Thank you for your question, what you say is true, the gravitational force exerted by the Earth on the Moon has to be equal to the centripetal force.
An interesting application of this principle is that it allows you to determine a relation between the period of an orbit and its size. Let us assume for simplicity the Moon's orbit as circular (it is not, but this is a good approximation for our purposes).
The gravitational acceleration that the Moon experience due to the gravitational attraction from the Earth is given by:
ag=G(MEarth+MMoon)/r2
Where G is the gravitational constant, M stands for mass, and r is the radius of the orbit. The centripetal acceleration is given by:
acentr=(4 pi2 r)/T2
Where T is the period. Since the two accelerations have to be equal, we obtain:
(4 pi2 r) /T2=G(MEarth+MMoon)/r2
Which implies:
r3/T2=G(MEarth+MMoon)/4 pi2=const.
This is the so-called third Kepler law, that states that the cube of the radius of the orbit is proportional to the square of the period.
This has interesting applications. In the Solar System, for example, if you know the period and the radius of one planet orbit, by knowing another planet's period you can determine its orbit radius. I hope that this answers your question.
The gravitational force between two object depends on their masses and on their distance.
Since the formula is

If the masses grow, the force also grows. But I'm assuming the two objects are fixed, so you can't enlarge their mass.
So, the only option remaining is to lower their distance: since it sits at the denominator, a smaller value of d results in a bigger value for F.
So, if you reduce the distance between two objects, the gravitational force between them will always result in an increase
Yea it’s called the Saffir-Simpson Hurricane scale, made in 1960s and further developed in 1970s
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300g is what that is on which is equal to 10.58oz