178 KANSAS CITY REVIEW OF SCIENCE. 
each, their number would be as above; while the diameter of each would be 2.21 
feet =2 feet 2^ inches. 
Problem: — How much matter must each of two spheres contain, whose centres 
ate one mile — ^,280 feet apart, to be able to attract each other with an intensity of i 
avoirdupois pound, both spheres to be isolated in infinite space ? 
Given the amount of matter in the earth, also its size, we learn its attractive 
force. It has been agreed to by some nations to call the attraction exerted by 
the mass of the earth on 27.727384 cubic inches of distilled water at 60° F.; or 
upon 3.559356 cubic inches pure iron, i pound avoirdupois. 
The diameter of a sphere of iron having this bulk is 1.9 inches. We now 
have given two spheres — one 7916. 3185 miles in diameter, and the other 1.9 
inches, together with the amount of attraction of each upon the other, which is i 
pound; also the distance between their centres which is 3958.15925 miles, from 
which data the question must be solved. And we proceed to discuss the case 
by means of the most elementary forms of analysis to avoid needless difficulties 
in the way of students. We have adopted the mean diameter and radius of the 
earth. 
It is proven in the higher mathematics that the force of gravity exerted by a 
sphere on a mass at any distance, is the same as would be if the entire mass of 
the sphere were concentrated at its centre. Now let us place enormous platform 
scales (Fairbank's ?) on the earth's surface, and place the little ball of iron on its 
platform and observe its weight, which will be seen to be i pound. We will see 
the necessity of using large scales to weigh the small ball as we proceed. Con- 
ceive the scales rigidly fixed in space by some power external to the earth, and 
let the earth become motionless upon its orbit. 
Imagine the earth to shrink and subside to a sphere only two miles in diam- 
eter, both its centre and also the scales remaining in space in their original posi- 
tions 3958.15925 miles apart. Then will the ball of iron still weigh i pound, 
because the earth, now two miles in diameter, will have its original mass, though 
of inconceivable density ; and will attract with the same intensity as when ex- 
panded. 
It is demonstrated by analysis that : Every particle of matter attracts every 
other with a force directly proportional to mass, but inversely proportional to the 
square of distance. That is : The greater the distance between two bodies the 
less is their mutual attraction. It follows then that the ball on the scales has a 
weight which we know to be diminished in the ratio of the square of the distance 
from it to the earth's centre. As our problem relates to a distance of one mile, 
we assume the earth to have dwindled to a sphere whose radius is one mile, 
which brings its surface to a distance of a mile from its centre. Thus : 3958. 15925 
less 1^3957.15925 miles, whose square is 15,659,110. Let us now move the 
scales to the surface of the sphere and weigh the little sphere of iron. Behold ! 
its weight is 15,659,110 pounds ! What scales must we have to determine its 
weight ? Indeed ! both ball and scales would be crushed by the appalling force 
-of gravity. We have now arrived at the fact that a sphere having a radius of 
