Page:Electricity (1912) Kapp.djvu/73

kinetic energy of the meteorite is the product of its weight multiplied by the radius of the earth. Adopting the engineer's unit of energy as the metre-kilogram, and the mass unit as that mass which weighs 9.81 kg., we must take the radius of the earth in metres, and shall get the velocity $v$ in metres per second. Since the mass of our meteorite is supposed to be unity, we have the equation

$E=9.81R$

from which we find ${\frac {1}{2}}v^{2}=9.81\times 636000$

$v=11150$

The meteorite will strike the earth with a velocity of 11.15 kilometres a second; in reality a little less, because of the resistance of the air.

This digression has been inserted to show the application of the potential theory to a purely mechanical problem. Let us now return to the electrical aspect of this theory. We have a large sphere, charged with $Q$ units of positive electricity, and suspended in the middle of a large room. The potential difference between any point of the wall and the surface of the sphere is numerically equal to the energy required to bring a unit of positive electricity from the wall to the