Some thirty years ago, soon after getting my Ph.D., I received a letter from a niece, my sister's youngest daughter. She was in high school and had been assigned (or chose) a science project consisting of presenting a talk on Newton's theory of gravity. Since I was the most educated person in the family and in physics to boot, she wrote me a letter requesting help in preparing that talk. So I wrote her back what I thought was a high school level presentation of Newton's theory.
When I saw the mess created when my mother-in-law's house burned, I was moved to clean out our attic. They had a lot of useless stuff in the attic which just created a burning pile of junk the fire department almost couldn't put out; in fact, two days after the firemen left we still found smoldering stuff in the bottoms of some piles of absolutely unidentifiable stuff.
At any rate, cleaning out my attic, I ran across my original work sheets for that letter to my niece. What follows here is essentially what I wrote to her.
The whole thing starts with Galilean relativity. Galilean relativity is essentially the idea that the rules of physics are the same in all frames of reference (Euclidean frames are all they had back then). Galileo really didn't worry much about “accelerating frames” so he was essentially talking about what Newton later called “inertial frames”: frames moving at a constant speed. Galileo pointed out that a rock dropped from the crow's nest on a ship hit the same place on the deck when the ship was moving as it hit when the ship was at rest. Essentially the fact that the ship was moving was completely hidden by the fact that the rock (when it started) was moving at exactly the same speed and in exactly the same direction as the ship.
Galileo also provided another deduction of great significance: everything fell at the same speed. People always say that Galileo dropped two objects from the leaning tower, but I have heard that he didn't; he simply used the idea of dropping two rocks from the leaning tower as a thought experiment and no one was able to show him wrong. From my understanding (which of course could be wrong) Galileo proposed the following thought experiment.
At the time, the authorities claimed that heavier objects fell faster. So suppose we drop two rocks from the top of the leaning tower at exactly the same time (one heavy and one light). The heavy one should hit the ground first. So now tie the two together with a string (loose enough so they don't actually touch one another) and drop them again. Since the heavy one falls faster, it must drag the light one by the string; right? Makes sense doesn't it? Ah, but it brings up another problem; the two rocks together weigh more so the combination (when they are tied together) must fall faster than the heavy rock when it is not tied to the light rock. So the existence of the light rock must cause the heavy rock to fall faster. So exactly how does it do this? It “pushes” the heavy rock through the string?
The only rational conclusion is that they fall at the same speed. Putting this together with his principle of relativity (for some reason no one ever refers to Galileo's theory of relativity; there is another paper right there), Galileo came up with the idea that the acceleration of gravity was constant (though I am not sure he referred to it as “gravity”). I don't read Italian and have never read what he actually said.
So now we come to Newton and his take on the principle of relativity and the constant acceleration of gravity. Essentially, what the falling apple pointed out to him was the idea that “the moon might be falling towards the earth” (an idea everyone else considered to be utterly ridiculous). But let us look at the idea carefully. When we fire a gun, the bullet (after it leaves the barrel) simply falls. This fact can be used to figure out exactly how far a bullet will go if the gun is perfectly level and there was no air to slow the bullet down. Just as with Galileo's rock dropped from the crows nest, it will hit exactly the same spot which would be hit by a bullet dropped from a platform moving horizontally at exactly the same speed as the bullet.
So let us put a little math to that. Let us start with a gun a distance “d” above a flat plane. We have an imaginary ship moving at exactly the same velocity and direction as the bullet. Let our imaginary ship consist of two important parts: a “crows nest” who's edge is exactly level with the end of the barrel and a flat deck exactly a distance “d” below that point. From that platform we drop a rock from a point on the crow's nest exactly at the end of the barrel at the very moment the gun is fired. What do we know? We know that both the bullet and the rock will, at exactly the same time, hit exactly the same spot on the deck of our imaginary ship. All we really need to know is, “how long does that take?” That together with the muzzle velocity of the gun and we have the fact that x=vt as the distance the bullet has gone.
Well, to figure that out you need calculus which you either know or you don't; therefore I will just give you the answer (it is quite easy if you know calculus). The distance the bullet drops as a function of its distance from the end of the barrel is given by

. If we know how far the bullet has to drop to hit the ground, we know exactly how far it will go.
Now, how about the fact that the earth is a sphere? As you go straight out in the direction the gun is pointed, the distance to the earth is further and further away. Suppose the muzzle velocity of the gun is high enough so that the distance the earth drops is exactly equal to the distance the bullet drops. Well, for that you need trigonometry so I will again just give you the answer: by similar triangles it is quite easy to show that x divided by the radius of the earth is identical to the distance the earth drops divided by half of x.

which implies
So, if the distance the earth drops (or rather a perfect sphere with no atmosphere) when the bullet moves a distance x is exactly the same as the distance the bullet drops, the height of the bullet above the sphere will remain exactly the same. (Don't worry about the acceleration; the force is perpendicular to the motion so the velocity is unchanged.) Putting the two equations above together,

, we can easily solve for that specific muzzle velocity. The critical velocity is exactly

. Since the radius of the earth is very close to 6.578x10
6 meters, if the muzzle velocity is exactly equal to 7,906 meters per second, you will shoot yourself in the back of the head. Or another way to view it, if you get out of the way, the bullet will be in orbit and it will stay at that altitude forever. A better measure of the critical velocity is the orbital period so we should look at the case where the circumference (or x) is exactly

: i.e.,

or

which is totally equivalent to

.
In other words, if the velocity is exactly equal to the velocity which will yield no change in the measured altitude above the surface, the period of the orbit will be given by

.
So it is entirely possible that the moon is just falling and the fact that its velocity is perpendicular to the direction to the earths center is just sufficient to yield a constant elevation is a quite possible circumstance. That would imply that the period of the moon (the length of time it takes to complete an orbit around the earth) given that it is at a radius of approximately 3.8x10
8 meters, would be roughly 10 hours, 55 minutes and 30 seconds. But that is the wrong answers! The correct answer is 27.3 days. So the idea that the moon is just falling towards the earth makes no sense at all.
But wait a moment, if you check it out, the answer is almost exactly one sixtieths of the correct answer! Put this together with the fact that the moon is 60 times as far away from the center of the earth as we are suggests the possibility that T is proportional to r divided by the radious of the earth or “g” (the acceleration of gravity at the moons orbit) is [imath]\left(\frac{6.578^6}{r}\right)^2 times that at the surface of the earth! Gee, that would yield exactly the correct period for the moon's orbit. That makes a lot of sense. It suggests that if we set the value of g to

we will get 9.8 meters per second per second at the earths surface and the correct period for the moon. Gee, that is pretty impressive.
So, let's check it. If that is the case, what would we expect the period of the earth around the sun to be? The distance to the sun is approximately 1.5x10
11 meters. Putting in the expected value of g we get the period (the length of time it takes the earth to go once around the sun) of about 570 years. That is certainly the wrong answer. So let's dump this whole idea.
But wait a moment. Suppose that the answer is again an incorrect value for g! The value of g is apparently too small by a facto of 3.25x10
4. Suppose that g is a function of the mass of the originating body. The assumption that the mass of the sun is 3.25x10
4 times the mass of the earth is not at all an unreasonable assumption. That would make the density of the sun roughly equal to the density of water and the fact that all evidence is that it is made of mostly hydrogen suggests that, if the gravity is considerably greater than the earth, that hydrogen might be sufficiently compressed to be roughly at the same density as water.
Best of all, setting the acceleration of gravity to

would yield the correct value of g at the earths surface, the correct period of for the moon and, in addition, the correct length of the year. That is one hell of a lot of support for the idea that they are all just falling toward one another. Furthermore, examination of the implied orbit periods of the rest of the planets in the solar system yield very little reason to disbelieve the theory. In fact, since then most everyone uses that very equation to determine the mass of the various planets.
And that, my dear, is the essence of Newton's theory of gravity. Actually it is pretty obvious if you think about it.
Have fun -- Uncle Dick