Comparing apples and oranges?

I have a post about Newton's brilliant insight that gravity as observed on earth could be extended to the moon. You can read about this lots of places, I like James Gleick's Isaac Newton (Amazon).

Many years later Newton told at least four people that he had been inspired by an apple in his Woolsthorpe garden--perhaps an apple actually falling from a tree, perhaps not. He never wrote of an apple. He recalled only:

I began to think of gravity extending to the orb of the Moon . . .

---gravity as a force, then, with an extended field of influence; no cutoff or boundary---

& computed the force requisite to kep the Moon in her Orb with the force of gravity at the surface of the earth . . . & found them answer pretty nearly. All this was in the two plague years of 1665-1666. For in those days I was in the prime of my age for invention & minded mathematicks and Philosophy more than at any time since.

The apple was nothing in itself. It was half of a couple---the moon's impish twin. As an apple falls toward the earth, so does the moon; falling away from a straight line, falling around the earth.

---Gleick
Isaac Newton (p. 55)

The force acting on the moon or the apple is proportional to the mass for each, but the acceleration imparted is in turn proportional to that force divided by the mass, which cancels the mass term (for apple and moon).

This leaves an equation relating the acceleration to a constant (G) times the mass of the earth (M) divided by the square of the distance to earth. Thus, we expect that the acceleration (and the distance moved) for the moon and the apple to be in the inverse ratio as the square of their distance from the earth.

a = GM/r2

The key relationship is shown in the diagram.

We draw the tangent to the orbital circle at time-zero at a right angle with the radius. Then the distance moved in one second, s, is the velocity times the time. The hypotenuse, H, is calculated from R and s by the Pythoagorean Theorem. And the distance "fallen" in the direction of the earth is the difference between H and the radius, R.

R = EM distance H = sqrt (R2 + (radial velocity * 1 sec)2) d = H - R

d is very small compared to R and H, but it still works. Using modern values for the distances:

mean distance from E to M (km) R (km) = 3.84x105 (range 3.63-4.06) circumference of M orbit = 2π 3.84x105 km mean radius of E (km) = 6.37x103 (range 6.36-6.38) ratio (EM dist/E radius) = 60.3 ratio2 = 3636 period of orbit of M T = 27.32 d = 2.36x106 seconds radial velocity of M C/T = 1.022 km/sec H = sqrt((3.84x105)2 + (1.022)2) = 384000.00000136002 M falls in 1 sec = 1.36 mm earth surface gravity = 9.8 m/s2 you fall in 1 sec = 1/2 g t2 = 4.9 m ratio = 4900 / 1.36 = 3603

I read somewhere that Newton first did this calculation with a value for the moon's orbital radius that was sufficiently off for him to think his result did not agree with it. Later he repeated his calculation with a more accurate value. But I can't find it now. One of these days I have to get a copy of the Principia (Amazon).