The application of Newton’s Law of Gravity in the above example
is a little more complicated than it might seem for the
following reasons:

- Mass m on the Earth’s
surface is attracted to every particle within the Earth.
Therefore, the attractive force W between object m and the
Earth (i.e., m’s weight) is a vector sum of all of the
m/particle forces.

In the above example, object m
and M1 (a random particle within the Earth’s sphere) have a
mutual attraction F. However, only the vector component of F.
(i.e. f) directed toward the Earth’s center contributes to the
weight of object m. The summation of all these vector forces
between mM1,
mM2, ...... will compute the accurate weight of m. This is a
calculus problem for a mathematician.

- Another condition which
complicates the calculation of m’s weight is the fact that
the earth is not a homogeneous sphere. The Earth’s interior
consists of the crust, asthenosphere, transition zone, lower
mantle, outer and inner core, all consisting of materials of
different densities.

- Also, the Earth is not a
perfect sphere. It is flattened at the poles and therefore,
the force of gravity is greater (i.e. an object’s weight is
greater) at the poles.

If object m could descend into
the Earth toward the Earth’s center, its weight would decrease
until it became weightless at the Earth’s center. As the object
descended from the Earth’s surface, the Earth’s mass above the
object would exert an attractive force tending to pull the
object toward the surface. The more powerful Earth centered
force would offset the surface directed force until the object
reached the Earth’s center.