Physics for Geologists, Second edition

10 Basic concepts
relating them by equating the exponents. It must be emphasized, though,
that if we omit any components, we may finish with the wrong answer. An
example will make this clear.
If we did not know the equation on page 1 relating depth to pressure, p,
in a static homogeneous liquid, we might assume that it is a function of mass
density, p, the acceleration due to gravity, g, and the volume, v, and write
p = f (p, g, v)
This can be written in dimensional form
from which we write the indicial equations
forM: 1 = a
forL: -I=-3a+b+3c
for T : - 2 = -2b.
(say, pressure is a function of p, g and v).
The solution of these simultaneous equations is a = 1, b = 1, and substitu-
tion of these into the indicial equation for L gives c = 4. This tells us that it
is not volume but length that is involved, and that that length is clearly the
depth at which p is required:
and the function is
where B is a dimensionless constant. A few measurements would establish
that the dimensionless constant
for a liquid of constant density.
Take another example, the results of which we will use later. The tension in
a string as we twirl a stone around in a horizontal orbit, or the gravitational
force that keeps a satellite in orbit, is called the centripetal force. What is
the equation for the centripetal force required to keep an artificial satellite
in a circular orbit? This force is clearly a function of the mass of the object,
Copyright 2002 by Richard E. Chapman