A COLOSSAL ENGINE AT WORK FOR MAN. 97 
let us calculate the strain on the rod of iron to keep the little ball from flying 
away. 
Proceeding as in the case of the cannon ball, we divide the square of the 
velocity by the radius in order to find the acceleration of centrifugal tendency. 
If we square the velocity per second of the revolving ball and divide the square 
by 92,882,917 the quotient will be .00000368 of a mile or .233285 of an inch per 
second. By rule 2d the total centrifugal tendency or strain on the iron rod, will 
be equal to the acceleration multiplied by the mass of the moving ball and the 
product divided by the force of gravity exerted by the mass of the large sphere 
upon the small. The question now — before we can proceed — is to find the mass 
and gravity of the two globes in space. 
We remark that our hypothetical spheres are none other than the Sun and 
Earth, as they actually exist in the solar system. 
The diameter of the Sun 866,394 Miles. 
The diameter of the Earth 75925 Miles. 
The diameter of the Sun 109.31823 Ratio. 
The diameter of the Earth i. Ratio. 
The mass of the Sun . ...... 333,426.2356 Ratio. 
The mass of the Earth i. Ratio, 
Distance between centres of Sun and 
Earth 92,882,917. Miles. 
The density of water. i. Ratio. 
The density of the Earth 5.66 Ratio. 
Weightof the Earth 6, 743, 6 10, 190,383,705,901,809. Tons. 
This weight of the Earth needs explanation, for considerable error is attached 
to its meaning. If we cut up the Earth into cubes, bring each, one at a time to 
the surface, weigh it and replace in original position, we would find, that if each 
block should weigh a ton, their number would be as given. The reason is, that 
each cube would be subjected to the attraction of the Earth's mass. Of course, 
if the Earth were the only mass in existence it would not "weigh" anything 
there being no matter to attract it. 
The mass of the Earth in tons as above is simply its mass in relation to itself, 
and cannot be used in a computation seeking centrifugal tendency. We must 
have its weight in reference to the Sun, before we can find the tension on a rod 
assumed to hold them together. And this weight in relation to the solar mass 
we shall find to be quite different; and we must also learn the intensity of gravity 
exerted by the Sun's mass at the Earth's distance, because we see by the law that 
attraction is a factor in these calculations. 
It is known that the force of gravity on the surface of a sphere is equal to 
its mass divided by the square of the radius. 
The mass, radius and square of the radius, and therefore the gravity of the 
Earth, are each equal to i. The mass of the Sun is 333,426, its radius, =109.- 
31823, the square of its radius,=i 1,950 whence, mass divided by the square of 
