,197 
miles; and 92,882,917 divided by 433,197=214.41248 whose square is 45,972.- 
71 158. Now we have seen that the Sun's attraction is 27.83 times stronger than 
the Earth's, hence one ton on the Earth taken to the Sun would weigh 27.83, 
tons. Suppose we now take the 27.83 tons into space 92,882,917 miles from 
the Sun, how much would it weigh then ? If we reduce 27.83 tons to ounces 
and divide by 45,972.71158 the quotient will be 19.37 ounces ! That is, if we 
carry a stone that weighs one ton on the Earth, to the Sun, it will weigh 27.83. 
tons, and then if we remove it 92,882,917 miles away from the Sun into space, 
the great weight will dwindle to 19.37 ounces or a little more than one pound. 
Great results follow from this method of mathematical induction. Let us see. 
Why not take the whole Earth to the Sun ? If we could, it would weigh 27.83 
times as much, or 187,718,612,962,379,075,475,000 tons. Now carry the Earth 
■ back into space from whence it came, then each ton of its weight on the Sun's- 
surface, would shrink to 19.37 ounces; the weight of the whole Earth dwindling 
in reference to its weight while on the Sun, to 4,083,261,651,590,510,370 tons. 
Then using rule 2d to find the tension exerted on our imaginary rod of cast- 
iron, we must multiply this newly determined mass of the Earth (in relation to 
the Sun) by the acceleration and divide the product by the force of solar gravity 
exerted by the Sun on the Earth. The acceleration we found to be .00000368 of 
a mile per second ,* but a curious circumstance comes in here, which renders the 
problem different from our cannon-ball computation. This is: the acceleration is 
just equal to gravity, i. e. the Earth does not fly away from the Sun; nor does it 
approach, hence we need not multiply one number by another and then divide 
the product by the same, we do not change values. Hence a cable or bar that 
would be able to hold the Earth on its orbit, must be strong enough to hold a 
weight of 4,083,261,651,590,510,370 tons. 
With what appalling force does the mighty mass of the Sun attract the Earth ; 
and what manner of engine is this, that toils night and day simply that such 
beings as men might live a i&w days and expire ? One-half of the Earth is turned 
toward the Sun all the time, and it is easy to find the number of square inches in 
the hemisphere. Suppose now that we attach the ends of a cast-iron bar to tlie 
Earth and Sun, and instantly let the Sun cease attracting, then the Earth would 
tend to fly away from its orbit on a straight line ; and since this centrifugal tend- 
ency is equal to its weight (referring to the Sun), we find the force or pressure 
exerted on each square inch of the entire half of the Earth's surface to be 20,619 
pounds, or 10.3096 tons. But cast-iron is capable of sustaining a pressure of only 
18,656 pounds, hence a soUd bar of this material 7,925 miles in diameter could 
