Newton's Theory of Gravity states that every object in the
universe pulls on every other object. Every object feels this
force, so it is a universal force. The force is always
attractive; it is always a pull, never a push. Picture the
force of gravity as the tension in an imaginary rope between two
objects. What factors determine the strength of the pull between
two objects?
The masses of the two objects are important factors. Remember
that mass is a measure of the amount of material in an object.
Thus, mass is a measure of the number of "pullers" in that
object. It makes sense that the more "pullers" there
are, the stronger the pull. If we double the mass of one of the
objects, we double the strength of the pull. If we double the
masses of both objects, we *quadruple* the
pull.

The other factor is not quite as easy to understand. The
farther apart two objects become, the weaker the pull between the two
becomes. Further, the relationship between distance and
gravitational pull is not a simple one, but rather what we call an *inverse
square* relationship. If, for example, we triple the
distance between two objects, the gravitational force between them
drops, not by three times, but by *three times three times*,
or nine times.

Newton summarized the relationship between the strength of
gravitation pull and these three factors in this equation:

**F**_{g}
= G M_{1} M_{2} / (D_{12})^{2}

where M_{1} and M_{2
}are the masses of the objects, and _{ }
D_{12}
is the distance between the **centers** of the two
objects. G is a number called Newton's Universal Constant of
Gravitation, which varies with the system of units we are using.

If we want to calculate absolute values of the gravitational force,
in units of dynes or pounds or Newtons, we need to know the value of
G. But that's not how Newton himself did it. We can get
around knowing the value of G by comparing gravitational forces.

Here's how it's done. Imagine that we have four objects,
numbered 1 through 4. How does the gravitational force between 1
and 2 compare to that between 3 and 4?

We must write out two different gravitational equations.
First, one for the force between 1 and 2:

**F**_{g12} = G M_{1} M_{2}
/ (D_{12})^{2}

And now an equation for the force between 3 and 4:

**F**_{g34} = G M_{3} M_{4}
/ (D_{34})^{2}

Notice the use of subscripts to tell one equation apart from the
other. Now, to compare the two forces, we divide one equation by
the other:

**F**_{g12}
G M_{1} M_{2} / (D_{12})^{2
-------- = --------------------------
}F_{g34}
G M_{3} M_{4} / (D_{34})^{2}

Notice how, since G is in both the numerator and the denominator,
it cancels out in this situation, and we don't have to worry about
it. We can rearrange some of the factors that remain like so:

Often, we will get a result that this ratio will equal some
number. This indicates how the gravitational force between 1 and
2 compares to that between 3 and 4. For example, if we get

**F**_{g12}
/ F_{g34} = 10

Then we can say "The gravitational force between
object 1 and object 2 is 10 times greater than the force between object
3 and object 4."

In many problems, we will make comparisons such that
object 3 is the same as object 1, making calculations even
simpler. We may also have a "before and after"
situation, where object 4 is object 2 after some change, and object 3
is object 1 after some change. Examples of both are given below.